question_id
stringlengths 8
35
| subject
stringclasses 1
value | chapter
stringclasses 32
values | topic
stringclasses 178
values | question
stringlengths 26
9.64k
| options
stringlengths 2
1.63k
| correct_option
stringclasses 5
values | answer
stringclasses 293
values | explanation
stringlengths 13
9.38k
| question_type
stringclasses 3
values | paper_id
stringclasses 149
values |
|---|---|---|---|---|---|---|---|---|---|---|
1l6dwd0i9 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the locus of the centre $$(\alpha, \beta), \beta>0$$, of the circle which touches the circle $$x^{2}+(y-1)^{2}=1$$ externally and also touches the $$x$$-axis be $$\mathrm{L}$$. Then the area bounded by $$\mathrm{L}$$ and the line $$y=4$$ is:</p> | [{"identifier": "A", "content": "$$\n\\frac{32 \\sqrt{2}}{3}\n$$"}, {"identifier": "B", "content": "$$\n\\frac{40 \\sqrt{2}}{3}\n$$"}, {"identifier": "C", "content": "$$\\frac{64}{3}$$"}, {"identifier": "D", "content": "$$\n\\frac{32}{3}\n$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97rzcly/4fa8a86d-de28-44c0-be52-7dde80194483/7a763360-4b5a-11ed-bfde-e1cb3fafe700/file-1l97rzclz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97rzcly/4fa8a86d-de28-44c0-be52-7dde80194483/7a763360-4b5a-11ed-bfde-e1cb3fafe700/file-1l97rzclz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Area Under The Curves Question 57 English Explanation 1"><br>
Radius of circle $S$ touching $x$-axis and centre $(\alpha, \beta)$ is $|\beta|$. According to given conditions
<br><br>
$$
\begin{aligned}
&\alpha^{2}+(\beta-1)^{2}=(|\beta|+1)^{2} \\\\
&\alpha^{2}+\beta^{2}-2 \beta+1=\beta^{2}+1+2|\beta| \\\\
&\alpha^{2}=4 \beta \text { as } \beta>0
\end{aligned}
$$
<br><br>
$\therefore \quad$ Required louse is $L: x^{2}=4 y$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97s0prs/1a087718-c94f-41f2-86cd-76314c7ef4ef/a0702f80-4b5a-11ed-bfde-e1cb3fafe700/file-1l97s0prt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97s0prs/1a087718-c94f-41f2-86cd-76314c7ef4ef/a0702f80-4b5a-11ed-bfde-e1cb3fafe700/file-1l97s0prt.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Area Under The Curves Question 57 English Explanation 2"><br>
The area of shaded region $=2 \int_{0}^{4} 2 \sqrt{y} d y$
<br><br>
$$
\begin{aligned}
&=4 . {\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{0}^{4} } \\\\
&=\frac{64}{3} \text { square units. }
\end{aligned}
$$
| mcq | jee-main-2022-online-25th-july-morning-shift |
1l6f3sq57 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area enclosed by the x-axis, and the tangent and normal drawn to the curve $$4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0$$ at the point ($$-$$2, 3) be A. Then 8A is equal to ______________.</p> | [] | null | 170 | $$
\begin{aligned}
& 4 x^3-3 x y^2+6 x^2-5 x y-8 y^2+9 x+14=0 \text { at } P(-2,3) \\\\
& 12 x^2-3\left(y^2+2 y x y^{\prime}\right)+12 x-5\left(x y^{\prime}+y\right)-16 y y^{\prime} + 9=0 \\\\
& 48-3\left(9-12 y^{\prime}\right)-24-5\left(-2 y^{\prime}+3\right)-48 y^{\prime}+9 =0 \\\\
& y^{\prime}=-9 / 2 \\\\
& \text { Tangent } y-3=-\frac{9}{2}(x+2) \Rightarrow 9 x+2 y=-12 \\\\
& \text { Normal : } y-3=\frac{2}{9}(x+2) \Rightarrow 9 y-2 x=31
\end{aligned}
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc3zqoek/4bbf8e50-e22c-45d7-936b-72d7ce06a9e2/55dcbdb0-84aa-11ed-bd29-d15f3a725cdf/file-1lc3zqoel.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc3zqoek/4bbf8e50-e22c-45d7-936b-72d7ce06a9e2/55dcbdb0-84aa-11ed-bd29-d15f3a725cdf/file-1lc3zqoel.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Evening Shift Mathematics - Area Under The Curves Question 56 English Explanation"><br>
$$
\begin{aligned}
& \text { Area }=\frac{1}{2}\left(\frac{31}{2}-4\right) \times 3=\frac{85}{4} \\\\
& 8 \mathrm{~A}=170
\end{aligned}
$$ | integer | jee-main-2022-online-25th-july-evening-shift |
1l6ggjssj | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The odd natural number a, such that the area of the region bounded by y = 1, y = 3, x = 0, x = y<sup>a</sup> is $${{364} \over 3}$$, is equal to :</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "9"}] | ["B"] | null | <p>$$\mathrm{a}$$ is a odd natural number and</p>
<p>$$\left| {\int\limits_1^3 {{y^a}dy} } \right| = {{364} \over 3}$$</p>
<p>$$ \Rightarrow \left| {{1 \over {a + 1}}\left( {{y^{a + 1}}} \right)_1^3} \right| = {{364} \over 3}$$</p>
<p>$$ \Rightarrow {{{3^{a + 1}} - 1} \over {a + 1}} = \, \pm \,{{364} \over 3}$$</p>
<p>Solving with $$( - )$$ sign,</p>
<p>$${{{3^{a + 1}} - 1} \over {a + 1}} = {{364} \over 3} \Rightarrow (a = 5)$$</p>
<p>Solving with $$( + )$$ sign,</p>
<p>$${{{3^{a + 1}} - 1} \over {a + 1}} = {{ - 364} \over 3}$$, No a exist</p>
<p>$$\therefore$$ $$(a = 5)$$</p> | mcq | jee-main-2022-online-26th-july-morning-shift |
1l6hzgs0d | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area bounded by the curves $$y=\left|x^{2}-1\right|$$ and $$y=1$$ is</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}(\\sqrt{2}+1)$$"}, {"identifier": "B", "content": "$$\\frac{4}{3}(\\sqrt{2}-1)$$"}, {"identifier": "C", "content": "$$2(\\sqrt{2}-1)$$"}, {"identifier": "D", "content": "$$\\frac{8}{3}(\\sqrt{2}-1)$$"}] | ["D"] | null | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nnzsdz/1978812c-886a-41da-a99e-ba9650919e28/fdc43870-2c7e-11ed-a18d-5933e4fde865/file-1l7nnzse0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nnzsdz/1978812c-886a-41da-a99e-ba9650919e28/fdc43870-2c7e-11ed-a18d-5933e4fde865/file-1l7nnzse0.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Evening Shift Mathematics - Area Under The Curves Question 54 English Explanation"> </p>
<p>Area $$ = 2\int\limits_0^{\sqrt 2 } {(1 - |{x^2} - 1|)dx} $$</p>
<p>$$2\left[ {\int\limits_0^1 {\left( {1 - (1 - {x^2})} \right)dx + \int\limits_1^{\sqrt 2 } {\left( {2 - {x^2}} \right)dx} } } \right]$$</p>
<p>$$ = 2\left[ {\left. {\left[ {{{{x^3}} \over 3}} \right]_0^1 + \left[ {2x - {{{x^3}} \over 3}} \right]_1^{\sqrt 2 }} \right]} \right]$$</p>
<p>$$ = 2\left( {{{4\sqrt 2 - 4} \over 3}} \right) = {8 \over 3}\left( {\sqrt 2 - 1} \right)$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6jbsxxf | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the smaller region enclosed by the curves $$y^{2}=8 x+4$$ and $$x^{2}+y^{2}+4 \sqrt{3} x-4=0$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{1}{3}(2-12 \\sqrt{3}+8 \\pi)$$"}, {"identifier": "B", "content": "$$\\frac{1}{3}(2-12 \\sqrt{3}+6 \\pi)$$"}, {"identifier": "C", "content": "$$\\frac{1}{3}(4-12 \\sqrt{3}+8 \\pi)$$"}, {"identifier": "D", "content": "$$\\frac{1}{3}(4-12 \\sqrt{3}+6 \\pi)$$"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ps543w/90436956-c0bb-4a4d-b417-ec2e224aeb75/c7ae54c0-2da8-11ed-8542-f96181a425b5/file-1l7ps543x.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ps543w/90436956-c0bb-4a4d-b417-ec2e224aeb75/c7ae54c0-2da8-11ed-8542-f96181a425b5/file-1l7ps543x.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Morning Shift Mathematics - Area Under The Curves Question 53 English Explanation"></p>
<p>$$\cos \theta = {{2\sqrt 3 } \over 4} = {{\sqrt 3 } \over 2} \Rightarrow \theta = 30^\circ $$</p>
<p>Area of the required region</p>
<p>$$ = {2 \over 3}\left( {4 \times {1 \over 2}} \right) + {4^2} \times {\pi \over 6} - {1 \over 2} \times 4 \times 2\sqrt 3 $$</p>
<p>$$ = {4 \over 3} + {{8\pi } \over 3} - 4\sqrt 3 = {1 \over 3}\left\{ {4 - 12\sqrt 3 + 8\pi } \right\}$$</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6kjv7su | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by $$y \leq 4 x^{2}, x^{2} \leq 9 y$$ and $$y \leq 4$$, is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{40}{3}$$"}, {"identifier": "B", "content": "$$\\frac{56}{3}$$"}, {"identifier": "C", "content": "$$\\frac{112}{3}$$"}, {"identifier": "D", "content": "$$\\frac{80}{3}$$"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7q9zzgg/929f9d9b-01c4-4a4d-8457-02912d2b4071/9d5df400-2dee-11ed-a744-1fb8f3709cfa/file-1l7q9zzgh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7q9zzgg/929f9d9b-01c4-4a4d-8457-02912d2b4071/9d5df400-2dee-11ed-a744-1fb8f3709cfa/file-1l7q9zzgh.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Evening Shift Mathematics - Area Under The Curves Question 51 English Explanation"></p>
<p>$$y \le 4{x^2},\,{x^2} \le 9y,\,y \le 4$$</p>
<p>So, required area</p>
<p>$$A = 2\int_0^4 {\left( {3\sqrt y - {1 \over 2}\sqrt y } \right)dy} $$</p>
<p>$$ = 2\,.\,{5 \over 2}\,\,\,\,\,\,\,\,\left[ {{2 \over 3}{y^{{3 \over 2}}}} \right]_0^4$$</p>
<p>$$ = {{10} \over 3}\,\,\,\,\,\,\,\,\,\,\left[ {{4^{{3 \over 2}}} - 0} \right] = {{80} \over 3}$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6kk0129 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Consider a curve $$y=y(x)$$ in the first quadrant as shown in the figure. Let the area $$\mathrm{A}_{1}$$ is twice the area $$\mathrm{A}_{2}$$. Then the normal to the curve perpendicular to the line $$2 x-12 y=15$$ does NOT pass through the point.</p>
<p><img src="data:image/png;base64,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"/></p> | [{"identifier": "A", "content": "(6, 21)"}, {"identifier": "B", "content": "(8, 9)"}, {"identifier": "C", "content": "(10, $$-$$4)"}, {"identifier": "D", "content": "(12, $$-$$15)"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qa2cbj/f16ef20a-d0a5-47f2-a99e-77b83471fd40/deec2e00-2dee-11ed-a744-1fb8f3709cfa/file-1l7qa2cbk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qa2cbj/f16ef20a-d0a5-47f2-a99e-77b83471fd40/deec2e00-2dee-11ed-a744-1fb8f3709cfa/file-1l7qa2cbk.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th July Evening Shift Mathematics - Area Under The Curves Question 52 English Explanation"></p>
<p>$${A_1} + {A_2} = xy - 8$$ & $${A_1} = 2{A_2}$$</p>
<p>$${A_1} + {{{A_1}} \over 2} = xy - 8$$</p>
<p>$${A_1} = {2 \over 3}(xy - 8)$$</p>
<p>$$\int\limits_4^x {f(x)dx = {2 \over 3}(xf(x) - 8)} $$</p>
<p>Differentiate w.r.t. x</p>
<p>$$f(x) = {2 \over 3}\{ xf'(x) + f(x)\} $$</p>
<p>$${2 \over 3}xf'(x) = {1 \over 3}f(x)$$</p>
<p>$$2\int {{{f'(x)} \over {f(x)}}dx = \int {{{dx} \over x}} } $$</p>
<p>$$2\ln f(x) = \ln x + \ln c$$</p>
<p>$${f^2}(x) = cx$$</p>
<p>Which passes through (4, 2)</p>
<p>$$4 = c \times 4 \Rightarrow c = 1$$</p>
<p>Equation of required curve</p>
<p>$${y^2} = x$$</p>
<p>Equation of normal having slope ($$-$$6) is</p>
<p>$$y = - 6x - 2\left( {{1 \over 4}} \right)( - 6) - {1 \over 4}{( - 6)^3}$$</p>
<p>$$y = - 6x + 57$$</p>
<p>Which does not pass through $$(10,\, - 4)$$</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nmba5c | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed by the curves $$y=\log _{e}\left(x+\mathrm{e}^{2}\right), x=\log _{e}\left(\frac{2}{y}\right)$$ and $$x=\log _{\mathrm{e}} 2$$, above the line $$y=1$$ is:</p> | [{"identifier": "A", "content": "$$2+\\mathrm{e}-\\log _{\\mathrm{e}} 2$$"}, {"identifier": "B", "content": "$$1+e-\\log _{e} 2$$"}, {"identifier": "C", "content": "$$e-\\log _{e} 2$$"}, {"identifier": "D", "content": "$$1+\\log _{e} 2$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97u3d0g/088f2998-8bd4-4720-8b55-fa1efe186501/bc448a00-4b62-11ed-80b9-4154b7faa509/file-1l97u3d0h.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97u3d0g/088f2998-8bd4-4720-8b55-fa1efe186501/bc448a00-4b62-11ed-80b9-4154b7faa509/file-1l97u3d0h.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Evening Shift Mathematics - Area Under The Curves Question 50 English Explanation"><br>
According to NTA, the required region $A_2$ which is shaded in crossed lines and comes out to be<br><br>
$$
A_2=\int_1^2\left(\ln \frac{2}{y}-e^y+e^2\right) d y=1+e-\ln 2
$$<br><br>
But according to us the required region $A_1$ comes out to be shaded in parallel lines, which can be obtained as<br><br>
$$
\begin{aligned}
A_1 &=\int_0^{\ln 2}\left(\ln \left(x+e^2\right)-2 e^{-x}\right) d x \\\\
&=\left.\left\{\left(x+e^2\right) \ln \left(x+e^2\right)-x+2 e^{-x}\right\}\right|_0 ^{\ln 2} \\\\
&=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2+1 \\\\
& \quad-2 e^2-2 \\\\
&=\left(\ln 2+e^2\right) \ln \left(\ln 2+e^2\right)-\ln 2-2 e^2-1
\end{aligned}
$$<br><br>
Not given in any option.<br><br>
The region asked in the question is bounded by three curves<br><br>
$$
\begin{aligned}
&y=\ln \left(x+e^2\right) \\\\
&x=\ln \left(\frac{2}{y}\right) \\\\
&x=\ln 2
\end{aligned}
$$<br><br>
There is only one region which satisfies above requirement and which also lies above line $y=1$<br><br> Line $y=1$ may or may not be the boundary of the region. | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6p2kie8 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region</p>
<p>$$\left\{(x, y):|x-1| \leq y \leq \sqrt{5-x^{2}}\right\}$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{5}{2} \\sin ^{-1}\\left(\\frac{3}{5}\\right)-\\frac{1}{2}$$"}, {"identifier": "B", "content": "$$\\frac{5 \\pi}{4}-\\frac{3}{2}$$"}, {"identifier": "C", "content": "$$\\frac{3 \\pi}{4}+\\frac{3}{2}$$"}, {"identifier": "D", "content": "$$\\frac{5 \\pi}{4}-\\frac{1}{2}$$"}] | ["D"] | null | <p>$$A = \int\limits_{ - 1}^1 {\left( {\sqrt {5 - {x^2}} - (1 - x)} \right)dx + \int\limits_1^2 {\left( {\sqrt {5 - {x^2}} - (x - 1)} \right)dx} } $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7ssldo3/57b8d8ee-cd1f-4f78-8605-15da9040ba0e/e5f60c30-2f50-11ed-85dd-19dc023e9ad1/file-1l7ssldo4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7ssldo3/57b8d8ee-cd1f-4f78-8605-15da9040ba0e/e5f60c30-2f50-11ed-85dd-19dc023e9ad1/file-1l7ssldo4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Morning Shift Mathematics - Area Under The Curves Question 49 English Explanation"></p>
<p>$$\left. {A = 2\left( {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }}} \right) - 2x} \right|_0^1 + \left. {{x \over 2}\sqrt {5 - {x^2}} + {5 \over 2}{{\sin }^{ - 1}}{x \over {\sqrt 5 }} - {{{x^2}} \over 2} + x} \right|_1^2$$</p>
<p>$$ = \left( {{{5\pi } \over 4} - {1 \over 2}} \right)$$ sq. units</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1ldo65ln2 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region given by $$\{ (x,y):xy \le 8,1 \le y \le {x^2}\} $$ is :</p> | [{"identifier": "A", "content": "$$16{\\log _e}2 - {{14} \\over 3}$$"}, {"identifier": "B", "content": "$$8{\\log _e}2 - {{13} \\over 3}$$"}, {"identifier": "C", "content": "$$16{\\log _e}2 + {7 \\over 3}$$"}, {"identifier": "D", "content": "$$8{\\log _e}2 + {7 \\over 6}$$"}] | ["A"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5fqsjf/90c628e8-42f9-49e3-ac52-f0f944e1d31e/2287dcb0-ad0e-11ed-a86d-8dfe0389db88/file-1le5fqsjg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5fqsjf/90c628e8-42f9-49e3-ac52-f0f944e1d31e/2287dcb0-ad0e-11ed-a86d-8dfe0389db88/file-1le5fqsjg.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Evening Shift Mathematics - Area Under The Curves Question 48 English Explanation">
<br><br>$\begin{aligned} & \text { Required area }=\int_1^2\left(x^2-1\right) d x+\int_2^8\left(\frac{8}{x}-1\right) d x \\\\ & =\left.\left(\frac{x^3}{3}-x\right)\right|_1 ^2+\left.(8 \ln x-x)\right|_2 ^8 \\\\ & =\left[\left(\frac{8}{3}-2\right)-\left(\frac{1}{3}-1\right)\right]+[8 \ln 8-8-(8 \ln 2-2)] \\\\ & =\frac{4}{3}+8 \ln 4-6 \\\\ & =8 \ln 4-\frac{14}{3} \\\\ & =16 \ln 2-\frac{14}{3}\end{aligned}$ | mcq | jee-main-2023-online-1st-february-evening-shift |
ldoah9qp | maths | area-under-the-curves | area-bounded-between-the-curves | Let the area of the region
<br/><br/>$\left\{(x, y):|2 x-1| \leq y \leq\left|x^{2}-x\right|, 0 \leq x \leq 1\right\}$ be $\mathrm{A}$.
