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 |
|---|---|---|---|---|---|---|---|---|---|---|
1lh00jkrz | maths | application-of-derivatives | maxima-and-minima | <p>If $$a_{\alpha}$$ is the greatest term in the sequence $$\alpha_{n}=\frac{n^{3}}{n^{4}+147}, n=1,2,3, \ldots$$, then $$\alpha$$ is equal to _____________.</p> | [] | null | 5 | $$
\begin{aligned}
& \text { Let } y=\frac{x^3}{x^4+147} \\\\
& \Rightarrow \frac{d y}{d x}=\frac{\left(x^4+147\right) \times 3 x^2-x^3\left(4 x^3\right)}{\left(x^4+147\right)^2} \\\\
& =\frac{3 x^6+441 x^2-4 x^6}{\left(x^4+147\right)^2}=\frac{441 x^2-x^6}{\left(x^4+147\right)^2}
\end{aligned}
$$
<br><br>For maxima/minima, put $\frac{d y}{d x}=0$
<br><br>$$
\begin{aligned}
& \Rightarrow 441 x^2-x^6=0 \Rightarrow x^4=441 \\\\
& \Rightarrow x= \pm \sqrt{21}, \pm \sqrt{21} i
\end{aligned}
$$
<br><br>Now, by descrates rule on number line we have
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lmls8gbc/7f8dbf18-7516-4da9-acc1-29bce97edc3b/409f8a70-546d-11ee-b10a-c342ea039081/file-6y3zli1lmls8gbd.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lmls8gbc/7f8dbf18-7516-4da9-acc1-29bce97edc3b/409f8a70-546d-11ee-b10a-c342ea039081/file-6y3zli1lmls8gbd.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 - Application of Derivatives Question 30 English Explanation">
<br><br>Since sign changes from negative to positive at 0 .
<br><br>$\therefore$ Maximum value of is at $x=\sqrt{21}=4.58$
<br><br>Now, $4<4.5<5$
<br><br>$$
\begin{aligned}
& \therefore \text { yat } x=4=\frac{64}{403}=0.159 \\\\
& y \text { at } x=5=\frac{125}{772}=0.162
\end{aligned}
$$
<br><br>So, $y$ is maximum at $x=5$
<br><br>$$
\therefore \alpha=5
$$ | integer | jee-main-2023-online-8th-april-morning-shift |
jaoe38c1lscngs3o | maths | application-of-derivatives | maxima-and-minima | <p>Let $$g(x)=3 f\left(\frac{x}{3}\right)+f(3-x)$$ and $$f^{\prime \prime}(x)>0$$ for all $$x \in(0,3)$$. If $$g$$ is decreasing in $$(0, \alpha)$$ and increasing in $$(\alpha, 3)$$, then $$8 \alpha$$ is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "20"}] | ["C"] | null | <p>$$g(x)=3 f\left(\frac{x}{3}\right)+f(3-x) \text { and } f^{\prime \prime}(x)>0 \forall x \in(0,3)$$</p>
<p>$$\Rightarrow \mathrm{f}^{\prime}(\mathrm{x})$$ is increasing function</p>
<p>$$\begin{aligned}
& g^{\prime}(x)=3 \times \frac{1}{3} \cdot f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x) \\
& =f^{\prime}\left(\frac{x}{3}\right)-f^{\prime}(3-x)
\end{aligned}$$</p>
<p>If $$\mathrm{g}$$ is decreasing in $$(0, \alpha)$$</p>
<p>$$\begin{aligned}
& \mathrm{g}^{\prime}(\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)-\mathrm{f}^{\prime}(3-\mathrm{x})<0 \\
& \mathrm{f}^{\prime}\left(\frac{\mathrm{x}}{3}\right)<\mathrm{f}^{\prime}(3-\mathrm{x}) \\
& \Rightarrow \frac{\mathrm{x}}{3}<3-\mathrm{x} \\
& \Rightarrow \mathrm{x}<\frac{9}{4}
\end{aligned}$$</p>
<p>Therefore $$\alpha=\frac{9}{4}$$</p>
<p>Then $$8 \alpha=8 \times \frac{9}{4}=18$$</p> | mcq | jee-main-2024-online-27th-january-evening-shift |
jaoe38c1lsf0sbol | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=2^x-x^2, x \in \mathbb{R}$$. If $$m$$ and $$n$$ are respectively the number of points at which the curves $$y=f(x)$$ and $$y=f^{\prime}(x)$$ intersect the $$x$$-axis, then the value of $$\mathrm{m}+\mathrm{n}$$ is ___________.</p> | [] | null | 5 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt307bt9/dd10aeba-887c-4185-a5fe-218e2ee9b2bf/09af1ae0-d4af-11ee-8384-811001421c41/file-1lt307bta.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt307bt9/dd10aeba-887c-4185-a5fe-218e2ee9b2bf/09af1ae0-d4af-11ee-8384-811001421c41/file-1lt307bta.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 - Application of Derivatives Question 19 English Explanation 1"></p>
<p>$$\begin{aligned}
& \therefore \mathrm{m}=3 \\
& \mathrm{f}^{\prime}(\mathrm{x})=2^{\mathrm{x}} \ln 2-2 \mathrm{x}=0 \\
& 2^{\mathrm{x}} \ln 2=2 \mathrm{x}
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt30872c/dd61fe66-bd37-4697-9d03-3f699eeb03a7/21d35f50-d4af-11ee-8384-811001421c41/file-1lt30872d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt30872c/dd61fe66-bd37-4697-9d03-3f699eeb03a7/21d35f50-d4af-11ee-8384-811001421c41/file-1lt30872d.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 - Application of Derivatives Question 19 English Explanation 2"></p>
<p>$$\begin{aligned}
& \therefore \mathrm{n}=2 \\
& \Rightarrow \mathrm{m}+\mathrm{n}=5
\end{aligned}$$</p> | integer | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfksmfv | maths | application-of-derivatives | maxima-and-minima | <p>The function $$f(x)=2 x+3(x)^{\frac{2}{3}}, x \in \mathbb{R}$$, has</p> | [{"identifier": "A", "content": "exactly one point of local minima and no point of local maxima\n"}, {"identifier": "B", "content": "exactly one point of local maxima and exactly one point of local minima\n"}, {"identifier": "C", "content": "exactly two points of local maxima and exactly one point of local minima\n"}, {"identifier": "D", "content": "exactly one point of local maxima and no point of local minima"}] | ["B"] | null | <p>$$\begin{aligned}
& f(x)=2 x+3(x)^{\frac{2}{3}} \\
& f^{\prime}(x)=2+2 x^{\frac{-1}{3}} \\
& =2\left(1+\frac{1}{x^{\frac{1}{3}}}\right) \\
& =2\left(\frac{x^{\frac{1}{3}}+1}{x^{\frac{1}{3}}}\right)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsr8t6nc/4c9643cb-de77-44b7-86fb-da033b7c270c/4abc4680-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8t6nd.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsr8t6nc/4c9643cb-de77-44b7-86fb-da033b7c270c/4abc4680-ce37-11ee-9412-cd4f9c6f2c40/file-6y3zli1lsr8t6nd.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 - Application of Derivatives Question 18 English Explanation"></p>
<p>So, maxima (M) at x = $$-$$1 & minima (m) at x = 0</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
1lsg4fvwc | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=(x+3)^2(x-2)^3, x \in[-4,4]$$. If $$M$$ and $$m$$ are the maximum and minimum values of $$f$$, respectively in $$[-4,4]$$, then the value of $$M-m$$ is</p> | [{"identifier": "A", "content": "108"}, {"identifier": "B", "content": "392"}, {"identifier": "C", "content": "608"}, {"identifier": "D", "content": "600"}] | ["C"] | null | <p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=(\mathrm{x}+3)^2 \cdot 3(\mathrm{x}-2)^2+(\mathrm{x}-2)^3 2(\mathrm{x}+3) \\
& =5(\mathrm{x}+3)(\mathrm{x}-2)^2(\mathrm{x}+1) \\
& \mathrm{f}^{\prime}(\mathrm{x})=0, \mathrm{x}=-3,-1,2
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsoxw4nw/e2162195-84ee-4126-95ca-5b36a9ae1ecf/06e90dc0-ccf3-11ee-a330-494dca5e9a63/file-6y3zli1lsoxw4nx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsoxw4nw/e2162195-84ee-4126-95ca-5b36a9ae1ecf/06e90dc0-ccf3-11ee-a330-494dca5e9a63/file-6y3zli1lsoxw4nx.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 - Application of Derivatives Question 16 English Explanation"></p>
<p>$$\begin{aligned}
& f(-4)=-216 \\
& f(-3)=0, f(4)=49 \times 8=392 \\
& M=392, m=-216 \\
& M-m=392+216=608 \\
& \text { Ans }=\text { '3' }
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-evening-shift |
1lsg91cac | maths | application-of-derivatives | maxima-and-minima | <p>The maximum area of a triangle whose one vertex is at $$(0,0)$$ and the other two vertices lie on the curve $$y=-2 x^2+54$$ at points $$(x, y)$$ and $$(-x, y)$$, where $$y>0$$, is :</p> | [{"identifier": "A", "content": "108"}, {"identifier": "B", "content": "122"}, {"identifier": "C", "content": "88"}, {"identifier": "D", "content": "92"}] | ["A"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqngokt/b3ef22bc-e563-4b45-a320-95a6d263c505/cf5ad1d0-cde3-11ee-a0d3-7b75c4537559/file-6y3zli1lsqngoku.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqngokt/b3ef22bc-e563-4b45-a320-95a6d263c505/cf5ad1d0-cde3-11ee-a0d3-7b75c4537559/file-6y3zli1lsqngoku.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 - Application of Derivatives Question 15 English Explanation"></p>
<p>Area of $$\Delta$$</p>
<p>$$\begin{aligned}
& =\frac{1}{2}\left|\begin{array}{ccc}
0 & 0 & 1 \\
x & y & 1 \\
-x & y & 1
\end{array}\right| \\
& \Rightarrow\left|\frac{1}{2}(x y+x y)\right|=|x y| \\
& \operatorname{Area}(\Delta)=|x y|=\left|x\left(-2 x^2+54\right)\right| \\
& \frac{d(\Delta)}{d x}=\left|\left(-6 x^2+54\right)\right| \Rightarrow \frac{d \Delta}{d x}=0 \text { at } x=3 \\
& \text { Area }=3(-2 \times 9+54)=108
\end{aligned}$$</p> | mcq | jee-main-2024-online-30th-january-morning-shift |
luy9clmh | maths | application-of-derivatives | maxima-and-minima | <p>Let the set of all positive values of $$\lambda$$, for which the point of local minimum of the function $$(1+x(\lambda^2-x^2))$$ satisfies $$\frac{x^2+x+2}{x^2+5 x+6}<0$$, be $$(\alpha, \beta)$$. Then $$\alpha^2+\beta^2$$ is equal to _________.</p> | [] | null | 39 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw3i47n9/78e3808f-b6b1-4774-ad57-06e016fc0e17/bd7fae50-1059-11ef-abcd-c333ada72a30/file-1lw3i47na.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw3i47n9/78e3808f-b6b1-4774-ad57-06e016fc0e17/bd7fae50-1059-11ef-abcd-c333ada72a30/file-1lw3i47na.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 - Application of Derivatives Question 13 English Explanation 1"></p>
<p>$$\begin{aligned}
& f(x)=1+x\left(\lambda^2-x^2\right) \\
& f(x)=-x^3+\left(\lambda^2 x+1\right) \\
& f^{\prime}(x)=-3 x^2+\lambda^2 \\
& x= \pm \frac{\lambda}{\sqrt{3}}
\end{aligned}$$</p>
<p>$$-\frac{\lambda}{\sqrt{3}}$$ should satisfy the given condition</p>
<p>$$\begin{aligned}
& \frac{x^2+x+2}{x^2+5 x+6}<6 \\
& \frac{1}{(x+2)(x+3)}<0
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw3i5n3l/0bfecb58-7297-4604-ad51-f617b9a67d6c/e53eda10-1059-11ef-abcd-c333ada72a30/file-1lw3i5n3m.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw3i5n3l/0bfecb58-7297-4604-ad51-f617b9a67d6c/e53eda10-1059-11ef-abcd-c333ada72a30/file-1lw3i5n3m.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 - Application of Derivatives Question 13 English Explanation 2"></p>
<p>$$\begin{aligned}
&\begin{aligned}
& x \in(-3,-2) \\
& -3<-\frac{\lambda}{\sqrt{3}}<-2 \\
& -3 \sqrt{3}<-\lambda<-2 \sqrt{3} \\
& 2 \sqrt{3}<\lambda<3 \sqrt{3} \\
& \left.\begin{array}{l}
\alpha=2 \sqrt{3} \\
\beta=2 \sqrt{3}
\end{array}\right\} \\
& (2 \sqrt{3})^2+(3 \sqrt{2})^2 \\
& 12+27 \\
\end{aligned}\\
&39
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-morning-shift |
lv0vxd5u | maths | application-of-derivatives | maxima-and-minima | <p>Let the sum of the maximum and the minimum values of the function $$f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}$$ be $$\frac{m}{n}$$, where $$\operatorname{gcd}(\mathrm{m}, \mathrm{n})=1$$. Then $$\mathrm{m}+\mathrm{n}$$ is equal to :</p> | [{"identifier": "A", "content": "217"}, {"identifier": "B", "content": "182"}, {"identifier": "C", "content": "201"}, {"identifier": "D", "content": "195"}] | ["C"] | null | <p>$$\begin{aligned}
& f(x)=\frac{2 x^2-3 x+8}{2 x^2+3 x+8}=y, 2 x^2+3 x+8>0 \quad \forall x \in \mathbb{R} \\
& \Rightarrow \quad x^2(2 y-2)+x(3 y+3)+8 y-8=0
\end{aligned}$$</p>
<p>Since $$x \in \mathbb{R}$$, the equation has real roots</p>
<p>$$\begin{aligned}
& \Rightarrow \quad D \geq 0 \\
& \Rightarrow(3 y+3)^2-4(2 y-2)(8 y-8) \geq 0 \\
& \Rightarrow 9(y+1)^2-64 y(y-1)^2 \geq 0 \\
& \Rightarrow(3 y+3)^2-(8 y-8)^2 \geq 0 \\
& \Rightarrow(11 y-5)(-5 y+11) \geq 0 \\
& \Rightarrow\left(y-\frac{5}{11}\right)\left(y-\frac{11}{5}\right) \leq 0 \\
& \Rightarrow y \in\left[\frac{5}{11}, \frac{11}{5}\right]
\end{aligned}$$</p>
<p>Sum of maximum and minimum value</p>
<p>$$\begin{aligned}
& y_{\max }+y_{\min }=\frac{5}{11}+\frac{11}{5}=\frac{25+121}{55} \\
& =\frac{146}{55}=\frac{m}{n} \Rightarrow m+n=201
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-morning-shift |
lv2er3i1 | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=3 \sqrt{x-2}+\sqrt{4-x}$$ be a real valued function. If $$\alpha$$ and $$\beta$$ are respectively the minimum and the maximum values of $$f$$, then $$\alpha^2+2 \beta^2$$ is equal to</p> | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "38"}, {"identifier": "C", "content": "24"}, {"identifier": "D", "content": "44"}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)=3 \sqrt{x-2}+\sqrt{4-x} \\
& \text { Let } x=2 \sin ^2 \theta+4 \cos ^2 \theta \\
& =3 \sqrt{2 \sin ^2 \theta+4 \cos ^2 \theta-2}+\sqrt{4-2 \sin ^2 \theta-4 \cos ^2 \theta} \\
& =3 \sqrt{2 \cos ^2 \theta}+\sqrt{2 \sin ^2 \theta} \\
& =3 \sqrt{2}|\cos \theta|+\sqrt{2}|\sin \theta|
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{18+2} \\
& \Rightarrow 3 \sqrt{2} \cos \theta+\sqrt{2} \sin \theta \leq \sqrt{20}
\end{aligned}$$</p>
<p>Minimum value exists when $$\theta=\frac{\pi}{2}$$</p>
<p>$$\begin{aligned}
& \text { So, minimum value }=\sqrt{2} \\
& \Rightarrow \alpha=\sqrt{2} \text { and } \beta=\sqrt{20} \\
& \Rightarrow \alpha^2+2 \beta^2=2+40 \\
& =42
\end{aligned}$$</p> | mcq | jee-main-2024-online-4th-april-evening-shift |
lv3ve65y | maths | application-of-derivatives | maxima-and-minima | <p>If the function $$f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a}> 0$$ has a local maximum at $$x=\alpha$$ and a local minimum at $$x=\alpha^2$$, then $$\alpha$$ and $$\alpha^2$$ are the roots of the equation :</p> | [{"identifier": "A", "content": "$$x^2-6 x+8=0$$\n"}, {"identifier": "B", "content": "$$8 x^2-6 x+1=0$$\n"}, {"identifier": "C", "content": "$$8 x^2+6 x-1=0$$\n"}, {"identifier": "D", "content": "$$x^2+6 x+8=0$$"}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)=6 x^2-18 a x+12 a^2 \\
& =6\left(x^2-3 a+2 a^2\right) \\
& =6(x-a)(x-2 a)=0 \\
& x=a, 2 a
\end{aligned}$$</p>
<p>$$a=\alpha, \quad 2 a=\alpha^2 \quad \Rightarrow \alpha=0,2$$</p>
<p>$$\begin{array}{lll}
a>0 & \therefore & \alpha=2 \\
& & \alpha^2=4
\end{array}$$</p>
<p>$$\therefore x^2-6 x+8=0$$ is the required quadratic equation.</p> | mcq | jee-main-2024-online-8th-april-evening-shift |
lv3vef38 | maths | application-of-derivatives | maxima-and-minima | <p>Let $$\mathrm{A}$$ be the region enclosed by the parabola $$y^2=2 x$$ and the line $$x=24$$. Then the maximum area of the rectangle inscribed in the region $$\mathrm{A}$$ is ________.</p> | [] | null | 128 | <p>$$\begin{aligned}
& y^2=2 x \\
& a=\left(\frac{1}{2}\right)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4uhftn/eaae4a9d-2d8a-4b86-93c7-b8b3dfc45e4c/e388a2b0-1116-11ef-b9cb-b5e0fe4ba33b/file-1lw4uhfto.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4uhftn/eaae4a9d-2d8a-4b86-93c7-b8b3dfc45e4c/e388a2b0-1116-11ef-b9cb-b5e0fe4ba33b/file-1lw4uhfto.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 - Application of Derivatives Question 9 English Explanation 1"></p>
<p>$$\begin{aligned}
& A(t)=2 t \times\left(24-\frac{t^2}{2}\right) \\
& A=48 t-t^3
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4uiqqc/051e5d6a-5dcd-41d3-92ba-260c0a74c2b1/07c4e940-1117-11ef-b9cb-b5e0fe4ba33b/file-1lw4uiqqd.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4uiqqc/051e5d6a-5dcd-41d3-92ba-260c0a74c2b1/07c4e940-1117-11ef-b9cb-b5e0fe4ba33b/file-1lw4uiqqd.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 - Application of Derivatives Question 9 English Explanation 2"></p>
<p>$$\begin{aligned}
& \frac{d A}{d t}=48-3 t^2 \\
& 48-3 t^2=0 \\
& 3 t^2=48 \\
& t^2=16 \\
& t= \pm 4
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4uomxr/b876dfc7-50c6-4c89-9214-81e07c07cec0/abb17af0-1117-11ef-b9cb-b5e0fe4ba33b/file-1lw4uomxs.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4uomxr/b876dfc7-50c6-4c89-9214-81e07c07cec0/abb17af0-1117-11ef-b9cb-b5e0fe4ba33b/file-1lw4uomxs.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 - Application of Derivatives Question 9 English Explanation 3"></p>
<p>$$\begin{aligned}
& A(4)=48 \times 4-4^3 \\
& =192-64 \\
& A(4)=128
\end{aligned}$$</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv5grwlv | maths | application-of-derivatives | maxima-and-minima | <p>Let $$f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10$$. The number of points of local maxima of $$f$$ in interval $$(0,2 \pi)$$ is</p> | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "2"}] | ["D"] | null | <p>$$\begin{aligned}
& f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10 \\
& f^{\prime}(x)=12 \cos ^2 x \cdot(-\sin x)+6 \sqrt{3} \cos x \cdot(-\sin x)=0 \\
& =-6 \sqrt{3} \cos x \cdot \sin x\left(1+\frac{2}{\sqrt{3}} \cos x\right)=0 \\
& \cos x=0, \sin x=0, \cos x=\frac{-\sqrt{3}}{2}
\end{aligned}$$</p>
<p>Sign of $$f(x)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8slgdc/e7ec72d2-9c20-4762-a251-c217bf88f1ea/a3884200-1342-11ef-9cb4-095599b9956f/file-1lw8slgdd.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8slgdc/e7ec72d2-9c20-4762-a251-c217bf88f1ea/a3884200-1342-11ef-9cb4-095599b9956f/file-1lw8slgdd.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 - Application of Derivatives Question 7 English Explanation"></p>
<p>$$\therefore \text { Maxima at } \frac{5 \pi}{6}, \frac{7 \pi}{6}$$</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv5grwiz | maths | application-of-derivatives | maxima-and-minima | <p>The number of critical points of the function $$f(x)=(x-2)^{2 / 3}(2 x+1)$$ is</p> | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "3"}] | ["A"] | null | <p>To find the number of critical points of the function $$f(x)=(x-2)^{2 / 3}(2 x+1)$$, we need to determine where its derivative $$f'(x)$$ is equal to zero or undefined. Critical points occur where the derivative is zero or does not exist.</p>
<p>First, let's find the derivative of the function:</p>
<p>$$f(x)=(x-2)^{2 / 3}(2 x+1)$$</p>
<p>We apply the product rule for differentiation, which states that $$(uv)' = u'v + uv'$$, where $$u = (x-2)^{2/3}$$ and $$v = 2x + 1$$.</p>
<p>We need the derivatives of $$u$$ and $$v$$:</p>
<p>For $$u = (x-2)^{2/3}$$, we use the chain rule:</p>
<p>$$u'(x) = \frac{d}{dx}[(x-2)^{2/3}] = \frac{2}{3}(x-2)^{-1/3} \cdot 1 = \frac{2}{3}(x-2)^{-1/3}$$</p>
<p>The derivative of $$v$$ is straightforward, as $$v = 2x + 1$$:</p>
<p>$$v'(x) = 2$$</p>
<p>Now we apply the product rule:</p>
<p>$$f'(x) = u'(x)v(x) + u(x)v'(x)$$</p>
<p>Substituting $$u$$, $$u'$$, and $$v$$, we get:</p>
<p>$$f'(x) = \left( \frac{2}{3}(x-2)^{-1/3} \right)(2x+1) + \left( (x-2)^{2/3} \right)(2)$$</p>
<p>This simplifies to:</p>
<p>$$f'(x) = \frac{2(2x + 1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3}$$</p>
<p>For critical points, we need to solve $$f'(x) = 0$$ or where it is undefined.</p>
<p>1. <strong>Solve for where the derivative is zero:</strong></p>
<p>$$\frac{2(2x + 1)}{3(x-2)^{1/3}} + 2(x-2)^{2/3} = 0$$</p>
<p>Combining like terms in a common denominator, we get:</p>
<p>$$\frac{2(2x + 1) + 6(x - 2)}{3(x-2)^{1/3}} = 0$$</p>
<p>Simplifying the numerator:</p>
<p>$$2(2x + 1) + 6(x-2) = 4x + 2 + 6x - 12 = 10x - 10 = 10(x-1)$$</p>
<p>So:</p>
<p>$$\frac{10(x-1)}{3(x-2)^{1/3}} = 0$$</p>
<p>The numerator is zero when:</p>
<p>$$10(x-1) = 0$$</p>
<p>Therefore:</p>
<p>$$x = 1$$</p>
<p>2. <strong>Solve for where the derivative is undefined:</strong></p>
<p>The denominator, $$3(x-2)^{1/3}$$, is undefined when $$(x-2)^{1/3} = 0$$, which happens at:</p>
<p>$$x = 2$$</p>
<p>From the above analysis, the critical points are at $$x = 1$$ and $$x = 2$$. Thus, there are 2 critical points.</p>
<p>Therefore, the number of critical points of the function $$f(x)=(x-2)^{2 / 3}(2 x+1)$$ is:</p>
<p><b>Option A</b></p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv7v3jz2 | maths | application-of-derivatives | maxima-and-minima | <p>Let a rectangle ABCD of sides 2 and 4 be inscribed in another rectangle PQRS such that the vertices of the rectangle ABCD lie on the sides of the rectangle PQRS. Let a and b be the sides of the rectangle PQRS when its area is maximum. Then (a+b)$$^2$$ is equal to :</p> | [{"identifier": "A", "content": "64"}, {"identifier": "B", "content": "80"}, {"identifier": "C", "content": "60"}, {"identifier": "D", "content": "72"}] | ["D"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwgejsnu/15565bcf-b8bf-4793-9f53-176aa69e81ff/facbac90-1771-11ef-910c-8da948f7ddd9/file-1lwgejsnv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwgejsnu/15565bcf-b8bf-4793-9f53-176aa69e81ff/facbac90-1771-11ef-910c-8da948f7ddd9/file-1lwgejsnv.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 Morning Shift Mathematics - Application of Derivatives Question 4 English Explanation"></p>
<p>$$\begin{aligned}
& P D=4 \cos \theta \Rightarrow P S=4 \cos \theta+2 \sin \theta \\
& D S=2 \sin \theta \\
& A P=4 \sin \theta \\
& Q A=2 \cos \theta \Rightarrow P Q=2 \cos \theta+4 \sin \theta \\
& \Rightarrow \text { Area of } P Q R S=4(2 \cos \theta+\sin \theta)(\cos \theta+2 \sin \theta) \\
&=4\left[2 \cos ^2 \theta+2 \sin ^2 \theta+5 \sin \theta \cos \theta\right] \\
&=8+10 \sin 2 \theta
\end{aligned}$$</p>
<p>Area is maximum when $$\sin 2 \theta=1 \Rightarrow \theta=45^{\circ}$$</p>
<p>$$\Rightarrow$$ Maximum area $$=8+10=18$$</p>
<p>$$\therefore P S=4 \times \frac{1}{\sqrt{2}}+2 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}}$$</p>
<p>$$\begin{aligned}
& P Q=2 \times \frac{1}{\sqrt{2}}+4 \times \frac{1}{\sqrt{2}}=\frac{6}{\sqrt{2}} \\
\therefore \quad & (a+b)^2=\left(\frac{6}{\sqrt{2}}+\frac{6}{\sqrt{2}}\right)^2=\left(\frac{12}{\sqrt{2}}\right)=(6 \sqrt{2})^2=72
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
lv9s20kb | maths | application-of-derivatives | maxima-and-minima | <p>Let the maximum and minimum values of $$\left(\sqrt{8 x-x^2-12}-4\right)^2+(x-7)^2, x \in \mathbf{R}$$ be $$\mathrm{M}$$ and $$\mathrm{m}$$, respectively. Then $$\mathrm{M}^2-\mathrm{m}^2$$ is equal to _________.</p> | [] | null | 1600 | <p>$$\begin{aligned}
& \text { Let } y=\sqrt{8 x-x^2-12} \Rightarrow(x-4)^2+y^2=2^2 \\
& \Rightarrow d=(y-4)^2+(x-7)^2
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lweqs4gr/cb92f82c-cbcf-4ca2-8ad5-d05029ed7bb2/3ea356b0-1688-11ef-9ee8-13752e98d8d6/file-1lweqs4gs.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lweqs4gr/cb92f82c-cbcf-4ca2-8ad5-d05029ed7bb2/3ea356b0-1688-11ef-9ee8-13752e98d8d6/file-1lweqs4gs.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 - Application of Derivatives Question 2 English Explanation"></p>
<p>$$\begin{aligned}
\Rightarrow & M=P A^2=16+25=41 \\
& m=P Q^2=(\sqrt{16+9}-2)^2=9 \\
\Rightarrow & M^2-m^2=1681-81=1600
\end{aligned}$$</p> | integer | jee-main-2024-online-5th-april-evening-shift |
WKbUFfnFg5soehXh | maths | application-of-derivatives | mean-value-theorem | If $$2a+3b+6c=0,$$ $$\left( {a,b,c \in R} \right)$$ then the quadratic equation $$a{x^2} + bx + c = 0$$ has | [{"identifier": "A", "content": "at least one root in $$\\left[ {0,1} \\right]$$"}, {"identifier": "B", "content": "at least one root in $$\\left[ {2,3} \\right]$$"}, {"identifier": "C", "content": "at least one root in $$\\left[ {4,5} \\right]$$"}, {"identifier": "D", "content": "none of these "}] | ["A"] | null | Let $$f\left( x \right) = {{a{x^3}} \over 3} + {{b{x^2}} \over 2} + cx \Rightarrow f\left( 0 \right) = 0$$ and $$f(1)$$
<br><br>$$ = {a \over 3} + {b \over 2} + c = {{2a + 3b + 6c} \over 6} = 0$$
<br><br>Also $$f(x)$$ is continuous and differentiable in $$\left[ {0,1} \right]$$ and
<br><br>$$\left[ {0,1\left[ {.\,\,} \right.} \right.$$ So by Rolle's theorem $$f'\left( x \right) = 0.$$
<br><br>i.e $$\,\,a{x^2} + bx + c = 0$$ has at least one root in $$\left[ {0,1} \right].$$ | mcq | aieee-2002 |
s1Rt0Q9slH5xcy55 | maths | application-of-derivatives | mean-value-theorem | If $$2a+3b+6c=0$$, then at least one root of the equation
<br/>$$a{x^2} + bx + c = 0$$ lies in the interval | [{"identifier": "A", "content": "$$(1, 3)$$ "}, {"identifier": "B", "content": "$$(1, 2)$$ "}, {"identifier": "C", "content": "$$(2, 3)$$ "}, {"identifier": "D", "content": "$$(0, 1)$$ "}] | ["D"] | null | Let us define a function
<br><br>$$f\left( x \right) = {{ax{}^3} \over 3} + {{b{x^2}} \over 2} + cx$$
<br><br>Being polynomial, it is continuous and differentiable, also,
<br><br>$$f\left( 0 \right) = 0\,$$ and $$\,\,f\left( 1 \right) = {a \over 3} + {b \over 2} + c$$
<br><br>$$ \Rightarrow f\left( 1 \right) = {{2a + 3b + 6c} \over 6} = 0$$ (given)
<br><br>$$\therefore$$ $$f\left( 0 \right) = f\left( 1 \right)$$
<br><br>$$\therefore$$ $$f(x)$$ satisfies all conditions of Rolle's theorem
<br><br>therefore $$f'\left( x \right) = 0$$ has a root in $$\left( {0,1} \right)$$
<br><br>i.e. $$a{x^2} + bx + c = 0$$ has at lease one root in $$(0, 1)$$ | mcq | aieee-2004 |
FvFBxKtfz58VB2r3 | maths | application-of-derivatives | mean-value-theorem | If the equation $${a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$
<br/>$${a_1} \ne 0,n \ge 2,$$ has a positive root $$x = \alpha $$, then the equation
<br/>$$n{a_n}{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + ........... + {a_1} = 0$$ has a positive root, which is | [{"identifier": "A", "content": "greater than $$\\alpha $$ "}, {"identifier": "B", "content": "smaller than $$\\alpha $$ "}, {"identifier": "C", "content": "greater than or equal to smaller than $$\\alpha $$ "}, {"identifier": "D", "content": "equal to smaller than $$\\alpha $$ "}] | ["B"] | null | Let $$f\left( x \right) = {a_n}{x^n} + {a_{n - 1}}{x^{n - 1}} + ........... + {a_1}x = 0$$
<br><br>The other given equation,
<br><br>$$na{}_n{x^{n - 1}} + \left( {n - 1} \right){a_{n - 1}}{x^{n - 2}} + .... + {a_1} = 0 = f'\left( x \right)$$
<br><br>Given $${a_1} \ne 0 \Rightarrow f\left( 0 \right) = 0$$
<br><br>Again $$f(x)$$ has root $$\alpha ,\,\,\, \Rightarrow f\left( \alpha \right) = 0$$
<br><br>$$\therefore$$ $$\,\,\,\,\,f\left( 0 \right) = f\left( \alpha \right)$$
<br><br>$$\therefore$$ By Roll's theorem $$f'(x)=0$$ has root be-
<br><br>tween $$\left( {0,\alpha } \right)$$
<br><br>Hence $$f'(x)$$ has a positive root smaller than $$\alpha .$$ | mcq | aieee-2005 |
UJUItPkT64O9QtSG | maths | application-of-derivatives | mean-value-theorem | Let f be differentiable for all x. If f(1) = -2 and f'(x) $$ \ge $$ 2 for
<br/>x $$ \in \left[ {1,6} \right]$$, then | [{"identifier": "A", "content": "f(6) $$ \\ge $$ 8"}, {"identifier": "B", "content": "f(6) < 8"}, {"identifier": "C", "content": "f(6) < 5"}, {"identifier": "D", "content": "f(6) = 5"}] | ["A"] | null | As $$\,\,f\left( 1 \right) = - 2\,\,\& \;\,f'\left( x \right) \ge 2\,\forall x \in \left[ {1,6} \right]$$
<br><br>Applying Lagrange's mean value theorem
<br><br>$${{f\left( 6 \right) - f\left( 1 \right)} \over 5} = f'\left( c \right) \ge 2$$
<br><br>$$ \Rightarrow f\left( 6 \right) \ge 10 + f\left( 1 \right)$$
<br><br>$$ \Rightarrow f\left( 6 \right) \ge 10 - 2$$
<br><br>$$ \Rightarrow f\left( 6 \right) \ge 8.$$ | mcq | aieee-2005 |
yVtBIed6SEB53zdc | maths | application-of-derivatives | mean-value-theorem | A value of $$c$$ for which conclusion of Mean Value Theorem holds for the function $$f\left( x \right) = {\log _e}x$$ on the interval $$\left[ {1,3} \right]$$ is | [{"identifier": "A", "content": "$${\\log _3}e$$ "}, {"identifier": "B", "content": "$${\\log _e}3$$"}, {"identifier": "C", "content": "$$2\\,\\,{\\log _3}e$$ "}, {"identifier": "D", "content": "$${1 \\over 2}{\\log _3}e$$ "}] | ["C"] | null | Using Lagrange's Mean Value Theorem
<br><br>Let $$f(x)$$ be a function defined on $$\left[ {a,b} \right]$$
<br><br>then, $$f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$
<br><br>$$c\,\, \in \left[ {a,b} \right]$$
<br><br>$$\therefore$$ Given $$f\left( x \right) = {\log _e}x$$
<br><br>$$\therefore$$ $$f'\left( x \right) = {1 \over x}$$
<br><br>$$\therefore$$ equation $$(i)$$ become $${1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}$$
<br><br>$$ \Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}$$
<br><br>$$ \Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e$$ | mcq | aieee-2007 |
9i0za95SuBx6OsEG | maths | application-of-derivatives | mean-value-theorem | If $$f$$ and $$g$$ are differentiable functions in $$\left[ {0,1} \right]$$ satisfying
<br/>$$f\left( 0 \right) = 2 = g\left( 1 \right),g\left( 0 \right) = 0$$ and $$f\left( 1 \right) = 6,$$ then for some $$c \in \left] {0,1} \right[$$ | [{"identifier": "A", "content": "$$f'\\left( c \\right) = g'\\left( c \\right)$$ "}, {"identifier": "B", "content": "$$f'\\left( c \\right) = 2g'\\left( c \\right)$$"}, {"identifier": "C", "content": "$$2f'\\left( c \\right) = g'\\left( c \\right)$$"}, {"identifier": "D", "content": "$$2f'\\left( c \\right) = 3g'\\left( c \\right)$$"}] | ["B"] | null | Since, $$f$$ and $$g$$ both are continuous function on $$\left[ {0,1} \right]$$