<br/><br/>Then $(6 \mathrm{~A}+11)^{2}$ is equal to | [] | null | 125 | For $B$,
<br><br>$$
\begin{aligned}
& x-x^{2}=2 x-1 \\\\
& x^{2}+x-1=0 \\\\
& x=\frac{-1+\sqrt{5}}{2}
\end{aligned}
$$
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leeiutst/47959be2-54bf-4c50-8327-bce6176bbe32/5bf957e0-b20d-11ed-9c9a-c3bd979bc154/file-1leeiutsu.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leeiutst/47959be2-54bf-4c50-8327-bce6176bbe32/5bf957e0-b20d-11ed-9c9a-c3bd979bc154/file-1leeiutsu.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Evening Shift Mathematics - Area Under The Curves Question 47 English Explanation">
<br><br>Area $=2($ area of $B C E)$
<br><br>$A=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}}\left(x-x^{2}\right)-(2 x-1) d x$
<br><br>$=2 \int_{\frac{1}{2}}^{\frac{\sqrt{5}-1}{2}} (1-x-x^{2}) d x=\left.2\left(x-\frac{x^{2}}{2}-\frac{x^{3}}{3}\right)\right|_{\frac{1}{2}} ^{]_{5}^{\frac{\sqrt{5}-1}{2}}}$
<br><br>$$
\begin{aligned}
=2\left\{\left(\frac{\sqrt{5}-1}{2}-\frac{1}{2}\right)-\left\{\left(\frac{\sqrt{5}-1}{2}\right)^{2}\right.\right. & \left.-\frac{1}{4}\right\} \frac{1}{2} \\\\
& \left.-\left[\left(\frac{\sqrt{5}-1}{2}\right)^{3}-\frac{1}{8}\right] \frac{1}{3}\right\}
\end{aligned}
$$
<br><br>$$
A=\frac{-11+5 \sqrt{5}}{6}
$$
<br><br>$\Rightarrow(6 A+11)^{2}=125$ | integer | jee-main-2023-online-31st-january-evening-shift |
1ldoocgjv | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$A$$ be the area bounded by the curve $$y=x|x-3|$$, the $$x$$-axis and the ordinates $$x=-1$$ and $$x=2$$. Then $$12 A$$ is equal to ____________.</p> | [] | null | 62 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le9xvce9/cce9188c-fc73-4ab5-a87d-33da42a11b53/28073f10-af88-11ed-bd02-1d2a2a7b6687/file-1le9xvcea.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le9xvce9/cce9188c-fc73-4ab5-a87d-33da42a11b53/28073f10-af88-11ed-bd02-1d2a2a7b6687/file-1le9xvcea.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 1st February Morning Shift Mathematics - Area Under The Curves Question 46 English Explanation">
<br><br>$=\int\limits_{-1}^{2}\left|3 x-x^{2}\right|$
<br><br>$A=\int\limits_{-1}^{0} x^{2}-3 x d x+\int\limits_{0}^{2} 3 x-x^{2} d x$
<br><br>$\left.\left.=\frac{x^{3}}{3}-\frac{3 x^{2}}{2}\right]_{-1}^{0}+\frac{3 x^{2}}{2}-\frac{x^{3}}{3}\right]_{0}^{2}$
<br><br>$=0-\left(\frac{-1}{3}-\frac{3}{2}\right)+\left(6-\frac{8}{3}\right)-0$
<br><br>$=\frac{31}{6}$
<br><br>$\therefore 12 A=62$ | integer | jee-main-2023-online-1st-february-morning-shift |
1ldptrjv7 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let for $$x \in \mathbb{R}$$,</p>
<p>$$
f(x)=\frac{x+|x|}{2} \text { and } g(x)=\left\{\begin{array}{cc}
x, & x<0 \\
x^{2}, & x \geq 0
\end{array}\right. \text {. }
$$</p>
<p>Then area bounded by the curve $$y=(f \circ g)(x)$$ and the lines $$y=0,2 y-x=15$$ is equal to __________.</p> | [] | null | 72 | $f(x)=\frac{x+|x|}{2}=\left[\begin{array}{ll}x & x \geq 0 \\ 0 & x<0\end{array}\right.$
<br><br>$g(x)=\left[\begin{array}{cc}x^{2} & x \geq 0 \\ x & x<0\end{array}\right.$
<br><br>$f(x)=f[g(x)]=\left[\begin{array}{cl}g(x), & g(x) \geq 0 \\ 0, & g(x)<0\end{array}\right.$
<br><br>$\operatorname{fog}(x)=\left[\begin{array}{cc}x^{2}, & x \geq 0 \\ 0, & x<0\end{array}\right.$
<br><br>$2 y-x=15$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lelnbcxl/c11e6b70-de76-422b-8148-ab95b681cbde/48c30290-b5f8-11ed-bc05-1be7a3b49a67/file-1lelnbcxm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lelnbcxl/c11e6b70-de76-422b-8148-ab95b681cbde/48c30290-b5f8-11ed-bc05-1be7a3b49a67/file-1lelnbcxm.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 31st January Morning Shift Mathematics - Area Under The Curves Question 45 English Explanation">
<br><br>$\mathrm{A}=\int_{0}^{3}\left(\frac{\mathrm{x}+15}{2}-\mathrm{x}^{2}\right) \mathrm{dx}+\frac{1}{2} \times \frac{15}{2} \times 15$
<br><br>= $\frac{x^{2}}{4}+\frac{15 x}{2}-\left.\frac{x^{3}}{3}\right|_{0} ^{3}+\frac{225}{4}$
<br><br>$=\frac{9}{4}+\frac{45}{2}-9+\frac{225}{4}=\frac{99-36+225}{4}$
<br><br>$=\frac{288}{4}=72$ | integer | jee-main-2023-online-31st-january-morning-shift |
ldqwiizg | maths | area-under-the-curves | area-bounded-between-the-curves | Let $q$ be the maximum integral value of $p$ in $[0,10]$ for which the roots of the equation $x^2-p x+\frac{5}{4} p=0$ are rational. Then the area of the region $\left\{(x, y): 0 \leq y \leq(x-q)^2, 0 \leq x \leq q\right\}$ is : | [{"identifier": "A", "content": "$\\frac{125}{3}$"}, {"identifier": "B", "content": "243"}, {"identifier": "C", "content": "164"}, {"identifier": "D", "content": "25"}] | ["B"] | null | <p>Given equation : $$4{x^2} - 4px + 5p = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leolev6m/9d2f1ed0-3ff3-4daa-b4d0-ae63d5dd71a4/31fa37e0-b797-11ed-b103-ed967fad3dff/file-1leolev6n.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leolev6m/9d2f1ed0-3ff3-4daa-b4d0-ae63d5dd71a4/31fa37e0-b797-11ed-b103-ed967fad3dff/file-1leolev6n.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Evening Shift Mathematics - Area Under The Curves Question 44 English Explanation"></p>
<p>for rational roots, D must be perfect square</p>
<p>$$D = 16{p^2} - 4 \times 4 \times 5p = 16p(p - 5)$$</p>
<p>So, max. Integral value of $$p = 9$$ for making D is perfect square</p>
<p>$$\therefore$$ $$q = 9$$</p>
<p>Area of shared region</p>
<p>$$ = \int\limits_0^9 {{{(x - 9)}^2}dx} $$</p>
<p>$$ = \left. {{{{{(x - 9)}^3}} \over 3}} \right|_0^9 = {{{9^3}} \over 3} = 243$$ sq. units</p> | mcq | jee-main-2023-online-30th-january-evening-shift |
ldr01u60 | maths | area-under-the-curves | area-bounded-between-the-curves | Let $A$ be the area of the region
<br/><br/>$\left\{(x, y): y \geq x^2, y \geq(1-x)^2, y \leq 2 x(1-x)\right\}$.
<br/><br/>Then $540 \mathrm{~A}$ is equal to : | [] | null | 25 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leol5gd1/d39e5f85-8fa2-457b-8f2b-aa98ad9bb499/2c3ee450-b796-11ed-b103-ed967fad3dff/file-1leol5gd2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leol5gd1/d39e5f85-8fa2-457b-8f2b-aa98ad9bb499/2c3ee450-b796-11ed-b103-ed967fad3dff/file-1leol5gd2.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Evening Shift Mathematics - Area Under The Curves Question 43 English Explanation"></p>
<p>$$y = {x^2}$$ and $$y = 2x(1 - x)$$</p>
<p>$$ \Rightarrow {x^2} = 2x - 2{x^2}$$</p>
<p>$$ \Rightarrow x = 0,x = {2 \over 3}$$</p>
<p>Now,</p>
<p>$$y = {(1 - x)^2}$$ and $$y = 2x(1 - x)$$</p>
<p>$$ \Rightarrow 1 + {x^2} - 2x = 2x - 2{x^2}$$</p>
<p>$$ \Rightarrow 3{x^2} - 4x + 1 = 0$$</p>
<p>$$x = 1,x = {1 \over 3}$$</p>
<p>$$\therefore$$ $$A = \int\limits_{{1 \over 3}}^{{2 \over 3}} {(2x - 2{x^2})dx - \left\{ {\int\limits_{{1 \over 3}}^{{1 \over 2}} {{{(1 - x)}^2}dx + \int\limits_{{1 \over 2}}^{{2 \over 3}} {{x^2}dx} } } \right\}} $$</p>
<p>= $$\left( {{x^2} - {{2{x^3}} \over 3}} \right)_{{1 \over 3}}^{{2 \over 3}} - \left\{ {\left( {{{{{\left( {x - 1} \right)}^3}} \over 3}} \right)_{{1 \over 3}}^{{1 \over 2}} + \left( {{{{x^3}} \over 3}} \right)_{{1 \over 2}}^{{2 \over 3}}} \right\}$$</p>
<p>$$ = {5 \over {108}}$$</p>
<p>$$\therefore$$ $$540A = 25$$</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldr7y2pf | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$\alpha$$ be the area of the larger region bounded by the curve $$y^{2}=8 x$$ and the lines $$y=x$$ and $$x=2$$, which lies in the first quadrant. Then the value of $$3 \alpha$$ is equal to ___________.</p> | [] | null | 22 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leq0fubs/f2abef4a-3124-4350-839e-3e7bfc1aa857/beb62c80-b85e-11ed-9fed-b1659a6c339b/file-1leq0fubt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leq0fubs/f2abef4a-3124-4350-839e-3e7bfc1aa857/beb62c80-b85e-11ed-9fed-b1659a6c339b/file-1leq0fubt.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 30th January Morning Shift Mathematics - Area Under The Curves Question 42 English Explanation"></p>
<p>$${A_1} = \int\limits_0^2 {2\sqrt 2 \sqrt x - xdx} $$</p>
<p>$$ = \left. {2\sqrt 2 \times {2 \over 3}\,.\,{x^{{3 \over 2}}} - {{{x^2}} \over 2}} \right]_0^2$$</p>
<p>$$ = {{4\sqrt 2 } \over 3} \times 2\sqrt 2 - 2$$</p>
<p>$$ = {{16} \over 3} - 2 = {{10} \over 3}$$</p>
<p>$${A_2} = \left. {\int\limits_2^8 {2\sqrt 2 \sqrt x - xdx = {{4\sqrt 2 } \over 3}\,.\,{x^{{3 \over 2}}} - {{{x^2}} \over 2}} } \right]_2^8$$</p>
<p>$$ = {{4\sqrt 2 } \over 3}(16\sqrt 2 - 2\sqrt 2 ) - 30$$</p>
<p>$$ = {{112} \over 3} - 30 = {{22} \over 3}$$</p>
<p>$${A_2} > {A_1} \Rightarrow 3\alpha = 22$$</p> | integer | jee-main-2023-online-30th-january-morning-shift |
1ldsf7k2m | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$A = \left\{ {(x,y):\left| {\cos x - \sin x} \right| \le y \le \sin x,0 \le x \le {\pi \over 2}} \right\}$$ is</p> | [{"identifier": "A", "content": "$$\\sqrt 5 + 2\\sqrt 2 - 4.5$$"}, {"identifier": "B", "content": "$$1 - {3 \\over {\\sqrt 2 }} + {4 \\over {\\sqrt 5 }}$$"}, {"identifier": "C", "content": "$$\\sqrt 5 - 2\\sqrt 2 + 1$$"}, {"identifier": "D", "content": "$${3 \\over {\\sqrt 5 }} - {3 \\over {\\sqrt 2 }} + 1$$"}] | ["C"] | null | <p>$$
|\cos x-\sin x| \leq y \leq \sin x
$$
<br><br>Intersection point of $\cos x-\sin x=\sin x$
<br><br>$$
\Rightarrow \tan x=\frac{1}{2}
$$
<br><br>Let $\psi=\tan ^{-1} \frac{1}{2}$
<br><br>So, $\tan \psi=\frac{1}{2}, \sin \psi=\frac{1}{\sqrt{5}}, \cos \psi=\frac{2}{\sqrt{5}}$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1let7mhvc/30704cc1-9134-44fb-84c5-b2ca232ae1f5/319af780-ba21-11ed-b1c7-2f0d4a78b053/file-1let7mhvd.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1let7mhvc/30704cc1-9134-44fb-84c5-b2ca232ae1f5/319af780-ba21-11ed-b1c7-2f0d4a78b053/file-1let7mhvd.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Evening Shift Mathematics - Area Under The Curves Question 41 English Explanation"></p>
<p>Area $$ = \int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{{\pi \over 4}} {(\sin x - (\cos x - \sin x))dx + \int\limits_{{\pi \over 4}}^{{\pi \over 2}} {(\sin x - (\sin x - \cos x))dx} } $$</p>
<p>$$ =\int\limits_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}(2 \sin x-\cos x) d x+\int\limits_{\pi / 4}^{\pi / 2} \cos x d x $$</p>
<p>$$ =[-2 \cos x-\sin x]_{{{\tan }^{ - 1}}{1 \over 2}}^{\pi / 4}+[\sin x]_{\pi / 4}^{\pi / 2} $$</p>
<p>$$ =-\sqrt{2}-\frac{1}{\sqrt{2}}+2 \cos ({{{\tan }^{ - 1}}{1 \over 2}}) +\sin ({{{\tan }^{ - 1}}{1 \over 2}})+\left(1-\frac{1}{\sqrt{2}}\right)$$</p>
<p>$=-\sqrt{2}-\frac{1}{\sqrt{2}}+2\left(\frac{2}{\sqrt{5}}\right)+\left(\frac{1}{\sqrt{5}}\right)+1-\frac{1}{\sqrt{2}}$</p>
<p>$$ = \sqrt 5 - 2\sqrt 2 + 1$$</p> | mcq | jee-main-2023-online-29th-january-evening-shift |
1ldsv2h4c | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$\Delta$$ be the area of the region $$\left\{ {(x,y) \in {R^2}:{x^2} + {y^2} \le 21,{y^2} \le 4x,x \ge 1} \right\}$$. Then $${1 \over 2}\left( {\Delta - 21{{\sin }^{ - 1}}{2 \over {\sqrt 7 }}} \right)$$ is equal to</p> | [{"identifier": "A", "content": "$$2\\sqrt 3 - {1 \\over 3}$$"}, {"identifier": "B", "content": "$$2\\sqrt 3 - {2 \\over 3}$$"}, {"identifier": "C", "content": "$$\\sqrt 3 - {4 \\over 3}$$"}, {"identifier": "D", "content": "$$\\sqrt 3 - {2 \\over 3}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1leksols2/a3491772-b199-41ab-9761-0d39a13abce7/7e1c5220-b580-11ed-b843-fd540edb80bf/file-1leksols3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1leksols2/a3491772-b199-41ab-9761-0d39a13abce7/7e1c5220-b580-11ed-b843-fd540edb80bf/file-1leksols3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 38 English Explanation"><br>
Required area $=2 \int\limits_{1}^{3} 2 \sqrt{x} d x+\int_{3}^{\sqrt{21}} \sqrt{\left(21-x^{2}\right)} d x$
<br><br>
$=2\left(\left[\left.2\left(\frac{x^{3 / 2}}{3 / 2}\right)\right|_{1} ^{3}\right]+\left[\frac{x}{2} \sqrt{21-x^{2}}+\frac{21}{2} \sin ^{-1}\left(\frac{x}{\sqrt{21}}\right)\right]_{3}^{\sqrt{21}}\right)$
<br><br>
$=2 \sqrt{3}+\frac{21 \pi}{2}-\frac{8}{3}-21 \sin ^{-1} \sqrt{\frac{3}{7}}=\Delta$
<br><br>
$\therefore \frac{1}{2}\left(\Delta-21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)\right)=\sqrt{3}-\frac{4}{3}$ | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldsvuk9z | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$[x]$$ denote the greatest integer $$\le x$$. Consider the function $$f(x) = \max \left\{ {{x^2},1 + [x]} \right\}$$. Then the value of the integral $$\int\limits_0^2 {f(x)dx} $$ is</p> | [{"identifier": "A", "content": "$${{5 + 4\\sqrt 2 } \\over 3}$$"}, {"identifier": "B", "content": "$${{4 + 5\\sqrt 2 } \\over 3}$$"}, {"identifier": "C", "content": "$${{8 + 4\\sqrt 2 } \\over 3}$$"}, {"identifier": "D", "content": "$${{1 + 5\\sqrt 2 } \\over 3}$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ldt0ucjn/f3408cd8-1181-40fe-aecc-1f2a69b3243c/74ae7e20-a63a-11ed-a341-61f046f8f7d7/file-1ldt0ucjo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ldt0ucjn/f3408cd8-1181-40fe-aecc-1f2a69b3243c/74ae7e20-a63a-11ed-a341-61f046f8f7d7/file-1ldt0ucjo.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 40 English Explanation 1"></p>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ldt0xhxy/b01c4c6d-ed94-4045-ae7d-51c181e8b9a2/cc476b60-a63a-11ed-8e98-03be028140e9/file-1ldt0xhxz.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ldt0xhxy/b01c4c6d-ed94-4045-ae7d-51c181e8b9a2/cc476b60-a63a-11ed-8e98-03be028140e9/file-1ldt0xhxz.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 40 English Explanation 2">
<p>Combining both the graph we get,</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1ldt1346q/f2c4d820-3596-4684-af86-b7ffd222c8ce/6881ed20-a63b-11ed-8e98-03be028140e9/file-1ldt1346r.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1ldt1346q/f2c4d820-3596-4684-af86-b7ffd222c8ce/6881ed20-a63b-11ed-8e98-03be028140e9/file-1ldt1346r.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 40 English Explanation 3"></p>
<p>Both graph cuts each other at $$y=2$$ and $$y=1$$</p>
<p>$$\therefore x^2=2$$</p>
<p>$$\Rightarrow x=\sqrt2$$</p>
<p>Two graphs cuts each other at $$x=1$$ and $$x=\sqrt2$$</p>
<p>In 0 to 1, $$f(x) = \max ({x^2},1 + [x]) = 1$$</p>
<p>In 1 to $$\sqrt2$$, $$f(x) = \max ({x^2},1 + [x]) = 2$$</p>
<p>In $$\sqrt2$$ to 2, $$f(x) = \max ({x^2},1 + [x]) = {x^2}$$</p>
<p>$$\therefore$$ $$\int\limits_0^2 {f(x)dx} $$</p>
<p>$$ = \int\limits_0^1 {1\,dx + \int\limits_1^{\sqrt 2 } {2\,dx + \int\limits_{\sqrt 2 }^2 {{x^2}\,dx} } } $$</p>
<p>$$ = \left[ x \right]_0^1 + 2\left[ x \right]_1^{\sqrt 2 } + \left[ {{{{x^3}} \over 3}} \right]_{\sqrt 2 }^2$$</p>
<p>$$ = 1 + 2(\sqrt 2 - 1) + {8 \over 3} - {{2\sqrt 2 } \over 3}$$</p>
<p>$$ = {8 \over 3} - 1 + 2\sqrt 2 - {{2\sqrt 2 } \over 3}$$</p>
<p>$$ = {5 \over 3} + {{6\sqrt 2 - 2\sqrt 2 } \over 3}$$</p>
<p>$$ = {5 \over 3} + {{4\sqrt 2 } \over 3}$$</p>
<p>$$ = {{5 + 4\sqrt 2 } \over 3}$$</p> | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldsvyln7 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$A=\left\{(x, y) \in \mathbb{R}^{2}: y \geq 0,2 x \leq y \leq \sqrt{4-(x-1)^{2}}\right\}$$ and<br/><br/> $$
B=\left\{(x, y) \in \mathbb{R} \times \mathbb{R}: 0 \leq y \leq \min \left\{2 x, \sqrt{4-(x-1)^{2}}\right\}\right\} \text {. }
$$.</p>
<p>Then the ratio of the area of A to the area of B is</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{\\pi+1}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi-1}{\\pi+1}$$"}, {"identifier": "C", "content": "$$\\frac{\\pi}{\\pi-1}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi+1}{\\pi-1}$$"}] | ["B"] | null | <p>$y^{2}+(x-1)^{2}=4$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lekro7yb/473e481f-1c3a-4e67-a338-a22cacc1490d/8a4d0930-b57c-11ed-9ad9-3b1cedbe69d8/file-1lekro7yc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lekro7yb/473e481f-1c3a-4e67-a338-a22cacc1490d/8a4d0930-b57c-11ed-9ad9-3b1cedbe69d8/file-1lekro7yc.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 39 English Explanation 1"></p>