<br><br>and differentiable on $$\left( {0,1} \right)$$ then $$\exists c \in \left( {0,1} \right)$$ such that
<br><br>$$f'\left( c \right) = {{f\left( 1 \right) - f\left( 0 \right)} \over 1} = {{6 - 2} \over 1} = 4$$
<br><br>and $$g'\left( c \right) = {{g\left( 1 \right) - g\left( 0 \right)} \over 1} = {{2 - 0} \over 1} = 2$$
<br><br>Thus, we get $$f'\left( c \right) = 2g'\left( c \right)$$ | mcq | jee-main-2014-offline |
RiuUixc2KvzB1Mn3kM7k9k2k5e32k5u | maths | application-of-derivatives | mean-value-theorem | Let the function, Ζ:[-7, 0]$$ \to $$R be continuous on [-7,0] and differentiable on (-7, 0). If Ζ(-7) = -
3 and Ζ'(x) $$ \le $$ 2, for all x $$ \in $$ (-7,0), then for all such functions Ζ, Ζ(-1) + Ζ(0) lies in the interval:
| [{"identifier": "A", "content": "$$\\left[ { - 6,20} \\right]$$"}, {"identifier": "B", "content": "$$\\left( { - \\infty ,\\left. {20} \\right]} \\right.$$"}, {"identifier": "C", "content": "$$\\left[ { - 3,11} \\right]$$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,\\left. {11} \\right]} \\right.$$"}] | ["B"] | null | Using Lagrangeβs Mean Value Theorem in [β7, β1]
<br><br>$${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}$$ = f'(c<sub>1</sub>)
<br><br>As Ζ'(x) $$ \le $$ 2 then f'(c<sub>1</sub>) $$ \le $$ 2
<br><br>$$ \therefore $$ $${{f\left( { - 1} \right) - f\left( { - 7} \right)} \over { - 1 - \left( { - 7} \right)}}$$ $$ \le $$ 2
<br><br>$$ \Rightarrow $$ $${{f\left( { - 1} \right) + 3} \over 6}$$ $$ \le $$ 2
<br><br>$$ \Rightarrow $$ f(-1) $$ \le $$ 9
<br><br>Using Lagrangeβs Mean Value Theorem in [β7, 0]
<br><br>$$ \Rightarrow $$ $${{f\left( 0 \right) - f\left( { - 7} \right)} \over {0 - \left( { - 7} \right)}}$$ $$ \le $$ 2
<br><br>$$ \Rightarrow $$ f(0) $$ \le $$ 11
<br><br>$$ \therefore $$ Ζ(-1) + Ζ(0) $$ \le $$ 11 + 9
<br><br>$$ \Rightarrow $$ Ζ(-1) + Ζ(0) $$ \le $$ 20 | mcq | jee-main-2020-online-7th-january-morning-slot |
SY1QQZrwAcg8toamm57k9k2k5fnb4sd | maths | application-of-derivatives | mean-value-theorem | The value of c in the Lagrange's mean value theorem for the function <br/>Ζ(x) = x<sup>3</sup>
- 4x<sup>2</sup>
+ 8x + 11,
when x $$ \in $$ [0, 1] is:
| [{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${{\\sqrt 7 - 2} \\over 3}$$"}, {"identifier": "C", "content": "$${{4 - \\sqrt 5 } \\over 3}$$"}, {"identifier": "D", "content": "$${{4 - \\sqrt 7 } \\over 3}$$"}] | ["D"] | null | Ζ(x) = x<sup>3</sup>
- 4x<sup>2</sup>
+ 8x + 11
<br><br>f(0) = 11
<br><br>f(1) = 16
<br><br>Using LMVT
<br><br>f'(c) = $${{f\left( 1 \right) - f\left( 0 \right)} \over {1 - 0}}$$
<br><br>$$ \Rightarrow $$ 3c<sup>2</sup>
β 8c + 8 = $${{16 - 11} \over {1 - 0}}$$
<br><br>$$ \Rightarrow $$ 3c<sup>2</sup>
β 8c + 3 = 0
<br><br>$$ \therefore $$ c = $${{8 \pm 2\sqrt 7 } \over 6}$$
<br><br>$$ \therefore $$ c = $${{4 - \sqrt 7 } \over 3}$$ as c $$ \in $$ [0, 1] | mcq | jee-main-2020-online-7th-january-evening-slot |
7HHI3mgBSQbnJgC9CJ7k9k2k5gyu0fm | maths | application-of-derivatives | mean-value-theorem | If c is a point at which Rolle's theorem holds
for the function,
<br/>f(x) = $${\log _e}\left( {{{{x^2} + \alpha } \over {7x}}} \right)$$ in the
interval [3, 4], where a $$ \in $$ R, then Ζ''(c) is equal
to | [{"identifier": "A", "content": "$${1 \\over {12}}$$"}, {"identifier": "B", "content": "$${{\\sqrt 3 } \\over 7}$$"}, {"identifier": "C", "content": "$$-{1 \\over {12}}$$"}, {"identifier": "D", "content": "$$-{1 \\over {24}}$$"}] | ["A"] | null | For Rolleβs theorem to be applicable in [3, 4]
<br><br>Ζ(3) = Ζ(4)
<br><br>$$ \Rightarrow $$ $${\log _e}\left( {{{9 + \alpha } \over {21}}} \right) = {\log _e}\left( {{{16 + \alpha } \over {28}}} \right)$$
<br><br>$$ \Rightarrow $$ $$\left( {{{9 + \alpha } \over {21}}} \right) = \left( {{{16 + \alpha } \over {28}}} \right)$$
<br><br>$$ \Rightarrow $$ 36 + 4$$\alpha $$ = 48 + 3$$\alpha $$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = 12
<br><br>According to Rolleβs theorem, f'(c) = 0
<br>where c $$ \in $$ (3, 4)
<br><br>f'(x) = $${{7x} \over {{x^2} + 12}}$$ $$ \times $$ $$\left( {{{2x\left( {7x} \right) - \left( {{x^2} + 12} \right)7} \over {{7^2}{x^2}}}} \right)$$
<br><br>$$ \Rightarrow $$ f'(x) = $${{{x^2} - 12} \over {\left( {{x^2} + 12} \right)x}}$$
<br><br>$$ \therefore $$ f'(c) = $${{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}$$
<br><br>From Rolle's theorem
<br><br>$${{{c^2} - 12} \over {\left( {{c^2} + 12} \right)c}}$$ = 0
<br><br>$$ \Rightarrow $$ c<sup>2</sup> = 12
<br><br>Now f''(c) = $${{2c\left( {{c^3} + 12c} \right) - \left( {{c^2} - 12} \right)\left( {3{c^2} + 12} \right)} \over {{{\left( {{c^2} + 12} \right)}^2}{c^2}}}$$
<br><br>= $${{2\left( {12} \right)\left( {24} \right) - 0} \over {{{\left( {24} \right)}^2} \times 12}}$$
<br><br>= $${1 \over {12}}$$ | mcq | jee-main-2020-online-8th-january-morning-slot |
hRKCTLC497lJlskzXd1kls4t2ij | maths | application-of-derivatives | mean-value-theorem | If Rolle's theorem holds for the function $$f(x) = {x^3} - a{x^2} + bx - 4$$, $$x \in [1,2]$$ with $$f'\left( {{4 \over 3}} \right) = 0$$, then ordered pair (a, b) is equal to : | [{"identifier": "A", "content": "($$-$$5, $$-$$8)"}, {"identifier": "B", "content": "(5, $$-$$8)"}, {"identifier": "C", "content": "($$-$$5, 8)"}, {"identifier": "D", "content": "(5, 8)"}] | ["D"] | null | $$f(1) = f(2)$$<br><br>$$ \Rightarrow 1 - a + b - 4 = 8 - 4a + 2b - 4$$<br><br>$$3a - b = 7$$ ..... (1)<br><br>$$f'(x) = 3{x^2} - 2ax + b$$<br><br>$$ \Rightarrow f'\left( {{4 \over 3}} \right) = 0 \Rightarrow 3 \times {{16} \over 9} - {8 \over 3}a + b = 0$$<br><br>$$ \Rightarrow - 8a + 3b = - 16$$ ..... (2)<br><br>$$ \therefore $$ $$a = 5,b = 8$$ | mcq | jee-main-2021-online-25th-february-morning-slot |
IvgThkO2TD7WaLN3jF1klui2sw9 | maths | application-of-derivatives | mean-value-theorem | Let f be any function defined on R and let it satisfy the condition : $$|f(x) - f(y)|\, \le \,|{(x - y)^2}|,\forall (x,y) \in R$$<br/><br/>If f(0) = 1, then : | [{"identifier": "A", "content": "f(x) can take any value in R"}, {"identifier": "B", "content": "$$f(x) < 0,\\forall x \\in R$$"}, {"identifier": "C", "content": "$$f(x) > 0,\\forall x \\in R$$"}, {"identifier": "D", "content": "$$f(x) = 0,\\forall x \\in R$$"}] | ["C"] | null | $$|f(x) - f(y)|\, \le \,|{(x - y)^2}|$$<br><br>$$ \Rightarrow \left| {{{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|x - y|$$<br><br>$$ \Rightarrow \left| {\mathop {\lim }\limits_{x \to y} {{f(x) - f(y)} \over {x - y}}} \right|\, \le \,|\mathop {\lim }\limits_{x \to y} (x - y)|$$<br><br>$$ \Rightarrow |f'(x)|\, \le 0$$<br><br>$$ \Rightarrow f'(x) = 0$$<br><br>$$ \Rightarrow f(x)$$ is constant function.<br><br>$$ \because $$ $$f(0)$$ = 1 then f(x) = 1 | mcq | jee-main-2021-online-26th-february-morning-slot |
1lguw40hj | maths | application-of-derivatives | mean-value-theorem | <p>Let $$f:[2,4] \rightarrow \mathbb{R}$$ be a differentiable function such that $$\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]$$ with $$f(2)=\frac{1}{2}$$ and $$f(4)=\frac{1}{4}$$.</p>
<p>Consider the following two statements :</p>
<p>(A) : $$f(x) \leq 1$$, for all $$x \in[2,4]$$</p>
<p>(B) : $$f(x) \geq \frac{1}{8}$$, for all $$x \in[2,4]$$</p>
<p>Then,</p> | [{"identifier": "A", "content": "Neither statement (A) nor statement (B) is true"}, {"identifier": "B", "content": "Only statement (A) is true"}, {"identifier": "C", "content": "Only statement (B) is true"}, {"identifier": "D", "content": "Both the statements $$(\\mathrm{A})$$ and (B) are true"}] | ["D"] | null | Given, $$\left(x \log _{e} x\right) f^{\prime}(x)+\left(\log _{e} x\right) f(x)+f(x) \geq 1, x \in[2,4]$$
<br/><br/>$$
\left(x\log _e x\right) f^{\prime}(x)+f(x)\left[\log _e x+1\right] \geq 1$$
<br/><br/>$$ \Rightarrow $$ $$
\frac{d}{d x}\left[x \log _e x f(x)\right] \geq 1$$
<br/><br/>$$
\begin{aligned}
&\Rightarrow \frac{d}{d x}\left[x \log _e x f(x)-x\right] \geq 0 \quad\left[\because \frac{d}{d x}(x)=1\right] \\\\
& \forall x \in[2,4]
\end{aligned}
$$
<br/><br/>Let $g(x)=x \log _e x f(x)-x$
<br/><br/>As $g(x) \geq 0, \forall x \in[2,4],
g(x)$ is an increasing function in $[2,4]$
<br/><br/>$$
\begin{aligned}
g(2) & =2 \log _e 2 f(2)-2 \\\\
& =\log _e 2-2 \quad\left[\because f(x)=\frac{1}{2}\right]
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& g(4)=4 \log _e 4 f(4)-4=\log _e 4-4 \\\\
&=2\left(\log _e 2-2\right)\\\\
& {\left[\therefore f(4)=\frac{1}{4}\right] }
\end{aligned}
$$
<br/><br/>As, $g(x)$ is an increasing function,
<br/><br/>$$
\begin{aligned}
& g(2) \leq g(x) \leq g(4) \\\\
& \log _e 2-2 \leq g(x) \leq 2\left(\log _e 2-2\right) \\\\
& \log _e 2-2 \leq x \log _e x f(x)-x \leq 2\left(\log _e 2-2\right)
\end{aligned}
$$
<br/><br/>$$
\frac{\log _e 2-2+x}{x \log _e x} \leq f(x) \leq \frac{2\left(\log _e 2-2\right)+x}{x \log _e x}
$$
<br/><br/>$$
\begin{aligned}
& \text {Now for } x \in[2,4] \\\\
& \begin{aligned}
\frac{\log _e 2-2+x}{x \log _e x} & \leq \frac{2\left(\log _e 2-2\right)+e^2}{2 \log _e 2} =1-\frac{1}{\log _e 2}<1
\end{aligned}
\end{aligned}
$$
<br/><br/>$$
\Rightarrow \mathrm{f}(\mathrm{x}) \leq 1 \text { for } \mathrm{x} \in[2,4]
$$
<br/><br/>$$
\text { Also for } \mathrm{x} \in[2,4] \text { : }
$$
<br/><br/>Now,
<br/><br/>$$
\begin{aligned}
\frac{2\left(\log _e 2-2\right)+2}{2 \log _e 2} & \geq \frac{\log _e 2-2+4}{4 \log _e 4} =\frac{1}{8}+\frac{1}{2 \log _e 2}>\frac{1}{8}
\end{aligned}
$$
<br/><br/>$$
\therefore f(x) \geq \frac{1}{8}, \forall x \in[2,4]
$$
<br/><br/>Hence, both statements $A$ and $B$ are true.
<br/><br/><b>Note :</b> LMVT on $(\mathrm{yx}(\ln \mathrm{x}))$ not satisfied.
<br/><br/>Hence no such function exists.
<br/><br/>Therefore it should be bonus. | mcq | jee-main-2023-online-11th-april-morning-shift |
lsblgir9 | maths | application-of-derivatives | mean-value-theorem | Let for a differentiable function $f:(0, \infty) \rightarrow \mathbf{R}, f(x)-f(y) \geqslant \log _{\mathrm{e}}\left(\frac{x}{y}\right)+x-y, \forall x, y \in(0, \infty)$. Then $\sum\limits_{n=1}^{20} f^{\prime}\left(\frac{1}{n^2}\right)$ is equal to ____________. | [] | null | 2890 | <p>$$\begin{aligned}
& f(x)-f(y) \geq \ln x-\ln y+x-y \\
& \frac{f(x)-f(y)}{x-y} \geq \frac{\ln x-\ln y}{x-y}+1
\end{aligned}$$</p>
<p>Let $$x>y$$</p>
<p>$$\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{-}\right) \geq \frac{1}{x}+1\quad\text{.... (1)}$$</p>
<p>Let $$x< y$$</p>
<p>$$\lim _\limits{y \rightarrow x} f^{\prime}\left(x^{+}\right) \leq \frac{1}{x}+1 \quad\text{.... (2)}$$</p>
<p>$$\begin{aligned}
& \mathrm{f}^1\left(\mathrm{x}^{-}\right)=\mathrm{f}^1\left(\mathrm{x}^{+}\right) \\
& \mathrm{f}^1(\mathrm{x})=\frac{1}{\mathrm{x}}+1 \\
& \mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}^2}\right)=\mathrm{x}^2+1
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \sum_{x=1}^{20}\left(x^2+1\right)=\sum_{x-1}^{20} x^2+20 \\
& =\frac{20 \times 21 \times 41}{6}+20 \\
& =2890
\end{aligned}$$</p> | integer | jee-main-2024-online-27th-january-morning-shift |
1ty77dtcxjHwzLQg | maths | application-of-derivatives | monotonicity | A function is matched below against an interval where it is supposed to be
increasing. Which of the following pairs is incorrectly matched? | [{"identifier": "A", "content": "<table class=\"tg\">\n <tbody><tr>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Interval</span></th>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Function</span></th>\n </tr>\n <tr>\n <td class=\"tg-s6z2\">(- $$\\infty $$, $$\\infty $$)</td>\n <td class=\"tg-s6z2\">x<sup>3</sup> - 3x<sup>2</sup> + 3x + 3</td>\n </tr>\n</tbody></table>"}, {"identifier": "B", "content": "<table class=\"tg\">\n <tbody><tr>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Interval</span></th>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Function</span></th>\n </tr>\n <tr>\n <td class=\"tg-s6z2\">[2, $$\\infty $$)</td>\n <td class=\"tg-s6z2\">2x<sup>3</sup> - 3x<sup>2</sup> - 12x + 6</td>\n </tr>\n</tbody></table>"}, {"identifier": "C", "content": "<table class=\"tg\">\n <tbody><tr>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Interval</span></th>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Function</span></th>\n </tr>\n <tr>\n <td class=\"tg-s6z2\">$$\\left( { - \\infty ,{1 \\over 3}} \\right]$$</td>\n <td class=\"tg-s6z2\">3x<sup>2</sup> - 2x + 1</td>\n </tr>\n</tbody></table>"}, {"identifier": "D", "content": "<table class=\"tg\">\n <tbody><tr>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Interval</span></th>\n <th class=\"tg-s6z2\"><span style=\"font-weight:bold\">Function</span></th>\n </tr>\n <tr>\n <td class=\"tg-s6z2\">($$ - \\infty $$, - 4 )</td>\n <td class=\"tg-s6z2\">x<sup>3</sup> + 6x<sup>2</sup> + 6</td>\n </tr>\n</tbody></table>"}] | ["C"] | null | Clearly function $$f\left( x \right) = 3{x^2} - 2x + 1$$ is increasing
<br><br> when $$f'\left( x \right) = 6x - 2 \ge 0 \Rightarrow \,\,\,\,\,x \in \left[ {1/3,\left. \infty \right)} \right.$$
<br><br>$$\therefore$$ $$f(x)$$ is incorrectly matched with $$\left( { - \infty ,{1 \over 3}} \right)$$ | mcq | aieee-2005 |
jaXMH0IezPGTzEA1 | maths | application-of-derivatives | monotonicity | The function $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$ is an incresing function in | [{"identifier": "A", "content": "$$\\left( {0,{\\pi \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - {\\pi \\over 2},{\\pi \\over 2}} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { {\\pi \\over 4},{\\pi \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {\\pi \\over 2},{\\pi \\over 4}} \\right)$$"}] | ["D"] | null | Given $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sin x + \cos x} \right)$$
<br><br>$$f'\left( x \right) = {1 \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}.\left( {\cos x - \sin x} \right)$$
<br><br>$$ = {{\sqrt 2 .\left( {{1 \over {\sqrt 2 }}\cos x - {1 \over {\sqrt 2 }}\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
<br><br>$$ = {{\left( {\cos {\pi \over 4}.\cos x - \sin {\pi \over 4}.\sin x} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
<br><br>$$\therefore$$ $$f'\left( x \right) = {{\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)} \over {1 + {{\left( {\sin x + \cos x} \right)}^2}}}$$
<br><br>if $$f'\left( x \right) > O$$ then $$f\left( x \right)$$ is increasing function.