shaded portion $=$ circular $(\mathrm{OABC})$
<br><br>
$$
\begin{aligned}
& -\operatorname{Ar}(\Delta \mathrm{OAB}) \\\\
& =\frac{\pi(4)}{4}-\frac{1}{2}(2)(1) \\\\
& \mathrm{A}=(\pi-1)
\end{aligned}
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lekrqfr8/c116d1fd-fb30-46f1-8aaa-33c69069d967/c7f29750-b57c-11ed-860d-ed3e7068b1ec/file-1lekrqfr9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lekrqfr8/c116d1fd-fb30-46f1-8aaa-33c69069d967/c7f29750-b57c-11ed-860d-ed3e7068b1ec/file-1lekrqfr9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 39 English Explanation 2"><br>
Area $B=\operatorname{Ar}(\triangle \mathrm{AOB})+$ Area of arc of circle $(\mathrm{ABC})$
<br><br>
$$
=\frac{1}{2}(1)(2)+\frac{\pi(2)^{2}}{4}=\pi+1
$$
<br><br>
$$
\frac{\mathrm{A}}{\mathrm{B}}=\frac{\pi-1}{\pi+1}
$$
| mcq | jee-main-2023-online-29th-january-morning-shift |
1ldv36uu9 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area enclosed by the parabolas $$\mathrm{P_1:2y=5x^2}$$ and $$\mathrm{P_2:x^2-y+6=0}$$ is equal to the area enclosed by $$\mathrm{P_1}$$ and $$\mathrm{y=\alpha x,\alpha > 0}$$, then $$\alpha^3$$ is equal to ____________.</p> | [] | null | 600 | $x^{2}+6=\frac{5}{2} x^{2} \Rightarrow x=\pm 2$
<br/><br/>
$$
\begin{aligned}
& \text { Area between } P_{1} \text { and } P_{2} \quad \text { [Say } \left.A_{1}\right] \\\\
& =\int\limits_{-2}^{2}\left(x^{2}+6\right)-\frac{5}{2} x^{2} d x \\\\
& =2 \int\limits_{0}^{2}\left(6-\frac{3}{2} x^{2}\right) d x=2\left[6 x-\frac{x^{3}}{2}\right]_{0}^{2}=16
\end{aligned}
$$<br/><br/>
$$
\begin{aligned}
& a x=\frac{5}{2} x^2 \Rightarrow x=0, \frac{2 a}{5} \\\\
& \text { Area between } P_1 \text { and } y=a x \quad \text { [Say } A_2 \text { ] } \\\\
& =\int\limits_0^{\frac{2 \alpha}{5}} a x-\frac{5}{2} x^2 d x \\\\
& \left.=\frac{a x^2}{2}-\frac{5}{6} x^3\right]_0^{\frac{2 a}{5}}: \frac{2 a^3}{75} \\\\
& A_1=A_2 \Rightarrow \frac{2 a^3}{75}=16 \\\\
& a^3=600
\end{aligned}
$$ | integer | jee-main-2023-online-25th-january-morning-shift |
1ldwxskv5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region bounded by the curves $$y^2-2y=-x,x+y=0$$ is A, then 8 A is equal to __________</p> | [] | null | 36 | Area enclosed by
<br><br>
$$
\begin{aligned}
& y^{2}-2 y=-x \\\\
& x+y=0
\end{aligned}
$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le5i5l5v/1c86c1a0-48dd-4e1e-9be0-6401a21b507d/903a5b30-ad17-11ed-8a8c-4d67f5492755/file-1le5i5l5w.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le5i5l5v/1c86c1a0-48dd-4e1e-9be0-6401a21b507d/903a5b30-ad17-11ed-8a8c-4d67f5492755/file-1le5i5l5w.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Evening Shift Mathematics - Area Under The Curves Question 36 English Explanation"><br>
Area $=\int_{0}^{3}\left(2 y-y^{2}\right)-(-y) d y$
<br><br>
$$
=\int_{0}^{3}\left(3 y-y^{2}\right) d y
$$
<br><br>
$$
\begin{aligned}
& \left.=\frac{3 y^{2}}{2}-\frac{y^{3}}{3}\right]_{0}^{3} \\\\
& =\frac{27}{2}-9 \\\\
& =\frac{27-18}{2}=\frac{9}{2}=A
\end{aligned}
$$
<br><br>
$8 A=\frac{9}{2} \times 8=36$ sq. units | integer | jee-main-2023-online-24th-january-evening-shift |
1ldyb9d7r | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed by the curves $${y^2} + 4x = 4$$ and $$y - 2x = 2$$ is :</p> | [{"identifier": "A", "content": "$${{22} \\over 3}$$"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "$${{23} \\over 3}$$"}, {"identifier": "D", "content": "$${{25} \\over 3}$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le1yf9t4/b18e1d13-83c3-4532-94a6-9b8efbcd5fcf/0d044770-ab24-11ed-b3ea-6525e5fb1f1e/file-1le1yf9t5.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le1yf9t4/b18e1d13-83c3-4532-94a6-9b8efbcd5fcf/0d044770-ab24-11ed-b3ea-6525e5fb1f1e/file-1le1yf9t5.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Morning Shift Mathematics - Area Under The Curves Question 35 English Explanation"><br>
Required area $=\int_{-4}^{2}\left(\frac{4-y^{2}}{4}-\frac{y-2}{2}\right) d y$
<br><br>
$=\int_{-4}^{2} \frac{8-2 y-y^{2}}{4} d y$
<br><br>
$=\frac{1}{4}\left\{8 y-y^{2}-\frac{y^{3}}{3}\right\}_{-4}^{2}$
<br><br>
$=9$ square units | mcq | jee-main-2023-online-24th-january-morning-shift |
lgnz10hf | maths | area-under-the-curves | area-bounded-between-the-curves | If the area bounded by the curve $2 y^{2}=3 x$, lines $x+y=3, y=0$ and outside the circle $(x-3)^{2}+y^{2}=2$ is $\mathrm{A}$, then $4(\pi+4 A)$ is equal to ____________. | [] | null | 42 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lgrn6cu8/fb34c18c-fa58-4f6a-a195-c84349a81234/e04c5800-e0dc-11ed-9ecd-e999028462e7/file-1lgrn6cu9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lgrn6cu8/fb34c18c-fa58-4f6a-a195-c84349a81234/e04c5800-e0dc-11ed-9ecd-e999028462e7/file-1lgrn6cu9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 15th April Morning Shift Mathematics - Area Under The Curves Question 34 English Explanation"><br>
$$
\begin{aligned}
& y^2=\frac{3 x}{2}, x+y=3, y=0 \\\\
& 2 y^2=3(3-y) \\\\
& 2 y^2+3 y-9=0 \\\\
& 2 y^2-3 y+6 y-9=0 \\\\
& (2 y-3)(y+2)=0 ; y=3 / 2 \\\\
& \text { Area }\left(\int_0^{\frac{3}{2}}\left(x_R-x_2\right) d y\right)-A_1 \\\\
& =\int_0^{\frac{3}{2}}\left((3-y)-\frac{2 y^2}{3}\right) d y-\frac{\pi}{8}(2) \\\\
& A=\left(3 y-\frac{y^2}{2}-\frac{2 y^3}{9}\right)_0^{\frac{3}{2}}-\frac{\pi}{4} \\\\
& 4 \mathrm{~A}+\pi=4\left[\frac{9}{2}-\frac{9}{8}-\frac{3}{4}\right]=\frac{21}{2}=10.50 \\\\
& \therefore 4(4 \mathrm{~A}+\pi)=42
\end{aligned}
$$ | integer | jee-main-2023-online-15th-april-morning-shift |
1lgoxjwx4 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$\left\{(x, y): x^{2} \leq y \leq\left|x^{2}-4\right|, y \geq 1\right\}$$ is</p> | [{"identifier": "A", "content": "$$\\frac{4}{3}(4 \\sqrt{2}+1)$$"}, {"identifier": "B", "content": "$$\\frac{3}{4}(4 \\sqrt{2}+1)$$"}, {"identifier": "C", "content": "$$\\frac{4}{3}(4 \\sqrt{2}-1)$$"}, {"identifier": "D", "content": "$$\\frac{3}{4}(4 \\sqrt{2}-1)$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh2jl21d/1eaa8b9f-d025-4054-8bc8-91786b0e9ad1/73b7e010-e6db-11ed-948f-4b963ec65c15/file-1lh2jl21e.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh2jl21d/1eaa8b9f-d025-4054-8bc8-91786b0e9ad1/73b7e010-e6db-11ed-948f-4b963ec65c15/file-1lh2jl21e.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 13th April Evening Shift Mathematics - Area Under The Curves Question 33 English Explanation">
<br><br>$$
\begin{aligned}
& \text { Required area }=2\left[\int_1^2 \sqrt{y} d y+\int_4^2 \sqrt{4-y} d y\right] \\\\
& \left.\left.=2\left[\frac{y^{3 / 2}}{\frac{3}{2}}\right]_1^2-\frac{2(4-y)^{3 / 2}}{3}\right]_2^4\right] \\\\
& =\frac{4}{3}(4 \sqrt{2}-1)
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-evening-shift |
1lgpxipld | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the curve
$$f(x)=\max \{\sin x, \cos x\},-\pi \leq x \leq \pi$$ and the $$x$$-axis is</p> | [{"identifier": "A", "content": "$$2 \\sqrt{2}(\\sqrt{2}+1)$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "$$2(\\sqrt{2}+1)$$"}, {"identifier": "D", "content": "$$4(\\sqrt{2})$$"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lh3urauz/df812b2b-41d3-412b-9523-6e99a63e5a07/ee7558b0-e793-11ed-9b1f-65ec4fb7911b/file-1lh3urav0.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lh3urauz/df812b2b-41d3-412b-9523-6e99a63e5a07/ee7558b0-e793-11ed-9b1f-65ec4fb7911b/file-1lh3urav0.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 13th April Morning Shift Mathematics - Area Under The Curves Question 32 English Explanation">
<br><br>
Area =
<br><br>= $$
\begin{aligned}
&\left|\int_{-\pi}^{-3 \pi / 4} \sin x d x\right|+\left|\int_{\pi / 4}^\pi \sin x d x\right|+\left|\int_{-3 \pi / 4}^{-\pi / 2} \cos x d x\right|
+\left|\int_{-\pi / 2}^{\pi / 4} \cos x d x\right|
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& =\left|\frac{1}{\sqrt{2}}-1\right|+\left|1+\frac{1}{\sqrt{2}}\right|+\left|-1+\frac{1}{\sqrt{2}}\right|+\left|\frac{1}{\sqrt{2}}+1\right| \\\\
& = 2+\frac{2}{\sqrt{2}}+2-\frac{2}{\sqrt{2}}=4
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgrgc23k | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the curve $$y=x^{3}$$ and its tangent at the point $$(-1,-1)$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{23}{4}$$"}, {"identifier": "B", "content": "$$\\frac{19}{4}$$"}, {"identifier": "C", "content": "$$\\frac{27}{4}$$"}, {"identifier": "D", "content": "$$\\frac{31}{4}$$<br/><br/>"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1li62ldz6/e3e1ec7b-7f4a-4711-90a5-23ea1380cf4d/5bb8a710-fc98-11ed-bb91-ad55392bec91/file-1li62ldz7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1li62ldz6/e3e1ec7b-7f4a-4711-90a5-23ea1380cf4d/5bb8a710-fc98-11ed-bb91-ad55392bec91/file-1li62ldz7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 12th April Morning Shift Mathematics - Area Under The Curves Question 31 English Explanation">
<br><br>Given $y=x^3$
<br><br>$$
\begin{aligned}
& \Rightarrow \frac{d y}{d x}=3 x^2 \\\\
& \left(\frac{d y}{d x}\right)_{(-1,-1)}=3
\end{aligned}
$$
<br><br>Equation of tangent at $(-1,-1)$
<br><br>$$
\begin{aligned}
& (y+1)=3(x+1) \\\\
& y=3 x+2
\end{aligned}
$$
<br><br>Solving (i) and (ii)
<br><br>$$
x^3=3 x+2
$$
<br><br>We can rewrite this as $x^3 - 3x - 2 = 0$. This equation will give us the x-coordinates of the points where the line $y = 3x + 2$ intersects the curve $y = x^3$. We already know that one point of intersection is $(-1, -1)$ (as it's a point of tangency).
<br><br>To find the other points of intersection, we can solve the cubic equation $x^3 - 3x - 2 = 0$.
<br><br>Let's factor this equation:
<br><br>$x^3 - 3x - 2 = 0 \Rightarrow (x + 1)(x + 1)(x - 2) = 0$.
<br><br>Setting each factor equal to zero gives the solutions $x = -1, -1, 2$.
<br><br>$$ \therefore $$ Other point is $Q(2,8)$.
<br><br>
We have the required area as
$$
\int_{-1}^{2} (3x + 2 - x^3) \, dx = \left[ \frac{3}{2}x^2 + 2x - \frac{1}{4}x^4 \right]_{-1}^{2}.
$$
<br><br>Evaluating this at $x = 2$ and $x = -1$ gives
<br><br>$$
\left[ \frac{3}{2} \cdot (2)^2 + 2 \cdot 2 - \frac{1}{4} \cdot (2)^4 \right] - \left[ \frac{3}{2} \cdot (-1)^2 + 2 \cdot (-1) - \frac{1}{4} \cdot (-1)^4 \right] $$
<br><br>$$
= \left[ 6 + 4 - 4 \right] - \left[ \frac{3}{2} - 2 - \frac{1}{4} \right] $$
<br><br>$$
= 6 - \left( \frac{1}{4} \right) $$
<br><br>$$
= \frac{27}{4}
$$
| mcq | jee-main-2023-online-12th-april-morning-shift |
1lgswa4aj | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If A is the area in the first quadrant enclosed by the curve $$\mathrm{C: 2 x^{2}-y+1=0}$$, the tangent to $$\mathrm{C}$$ at the point $$(1,3)$$ and the line $$\mathrm{x}+\mathrm{y}=1$$, then the value of $$60 \mathrm{~A}$$ is _________.</p> | [] | null | 16 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lie8i8gh/9671b86e-ee06-496d-abb6-609a695b7ace/c0004c10-0115-11ee-9f57-5de63d0f488b/file-1lie8i8gi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lie8i8gh/9671b86e-ee06-496d-abb6-609a695b7ace/c0004c10-0115-11ee-9f57-5de63d0f488b/file-1lie8i8gi.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 11th April Evening Shift Mathematics - Area Under The Curves Question 30 English Explanation">
<br><br>$$
y=2 x^2+1
$$
<br><br>Tangent at (1, 3) : $y-3=4(x-1)$
<br><br>$$
y=4 x-1
$$
<br><br>$\mathrm{A}=\int\limits_0^1\left(2 \mathrm{x}^2+1\right) \mathrm{dx}-$ area of $(\Delta \mathrm{QOT})-$ area of $(\Delta \mathrm{PQR})+$ area of $(\Delta \mathrm{QRS})$
<br><br>$$
\mathrm{A}=\left(\frac{2}{3}+1\right)-\frac{1}{2}-\frac{9}{8}+\frac{9}{40}=\frac{16}{60}
$$
<br><br>$$ \therefore $$ $$ 60 A=16 $$ | integer | jee-main-2023-online-11th-april-evening-shift |
1lguvhkpl | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Area of the region $$\left\{(x, y): x^{2}+(y-2)^{2} \leq 4, x^{2} \geq 2 y\right\}$$ is</p> | [{"identifier": "A", "content": "$$2 \\pi+\\frac{16}{3}$$"}, {"identifier": "B", "content": "$$\\pi-\\frac{8}{3}$$"}, {"identifier": "C", "content": "$$\\pi+\\frac{8}{3}$$"}, {"identifier": "D", "content": "$$2 \\pi-\\frac{16}{3}$$"}] | ["D"] | null | We have,
<br><br>$$
x^2+(y-2)^2 \leq 2^2 \text { and } x^2 \geq 2 y
$$
<br><br>On solving the given equation of parabola and circle, we get
<br><br>$$
\begin{aligned}
2 y+(y-2)^2 & =4 \\\\
\Rightarrow y =0 \text { or } 2
\end{aligned}
$$
<br><br>If $y=0$, then $x=0$ and
<br><br>If $y=2$, then $x= \pm 2$
<br><br>$\therefore$ The intersecting point of circle and parabola are $(0,0),(-2,2),(2,2)$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5indhw/a6c6d162-df13-4d2e-9e8e-b7936ef392f6/714f6750-5f47-11ee-8783-9144a2081c14/file-6y3zli1ln5indj9.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ln5indhw/a6c6d162-df13-4d2e-9e8e-b7936ef392f6/714f6750-5f47-11ee-8783-9144a2081c14/file-6y3zli1ln5indj9.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Area Under The Curves Question 29 English Explanation 1">
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ln5jlzv2/c746028e-399f-4439-a1c3-68cff047545d/34119ee0-5f4b-11ee-8f55-db3179c35b35/file-6y3zli1ln5jlzv3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ln5jlzv2/c746028e-399f-4439-a1c3-68cff047545d/34119ee0-5f4b-11ee-8f55-db3179c35b35/file-6y3zli1ln5jlzv3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 11th April Morning Shift Mathematics - Area Under The Curves Question 29 English Explanation 2">
<br><br>Area of the shaded region = $$
=2 \times 2-\frac{1}{4} \cdot \pi \cdot 2^2=4-\pi
$$
<br><br>$$
\begin{aligned}
\text { Required area } & =2\left[\int\limits_0^2 \frac{x^2}{2} d x-(4-\pi)\right] \\\\
& =2\left[\left.\frac{x^3}{6}\right|_0 ^2-4+\pi\right] \\\\
& =2\left[\frac{4}{3}+\pi-4\right] \\\\
& =2\left[\pi-\frac{8}{3}\right] \\\\
& =2 \pi-\frac{16}{6}
\end{aligned}
$$ | mcq | jee-main-2023-online-11th-april-morning-shift |
1lgvqkz5x | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\left\{(x, \mathrm{y}):\left|x^{2}-2\right| \leq y \leq x\right\}$$ is $$\mathrm{A}$$, then $$6 \mathrm{A}+16 \sqrt{2}$$ is equal to __________.</p> | [] | null | 27 | $$
\text { We have, }\left\{(x, y):\left|x^2-2\right| \leq y \leq x\right\}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnk7a0xn/c3e34f25-2aa9-4803-b342-54bfe7409450/bb60e0b0-675a-11ee-a06a-699a057b80c4/file-6y3zli1lnk7a0xo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnk7a0xn/c3e34f25-2aa9-4803-b342-54bfe7409450/bb60e0b0-675a-11ee-a06a-699a057b80c4/file-6y3zli1lnk7a0xo.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Evening Shift Mathematics - Area Under The Curves Question 28 English Explanation">
<br><br>On solving, $\left|x^2-2\right|=y$ and $y=x$, we get $(1,1)$ and $(2,2)$
<br><br>$$
\begin{aligned}
& A=\int\limits_1^{\sqrt{2}}\left(x-\left(2-x^2\right)\right) d x+\int\limits_{\sqrt{2}}^2\left(x-\left(x^2-2\right)\right) d x \\\\
& A=\left[\frac{x^2}{2}-2 x+\frac{x^3}{3}\right]_1^{\sqrt{2}}+\left[\frac{x^2}{2}-\frac{x^3}{3}+2 x\right]_{\sqrt{2}}^2
\end{aligned}
$$
<br><br>$$
\begin{aligned}
A=\left(1-2 \sqrt{2}+\frac{2 \sqrt{2}}{3}\right)-\left(\frac{1}{2}-2+\frac{1}{3}\right)+ & \left(2-\frac{8}{3}+4\right) \\
& -\left(1-\frac{2 \sqrt{2}}{3}+2 \sqrt{2}\right)
\end{aligned}
$$
<br><br>$$
\begin{aligned}
& A=-4 \sqrt{2}+\frac{4 \sqrt{2}}{3}+\frac{7}{6}+\frac{10}{3} \\\\
& A=\frac{-8 \sqrt{2}}{3}+\frac{9}{2} \Rightarrow A=\frac{-16 \sqrt{2}+27}{6} \\\\
& 6 A=-16 \sqrt{2}+27 \\\\
& 6 A+16 \sqrt{2}=27
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-evening-shift |
1lgxw9wy2 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let $$y = p(x)$$ be the parabola passing through the points $$( - 1,0),(0,1)$$ and $$(1,0)$$. If the area of the region $$\{ (x,y):{(x + 1)^2} + {(y - 1)^2} \le 1,y \le p(x)\} $$ is A, then $$12(\pi - 4A)$$ is equal to ___________.</p> | [] | null | 16 | Let, $y=p(x)$ be the parabola passing through the points $(-1,0)(0,1)(1,0)$.
<br><br>Now, to find the area of the region
<br><br>$$
\left\{(x, y) ;(x+1)^2+(y-1)^2 \leq 1, y \leq p(x)\right\}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnda9au3/d6943506-1c0a-4dc0-87a7-f279a28202cc/e60f2eb0-638c-11ee-9778-43ae0741eedc/file-6y3zli1lnda9au4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnda9au3/d6943506-1c0a-4dc0-87a7-f279a28202cc/e60f2eb0-638c-11ee-9778-43ae0741eedc/file-6y3zli1lnda9au4.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Area Under The Curves Question 27 English Explanation">
<br><br>Thus, area of shaded region is
<br><br>$$
\begin{aligned}
& A =\int\limits_{-1}^0\left\{\left(1-x^2\right)-\left(1-\sqrt{1-(x+1)^2}\right\} d x\right. \\\\
& =\int\limits_{-1}^0\left\{-x^2+\sqrt{1-(x+1)^2}\right\} d x \\\\
& =\int\limits_{-1}^0-x^2 d x+\int_{-1}^0 \sqrt{1-(x+1)^2} d x \\\\
& =-\left(\frac{x^3}{3}\right)_{-1}^0+\left[\frac{\sqrt{1-(x+1)^2}}{2}+\frac{1}{2} \sin ^{-1}(x+1)\right]_{-1}^0 \\\\
& =-\frac{1}{3}+\left[\frac{1}{2}+\frac{\pi}{4}-\frac{1}{2}\right] \\\\
& A =\frac{\pi}{4}-\frac{1}{3}
\end{aligned}
$$
<br><br>Now, $12(\pi-4 A)$
<br><br>$$
\begin{aligned}
& =12\left[\pi-4\left(\frac{\pi}{4}-\frac{1}{3}\right)\right] \\\\
& =12\left(\pi-\pi+\frac{4}{3}\right)=16
\end{aligned}
$$ | integer | jee-main-2023-online-10th-april-morning-shift |
1lgyom5cr | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area enclosed by the lines $$x+y=2, \mathrm{y}=0, x=0$$ and the curve $$f(x)=\min \left\{x^{2}+\frac{3}{4}, 1+[x]\right\}$$ where $$[x]$$
denotes the greatest integer $$\leq x$$, be $$\mathrm{A}$$. Then the value of $$12 \mathrm{~A}$$ is _____________.</p> | [] | null | 17 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmyue4ph/cd17397d-b57f-4fb7-a489-8f746007324f/e1368850-5b9b-11ee-83da-a3f80d422da4/file-6y3zli1lmyue4pi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lmyue4ph/cd17397d-b57f-4fb7-a489-8f746007324f/e1368850-5b9b-11ee-83da-a3f80d422da4/file-6y3zli1lmyue4pi.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 8th April Evening Shift Mathematics - Area Under The Curves Question 26 English Explanation">
<br><br>$$
\text { Required area }=\left[\int\limits_0^{\frac{1}{2}}\left(x^2+\frac{3}{4}\right) d x\right]+\left[\frac{1}{2}\left(\frac{3}{2}+\frac{1}{2}\right) \times 1\right]
$$
<br><br>$$
\begin{aligned}
& =\left[\frac{x^3}{3}+\frac{3 x}{4}\right]_0^{\frac{1}{2}}+1 \\\\
& =\frac{1}{24}+\frac{3}{8}-0+1=\frac{1+9+24}{24}=\frac{34}{24}=\frac{17}{12}
\end{aligned}
$$
<br><br>$$
\text { So, } 12 \mathrm{~A}=12 \times \frac{17}{12}=17
$$ | integer | jee-main-2023-online-8th-april-evening-shift |
1lgzyidm2 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$\left\{(x, y): x^{2} \leq y \leq 8-x^{2}, y \leq 7\right\}$$ is :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "21"}] | ["C"] | null | The given curves are
<br><br>$x^2 \leq y, y \leq 8-x^2 ; y \leq 7$
<br><br>On solving, we get $x^2=8-x^2$
<br><br>$$
\begin{aligned}
& \Rightarrow x^2=4 \\\\
& \Rightarrow x= \pm 2
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ljg19fl1/2d19d263-28f7-4e45-975b-04e64a691566/33ff3e40-15df-11ee-bf54-2dbadff89573/file-6y3zli1ljg19fl2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ljg19fl1/2d19d263-28f7-4e45-975b-04e64a691566/33ff3e40-15df-11ee-bf54-2dbadff89573/file-6y3zli1ljg19fl2.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 8th April Morning Shift Mathematics - Area Under The Curves Question 25 English Explanation">
<br><br>$$
\text { So, area }=2\left[\int_0^4 \sqrt{y} d y+\int_4^7 \sqrt{8-y} d y\right]
$$
<br><br>$$
=2\left\{\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_0^4+\left[\frac{-(8-y)^{\frac{3}{2}}}{\frac{3}{2}}\right]_4^7\right\}
$$
<br><br>$$
\begin{aligned}
& =2 \times \frac{2}{3}\left\{\left[4^{3 / 2}-0\right]+\left(-(1)^{3 / 2}+(4)^{3 / 2}\right)\right\} \\\\
& =\frac{4}{3}\{8-1+8\}=\frac{4}{3} \times 15=20 \text { sq. units }
\end{aligned}
$$ | mcq | jee-main-2023-online-8th-april-morning-shift |
1lh23quiq | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$S=\left\{(x, y): 2 y-y^{2} \leq x^{2} \leq 2 y, x \geq y\right\}$$ is equal to $$\frac{n+2}{n+1}-\frac{\pi}{n-1}$$, then the natural number $$n$$ is equal to ___________.</p> | [] | null | 5 | Given region,
<br><br>$$
S=\left\{(x, y): 2 y-y^2 \leq x^2 \leq 2 y, x \geq y\right\}
$$
<br><br>Here, we have three curves
<br><br>$$
\begin{aligned}
&2 y-y^2 =x^2 ..........(i)\\\\
&x^2 =2 y ..........(2)\\\\
&\text {and}~~ x = y ...........(3)
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnuiszjt/77b5ee73-13d9-423b-bb94-fc3ea2d14833/a598d780-6d07-11ee-a732-754c539ba542/file-6y3zli1lnuiszju.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnuiszjt/77b5ee73-13d9-423b-bb94-fc3ea2d14833/a598d780-6d07-11ee-a732-754c539ba542/file-6y3zli1lnuiszju.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Morning Shift Mathematics - Area Under The Curves Question 24 English Explanation">
<br><br>From Eq. (i),
<br><br>$$
x^2+y^2-2 y=0
$$
<br><br>$\Rightarrow x^2+(y-1)^2=1$ is a circle with centre $(0,1)$ and radius is 1.
<br><br>Intersection points of Equations (i) and (iii) are,
<br><br>$$
\begin{array}{rlrl}
& 2 x-x^2 =x^2 \\\\
&\Rightarrow 2 x^2 =2 x \\\\
&\Rightarrow x(x-1) =0 \\\\
&\Rightarrow x =0,1 \\\\
& \text { When, } x=0 \text {, then } y=0 \\\\
& \text { and when } x=1 \text {, then } y=1
\end{array}
$$
<br><br>$\therefore$ Required points are $(0,0)$ and $(1,1)$.
<br><br>Also, intersection points of (ii) and (iii)
<br><br>$$
\begin{aligned}
& \quad x^2=2 x \\\\
& \Rightarrow x(x-2)=0 \\\\
& \Rightarrow x=0,2 \\\\
& \text { When, } x=0 \text {, then } y=0 \\\\
& \text { and when } x=2 \text {, then } y=2 \\\\
& \therefore \text { Required points are }(0,0) \text { and }(2,2)
\end{aligned}
$$
<br><br>$\therefore$ Required area $=$ Area of triangle - Area of part I - Area of part II
<br><br>$$
=\frac{1}{2} \times 2 \times 2-\int\limits_0^2 \frac{x^2}{2}-\left(\frac{\pi}{4}-\frac{1}{2}\right)
$$
<br><br>$$
\begin{aligned}
& =2-\frac{1}{2}\left(\frac{x^3}{3}\right)_0^2-\frac{\pi}{4}+\frac{1}{2} \\\\
& =2-\frac{1}{2} \times \frac{8}{3}-\frac{\pi}{4}+\frac{1}{2} \\\\
& =\frac{7}{6}-\frac{\pi}{4} \\\\
& =\frac{5+2}{5+1}-\frac{\pi}{5-1} \\\\
& \therefore n =5
\end{aligned}
$$ | integer | jee-main-2023-online-6th-april-morning-shift |
1lh2xt7m5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area bounded by the curves $$y=|x-1|+|x-2|$$ and $$y=3$$ is equal to :</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "3"}] | ["B"] | null | Given equation of curve $y=|x-1|+|x-2|$ and $y=3$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lo8c5sm6/ed381416-4cb6-4a32-b174-452035f058cf/a2f26cd0-74a0-11ee-8723-4d48ce392782/file-6y3zli1lo8c5sm7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lo8c5sm6/ed381416-4cb6-4a32-b174-452035f058cf/a2f26cd0-74a0-11ee-8723-4d48ce392782/file-6y3zli1lo8c5sm7.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 6th April Evening Shift Mathematics - Area Under The Curves Question 23 English Explanation">
<br><br>When, $x-1=0 \Rightarrow x=1$, then $y=1$
<br><br>and when, $x-2=0 \Rightarrow x=2$, then $y=1$
<br><br>Equation of curve passing through point $(1,1)$ and $(2,1)$ and $(0,3),(3,3)$
<br><br>$\therefore$ Area bounded by given curves
<br><br>$$
=\frac{1}{2}(1+3) \times 2(\text { Area of trapezium })=4
$$ | mcq | jee-main-2023-online-6th-april-evening-shift |
lsan0p30 | maths | area-under-the-curves | area-bounded-between-the-curves | Three points $\mathrm{O}(0,0), \mathrm{P}\left(\mathrm{a}, \mathrm{a}^2\right), \mathrm{Q}\left(-\mathrm{b}, \mathrm{b}^2\right), \mathrm{a}>0, \mathrm{~b}>0$, are on the parabola $y=x^2$. Let $\mathrm{S}_1$ be the area of the region bounded by the line $\mathrm{PQ}$ and the parabola, and $\mathrm{S}_2$ be the area of the triangle $\mathrm{OPQ}$. If the minimum value of $\frac{\mathrm{S}_1}{\mathrm{~S}_2}$ is $\frac{\mathrm{m}}{\mathrm{n}}, \operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$, then $\mathrm{m}+\mathrm{n}$ is equal to __________. | [] | null | 7 | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsogvt13/49cb1ddc-aaf9-4143-8cc4-b6d4940befaa/82ba6c70-ccb0-11ee-aa98-13f456b8f7af/file-6y3zli1lsogvt14.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsogvt13/49cb1ddc-aaf9-4143-8cc4-b6d4940befaa/82ba6c70-ccb0-11ee-aa98-13f456b8f7af/file-6y3zli1lsogvt14.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Evening Shift Mathematics - Area Under The Curves Question 22 English Explanation">
<br><br>Equation of $P Q$
<br><br>$$
\begin{aligned}
& y+b=\frac{a^2-b^2}{a+b}\left(x-b^2\right) \\\\
& \Rightarrow y=(a-b) x+a b
\end{aligned}
$$
<br><br>$\begin{aligned} & S_1=\int\limits_{-b}^a\left((a-b) x+a b-x^2\right) d x \\\\ & =\left(\frac{(a-b)}{2} x^2+a b x-\frac{x^3}{3}\right)_{-b}^a \\\\ & =\frac{1}{6}(a+b)^3\end{aligned}$
<br><br>$S_2=\frac{1}{2}\left|\begin{array}{ccc}-b & b^2 & 1 \\ 0 & 0 & 1 \\ a & a^2 & 1\end{array}\right|=\frac{1}{2} a b(a+b)$
<br><br>$\begin{aligned} \frac{S_1}{S_2} & =\frac{\frac{1}{6}(a+b)^3}{\frac{1}{2} \cdot a b(a+b)} \\\\ & =\frac{1}{3} \frac{(a+b)^2}{a b}=\frac{1}{3}\left(\frac{a}{b}+\frac{b}{a}+2\right)\end{aligned}$
<br><br>$\begin{aligned} & \because \frac{b}{a}+\frac{a}{b} \geq 2 \\\\ & \Rightarrow\left(\frac{S_1}{S_2}\right)_{\min }=\frac{4}{3}=\frac{m}{n} \\\\ & \Rightarrow m+n=7\end{aligned}$ | integer | jee-main-2024-online-1st-february-evening-shift |
lsan1ar0 | maths | area-under-the-curves | area-bounded-between-the-curves | The sum of squares of all possible values of $k$, for which area of the region bounded by the parabolas $2 y^2=\mathrm{k} x$ and $\mathrm{ky}^2=2(y-x)$ is maximum, is equal to : | [] | null | 8 | Given $k y^2=2(y-x)$ .........(i)
<br/><br/>$$
2 y^2=k x
$$ .........(ii)
<br/><br/>Point of intersection of (i) and (ii)
<br/><br/>$$
\begin{aligned}
& k y^2=2\left(y-\frac{2 y^2}{k}\right) \\\\
& \Rightarrow y=0, k y=2\left(1-\frac{2 y}{k}\right)
\end{aligned}
$$
<br/><br/>$\begin{aligned} & k y+\frac{4 y}{k}=2 \\\\ & y=\frac{2}{k+\frac{4}{k}}=\frac{2 k}{k^2+4}\end{aligned}$
<br/><br/>$\begin{aligned} & A=\int\limits_6^{\frac{2 k}{k^2+4}}\left(\left(y-\frac{k y^2}{2}\right)-\frac{2 y^2}{k}\right) d y \\\\ & A=\left[\frac{y^2}{2}-\left(\frac{k}{2}+\frac{2}{k}\right) \frac{y^3}{3}\right]_0^{\frac{2 k}{k^2+4}}\end{aligned}$
<br/><br/>$\begin{aligned} & =\left(\frac{2 k}{k^2+4}\right)^2\left[\frac{1}{2}-\frac{k^2+4}{2 k}\left(\frac{1}{3}\right)\left(\frac{2 k}{k^2+4}\right)\right] \\\\ & =\frac{1}{6} \times 4 \times\left(\frac{1}{k+\frac{4}{k}}\right)^2\end{aligned}$
<br/><br/>$\begin{aligned} & \text { A.M. } \geq \text { G.M. } \\\\ & \frac{\left(k+\frac{4}{k}\right)}{2} \geq 2 \\\\ & k+\frac{4}{k} \geq 4\end{aligned}$
<br/><br/>$\therefore$ Area is maximum when $k=\frac{4}{k}$
<br/><br/>$$
\begin{aligned}
& \therefore k^2=4 \\\\
& k= \pm 2 \\\\
& k_1=2, k_2=-2 \\\\
& \therefore k_1^2+k_2^2=(+2)^2+(-2)^2 \\\\
& =4+4 \\\\
& =08
\end{aligned}
$$ | integer | jee-main-2024-online-1st-february-evening-shift |
lsaonhys | maths | area-under-the-curves | area-bounded-between-the-curves | The area enclosed by the curves $x y+4 y=16$ and $x+y=6$ is equal to : | [{"identifier": "A", "content": "$28-30 \\log _{\\mathrm{e}} 2$"}, {"identifier": "B", "content": "$30-28 \\log _{\\mathrm{e}} 2$"}, {"identifier": "C", "content": "$30-32 \\log _{\\mathrm{e}} 2$"}, {"identifier": "D", "content": "$32-30 \\log _{\\mathrm{e}} 2$"}] | ["C"] | null | <p>To find the enclosed area between the two curves $ x y+4 y=16 $ and $ x+y=6 $, we need to determine the region of intersection and integrate the difference of the functions over the interval where they intersect.</p>
<p>First, let's solve the equations simultaneously to find the points of intersection.</p>
<p>The second curve $ x+y=6 $ can be rewritten as $ y=6-x $.</p>
<p>Substitute $ y=6-x $ into the first equation:</p>
<p>$ x(6-x) + 4(6-x) = 16$</p>
<p>$ 6x - x^2 + 24 - 4x = 16 $</p>
<p>$ -x^2 + 2x + 8 = 0 $</p>
<p>This can be simplified to:</p>
<p>$ x^2 - 2x - 8 = 0 $</p>
<p>Using the quadratic formula, we can find the roots of this equation:</p>
<p>$ x = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-8)}}{2(1)} $</p>
<p>$ x = \frac{2 \pm \sqrt{4 + 32}}{2} $</p>
<p>$ x = \frac{2 \pm \sqrt{36}}{2} $</p>
<p>$ x = \frac{2 \pm 6}{2} $</p>
<p>So we have two solutions:</p>
<p>$ x_1 = \frac{2 + 6}{2} = 4 $</p>
<p>$ x_2 = \frac{2 - 6}{2} = -2 $</p>
<p>For $ x_1 = 4 $, substitute this back into the line equation $ x+y=6 $:</p>
<p>$ 4+y=6 $</p>
<p>$ y=2 $</p>
<p>For $ x_2 = -2 $, do the same:</p>
<p>$ -2+y=6 $</p>
<p>$ y=8 $</p>
<p>So we have two points of intersection: $ (4,2) $ and $ (-2,8) $.</p>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1ltaoieus/c5d8f2a6-f8e6-40bf-901f-208d94d79f63/95946740-d8e7-11ee-a868-79761dafd483/file-6y3zli1ltaoieut.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1ltaoieus/c5d8f2a6-f8e6-40bf-901f-208d94d79f63/95946740-d8e7-11ee-a868-79761dafd483/file-6y3zli1ltaoieut.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 1st February Morning Shift Mathematics - Area Under The Curves Question 20 English Explanation">
<p>Now, to find the area between $ x y + 4y = 16 $ and $ x + y = 6 $, we'll integrate the top function minus the bottom function from $ x = -2 $ to $ x = 4 $. First, we need to express $ y $ from both equations in terms of $ x $:</p>
<p>For $ x y + 4 y = 16 $, express $ y $ in terms of $ x $:</p>
<p>$ y(x+4) = 16 $</p>
<p>$ y = \frac{16}{x+4} $</p>
<p>For $ x + y = 6 $, we have:</p>
<p>$ y = 6 - x $</p>
<p>The area $ A $ is therefore given by:</p>
<p>$ A = \int\limits_{-2}^{4}\left( 6 - x - \frac{16}{x+4} \right) dx $</p>
<p>Now, we integrate:</p>
<p>$ A = \int\limits_{-2}^{4} (6 - x) dx - \int_{-2}^{4} \frac{16}{x + 4} dx $</p>
<p>$ A = \left[6x - \frac{x^2}{2}\right]_{-2}^{4} - \left[16 \ln|x + 4|\right]_{-2}^{4} $</p>
<p>$ A = \left( 6 \times 4 - \frac{1}{2} \times 4^2 \right) - \left( 6 \times -2 - \frac{1}{2} \times -2^2 \right) - 16 \left( \ln|4+4| - \ln|-2+4| \right) $</p>
<p>$ A = \left( 24 - 8 \right) - \left( -12 - 2 \right) - 16 \left( \ln 8 - \ln 2 \right) $</p>
<p>$ A = 16 + 14 - 16 \left( \ln (2^3) - \ln 2 \right) $</p>
<p>$ A = 30 - 16 \left( 3 \ln 2 - \ln 2 \right) $</p>
<p>$ A = 30 - 16 (2 \ln 2) $</p>
<p>$ A = 30- 32 \ln 2 $</p>
| mcq | jee-main-2024-online-1st-february-morning-shift |
lsblh7jy | maths | area-under-the-curves | area-bounded-between-the-curves | Let the area of the region $\left\{(x, y): x-2 y+4 \geqslant 0, x+2 y^2 \geqslant 0, x+4 y^2 \leq 8, y \geqslant 0\right\}$ be $\frac{\mathrm{m}}{\mathrm{n}}$, where $\mathrm{m}$ and $\mathrm{n}$ are coprime numbers. Then $\mathrm{m}+\mathrm{n}$ is equal to _____________. | [] | null | 119 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1g6xtj/f303b300-000e-4c93-be80-f4117fb2f938/ffa44560-d3d3-11ee-a874-bd0f61840084/file-1lt1g6xtk.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1g6xtj/f303b300-000e-4c93-be80-f4117fb2f938/ffa44560-d3d3-11ee-a874-bd0f61840084/file-1lt1g6xtk.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Morning Shift Mathematics - Area Under The Curves Question 19 English Explanation"></p>
<p>$$\begin{aligned}
& A=\int_\limits0^1\left[\left(8-4 y^2\right)-\left(-2 y^2\right)\right] d y+ \\
& \int_\limits1^{3 / 2}\left[\left(8-4 y^2\right)-(2 y-4)\right] d y \\
& =\left[8 y-\frac{2 y^3}{3}\right]_0^1+\left[12 y-y^2-\frac{4 y^3}{3}\right]_1^{3 / 2}=\frac{107}{12}=\frac{m}{n} \\
& \therefore m+n=119
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
jaoe38c1lscojzx5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\left\{(x, y): 0 \leq y \leq \min \left\{2 x, 6 x-x^2\right\}\right\}$$ is $$\mathrm{A}$$, then $$12 \mathrm{~A}$$ is equal to ________.</p> | [] | null | 304 | <p>We have</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1wj8og/d813b308-3b51-4315-93b6-2504383585b1/e7ce7b00-d413-11ee-b9d5-0585032231f0/file-1lt1wj8oh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1wj8og/d813b308-3b51-4315-93b6-2504383585b1/e7ce7b00-d413-11ee-b9d5-0585032231f0/file-1lt1wj8oh.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 27th January Evening Shift Mathematics - Area Under The Curves Question 18 English Explanation"></p>
<p>$$\begin{aligned}
& A=\frac{1}{2} \times 4 \times 8+\int_\limits4^6\left(6 x-x^2\right) d x \\
& A=\frac{76}{3} \\
& 12 A=304
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsd4h6fi | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the parabolas $$y=4 x-x^2$$ and $$3 y=(x-4)^2$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{32}{9}$$\n"}, {"identifier": "B", "content": "$$\\frac{14}{3}$$"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "6"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwhl5f/5e904568-fdcb-45b9-b0dc-a84be5bf8837/9e21fa30-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwhl5g.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwhl5f/5e904568-fdcb-45b9-b0dc-a84be5bf8837/9e21fa30-ca2d-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwhl5g.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Evening Shift Mathematics - Area Under The Curves Question 17 English Explanation"></p>
<p>$$\begin{aligned}
& \text { Area }=\left\lvert\, \int_\limits1^4\left[\left(4 x-x^2\right)-\frac{(x-4)^2}{3}\right] d x\right. \\
& \text { Area }=\left|\frac{4 x^2}{2}-\frac{x^3}{3}-\frac{(x-4)^3}{9}\right|_1^4 \\
& =\left|\left(\frac{64}{2}-\frac{64}{3}-\frac{4}{2}+\frac{1}{3}-\frac{27}{9}\right)\right| \\
& \Rightarrow(27-21)=6
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lse5f37p | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region $$\left\{(x, y): y^2 \leq 4 x, x<4, \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0, x \neq 3\right\}$$ is</p> | [{"identifier": "A", "content": "$$\\frac{32}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{16}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{8}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{64}{3}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& y^2 \leq 4 x, x<4 \\
& \frac{x y(x-1)(x-2)}{(x-3)(x-4)}>0 \\
& \text { Case - I}: y>0 \\
& \frac{x(x-1)(x-2)}{(x-3)(x-4)}>0 \\
& x \in(0,1) \cup(2,3) \\
& \text { Case }- \text { II : y<0 } \\