<br><br>Hence $$f(x)$$ is increasing, if $$ - {\pi \over 2} < x + {\pi \over 4} < {\pi \over 2}$$
<br><br>$$ \Rightarrow - {{3\pi } \over 4} < x < {\pi \over 4}$$
<br><br>Hence, $$f(x)$$ is increasing when $$n \in \left( { - {\pi \over 2},{\pi \over 4}} \right)$$ | mcq | aieee-2007 |
ecU4XGHCPdnGACZd | maths | application-of-derivatives | monotonicity | How many real solutions does the equation
<br/>$${x^7} + 14{x^5} + 16{x^3} + 30x - 560 = 0$$ have? | [{"identifier": "A", "content": "$$7$$ "}, {"identifier": "B", "content": "$$1$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$5$$ "}] | ["B"] | null | Let $$f\left( x \right) = {x^7} + 14{x^5} + 16{x^3} + 30x - 560$$
<br><br>$$ \Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\,\forall x \in R$$
<br><br>$$ \Rightarrow f$$ is an increasing function on $$R$$
<br><br>Also $$\mathop {\lim }\limits_{x \to \infty } \,\,f\left( x \right) = \infty $$ and $$\mathop {\lim }\limits_{x \to - \infty } \,\,f\left( x \right) = - \infty $$
<br><br>$$ \Rightarrow $$ The curve $$y = f\left( x \right)$$ crosses $$x$$-axis only once.
<br><br>$$\therefore$$ $$f\left( x \right) = 0$$ has exactly one real root. | mcq | aieee-2008 |
1hCumHByo0ZaOkXiKDLcf | maths | application-of-derivatives | monotonicity | Let f(x) = sin<sup>4</sup>x + cos<sup>4</sup> x. Then <i>f</i> is an increasing function in the interval : | [{"identifier": "A", "content": "$$] 0, \\frac{\\pi}{4}[$$"}, {"identifier": "B", "content": "$$] \\frac{\\pi}{4}, \\frac{\\pi}{2}[$$"}, {"identifier": "C", "content": "$$] \\frac{\\pi}{2}, \\frac{5 \\pi}{8}[$$"}, {"identifier": "D", "content": "$$] \\frac{5 \\pi}{8}, \\frac{3 \\pi}{4}[$$"}] | ["B"] | null | f(x) = sin<sup>4</sup>x + cos<sup>4</sup>x
<br><br>$$ \therefore $$ f'(x) = 4sin<sup>3</sup>x cosx + 4cos<sup>3</sup>x ($$-$$ sinx)
<br><br>= 4sinx cosx (sin<sup>2</sup>x $$-$$ cos<sup>2</sup>x)
<br><br>= $$-$$ 2sin2x cos2x
<br><br>= $$-$$ sin4x
<br><br>As, f(x) is increasing function when f'(x) > 0
<br><br>$$ \Rightarrow $$ $$-$$ sin4x > 0
<br><br>$$ \Rightarrow $$ sin4x < 0
<br><br>$$ \therefore $$ $$\pi $$ < 4x < 2$$\pi $$
<br><br>$${\pi \over 4} < x < {\pi \over 2}$$
<br><br>$$ \therefore $$ x $$ \in $$ $$\left( {{\pi \over 4},{\pi \over 2}} \right)$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
QxDDYKNpsGnt8PZt0K0rq | maths | application-of-derivatives | monotonicity | The function f defined by
<br/><br/>f(x) = x<sup>3</sup> $$-$$ 3x<sup>2</sup> + 5x + 7 , is : | [{"identifier": "A", "content": "increasing in <b>R</b>."}, {"identifier": "B", "content": "decreasing in <b>R</b>."}, {"identifier": "C", "content": "decreasing in (0, $$\\infty $$) and increasing in ($$-$$ $$\\infty $$, 0)"}, {"identifier": "D", "content": "increasing in (0, $$\\infty $$) and decreasing in ($$-$$ $$\\infty $$, 0)"}] | ["A"] | null | <p>The given function is</p>
<p>$$f(x) = {x^2} - 3{x^2} + 5x + 7$$</p>
<p>$$f'(x) = 3{x^2} - 6x + 5$$</p>
<p>The discriminant of the above quadratic equation is</p>
<p>$$\Delta = 36 - 4(3)(5) = 36 - 60 < 0$$</p>
<p>Therefore, $$f'(x) > 0\,\forall x \in {R^ + }$$</p>
<p>Also, $$f'(x) > 0\,\forall x \in {R^ - }$$</p>
<p>Therefore, the given function f is increasing in R.</p> | mcq | jee-main-2017-online-9th-april-morning-slot |
XB56wgM0Ps8wqSh3VSc4L | maths | application-of-derivatives | monotonicity | Let f(x) = $${x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }},\,\,$$ x $$\, \in $$ R, where a, b and d are non-zero real constants. Then : | [{"identifier": "A", "content": "f is an increasing function of x"}, {"identifier": "B", "content": "f is neither increasing nor decreasing function of x"}, {"identifier": "C", "content": "f ' is not a continuous function of x"}, {"identifier": "D", "content": "f is a decreasing function of x"}] | ["A"] | null | $$f\left( x \right) = {x \over {\sqrt {{a^2} + {x^2}} }} - {{d - x} \over {\sqrt {{b^2} + {{\left( {d - x} \right)}^2}} }}$$
<br><br>$$f'\left( x \right) = {{{a^2}} \over {{{\left( {{a^2} + {x^2}} \right)}^{3/2}}}} + {{{b^2}} \over {{{\left( {{b^2} + {{\left( {d - x} \right)}^2}} \right)}^{3/2}}}} > 0\forall x \in R$$
<br><br>f(x) is an increasing function. | mcq | jee-main-2019-online-11th-january-evening-slot |
G8bQuG05kI5WhxXcgg9bh | maths | application-of-derivatives | monotonicity | If the function f given by f(x) = x<sup>3</sup> β 3(a β 2)x<sup>2</sup> + 3ax + 7, for some a$$ \in $$R is increasing in (0, 1] and decreasing in [1, 5), then a root of the equation, $${{f\left( x \right) - 14} \over {{{\left( {x - 1} \right)}^2}}} = 0\left( {x \ne 1} \right)$$ is : | [{"identifier": "A", "content": "$$-$$ 7"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "6"}] | ["C"] | null | f '(x) = 3x<sup>2</sup> $$-$$ 6(a $$-$$ 2)x + 3a
<br><br>f '(x) $$ \ge $$ 0 $$\forall $$ x $$ \in $$ (0, 1]
<br><br>f '(x) $$ \le $$ 0 $$\forall $$ x $$ \in $$ [1, 5)
<br><br>$$ \Rightarrow $$ f '(x) = 0 at x = 1 $$ \Rightarrow $$ a = 5
<br><br>f(x) $$-$$ 14 = (x $$-$$ 1)<sup>2</sup> (x $$-$$ 7)
<br><br>$${{f(x) - 14} \over {{{\left( {x - 1} \right)}^2}}} = x - 7$$ | mcq | jee-main-2019-online-12th-january-evening-slot |
kHdxiJODeiP3oavxynlS5 | maths | application-of-derivatives | monotonicity | Let Ζ : [0, 2] $$ \to $$ R be a twice differentiable
function such that Ζ''(x) > 0, for all x $$ \in $$ (0, 2).
If $$\phi $$(x) = Ζ(x) + Ζ(2 β x), then $$\phi $$ is : | [{"identifier": "A", "content": "decreasing on (0, 2)"}, {"identifier": "B", "content": "decreasing on (0, 1) and increasing on (1, 2)"}, {"identifier": "C", "content": "increasing on (0, 2)"}, {"identifier": "D", "content": "increasing on (0, 1) and decreasing on (1, 2)"}] | ["B"] | null | $$\phi $$(x) = Ζ(x) + Ζ(2 β x)
<br><br>$$ \Rightarrow $$ $$\phi $$'(x) = Ζ'(x) - Ζ'(2 β x)
<br><br>Since Ζ''(x) > 0 for all x $$ \in $$ (0, 2)
<br><br>$$ \Rightarrow $$ Ζ'(x) is an increasing function for all x $$ \in $$ (0, 2).
<br><br><b>Case 1 : When $$\phi $$(x) is increasing function</b>
<br><br>So $$\phi $$'(x) > 0
<br><br>$$ \Rightarrow $$ Ζ'(x) - Ζ'(2 β x) > 0
<br><br>$$ \Rightarrow $$ Ζ'(x) > Ζ'(2 β x)
<br><br>$$ \Rightarrow $$ x > 2 β x
<br><br>$$ \Rightarrow $$ x > 1
<br><br> $$ \therefore $$ $$\phi $$(x) is increasing on (1, 2).
<br><br><b>Case 2 : When $$\phi $$(x) is decreasing function</b>
<br><br>So $$\phi $$'(x) < 0
<br><br>$$ \Rightarrow $$ Ζ'(x) - Ζ'(2 β x) < 0
<br><br>$$ \Rightarrow $$ Ζ'(x) < Ζ'(2 β x)
<br><br>$$ \Rightarrow $$ x < 2 β x
<br><br>$$ \Rightarrow $$ x < 1
<br><br> $$ \therefore $$ $$\phi $$(x) is decreasing on (0, 1).
| mcq | jee-main-2019-online-8th-april-morning-slot |
EJjlcBzmev4ZccS9XX3rsa0w2w9jx5ejv28 | maths | application-of-derivatives | monotonicity | If m is the minimum value of k for which the function f(x) = x$$\sqrt {kx - {x^2}} $$ is increasing in the interval [0,3]
and M is the maximum value of f in [0, 3] when k = m, then the ordered pair (m, M) is equal to : | [{"identifier": "A", "content": "$$\\left( {5,3\\sqrt 6 } \\right)$$"}, {"identifier": "B", "content": "$$\\left( {4,3\\sqrt 3 } \\right)$$"}, {"identifier": "C", "content": "$$\\left( {4,3\\sqrt 2 } \\right)$$"}, {"identifier": "D", "content": "$$\\left( {3,3\\sqrt 3 } \\right)$$"}] | ["B"] | null | $$f\left( x \right) = x\sqrt {kx - {x^2}} $$<br><br>
$$ \Rightarrow $$ $$f'\left( x \right) = \sqrt {kx - {x^2}} + {{(k - 2x)x} \over {2\sqrt {kx - {x^2}} }}$$<br><br>
$$ \Rightarrow {{2\left( {kx - {x^2}} \right) + kx - 2{x^2}} \over {2\sqrt {kx - {x^2}} }} = {{3kx - 4{x^2}} \over {2\sqrt {kx - {x^2}} }}$$<br><br>
$$ \Rightarrow {{x(3k - 4x)} \over {2\sqrt {kx - {x^2}} }}$$<br><br>
Now for increasing function<br><br>
for f'(x) $$ \ge $$ 0, $$\forall x \in [0,3]$$<br><br>
$$ \Rightarrow kx - {x^2} \ge 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(3k - 4x) \ge 0,\,\forall x \in \left[ {0,3} \right]\,$$<br><br>
$$ \Rightarrow x(x - k) \le 0,\forall x \in \left[ {0,3} \right]\,\,and\,x(4x - 3k) \le 0,\,\forall x \in \left[ {0,3} \right]\,$$<br><br>
k $$ \ge $$ 3 and k $$ \ge $$ 4 $$ \Rightarrow $$ k $$ \ge $$ 4<br><br>
$$ \Rightarrow $$ m = 4<br><br>
So f(x) is maximum when k = 4 then $$3\sqrt {4 \times 3 - {3^2}} = 3\sqrt 3 = M$$<br><br>
$$ \therefore $$ (m, M) = (4, $$3\sqrt 3 $$) | mcq | jee-main-2019-online-12th-april-morning-slot |
5qPzYxvBb1sYWL6hEVjgy2xukf0vzp5w | maths | application-of-derivatives | monotonicity | The function, f(x) = (3x β 7)x<sup>2/3</sup>, x $$ \in $$ R, is
increasing for all x lying in : | [{"identifier": "A", "content": "$$\\left( { - \\infty ,0} \\right) \\cup \\left( {{3 \\over 7},\\infty } \\right)$$"}, {"identifier": "B", "content": "$$\\left( { - \\infty ,0} \\right) \\cup \\left( {{{14} \\over {15}},\\infty } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\infty ,{{14} \\over {15}}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,{{14} \\over {15}}} \\right) \\cup \\left( {0,\\infty } \\right)$$"}] | ["B"] | null | f(x) = (3x β 7)x<sup>2/3</sup>
<br><br>fβ(x) = $$\left( {3x - 7} \right){2 \over {3{x^{1/3}}}} + {x^{{2 \over 3}}}.3$$
<br><br>= $${{6x - 14 - 9x} \over {3{x^{1/3}}}}$$
<br><br>= $${{15x - 14} \over {3{x^{1/3}}}}$$
<br><br>As f(x) increasing so f'(x) > 0
<br><br>$$ \therefore $$ $${{15x - 14} \over {3{x^{1/3}}}}$$ > 0
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264871/exam_images/dc3lrjubrnemgdlvwzox.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 3rd September Morning Slot Mathematics - Application of Derivatives Question 122 English Explanation">
<br>$$ \therefore $$ x $$ \in $$ $$\left( { - \infty ,0} \right) \cup \left( {{{14} \over {15}},\infty } \right)$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
FBiTnaSpw8Lymvky7rjgy2xukf8z47kh | maths | application-of-derivatives | monotonicity | Let f be a twice differentiable function on (1, 6). If f(2) = 8, fβ(2) = 5, fβ(x) $$ \ge $$ 1 and f''(x) $$ \ge $$ 4, for all x $$ \in $$ (1, 6), then :
| [{"identifier": "A", "content": "f(5) $$ \\le $$ 10"}, {"identifier": "B", "content": "f(5) + f'(5) $$ \\ge $$ 28"}, {"identifier": "C", "content": "f(5) + f'(5) $$ \\le $$ 26"}, {"identifier": "D", "content": "f'(5) + f''(5) $$ \\le $$ 20"}] | ["B"] | null | Given, $$f'(x) \ge 1$$<br><br>$$ \therefore $$ $$\int_2^5 {f'(x)} dx\, \ge \,\int_2^5 {dx} $$<br><br>$$ \Rightarrow f(5) - f(2) \ge 3$$<br><br>$$ \Rightarrow f(5) - 8 \ge 3$$<br><br>$$ \Rightarrow f(5) \ge 11$$ ...(1)<br><br>Also, $$f''(x) \ge 4$$<br><br>$$ \therefore $$ $$\int_2^5 {f''(x)} dx\, \ge \,\int_2^5 {4dx} $$<br><br>$$ \Rightarrow f'(5) - f'(2) \ge 4(3)$$<br><br>$$ \Rightarrow f'(5) - 5 \ge 12$$<br><br>$$ \Rightarrow f'(5) \ge 17$$ ...(2)<br><br>From (1) and (2),<br><br>$$f'(5) + f'(5) \ge 11 + 17$$<br><br>$$ \Rightarrow f'(5) + f'(5) \ge 28$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
R9J8afK3LpQfSZFxYhjgy2xukez6upin | maths | application-of-derivatives | monotonicity | Let f : (β1,
$$\infty $$)
$$ \to $$ R be defined by f(0) = 1 and
<br/>f(x) = $${1 \over x}{\log _e}\left( {1 + x} \right)$$, x $$ \ne $$ 0. Then the function f : | [{"identifier": "A", "content": "decreases in (\u20131, $$\\infty $$)"}, {"identifier": "B", "content": "decreases in (\u20131, 0) and increases in (0, $$\\infty $$)"}, {"identifier": "C", "content": "increases in (\u20131, $$\\infty $$)"}, {"identifier": "D", "content": "increases in (\u20131, 0) and decreases in (0, $$\\infty $$)"}] | ["A"] | null | $$f(x) = {1 \over x}{\log _e}\left( {1 + x} \right)$$<br><br>
$$ \Rightarrow f'(x) = {{x{1 \over {1 + x}} - 1{{\log }_e}\left( {1 + x} \right)} \over {{x^2}}}$$<br><br>
$$ \Rightarrow f'(x) = {{x - \left( {1 + x} \right){{\log }_e}\left( {1 + x} \right)} \over {{x^2}\left( {1 + x} \right)}}$$<br><br>
Let $$g(x) = x - \left( {1 + x} \right){\log _e}\left( {1 + x} \right)$$<br><br>
$$ \Rightarrow g'(x) = 1 - \left( {1 + x} \right){1 \over {1 + x}} - \left( 1 \right) \times {\log _e}\left( {1 + x} \right)$$<br><br>
$$ = 1 - 1 - {\log _e}\left( {1 + x} \right)$$<br><br>
$$ = - {\log _e}\left( {1 + x} \right)$$<br><br>
For $$x \in \left( { - 1,0} \right),g'(x) > 0$$<br>
<br>and for $$x \in \left( {0,\infty } \right),g'(x) < 0$$<br><br>
Also, $$g(0) = 0 - \left( {1 + 0} \right){\log _e}\left( {1 + 0} \right) = 0$$<br><br>
$$ \therefore g'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$$<br><br>
$$ \Rightarrow f'(x) < 0\,\,\forall\,\, x \in \left( { - 1,\infty } \right)$$<br><br>
$$ \Rightarrow $$ f(x) is a decreasing function for all $$x \in \left( { - 1,\infty } \right)$$ | mcq | jee-main-2020-online-2nd-september-evening-slot |
Zpp13XvDvvswz7dWtl7k9k2k5gzwf3e | maths | application-of-derivatives | monotonicity | Let Ζ(x) = xcos<sup>β1</sup>(βsin|x|), $$x \in \left[ { - {\pi \over 2},{\pi \over 2}} \right]$$, then
which of the following is true? | [{"identifier": "A", "content": "\u0192' is decreasing in $$\\left( { - {\\pi \\over 2},0} \\right)$$ and increasing\nin $$\\left( {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "B", "content": "\u0192 '(0) = $${ - {\\pi \\over 2}}$$"}, {"identifier": "C", "content": "\u0192 is not differentiable at x = 0"}, {"identifier": "D", "content": "\u0192' is increasing in $$\\left( { - {\\pi \\over 2},0} \\right)$$ and decreasing\nin $$\\left( {0,{\\pi \\over 2}} \\right)$$"}] | ["A"] | null | We know, cos<sup>-1</sup>(-x) = $$\pi $$ - cos<sup>-1</sup>x
<br><br>$$ \therefore $$ Ζ(x) = x($$\pi $$ - cos<sup>β1</sup>(sin|x|))
<br><br>= x($$\pi $$ - $${\pi \over 2}$$ + sin<sup>β1</sup>(sin|x|))
<br><br>= x($$\pi $$ - $${\pi \over 2}$$ + sin<sup>β1</sup>(sin|x|))
<br><br>= x$${\pi \over 2}$$ + x|x|
<br><br>$$ \therefore $$ f(x) = $$\left\{ {\matrix{
{x{\pi \over 2} - {x^2},} & {x < 0} \cr
{x{\pi \over 2} + {x^2},} & {x \ge 0} \cr
} } \right.$$
<br><br>Now f'(x) = $$\left\{ {\matrix{
{{\pi \over 2} - 2x,} & {x < 0} \cr
{{\pi \over 2} + 2x,} & {x \ge 0} \cr
} } \right.$$
<br><br>and f''(x) = $$\left\{ {\matrix{
{ - 2,} & {x < 0} \cr
{2,} & {x \ge 0} \cr
} } \right.$$
<br><br>$$ \therefore $$ Ζ' is decreasing in $$\left( { - {\pi \over 2},0} \right)$$ and increasing
in $$\left( {0,{\pi \over 2}} \right)$$ | mcq | jee-main-2020-online-8th-january-morning-slot |
iyKGHbJgw1CgsdT1FY1klrf64yg | maths | application-of-derivatives | monotonicity | The function
<br/>f(x) = $${{4{x^3} - 3{x^2}} \over 6} - 2\sin x + \left( {2x - 1} \right)\cos x$$ : | [{"identifier": "A", "content": "increases in $$\\left( { - \\infty ,{1 \\over 2}} \\right]$$"}, {"identifier": "B", "content": "decreases in $$\\left( { - \\infty ,{1 \\over 2}} \\right]$$"}, {"identifier": "C", "content": "increases in $$\\left[ {{1 \\over 2},\\infty } \\right)$$"}, {"identifier": "D", "content": "decreases in $$\\left[ {{1 \\over 2},\\infty } \\right)$$"}] | ["C"] | null | Given, $$f(x) = {{4{x^3} - 3{x^2}} \over 6} - 2\sin x + (2x - 1)\cos x$$<br/><br/>$$f'(x) = {{12{x^2} - 6x} \over 6} - 2\cos x + (2x - 1)( - \sin x) + \cos x(2)$$<br/><br/>$$ = (2{x^2} - x) - 2\cos x - 2x\sin x + \sin x + 2\cos x$$<br/><br/>$$ = 2{x^2} - x - 2x\sin x + \sin x$$<br/><br/>$$ = 2x(x - \sin x) - 1(x - \sin x)$$<br/><br/>$$f'(x) = (2x - 1)(x - \sin x)$$<br/><br/>for $$x > 0,x - \sin x > 0$$<br/><br/>$$x < 0,x - \sin x < 0$$<br/><br/>for $$x \in ( - \infty ,0] \cup \left[ {{1 \over 2},\infty } \right),f'(x) \ge 0$$<br/><br/>for $$x \in \left[ {0,{1 \over 2}} \right],f'(x) \le 0$$<br/><br/>Hence, f(x) increases in $$\left[ {{1 \over 2},\infty } \right)$$. | mcq | jee-main-2021-online-24th-february-morning-slot |
blzO06JKs65FBoRsOx1klrlnsa6 | maths | application-of-derivatives | monotonicity | Let $$f:R \to R$$ be defined as<br/><br/>$$f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.$$<br/><br/>Let A = {x $$ \in $$ R : f is increasing}. Then A is equal to : | [{"identifier": "A", "content": "$$( - 5,\\infty )$$"}, {"identifier": "B", "content": "$$( - \\infty , - 5) \\cup (4,\\infty )$$"}, {"identifier": "C", "content": "$$( - 5, - 4) \\cup (4,\\infty )$$"}, {"identifier": "D", "content": "$$( - \\infty , - 5) \\cup ( - 4,\\infty )$$"}] | ["C"] | null | $$f(x) = \left\{ {\matrix{
{ - 55x,} & {if\,x < - 5} \cr
{2{x^3} - 3{x^2} - 120x,} & {if\, - 5 \le x \le 4} \cr
{2{x^3} - 3{x^2} - 36x - 336,} & {if\,x > 4,} \cr
} } \right.$$
<br><br>Now, $$f'(x) = \left\{ {\matrix{
{ - 55} & ; & {x < - 5} \cr
{6({x^2} - x - 20)} & ; & { - 5 < x < 4} \cr
{6({x^2} - x - 6)} & ; & {x > 4} \cr
} } \right.$$<br><br>$$f'(x) = \left\{ {\matrix{
{ - 55} & ; & {x < - 5} \cr
{6(x - 5)(x + 4)} & ; & { - 5 < x < 4} \cr
{6(x - 3)(x + 2)} & ; & {x > 4} \cr
} } \right.$$<br><br>Hence, f(x) is monotonically increasing in interval $$( - 5, - 4) \cup (4,\infty )$$ | mcq | jee-main-2021-online-24th-february-evening-slot |
jnHf1y8tK5XeR4tPda1kluz1e2i | maths | application-of-derivatives | monotonicity | Let a be an integer such that all the real roots of the polynomial <br/>2x<sup>5</sup> + 5x<sup>4</sup> + 10x<sup>3</sup> + 10x<sup>2</sup> + 10x + 10 lie in the interval (a, a + 1). Then, |a| is equal to ___________. | [] | null | 2 | Let, $$f(x) = 2{x^5} + 5{x^4} + 10{x^3} + 10{x^2} + 10x + 10$$<br><br>$$ \Rightarrow f'(x) = 10({x^4} + 2{x^3} + 3{x^2} + 2x + 1)$$<br><br>$$ = 10\left( {{x^2} + {1 \over {{x^2}}} + 2\left( {x + {1 \over x}} \right) + 3} \right)$$<br><br>$$ = 10\left( {{{\left( {x + {1 \over x}} \right)}^2} + 2\left( {x + {1 \over x}} \right) + 1} \right)$$<br><br>$$ = 10{\left( {\left( {x + {1 \over x}} \right) + 1} \right)^2} > 0;\forall x \in R$$<br><br>$$ \therefore $$ f(x) is strictly increasing function. Since, it is an odd degree polynomial it will have exactly one real root.<br><br>Now, by observation.<br><br>$$f( - 1) = 3 > 0$$<br><br>$$f( - 2) = - 64 + 80 - 80 + 40 - 20 + 10$$<br><br>$$ = - 34 < 0$$<br><br>$$ \Rightarrow f(x)$$ has at least one root in $$( - 2, - 1) \equiv (a,a + 1)$$<br><br>$$ \Rightarrow a = - 2$$
<br><br>$$ \Rightarrow $$ |a| = - 2 | integer | jee-main-2021-online-26th-february-evening-slot |
mQLintGwdQfFf2qnw61kmiw0lr1 | maths | application-of-derivatives | monotonicity | Let f be a real valued function, defined on R $$-$$ {$$-$$1, 1} and given by <br/><br/>f(x) = 3 log<sub>e</sub> $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$.<br/><br/>Then in which of the following intervals, function f(x) is increasing? | [{"identifier": "A", "content": "($$-$$$$\\infty $$, $$-$$1) $$\\cup$$ $$\\left( {[{1 \\over 2},\\infty ) - \\{ 1\\} } \\right)$$"}, {"identifier": "B", "content": "($$-$$$$\\infty $$, $$\\infty $$) $$-$$ {$$-$$1, 1)"}, {"identifier": "C", "content": "($$-$$$$\\infty $$, $${{1 \\over 2}}$$] $$-$$ {$$-$$1}"}, {"identifier": "D", "content": "($$-$$1, $${{1 \\over 2}}$$]"}] | ["A"] | null | f(x) = 3 log<sub>e</sub> $$\left| {{{x - 1} \over {x + 1}}} \right| - {2 \over {x - 1}}$$
<br><br>$$f'(x) = {{3(x + 1)} \over {x - 1}} \times {{(x + 1) - (x - 1)} \over {{{(x + 1)}^2}}} + {2 \over {{{(x - 1)}^2}}} > 0$$<br><br>$$ = {6 \over {{x^2} - 1}} + {2 \over {{{(x - 1)}^2}}} > 0$$<br><br>$$ = {{2(3(x - 1) + (x + 1))} \over {{{(x - 1)}^2}(x + 1)}} = {{4(2x - 1)} \over {{{(x - 1)}^2}(x + 1)}} > 0$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264455/exam_images/u5uxawmfdatdnnbbfy6y.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 16th March Evening Shift Mathematics - Application of Derivatives Question 96 English Explanation"><br><br>$$ \Rightarrow x \in ( - \infty , - 1) \cup \left[ {{1 \over 2},\infty ) - \{ 1\} } \right.$$ | mcq | jee-main-2021-online-16th-march-evening-shift |
4M0JODSlLdB22Grotz1kmkn7lpe | maths | application-of-derivatives | monotonicity | Consider the function f : R $$ \to $$ R defined by
<br/><br/>$$f(x) = \left\{ \matrix{
\left( {2 - \sin \left( {{1 \over x}} \right)} \right)|x|,x \ne 0 \hfill \cr
0,\,\,x = 0 \hfill \cr} \right.$$. Then f is : | [{"identifier": "A", "content": "not monotonic on ($$-$$$$\\infty $$, 0) and (0, $$\\infty $$)"}, {"identifier": "B", "content": "monotonic on (0, $$\\infty $$) only"}, {"identifier": "C", "content": "monotonic on ($$-$$$$\\infty $$, 0) only"}, {"identifier": "D", "content": "monotonic on ($$-$$$$\\infty $$, 0) $$\\cup$$ (0, $$\\infty $$)"}] | ["A"] | null | $$f(x) = \left\{ {\matrix{
{ - \left( {2 - \sin {1 \over x}} \right)x} & , & {x < 0} \cr
0 & , & {x = 0} \cr
{\left( {2 - \sin {1 \over x}} \right)x} & , & {x > 0} \cr
} } \right.$$<br><br>$$f'(x) = \left\{ \matrix{
- x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) - \left( {2 - \sin {1 \over x}} \right),x < 0 \hfill \cr
x\left( { - \cos {1 \over x}} \right)\left( { - {1 \over {{x^2}}}} \right) + \left( {2 - \sin {1 \over x}} \right),x > 0 \hfill \cr} \right.$$<br><br>= $$\left\{ \matrix{
- {1 \over x}\cos {1 \over x} + \sin {1 \over x} - 2,x < 0 \hfill \cr
{1 \over x}\cos {1 \over x} - \sin {1 \over x} + 2,x > 0 \hfill \cr} \right.$$
<br><br>$$ \therefore $$ f'(x) is an oscillating function which is non-monotonic on ($$-$$$$\infty $$, 0) and (0, $$\infty $$). | mcq | jee-main-2021-online-17th-march-evening-shift |
1krtbxxgh | maths | application-of-derivatives | monotonicity | Let f : R $$\to$$ R be defined as<br/><br/>$$f(x) = \left\{ {\matrix{
{ - {4 \over 3}{x^3} + 2{x^2} + 3x,} & {x > 0} \cr
{3x{e^x},} & {x \le 0} \cr
} } \right.$$. Then f is increasing function in the interval | [{"identifier": "A", "content": "$$\\left( { - {1 \\over 2},2} \\right)$$"}, {"identifier": "B", "content": "(0,2)"}, {"identifier": "C", "content": "$$\\left( { - 1,{3 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "($$-$$3, $$-$$1)"}] | ["C"] | null | $$f'(x)\left\{ {\matrix{
{ - 4{x^2} + 4x + 3} & {x > 0} \cr
{3{e^x}(1 + x)} & {x \le 0} \cr
} } \right.$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266545/exam_images/qfalbkppdpmpdrbmbsgj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 22th July Evening Shift Mathematics - Application of Derivatives Question 89 English Explanation"><br>For x > 0, $$f'(x) = - 4{x^2} + 4x + 3$$<br><br>f(x) is increasing in $$\left( { - {1 \over 2},{3 \over 2}} \right)$$<br><br>For x $$\le$$ 0, f'(x) = 3e<sup>x</sup>(1 + x)<br><br>f'(x) > 0 $$\forall$$ x $$\in$$($$-$$1, 0)<br><br>$$\Rightarrow$$ f(x) is increasing in ($$-$$1, 0)<br><br>So, in complete domain, f(x) is increasing in $$\left( { - 1,{3 \over 2}} \right)$$ | mcq | jee-main-2021-online-22th-july-evening-shift |
1krvrw9s4 | maths | application-of-derivatives | monotonicity | Let $$f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3$$, $$x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]$$. Then, f is : | [{"identifier": "A", "content": "increasing in $$\\left( { - {\\pi \\over 6},{\\pi \\over 2}} \\right)$$"}, {"identifier": "B", "content": "decreasing in $$\\left( {0,{\\pi \\over 2}} \\right)$$"}, {"identifier": "C", "content": "increasing in $$\\left( { - {\\pi \\over 6},0} \\right)$$"}, {"identifier": "D", "content": "decreasing in $$\\left( { - {\\pi \\over 6},0} \\right)$$"}] | ["D"] | null | $$f(x) = 3{\sin ^4}x + 10{\sin ^3}x + 6{\sin ^2}x - 3,x \in \left[ { - {\pi \over 6},{\pi \over 2}} \right]$$<br><br>$$f'(x) = 12{\sin ^3}x\cos x + 30{\sin ^2}x\cos x + 12\sin x\cos x$$<br><br>$$ = 6\sin x\cos x(2{\sin ^2}x + 5\sin x + 2)$$<br><br>$$ = 6\sin x\cos x(2\sin x + 1)(\sin + 2)$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263470/exam_images/ogwasocqxensp73bghx1.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th July Morning Shift Mathematics - Application of Derivatives Question 88 English Explanation"><br>Decreasing in $$\left( { - {\pi \over 6},0} \right)$$ | mcq | jee-main-2021-online-25th-july-morning-shift |
1ktis8nus | maths | application-of-derivatives | monotonicity | If 'R' is the least value of 'a' such that the function f(x) = x<sup>2</sup> + ax + 1 is increasing on [1, 2] and 'S' is the greatest value of 'a' such that the function f(x) = x<sup>2</sup> + ax + 1 is decreasing on [1, 2], then <br/>the value of |R $$-$$ S| is ___________. | [] | null | 2 | f(x) = x<sup>2</sup> + ax + 1<br><br>f'(x) = 2x + a<br><br>when f(x) is increasing on [1, 2]<br><br>2x + a $$\ge$$ 0 $$\forall$$ x$$\in$$[1, 2]<br><br>a $$\ge$$ $$-$$2x $$\forall$$ x$$\in$$[1, 2]<br><br>R = $$-$$4<br><br>when f(x) is decreasing on [1, 2]<br><br>2x + a $$\le$$ 0 $$\forall$$ x$$\in$$[1, 2]<br><br>a $$\le$$ $$-$$2 $$\forall$$ x$$\in$$[1, 2]<br><br>S = $$-$$2<br><br>|R $$-$$ S| = | $$-$$4 + 2 | = 2 | integer | jee-main-2021-online-31st-august-morning-shift |