& \frac{x(x-1)(x-2)}{(x-3)(x-4)}<0, x \in(1,2) \cup(3,4)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsnjpfob/e2731f55-b6f3-45cf-bfd6-95a36d5f1043/c44b86b0-cc2e-11ee-b20d-39b621d226e3/file-6y3zli1lsnjpfoc.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsnjpfob/e2731f55-b6f3-45cf-bfd6-95a36d5f1043/c44b86b0-cc2e-11ee-b20d-39b621d226e3/file-6y3zli1lsnjpfoc.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 31st January Morning Shift Mathematics - Area Under The Curves Question 16 English Explanation"></p>
<p>$$\begin{aligned}
& \text { Area }=2 \int_\limits0^4 \sqrt{x} d x \\
& =2 \cdot \frac{2}{3}\left[x^{3 / 2}\right]_0^4=\frac{32}{3}
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
jaoe38c1lsf0hs1k | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in sq. units) of the part of the circle $$x^2+y^2=169$$ which is below the line $$5 x-y=13$$ is $$\frac{\pi \alpha}{2 \beta}-\frac{65}{2}+\frac{\alpha}{\beta} \sin ^{-1}\left(\frac{12}{13}\right)$$, where $$\alpha, \beta$$ are coprime numbers. Then $$\alpha+\beta$$ is equal to __________.</p> | [] | null | 171 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt30icn5/b76421b2-6e9f-4b5a-94e8-92add4114de7/3c39af10-d4b0-11ee-8384-811001421c41/file-1lt30icn6.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt30icn5/b76421b2-6e9f-4b5a-94e8-92add4114de7/3c39af10-d4b0-11ee-8384-811001421c41/file-1lt30icn6.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Morning Shift Mathematics - Area Under The Curves Question 14 English Explanation"></p>
<p>$$\begin{aligned}
& \text { Area }=\int_\limits{-13}^{12} \sqrt{169-y^2} d y-\frac{1}{2} \times 25 \times 5 \\
& =\frac{\pi}{2} \times \frac{169}{2}-\frac{65}{2}+\frac{169}{2} \sin ^{-1} \frac{12}{13} \\
& \therefore \alpha+\beta=171
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsf0no1u | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the points of intersection of two distinct conics $$x^2+y^2=4 b$$ and $$\frac{x^2}{16}+\frac{y^2}{b^2}=1$$ lie on the curve $$y^2=3 x^2$$, then $$3 \sqrt{3}$$ times the area of the rectangle formed by the intersection points is _________.</p> | [] | null | 432 | <p>Putting $$y^2=3 x^2$$ in both the conics</p>
<p>We get $$x^2=b$$ and $$\frac{b}{16}+\frac{3}{b}=1$$</p>
<p>$$\Rightarrow \mathrm{b}=4,12 \quad(\mathrm{b}=4$$ is rejected because curves coincide)</p>
<p>$$\therefore \mathrm{b}=12$$</p>
<p>Hence points of intersection are</p>
<p>$$( \pm \sqrt{12}, \pm 6) \Rightarrow \text { area of rectangle }=432$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfl0gg5 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area of the region $$\left\{(x, y): 0 \leq x \leq 3,0 \leq y \leq \min \left\{x^2+2,2 x+2\right\}\right\}$$ be A. Then $$12 \mathrm{~A}$$ is equal to __________.</p> | [] | null | 164 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8vkvl/77732020-1aec-484b-950e-aed3c54f53c2/8d586e10-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8vkvm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8vkvl/77732020-1aec-484b-950e-aed3c54f53c2/8d586e10-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8vkvm.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 29th January Evening Shift Mathematics - Area Under The Curves Question 13 English Explanation"></p>
<p>$$\begin{aligned}
& y=2 x+2 \\
& A=\int_\limits0^2\left(x^2+2\right) d x+\int_\limits2^3(2 x+2) d x \\
& A=\frac{41}{3} \\
& 12 A=41 \times 4=164
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-evening-shift |
1lsg57scj | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the parabola $$(y-2)^2=x-1$$, the line $$x-2 y+4=0$$ and the positive coordinate axes is _________.</p> | [] | null | 5 | <p>Solving the equations</p>
<p>$$\begin{array}{r}
(y-2)^2=x-1 \text { and } x-2 y+4=0 \\
x=2(y-2)
\end{array}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxplac/f4bdc645-97cb-4588-be1e-e674c4cd7e92/51166c40-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxplad.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxplac/f4bdc645-97cb-4588-be1e-e674c4cd7e92/51166c40-ccf2-11ee-a330-494dca5e9a63/file-6y3zli1lsoxplad.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Evening Shift Mathematics - Area Under The Curves Question 12 English Explanation"></p>
<p>$$\begin{aligned}
& \frac{x^2}{4}=x-1 \\
& x^2-4 x+4=0 \\
& (x-2)^2=0 \\
& x=2
\end{aligned}$$</p>
<p>Exclose area (w.r.t. y-axis) $$=\int_\limits0^3 x d y-\text { Area of } \Delta$$.</p>
<p>$$\begin{aligned}
& =\int_\limits0^3\left((y-2)^2+1\right) d y-\frac{1}{2} \times 1 \times 2 \\
& =\int_\limits0^3\left(y^2-4 y+5\right) d y-1 \\
& =\left[\frac{y^3}{3}-2 y^2+5 y\right]_0^3-1 \\
& =9-18+15-1=5
\end{aligned}$$</p> | integer | jee-main-2024-online-30th-january-evening-shift |
1lsgb2phy | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in square units) of the region bounded by the parabola $$y^2=4(x-2)$$ and the line $$y=2 x-8$$, is :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "9"}, {"identifier": "D", "content": "6"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqnpz3w/f37f9151-5383-482a-a0a5-619fb7183b76/d1c573c0-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnpz3x.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqnpz3w/f37f9151-5383-482a-a0a5-619fb7183b76/d1c573c0-cde4-11ee-a0d3-7b75c4537559/file-6y3zli1lsqnpz3x.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 30th January Morning Shift Mathematics - Area Under The Curves Question 11 English Explanation">
<br><br>Area along $y$-axis
<br><br>$$
\begin{aligned}
& \int\limits_{-2}^4\left[\left(\frac{y+8}{2}\right)-\left(\frac{y^2+8}{4}\right)\right] d y \\\\
& =\frac{y^2}{4}+4 y-\frac{y^3}{12}-\left.2 y\right|_{-2} ^4
\end{aligned}
$$
<br><br>$\begin{aligned} & =\left(4+16-\frac{16}{3}-8\right)-\left(1-8+\frac{8}{12}+4\right) \\\\ & =\left(12-\frac{16}{3}\right)-\left(\frac{8}{12}-3\right)=15-\left(\frac{64+8}{12}\right) \\\\ & =15-6=9 \text { sq. unit }\end{aligned}$ | mcq | jee-main-2024-online-30th-january-morning-shift |
luxweb0r | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in square units) of the region enclosed by the ellipse $$x^2+3 y^2=18$$ in the first quadrant below the line $$y=x$$ is</p> | [{"identifier": "A", "content": "$$\\sqrt{3} \\pi+1$$\n"}, {"identifier": "B", "content": "$$\\sqrt{3} \\pi$$\n"}, {"identifier": "C", "content": "$$\\sqrt{3} \\pi-\\frac{3}{4}$$\n"}, {"identifier": "D", "content": "$$\\sqrt{3} \\pi+\\frac{3}{4}$$"}] | ["B"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw1i51n7/b27b05e5-86d3-44de-bbce-888961f304b0/435f7e30-0f40-11ef-a754-b58ce0eab988/file-1lw1i51n8.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw1i51n7/b27b05e5-86d3-44de-bbce-888961f304b0/435f7e30-0f40-11ef-a754-b58ce0eab988/file-1lw1i51n8.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Evening Shift Mathematics - Area Under The Curves Question 10 English Explanation"></p>
<p>$$\begin{aligned}
& \text { Area }=\int_\limits0^{3 \sqrt{2}} x d x+\int_\limits{3 / \sqrt{2}}^{3 \sqrt{2}} \sqrt{\frac{18-x^2}{3}} d x \\
& =\frac{1}{2}\left(x^2\right)_0^{3 \sqrt{2}}+\frac{1}{\sqrt{3}}\left[\frac{x}{2} \sqrt{18-x^2}+9 \sin ^{-1}\left(\frac{x}{3 \sqrt{2}}\right)\right]_{\frac{3}{\sqrt{2}}}^{3 \sqrt{2}} \\
& =\frac{1}{2}\left(\frac{9}{2}\right)+\frac{1}{\sqrt{3}}\left[9 \sin ^{-1}(1)-\frac{3}{2 \sqrt{2}} \cdot \frac{3 \sqrt{3}}{\sqrt{2}}-9 \sin ^{-1}\left(\frac{1}{2}\right)\right] \\
& =\frac{9}{4}+\frac{1}{\sqrt{3}}\left(\frac{9 \pi}{2}-\frac{9 \sqrt{3}}{4}-\frac{9 \pi}{6}\right)=\sqrt{3} \pi
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-evening-shift |
luy6z4gc | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The parabola $$y^2=4 x$$ divides the area of the circle $$x^2+y^2=5$$ in two parts. The area of the smaller part is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{2}{3}+5 \\sin ^{-1}\\left(\\frac{2}{\\sqrt{5}}\\right)$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{3}+\\sqrt{5} \\sin ^{-1}\\left(\\frac{2}{\\sqrt{5}}\\right)$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{3}+5 \\sin ^{-1}\\left(\\frac{2}{\\sqrt{5}}\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{3}+\\sqrt{5} \\sin ^{-1}\\left(\\frac{2}{\\sqrt{5}}\\right)$$"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw36wee0/c97c824c-882e-4206-aafc-00ef66493663/dfd02e70-102d-11ef-a47a-5f0284b5aece/file-1lw36wee1.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw36wee0/c97c824c-882e-4206-aafc-00ef66493663/dfd02e70-102d-11ef-a47a-5f0284b5aece/file-1lw36wee1.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 9th April Morning Shift Mathematics - Area Under The Curves Question 9 English Explanation"></p>
<p>The point of intersection of $$y^2=4 x$$ and $$x^2+y^2=5$$ are $$(1,2)$$ and $$(1,-2)$$.</p>
<p>$$\because$$ Area of smaller region bounded by $$y^2=4 x$$ and
$$x^2+y^2=5$$</p>
<p>$$=2\{\text {area of } O A C O+\text { area of } C A B C\}$$</p>
<p>$$\begin{aligned}
& =2\left[\int_\limits0^1 2 \sqrt{x} d x+\int_\limits1^{\sqrt{5}} \sqrt{5-x^2} d x\right. \\
& =2\left[\left|\frac{4}{3} x^{\frac{3}{2}}\right|_0^1+\left(\frac{1}{2} x \sqrt{5-x^2}+\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right)\right]_1^{\sqrt{5}} \\
& =2\left[\left(\frac{4}{3}-0\right)+\left(0+\frac{5 \pi}{4}\right)-\left(1+\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right)\right] \\
& =2\left[\frac{1}{3}+\frac{5 \pi}{4}-\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}\right]=\frac{2}{3}+\frac{5 \pi}{2}-5 \sin ^{-1} \frac{1}{\sqrt{5}} \\
& =\frac{2}{3}+5 \cos ^{-1}\left(\frac{1}{\sqrt{5}}\right)=\frac{2}{3}+5 \sin ^{-1}\left(\frac{2}{\sqrt{5}}\right)
\end{aligned}$$</p> | mcq | jee-main-2024-online-9th-april-morning-shift |
lv0vxbvp | maths | area-under-the-curves | area-bounded-between-the-curves | <p>One of the points of intersection of the curves $$y=1+3 x-2 x^2$$ and $$y=\frac{1}{x}$$ is $$\left(\frac{1}{2}, 2\right)$$. Let the area of the region enclosed by these curves be $$\frac{1}{24}(l \sqrt{5}+\mathrm{m})-\mathrm{n} \log _{\mathrm{e}}(1+\sqrt{5})$$, where $$l, \mathrm{~m}, \mathrm{n} \in \mathbf{N}$$. Then $$l+\mathrm{m}+\mathrm{n}$$ is equal to</p> | [{"identifier": "A", "content": "30"}, {"identifier": "B", "content": "29"}, {"identifier": "C", "content": "31"}, {"identifier": "D", "content": "32"}] | ["A"] | null | <p>Solving curves $$y=1+3 x-2 x^2 ~\& ~y=\frac{1}{x}$$</p>
<p>$$\begin{aligned}
& 2 x^3-3 x^2-x+1=0 \\
& \Rightarrow \quad(2 x-1)\left(x^2-x-1\right)=0 \\
& \Rightarrow \quad x=\frac{1}{2}, x=\frac{1 \pm \sqrt{5}}{2}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lwk3rf0n/84c1aefe-3dc9-416a-bb2f-8c36b5773026/ece80d70-197a-11ef-8f7b-89e146e4f804/file-jaoe38c1lwk3rf0o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lwk3rf0n/84c1aefe-3dc9-416a-bb2f-8c36b5773026/ece80d70-197a-11ef-8f7b-89e146e4f804/file-jaoe38c1lwk3rf0o.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Morning Shift Mathematics - Area Under The Curves Question 8 English Explanation"></p>
<p>Area $$ = \int\limits_{{1 \over 2}}^{{{1 + \sqrt 5 } \over 2}} {\left( {1 + 3x - 2{x^2} - {1 \over x}dx} \right)} $$</p>
<p>$$ = \left[ {x + {{3{x^2}} \over 2} - {{2x} \over 3} - \ln x} \right]$$</p>
<p>$$ = {{\sqrt 5 + 1} \over 2} + {3 \over 8}{(\sqrt 5 + 1)^2} - {1 \over {12}}(\sqrt 5 + 1) - \ln \left( {{{\sqrt 5 + 1} \over 2}} \right) - \left( {{1 \over 2} + {3 \over 8} - {1 \over {12}} - \ln {1 \over 2}} \right)$$</p>
<p>$$ = {1 \over {24}}\left[ {12(\sqrt 5 + 1) + 9{{(\sqrt 5 + 1)}^2} - 2(\sqrt 5 + 1) - 12 - 9 + 2] - \ln \left( {{{\sqrt 5 + 1} \over 2} \times 2} \right)} \right]$$</p>
<p>$$\begin{aligned}
= & \frac{1}{24}[12(\sqrt{5}+1)+9(6+2 \sqrt{5})- \\
& 2(5 \sqrt{5}+1+3 \sqrt{5}(\sqrt{5}+1)-19)]-\ln (\sqrt{5}+1) \\
= & \frac{1}{24}[14 \sqrt{5}+15]-\ln (\sqrt{5}+1) \\
\therefore \quad & \quad l=14, m=15, n=1 \\
& \quad I+m+n=30
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2er3kw | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area (in sq. units) of the region described by $$
\left\{(x, y): y^2 \leq 2 x \text {, and } y \geq 4 x-1\right\}
$$ is</p> | [{"identifier": "A", "content": "$$\\frac{9}{32}$$\n"}, {"identifier": "B", "content": "$$\\frac{11}{12}$$\n"}, {"identifier": "C", "content": "$$\\frac{8}{9}$$\n"}, {"identifier": "D", "content": "$$\\frac{11}{32}$$"}] | ["A"] | null | <p>$$\text { Area }=\int_\limits{-\frac{1}{2}}^1\left(\frac{y+1}{4}-\frac{y^2}{2}\right) d y$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhbnwbo/8e96ed5b-9608-42df-8bc6-dc691e8b1a15/7a1fa240-17f3-11ef-b996-7dd2f40f5f82/file-1lwhbnwbp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhbnwbo/8e96ed5b-9608-42df-8bc6-dc691e8b1a15/7a1fa240-17f3-11ef-b996-7dd2f40f5f82/file-1lwhbnwbp.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Area Under The Curves Question 7 English Explanation"></p>
<p>$$\begin{aligned}
& =\left[\frac{y^2}{8}+\frac{y}{4}-\frac{y^3}{6}\right]_{-\frac{1}{2}}^1 \\
& =\left(\frac{1}{8}+\frac{1}{4}-\frac{1}{6}\right)-\left(\frac{1}{32}-\frac{1}{8}+\frac{1}{48}\right) \\
& =\frac{5}{24}+\frac{7}{96} \\
& =\frac{27}{96} \\
& =\frac{9}{32}
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv3ve4fr | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region in the first quadrant inside the circle $$x^2+y^2=8$$ and outside the parabola $$y^2=2 x$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{\\pi}{2}-\\frac{1}{3}$$\n"}, {"identifier": "B", "content": "$$\\pi-\\frac{1}{3}$$\n"}, {"identifier": "C", "content": "$$\\pi-\\frac{2}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{2}-\\frac{2}{3}$$"}] | ["C"] | null | <p>We have, $$x^2+y^2=8$$ and $$y^2=2 x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4imme9/55025775-792e-443b-a53e-cd7d760a7686/861b5510-10e8-11ef-8bc0-edb751127fbd/file-1lw4immea.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4imme9/55025775-792e-443b-a53e-cd7d760a7686/861b5510-10e8-11ef-8bc0-edb751127fbd/file-1lw4immea.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Evening Shift Mathematics - Area Under The Curves Question 6 English Explanation"></p>
<p>Area of shaded region</p>
<p>$$\begin{aligned}
& =\int_\limits0^2 \frac{y^2}{2} d y+\int_\limits2^{2 \sqrt{2}} \sqrt{8-y^2} d y \\
& =\frac{1}{6}\left[y^3\right]_0^2+\left[\frac{y}{2} \sqrt{8-y^2}+4 \sin ^{-1}\left(\frac{y}{2 \sqrt{2}}\right)\right]_2^{2 \sqrt{2}} \\
& =\frac{4}{3}+\pi-2=\pi-\frac{2}{3}
\end{aligned}$$</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv5gs18q | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area of the region enclosed by the curve $$y=\min \{\sin x, \cos x\}$$ and the $$x$$ axis between $$x=-\pi$$ to $$x=\pi$$ be $$A$$. Then $$A^2$$ is equal to __________.</p> | [] | null | 16 | <p>$$y=f(x)=\min \{\sin x, \cos x\}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8v4grs/3efd55a7-6040-4108-bf47-c4327f08c8f1/86796680-134c-11ef-8bbe-1b4949638519/file-1lw8v4grt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8v4grs/3efd55a7-6040-4108-bf47-c4327f08c8f1/86796680-134c-11ef-8bbe-1b4949638519/file-1lw8v4grt.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 8th April Morning Shift Mathematics - Area Under The Curves Question 5 English Explanation"></p>
<p>$$\begin{aligned}
& A=-\int_\limits{-\pi}^{\frac{-3 \pi}{4}} \cos x d x-\int_\limits{\frac{-3 \pi}{4}}^0 \sin x d x+\int_\limits0^{\frac{\pi}{4}} \sin x d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{2}} \cos x d x \\
& -\int_\limits{\frac{\pi}{2}}^\pi \cos x d x \\
& A=4 \\
& A^2=16
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-morning-shift |
lv7v3qu3 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area of the region enclosed by the parabolas $$y=x^2-5 x$$ and $$y=7 x-x^2$$ is ________.</p> | [] | null | 198 | <p>$$\begin{aligned}
y=x^2-5 x, y=7 x-x^2 & \Rightarrow \quad x^2-5 x=7 x-x^2 \\
& \Rightarrow \quad x=0, x=6
\end{aligned}$$</p>
<p>$$\text { Area }=\int_\limits0^6\left[\left(7 x-x^2\right)-(x-5 x)\right] d x$$</p>
<p>$$=\int_\limits0^6\left(12 x-2 x^2\right) d x=6 x-\left.\frac{2 x^3}{3}\right|_0 ^6$$</p>
<p>$$=216-144=72 \text { sq. unit }$$</p>
<p>But answer is 198 as per NTA.</p> | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s20di | maths | area-under-the-curves | area-bounded-between-the-curves | <p>The area enclosed between the curves $$y=x|x|$$ and $$y=x-|x|$$ is :</p> | [{"identifier": "A", "content": "$$\\frac{8}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{2}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{4}{3}$$"}, {"identifier": "D", "content": "1"}] | ["C"] | null | <p>$$y=x|x|$$ & $$y=x-|x|$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweeutim/25a3e531-aad5-453f-b0c9-bc32519834c1/9c0ef0e0-1659-11ef-b26b-2993e9de41b8/file-1lweeutin.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweeutim/25a3e531-aad5-453f-b0c9-bc32519834c1/9c0ef0e0-1659-11ef-b26b-2993e9de41b8/file-1lweeutin.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 5th April Evening Shift Mathematics - Area Under The Curves Question 3 English Explanation"></p>
<p>$$\begin{aligned}
& y=x|x|= \begin{cases}x^2, & x>0 \\
-x^2, & x<0\end{cases} \\
& y=x-|x|= \begin{cases}0, & x>0 \\
2 x, & x<0\end{cases}
\end{aligned}$$</p>
<p>Point of intersections are $$(0,0)$$ & $$(-2,4)$$.</p>
<p>$$\begin{aligned}
\therefore \quad & \text { Area }=\int_\limits{-2}^0\left(-x^2-2 x\right) d x \\
& =\left[\frac{-x^3}{3}-\frac{2 x^2}{2}\right]_{-2}^0 \\
& =-\left[\frac{8}{3}-4\right]=\frac{4}{3} \text { sq. unit }
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-evening-shift |
lvb294kx | maths | area-under-the-curves | area-bounded-between-the-curves | <p>If the area of the region $$\left\{(x, y): \frac{\mathrm{a}}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0<\mathrm{a}<1\right\}$$ is $$\left(\log _{\mathrm{e}} 2\right)-\frac{1}{7}$$ then the value of $$7 \mathrm{a}-3$$ is equal to :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$-$$1"}] | ["D"] | null | <p>$$\left\{(x, y): \frac{a}{x^2} \leq y \leq \frac{1}{x}, 1 \leq x \leq 2,0< a<1\right\}$$</p>
<p>$$ \Rightarrow \int\limits_1^2 {\left( {{1 \over x} - {a \over {{x^2}}}} \right)dx = \left| {\ln |x| + {a \over x}} \right|_1^2} $$</p>
<p>$$\begin{aligned}
& \left(\ln 2+\frac{a}{2}\right)-(\ln 1+a)=\ln 2-\frac{a}{2} \\
& \Rightarrow \quad \frac{a}{2}=\frac{1}{7} \\
& \Rightarrow \quad a=\frac{2}{7} \\
& \quad 7 a-3=\frac{2}{7} \times 7-3=-1
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-evening-shift |