1kto3gifb | maths | application-of-derivatives | monotonicity | The function $$f(x) = {x^3} - 6{x^2} + ax + b$$ is such that $$f(2) = f(4) = 0$$. Consider two statements :<br/><br/>Statement 1 : there exists x<sub>1</sub>, x<sub>2</sub> $$\in$$(2, 4), x<sub>1</sub> < x<sub>2</sub>, such that f'(x<sub>1</sub>) = $$-$$1 and f'(x<sub>2</sub>) = 0.<br/><br/>Statement 2 : there exists x<sub>3</sub>, x<sub>4</sub> $$\in$$ (2, 4), x<sub>3</sub> < x<sub>4</sub>, such that f is decreasing in (2, x<sub>4</sub>), increasing in (x<sub>4</sub>, 4) and $$2f'({x_3}) = \sqrt 3 f({x_4})$$.<br/><br/>Then | [{"identifier": "A", "content": "both Statement 1 and Statement 2 are true"}, {"identifier": "B", "content": "Statement 1 is false and Statement 2 is true"}, {"identifier": "C", "content": "both Statement 1 and Statement 2 are false"}, {"identifier": "D", "content": "Statement 1 is true and Statement 2 is false"}] | ["A"] | null | $$f(x) = {x^3} - 6{x^2} + ax + b$$<br><br>$$f(2) = 8 - 24 + 2a + b = 0$$<br><br>$$2a + b = 16$$ .... (1)<br><br>$$f(4) = 64 - 96 + 4a + b = 0$$<br><br>$$4a + b = 32$$ .... (2)<br><br>Solving (1) and (2)<br><br>a = 8, b = 0<br><br>$$f(x) = {x^3} - 6{x^2} + 8x$$<br><br>$$f'(x) = 3{x^2} - 12x + 8$$<br><br>$$f''(x) = 6x - 12$$<br><br>$$\Rightarrow$$ f'(x) is $$ \uparrow $$ for x > 2, and f'(x) is $$ \downarrow $$ for x < 2<br><br>$$f'(2) = 12 - 24 + 8 = - 4$$<br><br>$$f'(4) = 48 - 48 + 8 = 8$$<br><br>$$f'(x) = 3{x^2} - 12x + 8$$<br><br>vertex (2, $$-$$4)<br><br>f'(2) = $$-$$4, f'(4) = 8, f'(3) = 27 $$-$$ 36 + 8<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1kwopvl27/adfb9c3f-bff0-4d72-a03f-536e9ff2142c/c29c0ef0-534c-11ec-9cbb-695a838b20fb/file-1kwopvl28.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1kwopvl27/adfb9c3f-bff0-4d72-a03f-536e9ff2142c/c29c0ef0-534c-11ec-9cbb-695a838b20fb/file-1kwopvl28.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2021 (Online) 1st September Evening Shift Mathematics - Application of Derivatives Question 79 English Explanation"><br>f'(x<sub>1</sub>) = $$-$$1, then x<sub>1</sub> = 3<br><br>f'(x<sub>2</sub>) = 0<br><br>Again<br><br>f'(x) < 0 for x $$\in$$ (2, x<sub>4</sub>)<br><br>f'(x) > 0 for x $$\in$$ (x<sub>4</sub>, 4)<br><br>x<sub>4</sub> $$\in$$ (3, 4)<br><br>f(x) = x<sup>3</sup> $$-$$ 6x<sup>2</sup> + 8x<br><br>f(3) = 27 $$-$$ 54 + 24 = $$-$$3<br><br>f(4) = 64 $$-$$ 96 + 32 = 0<br><br>For x<sub>4</sub>(3, 4)<br><br>f(x<sub>4</sub>) < $$-$$3$$\sqrt 3 $$<br><br>and f'(x<sub>3</sub>) > $$-$$4<br><br>2f'(x<sub>3</sub>) > $$-$$8<br><br>So, 2f'(x<sub>3</sub>) = $$\sqrt 3 $$ f(x<sub>4</sub>)<br><br>Correct Ans. (a). | mcq | jee-main-2021-online-1st-september-evening-shift |
1l566qk0u | maths | application-of-derivatives | monotonicity | <p>The number of real solutions of <br/><br/>$${x^7} + 5{x^3} + 3x + 1 = 0$$ is equal to ____________.</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "5"}] | ["B"] | null | <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lc8ekncq/3bb86ec8-2e21-4aa5-b7aa-771daae2cbcc/79bb29a0-8717-11ed-b3ec-0bde88094e1e/file-1lc8ekncr.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lc8ekncq/3bb86ec8-2e21-4aa5-b7aa-771daae2cbcc/79bb29a0-8717-11ed-b3ec-0bde88094e1e/file-1lc8ekncr.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th June Morning Shift Mathematics - Application of Derivatives Question 76 English Explanation"><br>
$f^{\prime}(x)=7 x^{6}+15 x^{2}+3>0 \,\,\forall\,\, x \in R$
<br><br>
$f(x)$ is always increasing
<br><br>
So clearly it intersects
<br><br>
x-axis at only one point | mcq | jee-main-2022-online-28th-june-morning-shift |
1l58a7w47 | maths | application-of-derivatives | monotonicity | <p>Let $$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10$$, $$x \in [ - 1,1]$$. If [a, b] is the range of the function f, then 4a $$-$$ b is equal to :</p> | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "11 $$-$$ $$\\pi$$"}, {"identifier": "C", "content": "11 + $$\\pi$$"}, {"identifier": "D", "content": "15 $$-$$ $$\\pi$$"}] | ["B"] | null | <p>$$f(x) = 2{\cos ^{ - 1}}x + 4{\cot ^{ - 1}}x - 3{x^2} - 2x + 10\,\forall x \in [ - 1,1]$$</p>
<p>$$ \Rightarrow f'(x) = - {2 \over {\sqrt {1 - {x^2}} }} - {4 \over {1 + {x^2}}} - 6x - 2 < 0\,\forall x \in [ - 1,1]$$</p>
<p>So f(x) is decreasing function and range of f(x) is [f(1), f($$-$$1)], which is [$$\pi$$ + 5, 5$$\pi$$ + 9]</p>
<p>Now $$4a - b = 4(\pi + 5) - (5\pi + 9)$$</p>
<p>$$ = 11 - \pi $$</p> | mcq | jee-main-2022-online-26th-june-morning-shift |
1l5bavxs8 | maths | application-of-derivatives | monotonicity | <p>Let $$\lambda$$$$^ * $$ be the largest value of $$\lambda$$ for which the function $${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36x + 48$$ is increasing for all x $$\in$$ R. Then $${f_{{\lambda ^ * }}}(1) + {f_{{\lambda ^ * }}}( - 1)$$ is equal to :</p> | [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "48"}, {"identifier": "C", "content": "64"}, {"identifier": "D", "content": "72"}] | ["D"] | null | <p>$$\because$$ $${f_\lambda }(x) = 4\lambda {x^3} - 36\lambda {x^2} + 36\lambda + 48$$</p>
<p>$$\therefore$$ $$f{'_\lambda }(x) = 12(\lambda {x^2} - 6\lambda x + 3)$$</p>
<p>For $${f_\lambda }(x)$$ increasing : $${(6\lambda )^2} - 12\lambda \le 0$$</p>
<p>$$\therefore$$ $$\lambda \in \left[ {0,\,{1 \over 3}} \right]$$</p>
<p>$$\therefore$$ $$\lambda^ * = {1 \over 3}$$</p>
<p>Now, $$f_\lambda ^*(x) = {4 \over 3}{x^3} - 12{x^2} + 36x + 48$$</p>
<p>$$\therefore$$ $$f_\lambda ^*(1) + f_\lambda ^*( - 1) = 73{1 \over 2} - 1{1 \over 2} = 72$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5c19dx8 | maths | application-of-derivatives | monotonicity | <p>For the function <br/><br/>$$f(x) = 4{\log _e}(x - 1) - 2{x^2} + 4x + 5,\,x > 1$$, which one of the following is NOT correct?</p> | [{"identifier": "A", "content": "f is increasing in (1, 2) and decreasing in (2, $$\\infty$$)"}, {"identifier": "B", "content": "f(x) = $$-$$1 has exactly two solutions"}, {"identifier": "C", "content": "$$f'(e) - f''(2) < 0$$"}, {"identifier": "D", "content": "f(x) = 0 has a root in the interval (e, e + 1)"}] | ["C"] | null | Lets draw the curve $y=f(x)=4 \log _e(x-1)-2 x^2$ $+4 x+5, x>1$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lkavtohl/5b2f07ad-6815-4447-8b5a-bda6bf0cb5b3/3c6a7290-26d6-11ee-b52b-3728f15f4ced/file-6y3zli1lkavtohm.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lkavtohl/5b2f07ad-6815-4447-8b5a-bda6bf0cb5b3/3c6a7290-26d6-11ee-b52b-3728f15f4ced/file-6y3zli1lkavtohm.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2022 (Online) 24th June Morning Shift Mathematics - Application of Derivatives Question 65 English Explanation">
<br>$$
\begin{aligned}
&f(x)=4 \log _e(x-1)-2 x^2+4 x+5, x > 1 \\\\
&f^{\prime}(x)=\frac{4}{x-1}-4(x-1) \\\\
&f^{\prime \prime}(x)=\frac{-4}{(x-1)^2}-4\\\\
&\text {For} 1 < x< 2 \Rightarrow f^{\prime}(x) > 0 \\\\
&\text {For}~ x > 2 \Rightarrow f^{\prime}(x)<0 \text { (option } A \text { is correct) } \\\\
&f(x)=-1 \text { has two solution (option } B \text { is correct) } \\\\
&f(e) > 0 \\\\
&f(e+1) < 0 \\\\
&f(e) \cdot f(e+1)<0(\text { option } D \text { is correct) } \\\\
&f^{\prime}(e)-f^{\prime \prime}(2)=\frac{4}{e-1}-4(e-1)+8 > 0
\end{aligned}
$$
<br><br>(option C is incorrect) | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6gjnyuy | maths | application-of-derivatives | monotonicity | <p>Let the function $$f(x)=2 x^{2}-\log _{\mathrm{e}} x, x>0$$, be decreasing in $$(0, \mathrm{a})$$ and increasing in $$(\mathrm{a}, 4)$$. A tangent to the parabola $$y^{2}=4 a x$$ at a point $$\mathrm{P}$$ on it passes through the point $$(8 \mathrm{a}, 8 \mathrm{a}-1)$$ but does not pass through the point $$\left(-\frac{1}{a}, 0\right)$$. If the equation of the normal at $$P$$ is : $$\frac{x}{\alpha}+\frac{y}{\beta}=1$$, then $$\alpha+\beta$$ is equal to ________________.</p> | [] | null | 45 | <p>$$\delta '(x) = {{4{x^2} - 1} \over x}$$ so f(x) is decreasing in $$\left( {0,{1 \over 2}} \right)$$ and increasing in $$\left( {{1 \over 2},\infty } \right) \Rightarrow a = {1 \over 2}$$</p>
<p>Tangent at $${y^2} = 2x \Rightarrow y = ,x + {1 \over {2m}}$$</p>
<p>It is passing through $$(4,3)$$</p>
<p>$$3 = 4m + {1 \over {2m}} \Rightarrow m = {1 \over 2}$$ or $${1 \over 4}$$</p>
<p>So tangent may be</p>
<p>$$y = {1 \over 2}x + 1$$ or $$y = {1 \over 4}x + 2$$</p>
<p>But $$y = {1 \over 2}x + 1$$ passes through $$( - 2,0)$$ so rejected.</p>
<p>Equation of normal</p>
<p>$$y = - 4x - 2\left( {{1 \over 2}} \right)( - 4) - {1 \over 2}{( - 4)^3}$$</p>
<p>or $$y = - 4x + 4 + 32$$</p>
<p>or $${x \over 9} + {y \over {36}} = 1$$</p> | integer | jee-main-2022-online-26th-july-morning-shift |
1l6nm1791 | maths | application-of-derivatives | monotonicity | <p>The function $$f(x)=x \mathrm{e}^{x(1-x)}, x \in \mathbb{R}$$, is :</p> | [{"identifier": "A", "content": "increasing in $$\\left(-\\frac{1}{2}, 1\\right)$$"}, {"identifier": "B", "content": "decreasing in $$\\left(\\frac{1}{2}, 2\\right)$$"}, {"identifier": "C", "content": "increasing in $$\\left(-1,-\\frac{1}{2}\\right)$$"}, {"identifier": "D", "content": "decreasing in $$\\left(-\\frac{1}{2}, \\frac{1}{2}\\right)$$"}] | ["A"] | null | <p>$$f(x) = x{e^{x(1 - x)}},\,x \in R$$</p>
<p>$$f'(x) = x{e^{x(1 - x)}}\,.\,(1 - 2x) + {e^{x(1 - x)}}$$</p>
<p>$$ = {e^{x(1 - x)}}[x - 2{x^2} + 1]$$</p>
<p>$$ = - {e^{x(1 - x)}}[2{x^2} - x - 1]$$</p>
<p>$$ = - {e^{x(1 - x)}}(2x + 1)(x - 1)$$</p>
<p>$$\therefore$$ $$f(x)$$ is increasing in $$\left( { - {1 \over 2},1} \right)$$ and decreasing in $$\left( { - \infty ,\, - {1 \over 2}} \right) \cup \left( {1,\infty } \right)$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldv2lvfb | maths | application-of-derivatives | monotonicity | <p>Let $$f:(0,1)\to\mathbb{R}$$ be a function defined $$f(x) = {1 \over {1 - {e^{ - x}}}}$$, and $$g(x) = \left( {f( - x) - f(x)} \right)$$. Consider two statements</p>
<p>(I) g is an increasing function in (0, 1)</p>
<p>(II) g is one-one in (0, 1)</p>
<p>Then,</p> | [{"identifier": "A", "content": "Both (I) and (II) are true"}, {"identifier": "B", "content": "Neither (I) nor (II) is true"}, {"identifier": "C", "content": "Only (II) is true"}, {"identifier": "D", "content": "Only (I) is true"}] | ["A"] | null | $g(x)=f(-x)-f(x)$
<br/><br/>
$$
\begin{aligned}
& =\frac{1}{1-e^{x}}-\frac{1}{1-e^{-x}} \\\\
& =\frac{1}{1-e^{x}}-\frac{e^{x}}{e^{x}-1} \\\\
& =\frac{1+e^{x}}{1-e^{x}} \\\\
g^{\prime}(x) & =\frac{\left(1-e^{x}\right) e^{x}-\left(1+e^{x}\right)\left(-e^{x}\right)}{\left(1-e^{x}\right)^{2}} \\\\
& =\frac{e^{x}-2 e^{x}+e^{x}+2 e^{x}}{\left(1-e^{x}\right)^{2}}>0
\end{aligned}
$$<br/><br/>
<p>So both statements are correct</p> | mcq | jee-main-2023-online-25th-january-morning-shift |
1lgvpk0py | maths | application-of-derivatives | monotonicity | <p>Let $$\mathrm{g}(x)=f(x)+f(1-x)$$ and $$f^{\prime \prime}(x) > 0, x \in(0,1)$$. If $$\mathrm{g}$$ is decreasing in the interval $$(0, a)$$ and increasing in the interval $$(\alpha, 1)$$, then $$\tan ^{-1}(2 \alpha)+\tan ^{-1}\left(\frac{1}{\alpha}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right)$$ is equal to :</p> | [{"identifier": "A", "content": "$$\\frac{3 \\pi}{4}$$\n"}, {"identifier": "B", "content": "$$\\pi$$"}, {"identifier": "C", "content": "$$\\frac{5 \\pi}{4}$$"}, {"identifier": "D", "content": "$$\\frac{3 \\pi}{2}$$"}] | ["B"] | null | We have, $g(x)=f(x)+f(1-x)$
<br/><br/>Differentiating both side, we get
<br/><br/>$g^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$
<br/><br/>As $f^{\prime \prime}(x)>0, f^{\prime}(x)$ is an increasing function.
<br/><br/>Also, $g(x)=f(x)+f(2 a-x)$ is always symmetric about $x=a$
<br/><br/>So, $g(x)=f(x)+f(1-x)$ is also symmetric about $x=1 / 2$
<br/><br/>$\therefore g$ is decreasing in the interval $(0,1 / 2)$ and increasing in the interval $(1 / 2,1)$.