lvc57b84 | maths | area-under-the-curves | area-bounded-between-the-curves | <p>Let the area of the region enclosed by the curves $$y=3 x, 2 y=27-3 x$$ and $$y=3 x-x \sqrt{x}$$ be $$A$$. Then $$10 A$$ is equal to</p> | [{"identifier": "A", "content": "172"}, {"identifier": "B", "content": "154"}, {"identifier": "C", "content": "162"}, {"identifier": "D", "content": "184"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwcyt7aa/3fbcb7c5-4c8c-4aaa-876a-5ac80c7dce41/145c0910-158e-11ef-bb3a-95d759b5a950/file-1lwcyt7ab.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwcyt7aa/3fbcb7c5-4c8c-4aaa-876a-5ac80c7dce41/145c0910-158e-11ef-bb3a-95d759b5a950/file-1lwcyt7ab.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Morning Shift Mathematics - Area Under The Curves Question 1 English Explanation 1"></p>
<p>$$\begin{aligned}
& y^{\prime}=3-\frac{3}{2} \sqrt{x} \\
& y^{\prime}=3\left(1-\frac{\sqrt{x}}{2}\right)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwcyxj07/0b18439b-a031-4a45-9320-e6c48137f34f/8ca66870-158e-11ef-bb3a-95d759b5a950/file-1lwcyxj08.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwcyxj07/0b18439b-a031-4a45-9320-e6c48137f34f/8ca66870-158e-11ef-bb3a-95d759b5a950/file-1lwcyxj08.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Morning Shift Mathematics - Area Under The Curves Question 1 English Explanation 2"></p>
<p>$$\begin{aligned}
& \text { Area }=\int_\limits0^3\left(3 x-\left(3 x-x^{3 / 2}\right)\right) d x+\int_\limits3^9 \frac{27-3 x}{2}-\left(3 x-x^{3 / 2}\right) d x \\
& =\int_\limits0^3 x^{3 / 2} d x+\frac{1}{2} \int_\limits3^9\left(27-9 x+2 x^{3 / 2}\right) d x \\
& =\left[\frac{2}{5} x^{5 / 2}\right]_0^3+\frac{1}{2}\left[27 x-\frac{9 x^2}{2}+2 \times \frac{2}{5} x^{5 / 2}\right]_3^9 \\
& =\frac{2}{5} \times 9 \sqrt{3}+\frac{1}{2}\left[243-\frac{729}{2}+\frac{4}{5} \times 81 \times 3-81+\frac{81}{2}-\frac{4}{5} \times 9 \sqrt{3}\right] \\
& =\frac{1}{2}\left[\frac{486-729-81}{2}+\frac{972}{5}\right]=\frac{81}{5} \\
& A=\frac{81}{5} \\
& \therefore 10 A=10 \times \frac{81}{5}=162
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
cZPz7mFQ5ox2xoIa | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area enclosed between the curve $$y = {\log _e}\left( {x + e} \right)$$ and the coordinate axes is : | [{"identifier": "A", "content": "$$1$$"}, {"identifier": "B", "content": "$$2$$"}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$4$$"}] | ["A"] | null | The graph of the curve $$y = {\log _e}\left( {x + e} \right)$$ is as shown in the fig.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267502/exam_images/ssxskaas7h8ixpoi326x.webp" loading="lazy" alt="AIEEE 2005 Mathematics - Area Under The Curves Question 130 English Explanation">
<br><br>Required area
<br><br>$$A = \int\limits_{1 - e}^0 {ydx} = \int\limits_{1 - e}^0 {{{\log }_e}} \left( {x + e} \right)dx$$
<br><br>put $$x + e = t \Rightarrow dx = dt$$
<br><br>also At $$x = 1 - e,t = 1$$
<br><br>At $$x = 0,\,\,t = e$$
<br><br>$$\therefore$$ $$A = \int\limits_1^e {{{\log }_e}} \,tdt = \left[ {t\,{{\log }_e}t - t_1^e} \right]$$
<br><br>$$e - e - 0 + 1 = 1$$
<br><br>Hence the required area is $$1$$ square unit. | mcq | aieee-2005 |
TRI2HT1xJbgKZHyz | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let $$f(x)$$ be a non - negative continuous function such that the area bounded by the curve $$y=f(x),$$ $$x$$-axis and the ordinates $$x = {\pi \over 4}$$ and $$x = \beta > {\pi \over 4}$$ is $$\left( {\beta \sin \beta + {\pi \over 4}\cos \beta + \sqrt 2 \beta } \right).$$ Then $$f\left( {{\pi \over 2}} \right)$$ is | [{"identifier": "A", "content": "$$\\left( {{\\pi \\over 4} + \\sqrt 2 - 1} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{\\pi \\over 4} - \\sqrt 2 + 1} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {1 - {\\pi \\over 4} - \\sqrt 2 } \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {1 - {\\pi \\over 4} + \\sqrt 2 } \\right)$$ "}] | ["D"] | null | Given that
<br><br>$$\int\limits_{\pi /4}^\beta {f\left( x \right)} dx = \beta \sin \beta + {\pi \over 4}\cos \,\beta + \sqrt 2 \beta $$
<br><br>Differentiating $$w.r.t$$ $$\beta $$
<br><br>$$f\left( \beta \right) = \beta \cos \beta + \sin \beta - {\pi \over 4}\sin \beta + \sqrt 2 $$
<br><br>$$f\left( {{\pi \over 2}} \right) = \left( {1 - {\pi \over 4}} \right)\sin {\pi \over 2} + \sqrt 2 $$
<br><br>$$ = 1 - {\pi \over 4} + \sqrt 2 $$ | mcq | aieee-2005 |
55bGpDRnxxSxO83e | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let g(x) = cosx<sup>2</sup>, f(x) = $$\sqrt x $$ and $$\alpha ,\beta \left( {\alpha < \beta } \right)$$ be the roots of the quadratic equation 18x<sup>2</sup> - 9$$\pi $$x + $${\pi ^2}$$ = 0. Then the area (in sq. units) bounded by the curve
<br/>y = (gof)(x) and the lines $$x = \alpha $$, $$x = \beta $$ and y = 0 is : | [{"identifier": "A", "content": "$${1 \\over 2}\\left( {\\sqrt 2 - 1} \\right)$$"}, {"identifier": "B", "content": "$${1 \\over 2}\\left( {\\sqrt 3 - 1} \\right)$$"}, {"identifier": "C", "content": "$${1 \\over 2}\\left( {\\sqrt 3 + 1} \\right)$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\left( {\\sqrt 3 - \\sqrt 2 } \\right)$$"}] | ["B"] | null | Given quadratic equation,
<br><br>$$18{x^2} - 9\pi x + {\pi ^2} = 0$$
<br><br>$$ \Rightarrow \,\,\,18{x^2} - 6\pi x - 3\pi x + {\pi ^2} = 0$$
<br><br>$$ \Rightarrow \,\,\,\,6x\left( {3x - \pi } \right) - \pi \left( {3x - \pi } \right) = 0$$
<br><br>$$ \Rightarrow \,\,\,\,\left( {3x - \pi } \right)\left( {6x - \pi } \right) = 0$$
<br><br>$$\therefore$$ $$\,\,\,\,x = {\pi \over 3},{\pi \over 6}$$
<br><br>as $$\,\,\,\, \propto < B$$
<br><br>$$\therefore$$ $$\,\,\,\, \propto = {\pi \over 6}$$ $$\,\,\,\,$$ and $$\,\,\,\,$$ $$\beta = {\pi \over 3}$$
<br><br>Given, $$g\left( x \right) = \cos {x^2}$$ and $$ + \left( x \right) = \sqrt x $$
<br><br>$$y = \left( {gof} \right)x$$
<br><br>$$ = \,\,\,\,\,g\left( {f\left( x \right)} \right)$$
<br><br>$$ = \,\,\,\,\cos \left( {f{{\left( x \right)}^2}} \right)$$
<br><br>$$ = \,\,\,\,\cos {\left( {\sqrt x} \right)^2}$$
<br><br>$$ = \,\,\,\,\cos x$$
<br><br>So, the required area in the curve is
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267685/exam_images/if6lbcuycldilm9eq8ed.webp" loading="lazy" alt="JEE Main 2018 (Offline) Mathematics - Area Under The Curves Question 117 English Explanation">
<br><br>Area $$ = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {\cos \,\,dx} $$
<br><br>$$ = \left[ {\sin x} \right]_{{\pi \over 6}}^{{\pi \over 3}}$$
<br><br>$$ = \sin {\pi \over 3} - \sin {\pi \over 6}$$
<br><br>$$ = {{\sqrt 3} \over 2} - {1 \over 2}$$
<br><br>$$ = {{\sqrt 3 - 1} \over 2}$$ | mcq | jee-main-2018-offline |
K7S3uSiqOCZZI0fYX9H7G | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area of the region
<br/><br/>A = {(x, y) : 0 $$ \le $$ y $$ \le $$x |x| + 1 and $$-$$1 $$ \le $$ x $$ \le $$1} in sq. units, is : | [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 3}$$"}] | ["B"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265519/exam_images/gatckweagely4rxump82.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th January Evening Slot Mathematics - Area Under The Curves Question 110 English Explanation">
<br><br>Required area
<br><br>$$ = \int\limits_{ - 1}^1 {\left( {x\left| x \right| + 1} \right)} dx$$
<br><br>$$ = 0 + \left( x \right)_{ - 1}^1 = 2$$ | mcq | jee-main-2019-online-9th-january-evening-slot |
6Oe5Nkd1BXdoY6FgrjK1p | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area (in sq. units) of the region<br/>
A = { (x, y) $$ \in $$ R × R| 0 $$ \le $$ x $$ \le $$ 3, 0 $$ \le $$ y $$ \le $$ 4,
y $$ \le $$ x<sup>2</sup> + 3x} is : | [{"identifier": "A", "content": "$${{59} \\over 6}$$"}, {"identifier": "B", "content": "$${{26} \\over 3}$$"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "$${{53} \\over 6}$$"}] | ["A"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264987/exam_images/bujwhpy9e56yw8c0s7ne.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267785/exam_images/zjv3gmgivbhrqjdtzjj2.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266159/exam_images/ttruyt20dueawjkbojg4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Morning Slot Mathematics - Area Under The Curves Question 105 English Explanation"></picture>
<br><br>When y = 4 then,
<br><br>x<sup>2</sup> + 3x = 4
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + 3x - 4 = 0
<br><br>$$ \Rightarrow $$ x<sup>2</sup> + 4x - x - 4 = 0
<br><br>$$ \Rightarrow $$ x(x + 4) - (x + 4) = 0
<br><br>$$ \Rightarrow $$ (x + 4)(x - 1) = 0
<br><br>$$ \Rightarrow $$ x = 1, - 4
<br><br>As 0 $$ \le $$ x $$ \le $$ 3,
<br><br>so possible value of x = 1.
<br><br>$$ \therefore $$ y = x<sup>2</sup> + 3x parabola cut the line y = 4 at x = 1.
<br><br>Required area
<br><br>= $$\int\limits_0^1 {\left( {{x^2} + 3x} \right)} dx$$ + 2 $$ \times $$ 4
<br><br>= $$\left[ {{{{x^3}} \over 3} + 3\left( {{{{x^2}} \over 2}} \right)} \right]_0^1$$ + 8
<br><br>= $${\left( {{1 \over 3} + {3 \over 2}} \right)}$$ + 8
<br><br>= $${{11} \over 6} + 8$$
<br><br>= $${{59} \over 6}$$ | mcq | jee-main-2019-online-8th-april-morning-slot |
bauLwyzOpatnYqvKltGUV | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let S($$\alpha $$) = {(x, y) : y<sup>2</sup>
$$ \le $$ x, 0 $$ \le $$ x $$ \le $$ $$\alpha $$} and A($$\alpha $$)
is area of the region S($$\alpha $$). If for a $$\lambda $$, 0 < $$\lambda $$ < 4,
A($$\lambda $$) : A(4) = 2 : 5, then $$\lambda $$ equals | [{"identifier": "A", "content": "$$2{\\left( {{4 \\over {25}}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "B", "content": "$$2{\\left( {{2 \\over {5}}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "C", "content": "$$4{\\left( {{4 \\over {25}}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "D", "content": "$$4{\\left( {{2 \\over {5}}} \\right)^{{1 \\over 3}}}$$"}] | ["C"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267162/exam_images/qcblehdbr1lbbocet0ni.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263555/exam_images/rfv4vtuykmtleeqymrfe.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266675/exam_images/bitcqno4ggj6nfusp1fi.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 8th April Evening Slot Mathematics - Area Under The Curves Question 104 English Explanation"></picture>
<br>A($$\lambda $$) = $$2\int\limits_0^\lambda {\sqrt x } dx$$<br><br>= $$2\left[ {{{{x^{{3 \over 2}}}} \over {{3 \over 2}}}} \right]_0^\lambda $$
<br><br>= $${4 \over 3}{\lambda ^{{3 \over 2}}}$$
<br><br>$$ \therefore $$ A(4) = $${4 \over 3}{\left( 4 \right)^{{3 \over 2}}}$$
<br><br>Given, $${{A\left( \lambda \right)} \over {A\left( 4 \right)}} = {2 \over 5}$$
<br><br>$$ \Rightarrow $$ $${{{4 \over 3}{{\left( \lambda \right)}^{{3 \over 2}}}} \over {{4 \over 3}{{\left( 4 \right)}^{{3 \over 2}}}}} = {2 \over 5}$$
<br><br>$$ \Rightarrow $$ $${\lambda ^{{3 \over 2}}} = {2 \over 5} \times 8$$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${\left( {{{16} \over 5}} \right)^{{2 \over 3}}}$$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${\left( {{{256} \over {25}}} \right)^{{1 \over 3}}}$$
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${\left( {{{{4^3} \times 4} \over {25}}} \right)^{{1 \over 3}}}$$
<br><br>= $$4{\left( {{4 \over {25}}} \right)^{{1 \over 3}}}$$ | mcq | jee-main-2019-online-8th-april-evening-slot |
nVJQQ0RIj7eHctBMc91klux59dv | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | Let A<sub>1</sub> be the area of the region bounded by the curves y = sinx, y = cosx and y-axis in the first quadrant. Also, let A<sub>2</sub> be the area of the region bounded by the curves y = sinx, y = cosx, x-axis and x = $${\pi \over 2}$$ in the first quadrant. Then, | [{"identifier": "A", "content": "$${A_1}:{A_2} = 1:\\sqrt 2 $$ and $${A_1} + {A_2} = 1$$"}, {"identifier": "B", "content": "$${A_1} = {A_2}$$ and $${A_1} + {A_2} = \\sqrt 2 $$"}, {"identifier": "C", "content": "$$2{A_1} = {A_2}$$ and $${A_1} + {A_2} = 1 + \\sqrt 2 $$"}, {"identifier": "D", "content": "$${A_1}:{A_2} = 1:2$$ and $${A_1} + {A_2} = 1$$"}] | ["A"] | null | $${A_1} + {A_2} = \int\limits_0^{\pi /2} {\cos x.\,dx = \left. {\sin x} \right|_0^{\pi /2}} = 1$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264467/exam_images/ar9w3rtouhitqkvqu61e.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th February Evening Shift Mathematics - Area Under The Curves Question 83 English Explanation"><br><br>$${A_1} = \left. {\int\limits_0^{\pi /4} {(\cos x - \sin x)dx = (\sin x + \cos x)} } \right|_0^{\pi /4} = \sqrt 2 - 1$$<br><br>$$ \therefore $$ $${A_2} = 1 - \left( {\sqrt 2 - 1} \right) = 2 - \sqrt 2 $$<br><br>$$ \therefore $$ $${{{A_1}} \over {{A_2}}} = {{\sqrt 2 - 1} \over {\sqrt 2 \left( {\sqrt 2 - 1} \right)}} = {1 \over {\sqrt 2 }}$$ | mcq | jee-main-2021-online-26th-february-evening-slot |
aqpkRU5oUGyE2Sx1Fh1kmm2mc49 | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | The area bounded by the curve 4y<sup>2</sup> = x<sup>2</sup>(4 $$-$$ x)(x $$-$$ 2) is equal to : | [{"identifier": "A", "content": "$${\\pi \\over {16}}$$"}, {"identifier": "B", "content": "$${\\pi \\over {8}}$$"}, {"identifier": "C", "content": "$${3\\pi \\over {2}}$$"}, {"identifier": "D", "content": "$${3\\pi \\over {8}}$$"}] | ["C"] | null | Given,<br><br>4y<sup>2</sup> = x<sup>2</sup>(4 $$-$$ x)(x $$-$$ 2) ..... (1)<br><br>Here, Left hand side 4y<sup>2</sup> is always positive. So Right hand side should also be positive.<br><br>In x$$\in$$ [2, 4] Right hand side is positive.<br><br>By putting y = $$-$$y in equation (1), equation remains same. So, graph is symmetric about x axis.<br><br> <picture><source media="(max-width: 1644px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264611/exam_images/blvaykyw9hc3nczukh6r.webp"><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265519/exam_images/ncpxdczukmcwb1qlwiv0.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263617/exam_images/rj0bq9zhrsnt0msr5oq8.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266073/exam_images/naqcmdynvfxkte8guctu.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266697/exam_images/h0wgevig3gqfqwtgkrbv.webp"><source media="(max-width: 1040px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264331/exam_images/jxemrftx9ukh0w7x4q1h.webp"><source media="(max-width: 1220px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266276/exam_images/lswda8kb5augjc20sbwh.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263264/exam_images/ejh1nelzcfnzif29c6zv.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 18th March Evening Shift Mathematics - Area Under The Curves Question 81 English Explanation"></picture> <br><br>Required Area = 2A<sub>1</sub><br><br>$$ = 2\int_2^4 y dx$$<br><br>$$ = \int_2^4 {x\sqrt {(x - 2)(4 - x)} } dx$$<br><br>put $$x = 4{\sin ^2}\theta + 2{\cos ^2}\theta $$<br><br>$$ \Rightarrow dx = \left[ {4(2\sin \theta \cos \theta - 4\sin \theta \cos \theta )} \right]d\theta $$<br><br>$$ \Rightarrow dx = 4\sin \theta \cos \theta d\theta $$<br><br>When lower limit = 2 then <br><br>$$2 = 4{\sin ^2}\theta + 2{\cos ^2}\theta $$<br><br>$$ \Rightarrow 4(1 - {\cos ^2}\theta ) + 2{\cos ^2}\theta = 2$$<br><br>$$ \Rightarrow 4 - 4{\cos ^2}\theta + 2{\cos ^2}\theta = 2$$<br><br>$$ \Rightarrow 2{\cos ^2}\theta = 2$$<br><br>$$ \Rightarrow \cos \theta = \pm 1$$<br><br>$$ \Rightarrow \theta = 0,\pi {} $$<br><br>When upper limit = 4 then<br><br>$$4 = 4{\sin ^2}\theta + 2{\cos ^2}\theta $$<br><br>$$ \Rightarrow - 2{\cos ^2}\theta = 0$$<br><br>$$ \Rightarrow \theta = {{\pi {} } \over 2}$$<br><br>$$ \therefore $$ Range of $$\theta$$ = 0 to $${{{\pi {} } \over 2}}$$<br><br>$$ \therefore $$ Area = $$ \int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )\sqrt {(2{{\sin }^2}\theta )(2{{\cos }^2}\theta )} (4\sin \theta \cos \theta )d\theta } $$<br><br>$$ = \int_0^{{{\pi {} } \over 2}} {(4{{\sin }^2}\theta + 2{{\cos }^2}\theta )8{{\sin }^2}\theta {{\cos }^2}\theta d\theta } $$<br><br>$$ = 32\int_0^{{{\pi {} } \over 2}} {{{\sin }^4}\theta {{\cos }^2}\theta d\theta } + 16\int_0^{{{\pi {} } \over 2}} {{{\sin }^2}\theta {{\cos }^2}\theta d\theta } $$<br><br>Using Wallis formula,<br><br>$$ = 32.{{3.1.1} \over {6.4.2}}.{{\pi {} } \over 2} + 16.{{1.3.1} \over {6.4.2}}.{{\pi {} } \over 2}$$<br><br>$$ = \pi {} + {{\pi {} } \over 2}$$<br><br>$$ = {{3\pi {} } \over 2}$$ | mcq | jee-main-2021-online-18th-march-evening-shift |
1ktkeocfd | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | If the line y = mx bisects the area enclosed by the lines x = 0, y = 0, x = $${3 \over 2}$$ and the curve y = 1 + 4x $$-$$ x<sup>2</sup>, then 12 m is equal to _____________. | [] | null | 26 | <p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kxvqwyud/3b08f599-748b-4dc3-95b3-f0ac0c10ea2e/861eb060-6af6-11ec-b350-33e20cd86462/file-1kxvqwyue.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kxvqwyud/3b08f599-748b-4dc3-95b3-f0ac0c10ea2e/861eb060-6af6-11ec-b350-33e20cd86462/file-1kxvqwyue.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 60vh" alt="JEE Main 2021 (Online) 31st August Evening Shift Mathematics - Area Under The Curves Question 71 English Explanation"> </p>
<p>According to the question,</p>
<p>$${1 \over 2}\int_0^{3/2} {(1 + 4x - {x^2})dx = \int_0^{3/2} {mx\,dx} } $$</p>
<p>$$ \Rightarrow {1 \over 2}\left[ {\left( {x + 2{x^2} - {{{x^3}} \over 3}} \right)} \right]_0^{3/2} = {m \over 2}[x]_0^{3/2} \Rightarrow {3 \over 2} + {9 \over 2} - {9 \over 8} = {{9m} \over 4}$$</p>
<p>$$\Rightarrow$$ m = 39/18 $$\Rightarrow$$ 12m = 26</p> | integer | jee-main-2021-online-31st-august-evening-shift |
1l55hj2ov | maths | area-under-the-curves | area-under-simple-curves-in-standard-forms | <p>The area of the bounded region enclosed by the curve <br/><br/>$$y = 3 - \left| {x - {1 \over 2}} \right| - |x + 1|$$ and the x-axis is :</p> | [{"identifier": "A", "content": "$${9 \\over 4}$$"}, {"identifier": "B", "content": "$${45 \\over 16}$$"}, {"identifier": "C", "content": "$${27 \\over 8}$$"}, {"identifier": "D", "content": "$${63 \\over 16}$$"}] | ["C"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l99thyns/e3d7926f-e2f7-4b54-82b0-67049b426481/fa7cf380-4c79-11ed-b94d-45a8040c2a81/file-1l99thynt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l99thyns/e3d7926f-e2f7-4b54-82b0-67049b426481/fa7cf380-4c79-11ed-b94d-45a8040c2a81/file-1l99thynt.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Evening Shift Mathematics - Area Under The Curves Question 68 English Explanation"><br>
$$
y=\left\{\begin{array}{cc}
2 x-\frac{7}{2} & x<-1 \\\\
\frac{3}{2} & -1 \leq x \leq \frac{1}{2} \\\\
\frac{5}{2}-2 x & x>\frac{1}{2}
\end{array}\right.