<br/><br/>Now,
<br/><br/>$$
\begin{aligned}
& \tan ^{-1} 2 \alpha+\tan ^{-1}\left(\frac{1}{2}\right)+\tan ^{-1}\left(\frac{\alpha+1}{\alpha}\right) \\\\
& =\tan ^{-1} 1+\tan ^{-1}(2)+\tan ^{-1} 3=\pi
\end{aligned}
$$ | mcq | jee-main-2023-online-10th-april-evening-shift |
lsaps6wb | maths | application-of-derivatives | monotonicity | If $5 f(x)+4 f\left(\frac{1}{x}\right)=x^2-2, \forall x \neq 0$ and $y=9 x^2 f(x)$, then $y$ is strictly increasing in : | [{"identifier": "A", "content": "$\\left(0, \\frac{1}{\\sqrt{5}}\\right) \\cup\\left(\\frac{1}{\\sqrt{5}}, \\infty\\right)$"}, {"identifier": "B", "content": "$\\left(-\\frac{1}{\\sqrt{5}}, 0\\right) \\cup\\left(\\frac{1}{\\sqrt{5}}, \\infty\\right)$"}, {"identifier": "C", "content": "$\\left(-\\frac{1}{\\sqrt{5}}, 0\\right) \\cup\\left(0, \\frac{1}{\\sqrt{5}}\\right)$"}, {"identifier": "D", "content": "$\\left(-\\infty, \\frac{1}{\\sqrt{5}}\\right) \\cup\\left(0, \\frac{1}{\\sqrt{5}}\\right)$"}] | ["B"] | null | $$
5 f(x)+4 f(1 / x)=x^2-2
$$ ........(1)
<br><br>Replace $x$ by $1 / x$
<br><br>$$
5 f(1 / x)+4 f(x)=\frac{1}{x^2}-2
$$ ..........(2)
<br><br>Multiply equation (1) by 5 and multiply equation (2) by 4 and then subtract equation (2) from (1)
<br><br>$\begin{aligned} & 25 f(x)-16 f(x)=5 x^2-10-\frac{4}{x^2}+8 \\\\ & 9 f(x)=5 x^2-\frac{4}{x^2}-2 \\\\ & 9 f(x)=\frac{5 x^4-4-2 x^2}{x^2}\end{aligned}$
<br><br>$\begin{aligned} & y=9 x^2 f(x) \\\\ & y=5 x^4-2 x^2-4 \\\\ & y^{\prime}=20 x^3-4 x \\\\ & \text { Put } y^{\prime}>0 \\\\ & 20 x^3-4 x>0 \\\\ & 5 x^3-x>0 \\\\ & x\left(5 x^2-1\right)>0\end{aligned}$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lt4ueirs/bdaf4cf1-dcad-4fa2-b9ed-fb23b230da96/ec3f4080-d5b1-11ee-954c-abbae7354768/file-6y3zli1lt4ueirt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lt4ueirs/bdaf4cf1-dcad-4fa2-b9ed-fb23b230da96/ec3f4080-d5b1-11ee-954c-abbae7354768/file-6y3zli1lt4ueirt.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 - Application of Derivatives Question 26 English Explanation">
<br><br>$x \in\left(-\frac{1}{\sqrt{5}}, 0\right) \cup\left(\frac{1}{\sqrt{5}}, \infty\right)$ | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lsd4jja0 | maths | application-of-derivatives | monotonicity | <p>Let $$f: \rightarrow \mathbb{R} \rightarrow(0, \infty)$$ be strictly increasing function such that $$\lim _\limits{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1$$. Then, the value of $$\lim _\limits{x \rightarrow \infty}\left[\frac{f(5 x)}{f(x)}-1\right]$$ is equal to</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "7/5"}] | ["A"] | null | <p>$$\begin{aligned}
& f: R \rightarrow(0, \infty) \\
& \lim _{x \rightarrow \infty} \frac{f(7 x)}{f(x)}=1
\end{aligned}$$</p>
<p>$$\because \mathrm{f}$$ is increasing</p>
<p>$$\begin{aligned}
& \therefore \mathrm{f}(\mathrm{x})<\mathrm{f}(5 \mathrm{x})<\mathrm{f}(7 \mathrm{x}) \\
& \because \frac{\mathrm{f}(\mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<\frac{\mathrm{f}(7 \mathrm{x})}{\mathrm{f}(\mathrm{x})} \\
& 1<\lim _{\mathrm{x} \rightarrow \infty} \frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}<1 \\
& \therefore\left[\frac{\mathrm{f}(5 \mathrm{x})}{\mathrm{f}(\mathrm{x})}-1\right] \\
& \Rightarrow 1-1=0
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsd4mvvi | maths | application-of-derivatives | monotonicity | <p>If the function $$f:(-\infty,-1] \rightarrow(a, b]$$ defined by $$f(x)=e^{x^3-3 x+1}$$ is one - one and onto, then the distance of the point $$P(2 b+4, a+2)$$ from the line $$x+e^{-3} y=4$$ is :</p> | [{"identifier": "A", "content": "$$2 \\sqrt{1+e^6}$$\n"}, {"identifier": "B", "content": "$$\\sqrt{1+e^6}$$\n"}, {"identifier": "C", "content": "$$3 \\sqrt{1+e^6}$$\n"}, {"identifier": "D", "content": "$$4 \\sqrt{1+e^6}$$"}] | ["A"] | null | <p>$$\begin{aligned}
& f(x)=e^{x^3-3 x+1} \\
& f^{\prime}(x)=e^{x^3-3 x+1} \cdot\left(3 x^2-3\right) \\
& =e^{x^3-3 x+1} \cdot 3(x-1)(x+1)
\end{aligned}$$</p>
<p>For $$\mathrm{f}^{\prime}(\mathrm{x}) \geq 0$$</p>
<p>$$\therefore \mathrm{f}(\mathrm{x})$$ is increasing function</p>
<p>$$\begin{aligned}
& \therefore \mathrm{a}=\mathrm{e}^{-\infty}=0=\mathrm{f}(-\infty) \\
& \mathrm{b}=\mathrm{e}^{-1+3+1}=\mathrm{e}^3=\mathrm{f}(-1) \\
& \mathrm{P}(2 \mathrm{~b}+4, \mathrm{a}+2) \\
& \therefore \mathrm{P}\left(2 \mathrm{e}^3+4,2\right)
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjw7x98/08591748-2ee9-41c3-9af0-f822bc19a883/916485c0-ca2c-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjw7x99.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjw7x98/08591748-2ee9-41c3-9af0-f822bc19a883/916485c0-ca2c-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjw7x99.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 - Application of Derivatives Question 22 English Explanation"></p>
<p>$$\mathrm{d}=\frac{\left(2 \mathrm{e}^3+4\right)+2 \mathrm{e}^{-3}-4}{\sqrt{1+\mathrm{e}^{-6}}}=2 \sqrt{1+\mathrm{e}^6}$$</p> | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsf073t9 | maths | application-of-derivatives | monotonicity | <p>Consider the function $$f:\left[\frac{1}{2}, 1\right] \rightarrow \mathbb{R}$$ defined by $$f(x)=4 \sqrt{2} x^3-3 \sqrt{2} x-1$$. Consider the statements</p>
<p>(I) The curve $$y=f(x)$$ intersects the $$x$$-axis exactly at one point.</p>
<p>(II) The curve $$y=f(x)$$ intersects the $$x$$-axis at $$x=\cos \frac{\pi}{12}$$.</p>
<p>Then</p> | [{"identifier": "A", "content": "Both (I) and (II) are correct.\n"}, {"identifier": "B", "content": "Only (I) is correct.\n"}, {"identifier": "C", "content": "Both (I) and (II) are incorrect.\n"}, {"identifier": "D", "content": "Only (II) is correct."}] | ["A"] | null | <p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=12 \sqrt{2} \mathrm{x}^2-3 \sqrt{2} \geq 0 \text { for }\left[\frac{1}{2}, 1\right] \\
& \mathrm{f}\left(\frac{1}{2}\right)<0
\end{aligned}$$</p>
<p>$$\mathrm{f}(1)>0 \Rightarrow(\mathrm{A})$$ is correct.</p>
<p>$$f(x)=\sqrt{2}\left(4 x^3-3 x\right)-1=0$$</p>
<p>Let $$\cos \alpha=\mathrm{x}$$,</p>
<p>$$\cos 3 \alpha=\cos \frac{\pi}{4} \Rightarrow \alpha=\frac{\pi}{12}$$</p>
<p>$$\mathrm{x}=\cos \frac{\pi}{12}$$</p>
<p>(4) is correct.</p> | mcq | jee-main-2024-online-29th-january-morning-shift |
jaoe38c1lsfkgqg0 | maths | application-of-derivatives | monotonicity | <p>The function $$f(x)=\frac{x}{x^2-6 x-16}, x \in \mathbb{R}-\{-2,8\}$$</p> | [{"identifier": "A", "content": "decreases in $$(-\\infty,-2) \\cup(-2,8) \\cup(8, \\infty)$$\n"}, {"identifier": "B", "content": "increases in $$(-\\infty,-2) \\cup(-2,8) \\cup(8, \\infty)$$\n"}, {"identifier": "C", "content": "decreases in $$(-2,8)$$ and increases in $$(-\\infty,-2) \\cup(8, \\infty)$$\n"}, {"identifier": "D", "content": "decreases in $$(-\\infty,-2)$$ and increases in $$(8, \\infty)$$"}] | ["A"] | null | <p>$$f(x)=\frac{x}{x^2-6 x-16}$$</p>
<p>Now,</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{-\left(\mathrm{x}^2+16\right)}{\left(\mathrm{x}^2-6 \mathrm{x}-16\right)^2} \\
& \mathrm{f}^{\prime}(\mathrm{x})<0
\end{aligned}$$</p>
<p>Thus $$f(x)$$ is decreasing in</p>
<p>$$(-\infty,-2) \cup(-2,8) \cup(8, \infty)$$</p> | mcq | jee-main-2024-online-29th-january-evening-shift |
luxwdep3 | maths | application-of-derivatives | monotonicity | <p>Let the set of all values of $$p$$, for which $$f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right)+2(2-p) x+7$$ does not have any critical point, be the interval $$(a, b)$$. Then $$16 a b$$ is equal to _________.</p> | [] | null | 252 | <p>$$\begin{aligned}
& f(x)=\left(p^2-6 p+8\right)\left(\sin ^2 2 x-\cos ^2 2 x\right) +2(2-p) x+7 \\
& f(x)=-\cos 4 x\left(p^2-6 p+8\right)+2(2-p) x+7 \\
& f^{\prime}(x)=4 \sin 4 x\left(p^2-6 p+8\right)+2(2-p) \neq 0 \\
& 2(2-p)+\left[-4\left(p^2-6 p+8\right), 4\left(p^2-6 p+8\right)\right] \\
& \Rightarrow\left[-4 p^2+24 p-32,4 p^2-24 p+32\right]+(4-2 p) \\
& {\left[-4 p^2+22 p-28,4 p^2-26 p+36\right]} \\
& {[(p-2)(-4 p+14),(p-2)(4 p-18)]} \\
& \Rightarrow(p-2)[(-4 p+14), 4 p-18] \Rightarrow p \in\left(\frac{7}{2}, \frac{9}{2}\right) \\
& \Rightarrow a=\frac{7}{2}, b=\frac{9}{2} \\
& \Rightarrow 16 a b=4 \times 63=252
\end{aligned}$$</p> | integer | jee-main-2024-online-9th-april-evening-shift |
lv5gt1vv | maths | application-of-derivatives | monotonicity | <p>For the function $$f(x)=(\cos x)-x+1, x \in \mathbb{R}$$, between the following two statements</p>
<p>(S1) $$f(x)=0$$ for only one value of $$x$$ in $$[0, \pi]$$.</p>
<p>(S2) $$f(x)$$ is decreasing in $$\left[0, \frac{\pi}{2}\right]$$ and increasing in $$\left[\frac{\pi}{2}, \pi\right]$$.</p> | [{"identifier": "A", "content": "Both (S1) and (S2) are incorrect.\n"}, {"identifier": "B", "content": "Only (S1) is correct.\n"}, {"identifier": "C", "content": "Only (S2) is correct.\n"}, {"identifier": "D", "content": "Both (S1) and (S2) are correct."}] | ["B"] | null | <p>Let's analyze the function $$f(x) = (\cos x) - x + 1$$ over the interval $$[0, \pi]$$ and the statements provided.</p>
<p>First, let's consider statement (S1):</p>
<p>(S1) $$f(x)=0$$ for only one value of $$x$$ in $$[0, \pi]$$.</p>
<p>To examine this statement, we need to explore the zeros of the function $$f(x)$$ within the given interval. Let's define and analyze the function:</p>
<p>$$f(x) = \cos x - x + 1$$</p>
<p>We seek to determine if $$f(x) = 0$$ has only one solution in the interval $$[0, \pi]$$. To do this, we can use the Intermediate Value Theorem and the behavior of the function's derivative. First, compute the derivative of $$f(x)$$:</p>
<p>$$f'(x) = \frac{d}{dx}(\cos x - x + 1) = -\sin x - 1$$</p>
<p>The critical points occur when $$f'(x) = 0$$:
<p>$$- \sin x - 1 = 0 \Rightarrow \sin x = -1$$.</p></p>
<p>The equation $$\sin x = -1$$ does not hold for any $$x$$ in $$[0, \pi]$$. Note that:
<ul>
<li>For $$x \in [0, \frac{\pi}{2}]$$, $$\sin x$$ ranges from 0 to 1.</li><br>
<li>For $$x \in [\frac{\pi}{2}, \pi]$$, $$\sin x$$ ranges from 1 to 0.</p></li>
</ul>
<p>Since $$f'(x)$$ is always negative (i.e., $$f'(x) < 0$$), $$f(x)$$ is a strictly decreasing function in $$[0, \pi]$$. Moreover, we evaluate:</p>
<p>$$f(0) = \cos 0 - 0 + 1 = 1 + 1 = 2$$</p>
<p>$$f(\pi) = \cos \pi - \pi + 1 = -1 - \pi + 1 = -\pi$$</p>
<p>Given the continuous and strictly decreasing nature of $$f(x)$$ in $$[0, \pi]$$, by the Intermediate Value Theorem, there is exactly one value of $$x$$ in the interval $$[0, \pi]$$ where $$f(x) = 0$$, confirming (S1).</p>
<p>Now, let's consider statement (S2):</p>
<p>(S2) $$f(x)$$ is decreasing in $$\left[0, \frac{\pi}{2}\right]$$ and increasing in $$\left[\frac{\pi}{2}, \pi\right]$$.</p>
<p>We already analyzed that $$f'(x)$$ shows that $$f'(x) = -\sin x - 1$$ is always less than zero in $$[0, \pi]$$. No interval exists where the derivative is positive. This means that $$f(x)$$ is strictly decreasing throughout the entire interval of $$[0, \pi]$$, invalidating (S2).</p>
<p>Therefore, the correct option is:</p>
<p><b>Option B</b>: Only (S1) is correct.</p> | mcq | jee-main-2024-online-8th-april-morning-shift |
lv7v3jxe | maths | application-of-derivatives | monotonicity | <p>For the function</p>
<p>$$f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right],$$</p>
<p>consider the following two statements :</p>
<p>(I) $$f$$ is increasing in $$\left(0, \frac{\pi}{2}\right)$$.</p>
<p>(II) $$f^{\prime}$$ is decreasing in $$\left(0, \frac{\pi}{2}\right)$$.</p>
<p>Between the above two statements,</p> | [{"identifier": "A", "content": "only (I) is true.\n"}, {"identifier": "B", "content": "both (I) and (II) are true.\n"}, {"identifier": "C", "content": "only (II) is true.\n"}, {"identifier": "D", "content": "neither (I) nor (II) is true."}] | ["B"] | null | <p>$$\begin{aligned}
& f(x)=\sin x+3 x-\frac{2}{\pi}\left(x^2+x\right), \text { where } x \in\left[0, \frac{\pi}{2}\right] \\
& f^{\prime}(x)=\cos x+3-\frac{2}{\pi}(2 x+1) \\
& =\cos x-\frac{4 x}{\pi}-\frac{2}{\pi}+3 \\
& \text { as } x \in\left[0, \frac{\pi}{2}\right] \\
& \frac{4 x}{\pi} \in[0,2]
\end{aligned}$$</p>
<p>$$\Rightarrow 3-\frac{2}{\pi}-\frac{4 x}{\pi}>0$$</p>
<p>and also $$\cos x>0$$ when $$x \in\left[0, \frac{\pi}{2}\right]$$</p>
<p>$$\Rightarrow f^{\prime}(x)>0$$</p>
<p>$$\Rightarrow f(x)$$ is increasing</p>
<p>Now, $$f^{\prime \prime}(x)=-\sin x-\frac{4}{\pi}<0 \forall x \in\left[0, \frac{\pi}{2}\right]$$</p>
<p>Hence, $$f^{\prime}(x)$$ is decreasing</p>
<p>$$\therefore \quad$$ Both statements (I) and (II) are true</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
lvc57uea | maths | application-of-derivatives | monotonicity | <p>The interval in which the function $$f(x)=x^x, x>0$$, is strictly increasing is</p> | [{"identifier": "A", "content": "$$(0, \\infty)$$\n"}, {"identifier": "B", "content": "$$\\left(0, \\frac{1}{e}\\right]$$\n"}, {"identifier": "C", "content": "$$\\left[\\frac{1}{e^2}, 1\\right)$$\n"}, {"identifier": "D", "content": "$$\\left[\\frac{1}{e}, \\infty\\right)$$"}] | ["D"] | null | <p>$$\begin{aligned}
& f(x)=x^x \\
& f(x)=x^x(\log x+1) \\
& f(x) \geq 0 \\
& \Rightarrow 1+\log x \geq 0 \\
& \Rightarrow \log x \geq-1 \\
& \Rightarrow x \geq e^{-1} \\
& \therefore x \in\left[\frac{1}{e^{\prime}}, \infty\right)
\end{aligned}$$</p> | mcq | jee-main-2024-online-6th-april-morning-shift |
5hg8moK4Iyz89JC7 | maths | application-of-derivatives | rate-of-change-of-quantity | A point on the parabola $${y^2} = 18x$$ at which the ordinate increases at twice the rate of the abscissa is | [{"identifier": "A", "content": "$$\\left( {{9 \\over 8},{9 \\over 2}} \\right)$$ "}, {"identifier": "B", "content": "$$(2, -4)$$ "}, {"identifier": "C", "content": "$$\\left( {{-9 \\over 8},{9 \\over 2}} \\right)$$"}, {"identifier": "D", "content": "$$(2, 4)$$ "}] | ["A"] | null | $${y^2} = 18x \Rightarrow 2y{{dy} \over {dx}} = 18 \Rightarrow {{dy} \over {dx}} = {9 \over y}$$
<br><br>Given $${{dy} \over {dx}} = 2 \Rightarrow {9 \over 2} = 2 \Rightarrow y = {9 \over 2}$$
<br><br>Puting in $${y^2} = 18x \Rightarrow x = {9 \over 8}$$
<br><br>$$\therefore$$ Required point is $$\left( {{9 \over 8},{9 \over 2}} \right)$$ | mcq | aieee-2004 |
eSua2dp9jsk4cv3A | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical iron ball $$10$$ cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of $$50$$ cm$$^3$$ /min. When the thickness of ice is $$5$$ cm, then the rate at which the thickness of ice decreases is
| [{"identifier": "A", "content": "$${1 \\over {36\\pi }}$$ cm/min"}, {"identifier": "B", "content": "$${1 \\over {18\\pi }}$$ cm/min"}, {"identifier": "C", "content": "$${1 \\over {54\\pi }}$$ cm/min"}, {"identifier": "D", "content": "$${5 \\over {6\\pi }}$$ cm/min"}] | ["B"] | null | Given that
<br><br>$${{dv} \over {dt}} = 50\,c{m^3}/\min $$
<br><br>$$ \Rightarrow {d \over {dt}}\left( {{4 \over 3}\pi {r^3}} \right) = 50$$
<br><br>$$ \Rightarrow 4\pi {r^2}{{dr} \over {dt}} = 50$$
<br><br>$$ \Rightarrow {{dr} \over {dt}} = {{50} \over {4\pi {{\left( {15} \right)}^2}}} = {1 \over {18\pi }}\,\,cm/\min \,\,$$
<br><br>(here $$r=10+5)$$ | mcq | aieee-2005 |
ljle24hk | maths | application-of-derivatives | rate-of-change-of-quantity | A lizard, at an initial distance of 21 cm behind an insect moves from rest with an acceleration of $2 \mathrm{~cm} / \mathrm{s}^2$ and pursues the insect which is crawling uniformly along a straight line at a speed of $20 \mathrm{~cm} / \mathrm{s}$. Then the lizard will catch the insect after : | [{"identifier": "A", "content": "20 s"}, {"identifier": "B", "content": "1 s"}, {"identifier": "C", "content": "21 s"}, {"identifier": "D", "content": "24 s"}] | ["C"] | null | <p>The motion of the lizard, which starts from rest and accelerates at a rate of $a = 2 \, \text{cm/s}^2$, can be described by the equation of motion :</p>
<p>$D_l = \frac{1}{2} a t^2$</p>
<p>where $D_l$ is the distance the lizard travels, $a$ is its acceleration, and $t$ is the time.</p>
<p>The insect, moving at a constant speed of $v = 20 \, \text{cm/s}$, has a motion that can be simply described as:</p>
<p>$D_i = v t$</p>
<p>where $D_i$ is the distance the insect travels, $v$ is its velocity, and $t$ is the time.</p>
<p>Since the lizard starts 21 cm behind the insect and needs to catch up, it must travel the distance the insect travels plus an additional 21 cm. Equating these two distances gives us :</p>