$$<br><br>
$$
y=3-\left|x-\frac{1}{2}\right|-|x+1|
$$
<br><br>
Area of shaded region (required area)
<br><br>
$$
=\frac{1}{2}\left(3+\frac{3}{2}\right) \cdot \frac{3}{2}=\frac{27}{8}
$$ | mcq | jee-main-2022-online-28th-june-evening-shift |
5d6Ld5edVvT7r0SH | maths | binomial-theorem | binomial-theorem-for-any-index | The positive integer just greater than $${\left( {1 + 0.0001} \right)^{10000}}$$ is | [{"identifier": "A", "content": "4 "}, {"identifier": "B", "content": "5 "}, {"identifier": "C", "content": "2 "}, {"identifier": "D", "content": "3 "}] | ["D"] | null | $${\left( {1 + 0.0001} \right)^{10000}}$$ = $${\left( {1 + {1 \over {{{10}^4}}}} \right)^{10000}}$$
<br><br>= 1 + 10000$${ \times {1 \over {{{10}^4}}}}$$ + $${{10000\left( {9999} \right)} \over {2!}} \times {\left( {{1 \over {{{10}^4}}}} \right)^2}$$+......$$\infty $$
<br><br>< 1 + 1 + $${1 \over {2!}}$$ + $${1 \over {3!}}$$ + ...... $$\infty $$ = e = 2.71828 < 3 | mcq | aieee-2002 |
1krzmg688 | maths | binomial-theorem | binomial-theorem-for-any-index | The lowest integer which is greater <br/><br/>than $${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$ is ______________. | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["A"] | null | Let $$P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$<br><br>Let $$x = {10^{100}}$$<br><br>$$ \Rightarrow P = {\left( {1 + {1 \over x}} \right)^x}$$<br><br>$$ \Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\,
2 \,}} \right. }}.{1 \over {{x^2}}} + {{(x)(x - 1)(x - 2)} \over {\left| \!{\underline {\,
3 \,}} \right. }}.{1 \over {{x^3}}} + ....$$<br><br>(upto 10<sup>100</sup> + 1 terms)<br><br>$$ \Rightarrow P = 1 + 1 + \left( {{1 \over {\left| \!{\underline {\,
2 \,}} \right. }} - {1 \over {\left| \!{\underline {\,
2 \,}} \right. {x^2}}}} \right) + \left( {{1 \over {\left| \!{\underline {\,
3 \,}} \right. }} - ...} \right) + ...$$ so on<br><br>$$ \Rightarrow P = 2 + \left( {Positive\,value\,less\,than\,{1 \over {\left| \!{\underline {\,
2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
4 \,}} \right. }} + ...} \right)$$<br><br>Also $$e = 1 + {1 \over {\left| \!{\underline {\,
1 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
4 \,}} \right. }} + ...$$<br><br>$$ \Rightarrow {1 \over {\left| \!{\underline {\,
2 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
3 \,}} \right. }} + {1 \over {\left| \!{\underline {\,
4 \,}} \right. }} + ... = e - 2$$<br><br>$$\Rightarrow$$ P = 2 + (positive value less than e $$-$$ 2)<br><br>$$\Rightarrow$$ P $$\in$$ (2, 3)<br><br>$$\Rightarrow$$ least integer value of P is 3 | mcq | jee-main-2021-online-25th-july-evening-shift |
lsamwj6p | maths | binomial-theorem | binomial-theorem-for-any-index | Let $m$ and $n$ be the coefficients of seventh and thirteenth terms respectively<br/><br/> in the expansion of $\left(\frac{1}{3} x^{\frac{1}{3}}+\frac{1}{2 x^{\frac{2}{3}}}\right)^{18}$. Then $\left(\frac{\mathrm{n}}{\mathrm{m}}\right)^{\frac{1}{3}}$ is : | [{"identifier": "A", "content": "$\\frac{1}{9}$"}, {"identifier": "B", "content": "$\\frac{1}{4}$"}, {"identifier": "C", "content": "$\\frac{4}{9}$"}, {"identifier": "D", "content": "$\\frac{9}{4}$"}] | ["D"] | null | $\begin{aligned} & \mathrm{t}_7={ }^{18} \mathrm{C}_6\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^{12}\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^6={ }^{18} \mathrm{C}_6 \frac{1}{(3)^{12}} \cdot \frac{1}{2^6} \\\\ & \mathrm{t}_{13}={ }^{18} \mathrm{C}_{12}\left(\frac{\mathrm{x}^{\frac{1}{3}}}{3}\right)^6\left(\frac{\mathrm{x}^{\frac{-2}{3}}}{2}\right)^{12}={ }^{18} \mathrm{C}_{12} \frac{1}{(3)^6} \cdot \frac{1}{2^{12}} \cdot \mathrm{x}^{-6}\end{aligned}$
<br/><br/>$$ \therefore $$ $m={ }^{18} C_6\left(\frac{1}{3}\right)^{12}\left(\frac{1}{2}\right)^6$
<br/><br/>$n={ }^{18} C_{12}\left(\frac{1}{3}\right)^6\left(\frac{1}{2}\right)^{12}$
<br/><br/>$\begin{aligned}\left(\frac{m}{n}\right)^{\frac{1}{3}} & =\left(\frac{{ }^{18} C_6\left(\frac{1}{3}\right)^{12}\left(\frac{1}{2}\right)^6}{{ }^{18} C_{12}\left(\frac{1}{3}\right)^6\left(\frac{1}{2}\right)^{12}}\right)^{\frac{1}{3}} \\\\ & =\left(\frac{\left(\frac{1}{3}\right)^6}{\left(\frac{1}{2}\right)^6}\right)^{\frac{1}{3}}=\left(\left(\frac{2}{3}\right)^6\right)^{\frac{1}{3}}=\frac{4}{9}\end{aligned}$
<br/><br/>$\therefore\left(\frac{n}{m}\right)^{\frac{1}{3}}=\frac{9}{4}$ | mcq | jee-main-2024-online-1st-february-evening-shift |
1lsgcji8h | maths | binomial-theorem | binomial-theorem-for-any-index | <p>$$\text { Number of integral terms in the expansion of }\left\{7^{\left(\frac{1}{2}\right)}+11^{\left(\frac{1}{6}\right)}\right\}^{824} \text { is equal to _________. }$$</p> | [] | null | 138 | <p>General term in expansion of $$\left((7)^{1 / 2}+(11)^{1 / 6}\right)^{824}$$ is $$\mathrm{t}_{\mathrm{r}+1}={ }^{824} \mathrm{C}_{\mathrm{r}}(7)^{\frac{824-\mathrm{r}}{2}}(11)^{\mathrm{r} / 6}$$</p>
<p>For integral term, $$r$$ must be multiple of 6.</p>
<p>Hence $$r=0,6,12, ....... 822$$</p> | integer | jee-main-2024-online-30th-january-morning-shift |
lv0vxd0y | maths | binomial-theorem | binomial-theorem-for-any-index | <p>The sum of all rational terms in the expansion of $$\left(2^{\frac{1}{5}}+5^{\frac{1}{3}}\right)^{15}$$ is equal to :</p> | [{"identifier": "A", "content": "633"}, {"identifier": "B", "content": "6131"}, {"identifier": "C", "content": "3133"}, {"identifier": "D", "content": "931"}] | ["C"] | null | <p>$$\begin{aligned}
& T_{r+1}={ }^{15} \mathrm{C}_r\left(2^{1 / 5}\right)^{15-r}\left(5^{1 / 3}\right)^r \\
& ={ }^{15} C_r 5^{r / 3} 2^{\left(3-\frac{r}{5}\right)}
\end{aligned}$$</p>
<p>For rational terms,</p>
<p>$$\frac{r}{3}$$ and $$\frac{r}{5}$$ must be integer</p>
<p>3 and 5 divide $$r \Rightarrow 15$$ divides $$r \Rightarrow r=0$$ and $$r=15$$</p>
<p>$${ }^{15} C_0 5^0 2^3+{ }^{15} C_{15} 5^5 2^{(0)}$$</p>
<p>$$\begin{aligned}
& =8+3125 \\
& =3133
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
PmsYoXcOZSnaCuzH | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The remainder left out when $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$ is divided by 9 is : | [{"identifier": "A", "content": "2 "}, {"identifier": "B", "content": "7 "}, {"identifier": "C", "content": "8 "}, {"identifier": "D", "content": "0"}] | ["A"] | null | $${8^{2n}} - {\left( {62} \right)^{2n + 1}}$$
<br><br>= $${\left( {{8^2}} \right)^n} - {\left( {62} \right)^{2n + 1}}$$
<br><br>= $${\left( {1 + 63} \right)^n} - {\left( {1 - 63} \right)^{2n + 1}}$$
<br><br> = $$\left( {1 + n.63 + {}^n{C_2}{{.63}^2} + ......} \right)$$
<br> + $$\left( {1 + {}^{2n + 1}{C_1}.\left( { - 63} \right) + {}^{2n + 1}{C_2}.{{\left( { - 63} \right)}^2} + ......} \right)$$
<br><br>= 2 + 63$$\left[ {\left( {n + {}^n{C_2} + ....} \right) + \left( { - {}^{2n + 1}{C_1} + {}^{2n + 1}{C_2}.63 + ......} \right)} \right]$$
<br><br>= 63$$ \times $$[Some integral value] + 2
<br><br>63$$ \times $$[Some integral value] + 2 by dividing with 9 we will get 2 as remainder as 63 is multiple of 9. | mcq | aieee-2009 |
njUyy7DYscXbuI3Jxazra | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If (27)<sup>999</sup> is divided by 7, then the remainder is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["D"] | null | We have,
<br><br>$${{{{\left( {27} \right)}^{999}}} \over 7}$$
<br><br>= $${{{{\left( {28 - 1} \right)}^{999}}} \over 7}$$
<br><br>= $${{28\,\lambda - 1} \over 7}$$
<br><br>= $${{28\,\lambda - 7 + 7 - 1} \over \lambda }$$
<br><br>= $${{7\left( {4\lambda - 1} \right) + 6} \over 7}$$
<br><br>$$\therefore\,\,\,$$ Remainder = 6 | mcq | jee-main-2017-online-8th-april-morning-slot |
y6YAjngUyXiNUzKKBU7k9k2k5e343pn | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The greatest positive integer k, for which 49<sup>k</sup> + 1 is a factor of the sum <br/>49<sup>125</sup> + 49<sup>124</sup> + ..... + 49<sup>2</sup> + 49 + 1, is: | [{"identifier": "A", "content": "32"}, {"identifier": "B", "content": "60"}, {"identifier": "C", "content": "63"}, {"identifier": "D", "content": "65"}] | ["C"] | null | 1 + 49 + 49<sup>2</sup>
+ ..... + 49<sup>125</sup>
<br><br>sum of G.P. = $${{1.\left( {{{49}^{126}} - 1} \right)} \over {49 - 1}}$$
<br><br>= $${{\left( {{{49}^{63}} + 1} \right)\left( {{{49}^{63}} - 1} \right)} \over {48}}$$
<br><br>Also 49<sup>63</sup> - 1
<br><br>= (1 + 48)<sup>63</sup> - 1
<br><br>= [<sup>63</sup>C<sub>0</sub>$$ \times $$1 + <sup>63</sup>C<sub>1</sub>$$ \times $$ 48 + <sup>63</sup>C<sub>2</sub>$$ \times $$ (48)<sup>2</sup> + .... ] - 1
<br><br>= [1 + 48$$\lambda $$] - 1 = 48$$\lambda $$
<br><br>So $${{\left( {{{49}^{63}} - 1} \right)} \over {48}}$$ = integer
<br><br>$$ \therefore $$ 49<sup>63</sup> + 1 is a factor.
<br><br>So k = 63. | mcq | jee-main-2020-online-7th-january-morning-slot |
aFW8brFlz4zwZsvzfZ1klt9oix2 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If the remainder when x is divided by 4 is 3, then the remainder when (2020 + x)<sup>2022</sup> is divided by 8 is __________. | [] | null | 1 | Let x = 4k + 3<br><br>(2020 + x)<sup>2022</sup><br><br>= (2020 + 4k + 3)<sup>2022</sup><br><br>= (4(505) + 4k + 3)<sup>2022</sup>
<br><br>= (4P + 3)<sup>2022</sup><br><br>= (4P + 4 $$-$$ 1)<sup>2022</sup><br><br>= (4A $$-$$ 1)<sup>2022</sup><br><br><sup>2022</sup>C<sub>0</sub>(4A)<sup>0</sup>($$-$$1)<sup>2022</sup> + <sup>2022</sup>C<sub>1</sub>(4A)<sup>1</sup>($$-$$1)<sup>2021</sup> + ......
<br><br>= 1 + 2022(4A)(-1) + .....
<br><br>= 1 + 8$$\lambda$$<br><br>$$ \therefore $$ Reminder is 1. | integer | jee-main-2021-online-25th-february-evening-slot |
ofQvn9MZjWjzpi3QoA1klt9p75z | maths | binomial-theorem | divisibility-concept-and-remainder-concept | The total number of two digit numbers 'n', such that 3<sup>n</sup> + 7<sup>n</sup> is a multiple of 10, is __________. | [] | null | 45 | $$ \because $$ $${7^n} = {(10 - 3)^n} = 10k + {( - 3)^n}$$<br><br>$${7^n} + {3^n} = 10k + {( - 3)^n} + {3^n}$$<br><br><picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264628/exam_images/ukvxzlftpfikzghxsw5c.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266742/exam_images/af4pahqgj1sbdtklrnkn.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266434/exam_images/tqd7mtwk97wryl4bzy6g.webp"><source media="(max-width: 860px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263287/exam_images/nsnzomfk6ungv89yxu7j.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266340/exam_images/jcx1zygqvh9prqa0yh9r.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Binomial Theorem Question 134 English Explanation"></picture><br><br>$$ \therefore $$ 3<sup>n</sup> = 3<sup>2t</sup> = (10 $$-$$ 1)<sup>t</sup><br><br>= 10p + ($$-$$1)<sup>t</sup><br><br>= 10p $$\pm$$ 1<br><br>$$ \therefore $$ if n = even then 7<sup>n</sup> + 3<sup>n</sup> will not be multiply of 10<br><br>So if n is odd then only 7<sup>n</sup> + 3<sup>n</sup> will be multiply of 10<br><br>$$ \therefore $$ n = 11, 13, 15, ..........., 99<br><br>$$ \therefore $$ Ans : 45 | integer | jee-main-2021-online-25th-february-evening-slot |
5UlwtPdmE6ETPIebwf1kmjbieu8 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | If (2021)<sup>3762</sup> is divided by 17, then the remainder is __________. | [] | null | 4 | 2021 = 17m - 2
<br><br>(2021)<sup>3762</sup> = (17m $$-$$ 2)<sup>3762</sup> = multiple of 17 + 2<sup>3762</sup><br><br>= 17$$\lambda$$ + 2<sup>2</sup> (2<sup>4</sup>)<sup>940</sup><br><br>= 17$$\lambda$$ + 4 (17 $$-$$ 1)<sup>940</sup><br><br>= 17$$\lambda$$ + 4 (17$$\mu$$ + 1)<br><br>= 17k + 4; (k $$\in$$ I)<br><br>$$ \therefore $$ Remainder = 4 | integer | jee-main-2021-online-17th-march-morning-shift |
1ktgq16qr | maths | binomial-theorem | divisibility-concept-and-remainder-concept | 3 $$\times$$ 7<sup>22</sup> + 2 $$\times$$ 10<sup>22</sup> $$-$$ 44 when divided by 18 leaves the remainder __________. | [] | null | 15 | 3(1 + 6)<sup>22</sup> + 2 . (1 + 9)<sup>22</sup> $$-$$ 44 = (3 + 2 $$-$$ 44) = 18 . I<br><br>= $$-$$ 39 + 18 . I<br><br>= (54 $$-$$ 39) + 18(I $$-$$ 3)<br><br>= 15 + 18I<sub>1</sub><br><br>$$\Rightarrow$$ Remainder = 15 | integer | jee-main-2021-online-27th-august-evening-shift |
1l587i0nd | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when (2021)<sup>2023</sup> is divided by 7 is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["C"] | null | (2021)<sup>2023</sup></p>
<p>= (2016 + 5)<sup>2023</sup> [here 2016 is divisible by 7]</p>
<p>= <sup>2023</sup>C<sub>0</sub> (2016)<sup>2023</sup> + .......... + <sup>2023</sup>C<sub>2022</sub> (2016) (5)<sup>2022</sup> + <sup>2023</sup>C<sub>2023</sub> (5)<sup>2023</sup></p>
<p>= 2016 [<sup>2023</sup>C<sub>0</sub> . (2016)<sup>2022</sup> + ....... + <sup>2023</sup>C<sub>2022</sub> . (5)<sup>2022</sup>] + (5)<sup>2023</sup></p>
<p>= 2016$$\lambda$$ + (5)<sup>2023</sup></p>
<p>= 7 $$\times$$ 288$$\lambda$$ + (5)<sup>2023</sup></p>
<p>= 7K + (5)<sup>2023</sup> ...... (1)</p>
<p>Now, (5)<sup>2023</sup></p>
<p>= (5)<sup>2022</sup> . 5</p>
<p>= (5<sup>3</sup>)<sup>674</sup> . 5</p>
<p>= (125)<sup>674</sup> . 5</p>
<p>= (126 $$-$$ 1)<sup>674</sup> . 5</p>
<p>= 5[<sup>674</sup>C<sub>0</sub> (126)<sup>674</sup> + ......... $$-$$ <sup>674</sup>C<sub>673</sub> (126) + <sup>674</sup>C<sub>674</sub>]</p>
<p>= 5 $$\times$$ 126 [<sup>674</sup>C<sub>0</sub>(126)<sup>673</sup> + ....... $$-$$ <sup>674</sup>C<sub>673</sub>] + 5</p>
<p>= 5 . 7 . 18 [<sup>674</sup>C<sub>0</sub>(126)<sup>673</sup> + ....... $$-$$ <sup>674</sup>C<sub>673</sub>] + 5</p>
<p>= 7$$\lambda$$ + 5</p>
<p>Replacing (5)<sup>2023</sup> in equation (1) with 7$$\lambda$$ + 5, we get,</p>
<p>(2021)<sup>2023</sup> = 7K + 7$$\lambda$$ + 5</p>
<p>= 7(K + $$\lambda$$) + 5</p>
<p>$$\therefore$$ Remainer = 5 | mcq | jee-main-2022-online-26th-june-morning-shift |
1l5ai3wa2 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>If $${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,.....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$, then the remainder when K is divided by 6 is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["D"] | null | <p>$${1 \over {2\,.\,{3^{10}}}} + {1 \over {{2^2}\,.\,{3^9}}} + \,\,....\,\, + \,\,{1 \over {{2^{10}}\,.\,3}} = {K \over {{2^{10}}\,.\,{3^{10}}}}$$</p>
<p>$$ \Rightarrow {1 \over {2\,.\,{3^{10}}}}\left[ {{{{{\left( {{3 \over 2}} \right)}^{10}} - 1} \over {{3 \over 2} - 1}}} \right] = {K \over {{2^{10}}\,.\,{3^{10}}}}$$</p>
<p>$$ = {{{3^{10}} - {2^{10}}} \over {{2^{10}}\,.\,{3^{10}}}} = {K \over {{2^{10}}\,.\,{3^{10}}}} \Rightarrow K = {3^{10}} - {2^{10}}$$</p>
<p>Now $$K = {(1 + 2)^{10}} - {2^{10}}$$</p>
<p>$$ = {}^{10}{C_0} + {}^{10}{C_1}2 + {}^{10}{C_2}{2^3} + \,\,....\,\, + \,\,{}^{10}{C_{10}}{2^{10}} - {2^{10}}$$</p>
<p>$$ = {}^{10}{C_0} + {}^{10}{C_1}2 + 6\lambda + {}^{10}{C_9}\,.\,{2^9}$$</p>
<p>$$ = 1 + 20 + 5120 + 6\lambda $$</p>
<p>$$ = 5136 + 6\lambda + 5$$</p>
<p>$$ = 6\mu + 5$$</p>
<p>$$\lambda ,\,\mu \in N$$</p>
<p>$$\therefore$$ remainder = 5</p> | mcq | jee-main-2022-online-25th-june-morning-shift |
1l5bb3q9d | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder on dividing 1 + 3 + 3<sup>2</sup> + 3<sup>3</sup> + ..... + 3<sup>2021</sup> by 50 is _________.</p> | [] | null | 4 | <p>Given,</p>
<p>$$1 + 3 + {3^2} + {3^3} + \,\,.....\,\, + \,\,{3^{2021}}$$</p>
<p>$$ = {3^0} + {3^1} + {3^2} + {3^3} + \,\,....\,\, + \,\,{3^{2021}}$$</p>
<p>This is a G.P with common ratio = 3</p>
<p>$$\therefore$$ Sum $$ = {{1({3^{2022}} - 1)} \over {3 - 1}}$$</p>
<p>$$ = {{{3^{2022}} - 1} \over 2}$$</p>
<p>$$ = {{{{({3^2})}^{2011}} - 1} \over 2}$$</p>
<p>$$ = {{{{(10 - 1)}^{1011}} - 1} \over 2}$$</p>
<p>$$ = {{\left[ {{}^{1011}{C_0}\,.\,{{10}^{1011}} - {}^{1011}{C_1}\,.\,{{10}^{1010}} + \,\,.....\,\, - \,\,{}^{1011}{C_{1009}}\,.\,{{(10)}^2} + {}^{1011}{C_{1010}}\,.\,10 - {}^{1011}{C_{1011}}} \right] - 1} \over 2}$$</p>
<p>$$ = {{{{10}^2}\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1009}} - {}^{1011}{C_1}\,.\,(1008) + \,\,.....\,\,{}^{1011}{C_{1009}}} \right] + 10110 - 1 - 1} \over 2}$$</p>
<p>$$ = {{100k + 10110 - 2} \over 2}$$</p>
<p>$$ = {{100k + 10108} \over 2}$$</p>
<p>$$ = 50k + 5054$$</p>
<p>$$ = 50k + 50 \times 101 + 4$$</p>
<p>$$ = 50[k + 101] + 4$$</p>
<p>$$ = 50k' + 4$$</p>
<p>$$\therefore$$ By dividing 50 we get remainder as 4.</p> | integer | jee-main-2022-online-24th-june-evening-shift |
1l5c0s15h | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when 3<sup>2022</sup> is divided by 5 is :</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["D"] | null | <p>$${3^{2022}}$$</p>
<p>$$ = {({3^2})^{1011}}$$</p>
<p>$$ = {(9)^{1011}}$$</p>
<p>$$ = {(10 - 1)^{1011}}$$</p>
<p>$$ = {}^{1011}{C_0}{(10)^{1011}} + \,\,.....\,\, + \,\,{}^{1011}{C_{1010}}\,.\,{(10)^1} - {}^{1011}{C_{1011}}$$</p>
<p>$$ = 10\left[ {{}^{1011}{C_0}{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right] - 1$$</p>
<p>$$ = 10\,K - 1$$</p>
<p>[As $$10\left[ {{}^{1011}{C_0}\,.\,{{(10)}^{1010}} + \,\,......\,\, + \,\,{}^{1011}{C_{1010}}} \right]$$ is multiple of 10]</p>
<p>$$ = 10K + 5 - 5 - 1$$</p>
<p>$$ = 10K - 5 + 5 - 1$$</p>
<p>$$ = 5(2K - 1) + 4$$</p>
<p>$$\therefore$$ Unit digit = 4 when divided by 5.</p> | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6f0uwiw | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$(11)^{1011}+(1011)^{11}$$ is divided by 9 is</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["D"] | null | <p>$${\mathop{\rm Re}\nolimits} \left( {{{{{(11)}^{1011}} + {{(1011)}^{11}}} \over 9}} \right) = {\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}} + {3^{11}}} \over 9}} \right)$$</p>
<p>For $${\mathop{\rm Re}\nolimits} \left( {{{{2^{1011}}} \over 9}} \right)$$</p>
<p>$${2^{1011}} = {(9 - 1)^{337}} = {}^{337}{C_0}{9^{337}}{( - 1)^0} + {}^{337}{C_1}{9^{336}}{( - 1)^1} + {}^{337}{C_2}{9^{335}}{( - 1)^2} + \,\,.....\,\, + \,\,{}^{337}{C_{337}}{9^0}{( - 1)^{337}}$$</p>
<p>So, remainder is 8 </p>
<p>and $${\mathop{\rm Re}\nolimits} \left( {{{{3^{11}}} \over 9}} \right) = 0$$</p>
<p>So, remainder is 8</p> | mcq | jee-main-2022-online-25th-july-evening-shift |
1l6jb84eh | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$(2021)^{2022}+(2022)^{2021}$$ is divided by 7 is</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "6"}] | ["A"] | null | <p>$${(2021)^{2022}} + {(2022)^{2021}}$$</p>
<p>$$ = {(7k - 2)^{2022}} + {(7{k_1} - 1)^{2021}}$$</p>
<p>$$ = {\left[ {{{(7k - 2)}^3}} \right]^{674}} + {(7{k_1})^{2021}} - 2021{(7{k_1})^{2020}}\, + \,....\, - 1$$</p>
<p>$$ = {(7{k_2} - 1)^{674}} + (7m - 1)$$</p>
<p>$$ = (7n + 1) + (7m - 1) = 7(m + n)$$ (multiple of 7)</p>
<p>$$\therefore$$ Remainder = 0</p> | mcq | jee-main-2022-online-27th-july-morning-shift |
1l6m5vk3w | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$7^{2022}+3^{2022}$$ is divided by 5 is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}] | ["C"] | null | $$
\begin{aligned}
& 7^{2022}+3^{2022} \\\\
& =\left(7^2\right)^{1011}+\left(3^2\right)^{1011} \\\\
&=(50-1)^{1011}+(10-1)^{1011} \\\\
&= (50^{1011}-1011.50^{1010}+\ldots-1) \\\\
& + (10^{1011}-1011.10^{1010}+\ldots . .-1) \\\\
&= 5 m-1+5 n-1=5(m+n)-2 \\\\
&= 5(m+n)-5+3=5(m+n-1)+3 \\\\
&= 5 k+3 \\\\
& \therefore \text { Remainder }=3
\end{aligned}
$$ | mcq | jee-main-2022-online-28th-july-morning-shift |
1ldoo7i54 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder, when $$19^{200}+23^{200}$$ is divided by 49 , is ___________.</p> | [] | null | 29 | $19^{200}+23^{200}$
<br/><br/>= $(21-2)^{200}+(21+2)^{200}=49 \lambda+2^{201}$
<br/><br/>Now, $2^{201}=8^{67}=(7+1)^{67}=49 \lambda+7 \times 67+1$
<br/><br/>$=49 \lambda+470$
<br/><br/>$=49(\lambda+9)+29$
<br/><br/>$$ \therefore $$ Remainder $=29$ | integer | jee-main-2023-online-1st-february-morning-shift |
1ldptern1 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder on dividing $$5^{99}$$ by 11 is ____________.</p> | [] | null | 9 | $5^{99}=5^{4} .5^{95}$
<br/><br/>$=625\left[5^{5}\right]^{19}$
<br/><br/>$=625[3125]^{19}$
<br/><br/>$=625[3124+1]^{19}$
<br/><br/>$=625[11 \mathrm{k} \times 19+1]$
<br/><br/>$=625 \times 11 \mathrm{k} \times 19+625$
<br/><br/>$=11 \mathrm{k}_{1}+616+9$
<br/><br/>$=11\left(\mathrm{k}_{2}\right)+9$
<br/><br/>Remainder $=9$ | integer | jee-main-2023-online-31st-january-morning-shift |
ldr0sc3g | maths | binomial-theorem | divisibility-concept-and-remainder-concept | $50^{\text {th }}$ root of a number $x$ is 12 and $50^{\text {th }}$ root of another number $y$ is 18 . Then the remainder obtained on dividing $(x+y)$ by 25 is ____________. | [] | null | 23 | <p>Given $${x^{{1 \over {50}}}} = 12 \Rightarrow x = {12^{50}}$$</p>
<p>$${y^{{1 \over {50}}}} = 18 \Rightarrow y = {18^{50}}$$</p>
<p>$$12\equiv13$$ (Mod 25)</p>
<p>$$12^2\equiv19$$ (Mod 25)</p>
<p>$$12^3\equiv-3$$ (Mod 25)</p>
<p>$$12^9\equiv-2$$ (Mod 25)</p>
<p>$$12^{10}\equiv-1$$ (Mod 25)</p>
<p>$$12^{50}\equiv-1$$ (Mod 25) ..... (i)</p>
<p>Now</p>
<p>$$18\equiv7$$ (Mod 25)</p>
<p>$$18^2\equiv-1$$ (Mod 25)</p>
<p>$$18^{-50}\equiv-1$$ (Mod 25) ..... (ii)</p>
<p>$$\therefore$$ $$12^{50}+18^{50}\equiv-2$$ (Mod 25)</p>
<p>$$\equiv23$$ (Mod 25)</p>
<p>$$\therefore$$ Answer = 23</p> | integer | jee-main-2023-online-30th-january-evening-shift |
1ldu60obo | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when (2023)$$^{2023}$$ is divided by 35 is __________.</p> | [] | null | 7 | $$
\begin{aligned}
& (2023)^{2023} \\\\
& =(2030-7)^{2023} \\\\
& =(35 \mathrm{~K}-7)^{2023} \\\\
& ={ }^{2023} \mathrm{C}_0(35 \mathrm{~K})^{2023}(-7)^0+{ }^{2023} \mathrm{C}_1(35 \mathrm{~K})^{2022}(-7)+ \\\\
& \ldots . .+\ldots \ldots .+{ }^{2023} \mathrm{C}_{2023}(-7)^{2023} \\\\
& =35 \mathrm{~N}-7^{2023} \\\\
& \text { Now },-7^{2023}=-7 \times 7^{2022}=-7\left(7^2\right)^{1011} \\\\
& =-7(50-1)^{1011} \\\\
& =-7\left({ }^{1011} \mathrm{C}_0 50^{1011}-{ }^{1011} \mathrm{C}_1(50)^{1010}+\ldots \ldots{ }^{1011} \mathrm{C}_{1011}\right) \\\\
& =-7(5 \lambda-1) \\\\
& =-35 \lambda+7
\end{aligned}
$$<br/><br/>
$\therefore$ when $(2023)^{2023}$ is divided by 35 remainder is 7 | integer | jee-main-2023-online-25th-january-evening-shift |
1lgoy2e8y | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder, when $$7^{103}$$ is divided by 17, is __________</p> | [] | null | 12 | $7^{103}=7 \times 7^{102}$
<br/><br/>$$
\begin{aligned}
& =7 \times(49)^{51} \\\\
& =7 \times(51-2)^{51}
\end{aligned}
$$
<br/><br/>Remainder = $7 \times(-2)^{51}$
<br/><br/>$$
\begin{aligned}
& =-7\left(2^3 \cdot(16)^{12}\right) \\\\
& =-56(17-1)^{12}
\end{aligned}
$$
<br/><br/>Remainder $=-56 \times(-1)^{12}=-56+68=12$ | integer | jee-main-2023-online-13th-april-evening-shift |
1lgpy5uw8 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Fractional part of the number $$\frac{4^{2022}}{15}$$ is equal to</p> | [{"identifier": "A", "content": "$$\\frac{8}{15}$$"}, {"identifier": "B", "content": "$$\\frac{4}{15}$$"}, {"identifier": "C", "content": "$$\\frac{1}{15}$$"}, {"identifier": "D", "content": "$$\\frac{14}{15}$$"}] | ["C"] | null | $$
\begin{aligned}
& \left\{\frac{4^{2022}}{15}\right\}=\left\{\frac{2^{4044}}{15}\right\} \\\\
& =\left\{\frac{(1+15)^{1011}}{15}\right\} \\\\
& =\frac{1}{15}
\end{aligned}
$$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lgvpgm7l | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Let the number $$(22)^{2022}+(2022)^{22}$$ leave the remainder $$\alpha$$ when divided by 3 and $$\beta$$ when divided by 7. Then $$\left(\alpha^{2}+\beta^{2}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "13"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "20"}, {"identifier": "D", "content": "5"}] | ["D"] | null | We have, $(22)^{2022}+(2022)^{22}$
<br/><br/>As 2022 is completely divisible by 3
<br/><br/>So, $(2022)^{22}$ is also divisible by 3
<br/><br/>$(22)^{2022}=(21+1)^{2022}=(3 \times 7+1)^{2022}=7 m+1$
<br/><br/>$\Rightarrow(22)^{2022}$ leave a remainder 1 , when divisible by 3 .
<br/><br/>$\therefore(22)^{2022}+(2022)^{22}$ leave a remainder when divisible by 3
<br/><br/>$\therefore \alpha=1$
<br/><br/>$$
\begin{aligned}
(22)^{2022}+(2022)^{22} & =(21+1)^{2022}+(2023-1)^{22} \\\\
& =7 K+1+7 \mu+1=7(K+\mu)+2
\end{aligned}
$$
<br/><br/>$\Rightarrow(22)^{2022}+(2022)^{22}$ leave a remainder 2 when divisible by 7
<br/><br/>$$
\begin{aligned}
& \therefore \beta=2 \\\\
& \text { Hence, } \alpha^2+\beta^2=1^2+2^2=5
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
1lgyl43d6 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>$$25^{190}-19^{190}-8^{190}+2^{190}$$ is divisible by :</p> | [{"identifier": "A", "content": "14 but not by 34"}, {"identifier": "B", "content": "neither 14 nor 34"}, {"identifier": "C", "content": "both 14 and 34"}, {"identifier": "D", "content": "34 but not by 14"}] | ["D"] | null | The given expression is divisible by 6 and 17 .
<br/><br/>Also, $25^{190}-8^{190}$ is not divisible by 7
<br/><br/>but $19^{190}-2^{190}$ is divisible by 7 ,
<br/><br/>So, $25^{190}-19^{190}-8^{190}+2^{190}$ is divisible by 34 but not by 14 . | mcq | jee-main-2023-online-8th-april-evening-shift |
1lh00oqex | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The largest natural number $$n$$ such that $$3^{n}$$ divides $$66 !$$ is ___________.</p> | [] | null | 31 | We have,
<br/><br/>$$
\begin{aligned}
& {\left[\frac{66}{3}\right]=22} \\\\
& {\left[\frac{66}{3^2}\right]=7} \\\\
& {\left[\frac{66}{3^3}\right]=2}
\end{aligned}
$$
<br/><br/>Highest powers of 3 is greater than 66. So, their g.i.f. is always 0.
<br/><br/>$\therefore$ Required natural number $=22+7+2=31$ | integer | jee-main-2023-online-8th-april-morning-shift |
1lh2y2hd7 | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Among the statements :</p>
<p>(S1) : $$2023^{2022}-1999^{2022}$$ is divisible by 8</p>
<p>(S2) : $$13(13)^{n}-12 n-13$$ is divisible by 144 for infinitely many $$n \in \mathbb{N}$$</p> | [{"identifier": "A", "content": "both (S1) and (S2) are incorrect"}, {"identifier": "B", "content": "only (S1) is correct"}, {"identifier": "C", "content": "only (S2) is correct"}, {"identifier": "D", "content": "both (S1) and (S2) are correct"}] | ["D"] | null | We have, $S_1$ : $(2023)^{2022}-(1999)^{2022}$
<br/><br/>$$
\begin{aligned}
& =(1999+24)^{2022}-(1999)^{2022}={ }^{2022} C_0(1999)^{2022}(24)^0 \\\\
& +{ }^{2022} C_1(1999)^{2021}(24)^1+{ }^{2022} C_2(1999)^{2020}(24)^2 \\\\
& +\ldots-(1999)^{2022} \\\\
& ={ }^{2022} C_1(1999)^{2021}(24)+{ }^{2022} C_2(1999)^{2022}(24)^2 \\\\
& =24\left({ }^{2022} C_1(1999)^{2021}+{ }^{2022} C_2(1999)^{2022}(24)+\ldots+\ldots\right) \\\\
& \Rightarrow S_1 \text { is divisible by } 24
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { Now, } S_2: 13(13)^n-12 n-13 \\\\
& \text { Here, } 13^n=(1+12)^n \\\\
& \quad=1+{ }^n C_1 12+{ }^n C_2(12)^2+{ }^n C_3(12)^3 \\\\
& \begin{aligned}
\therefore S_2: & 13\left(1+{ }^n C_1(12)+{ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\
\quad= & 13+156 n+13\left({ }^n C_2(12)^2+{ }^n C_3(12)^3+\ldots\right)-12 n-13 \\\\
\quad= & 144 \times 13\left({ }^n C_2+{ }^n C_3(12)+\ldots\right)
\end{aligned}
\end{aligned}
$$
<br/><br/>$\Rightarrow S_2$ is divisible by 144 for infinitely many $n \in N$ | mcq | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lsfl8i9p | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>Remainder when $$64^{32^{32}}$$ is divided by 9 is equal to ________.</p> | [] | null | 1 | <p>Let $$32^{32}=\mathrm{t}$$</p>
<p>$$\begin{aligned}
& 64^{32^{32}}=64^t=8^{2 t}=(9-1)^{2 t} \\
& =9 \mathrm{k}+1
\end{aligned}$$</p>
<p>Hence remainder $$=1$$</p> | integer | jee-main-2024-online-29th-january-evening-shift |
luy9clej | maths | binomial-theorem | divisibility-concept-and-remainder-concept | <p>The remainder when $$428^{2024}$$ is divided by 21 is __________.</p> | [] | null | 1 | <p>$$\begin{aligned}
& 428=21 \times 20+8 \\
\Rightarrow \quad & (428)^{2024} \equiv(20 \times 21+8)^{2024} \equiv 8^{2024}(\bmod 21) \\
& 8^2=21 \times 3+1 \\
& 8^{2024}=(21 \times 3+1)^{1012} \\
\Rightarrow \quad & 8^{2024} \equiv(21 \times 3+1)^{1012}(\bmod 21) \\
& \equiv 1^{2012}(\bmod 21) \\
& 428^{2024} \equiv 1(\bmod 21)
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-morning-shift |
v11RSlXUNBj0iZzn | maths | binomial-theorem | general-term | The coefficients of $${x^p}$$ and $${x^q}$$ in the expansion of $${\left( {1 + x} \right)^{p + q}}$$ are | [{"identifier": "A", "content": "equal "}, {"identifier": "B", "content": "equal with opposite signs "}, {"identifier": "C", "content": "reciprocals of each other "}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | Here in this expansion $${\left( {1 + x} \right)^{p + q}}$$
<br><br>The general term = $${T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}$$
<br><br>$$\therefore$$ $${x^p}$$ will be present in the term = $${}^{p + q}{C_p}.{\left( x \right)^p}$$
<br><br>So coefficient of $${x^p}$$ = $${}^{p + q}{C_p}$$
<br><br>And $${x^q}$$ will be present in the term = $${}^{p + q}{C_q}.{\left( x \right)^q}$$
<br><br>$$\therefore$$ coefficient of $${x^q}$$ = $${}^{p + q}{C_q}$$
<br><br>We know $${}^n{C_r}$$ = $${}^n{C_{n - r}}$$
<br><br>$$\therefore$$ $${}^{p + q}{C_q}$$ = $${}^{p + q}{C_{\left( {p + q} \right) - q}}$$ = $${}^{p + q}{C_p}$$
<br><br>So coefficients of $${x^p}$$ and $${x^q}$$ are equal. | mcq | aieee-2002 |
AuQk01qTdDzVQGQ4 | maths | binomial-theorem | general-term | The number of integral terms in the expansion of $${\left( {\sqrt 3 + \root 8 \of 5 } \right)^{256}}$$ is | [{"identifier": "A", "content": "35 "}, {"identifier": "B", "content": "32 "}, {"identifier": "C", "content": "33 "}, {"identifier": "D", "content": "34 "}] | ["C"] | null | General term = $${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$$
<br>= $${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$$
<br><br>When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.
<br>And when $${{r \over 8}}$$ is integer then $${\left( 5 \right)^{{r \over 8}}}$$ is integer.
<br><br>Entire general term will be integer when $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ both are integer.
<br><br>$${{{256 - r} \over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256
<br><br>$${{r \over 8}}$$ is integer when r = 0, 8, 16 ,......., 256
<br><br>Now both $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)
<br><br>$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, t<sub>n</sub> = a + (n - 1)d
<br><br>$$\therefore$$ n = $${{256} \over 8} + 1$$ = 32 + 1 = 33 | mcq | aieee-2003 |
Subsets and Splits