<p>$\frac{1}{2} a t^2 = v t + 21$</p>
<p>Substituting the given values into this equation gives :</p>
<p>$t^2 = \frac{(v t + 21) \cdot 2}{a}$</p>
<p>Substituting $a = 2 \, \text{cm/s}^2$ and $v = 20 \, \text{cm/s}$, we simplify to :</p>
<p>$t^2 = 20t + 21$</p>
<p>This simplifies to a quadratic equation :</p>
<p>$t^2 - 20t - 21 = 0$</p>
<p>Solving this quadratic equation for $t$ gives the solutions $t = 21 \, \text{s}$ and $t = -1 \, \text{s}$. Since time cannot be negative, we discard the -1 s solution. Therefore, the lizard will catch the insect after 21 seconds.</p>
<p>So, the correct answer is <strong>Option C : 21 s</strong>.</p>
| mcq | aieee-2005 |
ZlHAOfk1vBZtgBRm | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical balloon is filled with $$4500\pi $$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $$72\pi $$ cubic meters per minute, then the rate (in meters per minute) at which the radius of the balloon decreases $$49$$ minutes after the leakage began is : | [{"identifier": "A", "content": "$${{9 \\over 7}}$$"}, {"identifier": "B", "content": "$${{7 \\over 9}}$$"}, {"identifier": "C", "content": "$${{2 \\over 9}}$$"}, {"identifier": "D", "content": "$${{9 \\over 2}}$$"}] | ["C"] | null | Volume of spherical balloon $$ = V = {4 \over 3}\pi {r^3}$$
<br><br>$$ \Rightarrow 4500\pi = {{4\pi {r^3}} \over 3}$$
<br><br>( as Given, volume $$ = 4500\pi {m^3}$$ )
<br><br>Differentiating both the sides, $$w.r.t't'$$ we get,
<br><br>$${{dV} \over {dt}} = 4\pi {r^2}\left( {{{dr} \over {dt}}} \right)$$
<br><br>Now, it is given that $${{dV} \over {dt}} = 72\pi $$
<br><br>$$\therefore$$ After $$49$$ min, Volume -
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 49 \times 72} \right)\pi $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {4500 - 3528} \right)\pi $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 972\pi {m^3}$$
<br><br>$$ \Rightarrow V = 972\,\,\pi {m^3}$$
<br><br>$$\therefore$$ $$972\pi = {4 \over 3}\pi r{}^3$$
<br><br>$$ \Rightarrow {r^3} = 3 \times 243 = 3 \times {3^5} = {3^6} = {\left( {{3^2}} \right)^3} \Rightarrow r = 9$$
<br><br>Also, we have $${{dV} \over {dt}} = 72\pi $$
<br><br>$$\therefore$$ $$72\pi = 4\pi \times 9 \times 9\left( {{{dr} \over {dt}}} \right) \Rightarrow {{dr} \over {dt}} = \left( {{2 \over 9}} \right)$$ | mcq | aieee-2012 |
j6dKprjfuC9iNtm3Qv18hoxe66ijvwuw2ce | maths | application-of-derivatives | rate-of-change-of-quantity | A water tank has the shape of an inverted right
circular cone, whose semi-vertical angle is
$${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$. Water is poured into it at a constant
rate of 5 cubic meter per minute. The the rate
(in m/min.), at which the level of water is rising
at the instant when the depth of water in the tank
is 10m; is :- | [{"identifier": "A", "content": "$${1 \\over {15\\pi }}$$"}, {"identifier": "B", "content": "$${1 \\over {5\\pi }}$$"}, {"identifier": "C", "content": "$${1 \\over {10\\pi }}$$"}, {"identifier": "D", "content": "$${2 \\over \\pi }$$"}] | ["B"] | null | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265192/exam_images/spujcpaesz26bmcciwz1.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264494/exam_images/cmldsgcfwufq2oz2bpff.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266702/exam_images/ympsmpobjl8mibz9vqjo.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265845/exam_images/dk7w8bditkgxgia8lokx.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 9th April Evening Slot Mathematics - Application of Derivatives Question 141 English Explanation"></picture>
<br><br>Given $${{dv} \over {dt}}$$ = 5 cm<sup>3</sup>/min
<br><br> and $$\theta $$ = $${\tan ^{ - 1}}\left( {{1 \over 2}} \right)$$
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = $${1 \over 2}$$ = $${r \over h}$$
<br><br>$$ \Rightarrow $$ h = 2r
<br><br>Volume of the cone,
<br><br>v = $${1 \over 3}\pi {r^2}h$$ = $${1 \over 3}\pi {\left( {{h \over 2}} \right)^2}h$$ = $${{\pi {h^3}} \over {12}}$$
<br><br>$$ \therefore $$ $${{dv} \over {dt}} = {\pi \over {12}}\left( {3{h^2}} \right){{dh} \over {dt}}$$
<br><br>$$ \Rightarrow $$ 5 = $${\pi \over 4}{h^2}{{dh} \over {dt}}$$
<br><br>$$ \Rightarrow $$ 5 = $${\pi \over 4}{\left( {10} \right)^2}{{dh} \over {dt}}$$
<br><br>$$ \Rightarrow $$ $${{dh} \over {dt}}$$ = $${1 \over {5\pi }}$$ | mcq | jee-main-2019-online-9th-april-evening-slot |
kaw0SkAcvfOydLXkka3rsa0w2w9jx1zj510 | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical iron ball of radius 10 cm is coated with a layer of ice of uniform thickness that melts at a rate of
50 cm<sup>3</sup>
/min. When the thickness of the ice is 5 cm, then the rate at which the thickness (in cm/min) of the ice
decreases, is : | [{"identifier": "A", "content": "$${5 \\over {6\\pi }}$$"}, {"identifier": "B", "content": "$${1 \\over {9\\pi }}$$"}, {"identifier": "C", "content": "$${1 \\over {36\\pi }}$$"}, {"identifier": "D", "content": "$${1 \\over {18\\pi }}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263695/exam_images/a99ifvlxnsbj5wrmjdqh.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th April Evening Slot Mathematics - Application of Derivatives Question 140 English Explanation"><br>
$$V = {4 \over 3}\pi \left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)$$<br><br>
$$ \Rightarrow {{dV} \over {dt}} = 4\pi {(10 + h)^2}{{dh} \over {dt}}$$<br><br>
$$ \Rightarrow - 50 = 4\pi {\left( {10 + 5} \right)^2}{{dh} \over {dt}}$$<br><br>
$$ \Rightarrow {{dh} \over {dt}} = - {1 \over {18\pi }}$$ | mcq | jee-main-2019-online-10th-april-evening-slot |
D5MTcIGTKq9EGcY3D23rsa0w2w9jx65rjtw | maths | application-of-derivatives | rate-of-change-of-quantity | A 2 m ladder leans against a vertical wall. If the top of the ladder begins to slide down the wall at the rate
25 cm/sec, then the rate (in cm/sec.) at which the bottom of the ladder slides away from the wall on the
horizontal ground when the top of the ladder is 1 m above the ground is : | [{"identifier": "A", "content": "$${{25} \\over 3}$$"}, {"identifier": "B", "content": "25"}, {"identifier": "C", "content": "25$$\\sqrt 3 $$"}, {"identifier": "D", "content": "$${{25} \\over {\\sqrt 3 }}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266309/exam_images/axa7ph2cq7yfizyo5afc.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th April Morning Slot Mathematics - Application of Derivatives Question 137 English Explanation"><br>
x<sup>2</sup> + y<sup>2</sup> = 4 $$ \Rightarrow $$ $$2x{{dx} \over {dt}} + 2y{{dy} \over {dt}} = 0$$<br><br>
$$ \Rightarrow {{dx} \over {dt}} = - {y \over x}.{{dy} \over {dt}}$$<br><br>
When upper end is 1m above the ground, $${{dx} \over {dt}} = - {1 \over {\sqrt 3 }}.25 = - {{25} \over {\sqrt 3 }}$$ cm/sec. | mcq | jee-main-2019-online-12th-april-morning-slot |
WdFaC9JydJYDDwFq5b7k9k2k5ioxa2s | maths | application-of-derivatives | rate-of-change-of-quantity | A spherical iron ball of 10 cm radius is
coated with a layer of ice of uniform
thickness the melts at a rate of 50 cm<sup>3</sup>/min.
When the thickness of ice is 5 cm, then the rate
(in cm/min.) at which of the thickness of ice
decreases, is : | [{"identifier": "A", "content": "$${1 \\over {18\\pi }}$$"}, {"identifier": "B", "content": "$${1 \\over {36\\pi }}$$"}, {"identifier": "C", "content": "$${1 \\over {54\\pi }}$$"}, {"identifier": "D", "content": "$${5 \\over {6\\pi }}$$"}] | ["A"] | null | Let the thickness = h cm
<br><br>Volume of ice = v = $${{4\pi } \over 3}\left( {{{\left( {10 + h} \right)}^3} - {{10}^3}} \right)$$
<br><br>$$ \Rightarrow $$ $${{dv} \over {dt}} = {{4\pi } \over 3}\left( {3{{\left( {10 + h} \right)}^2}} \right).{{dh} \over {dt}}$$
<br><br>Given $${{dv} \over {dt}} = $$ 50 cm<sup>3</sup>/min and h = 5 cm
<br><br>$$ \therefore $$ 50 = $${{4\pi } \over 3}\left( {3{{\left( {10 + 5} \right)}^2}} \right).{{dh} \over {dt}}$$
<br><br>$$ \Rightarrow $$ $${{dh} \over {dt}} = {{50} \over {4\pi \times {{15}^2}}}$$ = $${1 \over {18\pi }}$$ cm/min | mcq | jee-main-2020-online-9th-january-morning-slot |
R9XIGlr0aLF6KGHJEFjgy2xukf46a92x | maths | application-of-derivatives | rate-of-change-of-quantity | If the surface area of a cube is increasing at a
rate of 3.6 cm<sup>2</sup>/sec, retaining its shape; then
the rate of change of its volume (in cm<sup>3</sup>/sec),
when the length of a side of the cube is
10 cm, is : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "18"}, {"identifier": "D", "content": "20"}] | ["A"] | null | For cube of side 'a'<br><br>A = 6a<sup>2</sup> and V = a<sup>3</sup><br><br>Given $${{dA} \over {dt}} = 3.6$$<br><br>$$ \Rightarrow $$$$ 12a{{da} \over {dt}}$$ = 3.6<br><br>$${{dV} \over {dt}} = 3{a^2}.{{da} \over {dt}} = 3{a^2}\left( {{{3.6} \over {12a}}} \right)$$<br><br>at a = 10<br><br>$${{dV} \over {dt}} = 9$$ | mcq | jee-main-2020-online-3rd-september-evening-slot |
1l59ke5cr | maths | application-of-derivatives | rate-of-change-of-quantity | <p>Water is being filled at the rate of 1 cm<sup>3</sup> / sec in a right circular conical vessel (vertex downwards) of height 35 cm and diameter 14 cm. When the height of the water level is 10 cm, the rate (in cm<sup>2</sup> / sec) at which the wet conical surface area of the vessel increases is</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "$${{\\sqrt {21} } \\over 5}$$"}, {"identifier": "C", "content": "$${{\\sqrt {26} } \\over 5}$$"}, {"identifier": "D", "content": "$${{\\sqrt {26} } \\over {10}}$$"}] | ["C"] | null | <p>$$\because$$ $$V = {1 \over 3}\pi {r^2}h$$ and $${r \over h} = {7 \over {35}} = {1 \over 5}$$</p>
<p>$$ \Rightarrow V = {1 \over {75}}\pi {h^3}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5th5qtk/a97fb8bf-1c8b-4d95-accc-6e78d7642980/88a53880-0818-11ed-98aa-f9038709a939/file-1l5th5qtl.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5th5qtk/a97fb8bf-1c8b-4d95-accc-6e78d7642980/88a53880-0818-11ed-98aa-f9038709a939/file-1l5th5qtl.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th June Evening Shift Mathematics - Application of Derivatives Question 68 English Explanation"></p>
<p>$${{dV} \over {dt}} = {1 \over {25}}\pi {h^2}{{dh} \over {dt}} = 1$$</p>
<p>$$ \Rightarrow {{dh} \over {dt}} = {{25} \over {\pi {h^2}}}$$</p>
<p>Now, $$S = \pi rl = \pi \left( {{h \over 5}} \right)\sqrt {{h^2} + {{{h^2}} \over {25}}} = {\pi \over {25}}\sqrt {26} {h^2}$$</p>
<p>$$ \Rightarrow {{dS} \over {dt}} = {{2\sqrt {26} \pi h} \over {25}}\,.\,{{dh} \over {dt}} = {{2\sqrt {26} } \over h}$$</p>
<p>$$ \Rightarrow {{dS} \over {d{t_{(h = 10)}}}} = {{\sqrt {26} } \over 5}$$</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l5c0t4qw | maths | application-of-derivatives | rate-of-change-of-quantity | <p>The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially, the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is :</p> | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "12"}] | ["A"] | null | <p>We know,</p>
<p>Surface area of balloon (s) = 4$$\pi$$r<sup>2</sup></p>
<p>$$\therefore$$ $${{ds} \over {dt}} = {d \over {dt}}(4\pi {r^2})$$</p>
<p>$$ \Rightarrow {{ds} \over {dt}} = 4\pi (2r) \times {{dr} \over {dt}}$$</p>
<p>$$ \Rightarrow {{ds} \over {dt}} = 8\pi r \times {{dr} \over {dt}}$$</p>
<p>Given that, surface area of balloon is increasing in constant rate.</p>
<p>$$\therefore$$ $${{ds} \over {dt}}$$ = constant = k (Assume)</p>
<p>$$ \Rightarrow k = 8\pi r\,.\,{{dr} \over {dt}}$$</p>
<p>$$ \Rightarrow \int {k\,dt = \int {8\pi r\,dr} } $$</p>
<p>$$ \Rightarrow kt = 8\pi \times {{{r^2}} \over 2} + c$$</p>
<p>$$ \Rightarrow kt = 4\pi {r^2} + c$$ ..... (1)</p>
<p>Given at t = 0, radius r = 3</p>
<p>So, $$0 = 4\pi ({3^2}) + c$$</p>
<p>$$ \Rightarrow c = - 36\pi $$</p>
<p>$$\therefore$$ Equation (1) becomes</p>
<p>$$kt = 4\pi {r^2} - 36\pi $$</p>
<p>Also given, at t = 5, radius r = 7</p>
<p>$$\therefore$$ $$k(5) = 4\pi {(7)^2} - 36$$</p>
<p>$$ \Rightarrow k = 32\pi $$</p>
<p>$$\therefore$$ Equation (1) is</p>
<p>$$32\pi t = 4\pi {r^2} - 36\pi $$</p>
<p>Now at $$t = 9$$</p>
<p>$$ \Rightarrow 32\pi (9) = 4\pi {r^2} - 36\pi $$</p>
<p>$$ \Rightarrow 8 \times 9 = {r^2} - 9$$</p>
<p>$$ \Rightarrow {r^2} = 81 \Rightarrow r = 9$$</p> | mcq | jee-main-2022-online-24th-june-morning-shift |
1l5w1f9pv | maths | application-of-derivatives | rate-of-change-of-quantity | <p>A hostel has 100 students. On a certain day (consider it day zero) it was found that two students are infected with some virus. Assume that the rate at which the virus spreads is directly proportional to the product of the number of infected students and the number of non-infected students. If the number of infected students on 4<sup>th</sup> day is 30, then number of infected students on 8<sup>th</sup> day will be __________.</p> | [] | null | 90 | <p>Total students = 100</p>
<p>At t = 0 (zero day), infected student = 2</p>
<p>Let at t = t day infected student = x</p>
<p>$$\therefore$$ At t = t day non infected student = (100 $$-$$ x)</p>
<p>Rate of infection $$ = {{dx} \over {dt}}$$</p>
<p>Given, $${{dx} \over {dt}} \propto x(100 - x)$$</p>
<p>$$ \Rightarrow \int\limits_{}^{} {{{dx} \over {x(100 - x)}} = \int\limits_{}^{} {k\,dt} } $$</p>
<p>$$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {{{100 - x + x} \over {x(100 - x)}}dx = k\,t + c} $$</p>
<p>$$ \Rightarrow {1 \over {100}}\int\limits_{}^{} {\left( {{1 \over x} + {1 \over {100 - x}}} \right)dx = k\,t + c} $$</p>
<p>$$ \Rightarrow {1 \over {100}}\left[ {\ln x - \ln (100 - x)} \right] = k\,t + c$$</p>
<p>$$ \Rightarrow {1 \over {100}}\ln {x \over {100 - x}} = k\,t + c$$ ...... (1)</p>
<p>Given, At, t = 0, x = 2</p>
<p>$$\therefore$$ $${1 \over {100}}\ln {2 \over {98}} = c$$</p>
<p>Putting value of c in equation (1), we get</p>
<p>$${1 \over {100}}\ln {x \over {100 - x}} = kt + {1 \over {100}}\ln {2 \over {98}}$$</p>
<p>$$ \Rightarrow {1 \over {100}}\ln {x \over {100 - x}} - {1 \over {100}}\ln {2 \over {98}} = kt$$</p>
<p>$$ \Rightarrow {1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = kt$$</p>
<p>Given, At t = 4, x = 30</p>
<p>$$\therefore$$ $${1 \over {100}}\ln {{30 \times 98} \over {2(70)}} = k \times 4$$</p>
<p>$$ \Rightarrow k = {1 \over {400}}\ln 21$$</p>
<p>$$\therefore$$ $${1 \over {100}}\ln {{x \times 98} \over {2(100 - x)}} = t \times {1 \over {400}} \times \ln 21$$</p>
<p>Now, when t = 8, then r = ?</p>
<p>$${1 \over {100}}\ln {{49x} \over {(100 - x)}} = 8 \times {1 \over {400}} \times \ln 21$$</p>
<p>$$ \Rightarrow \ln {{49x} \over {(100 - x)}} = 2\ln 21$$</p>
<p>$$ \Rightarrow {{49x} \over {100 - x}} = {21^2}$$</p>
<p>$$ \Rightarrow {x \over {100 - x}} = {{21 \times 21} \over {49}}$$</p>
<p>$$ \Rightarrow {x \over {100 - x}} = 9$$</p>
<p>$$ \Rightarrow x = 900 - 9x$$</p>
<p>$$ \Rightarrow 10x = 900$$</p>
<p>$$ \Rightarrow x = 90$$</p> | integer | jee-main-2022-online-30th-june-morning-shift |
1l6kljqrp | maths | application-of-derivatives | rate-of-change-of-quantity | <p>A water tank has the shape of a right circular cone with axis vertical and vertex downwards. Its semi-vertical angle is $$\tan ^{-1} \frac{3}{4}$$. Water is poured in it at a constant rate of 6 cubic meter per hour. The rate (in square meter per hour), at which the wet curved surface area of the tank is increasing, when the depth of water in the tank is 4 meters, is ______________.</p> | [] | null | 5 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7qaiwsv/317e788b-9fe5-4c30-acd5-40a1d96c242d/abb0b900-2df0-11ed-a744-1fb8f3709cfa/file-1l7qaiwsw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7qaiwsv/317e788b-9fe5-4c30-acd5-40a1d96c242d/abb0b900-2df0-11ed-a744-1fb8f3709cfa/file-1l7qaiwsw.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 - Application of Derivatives Question 51 English Explanation"></p>
<p>$$v = {1 \over 3}\pi {r^2}h$$ ..... (i)</p>
<p>And $$\tan \theta = {3 \over 4} = {r \over h}$$ ...... (ii)</p>
<p>i.e. if $$h = 4,\,r = 3$$</p>
<p>$$v = {1 \over 3}\pi {r^2}\left( {{{4r} \over 3}} \right)$$</p>
<p>$${{dv} \over {dt}} = {{4\pi } \over 9}3{r^2}{{dr} \over {dt}} \Rightarrow 6 = {{4\pi } \over 3}(9){{dr} \over {dt}}$$</p>
<p>$$ \Rightarrow {{dr} \over {dt}} = {1 \over {2\pi }}$$</p>
<p>Curved area $$ = \pi r\sqrt {{r^2} + {h^2}} $$</p>
<p>$$ = \pi r\sqrt {{r^2} + {{16{r^2}} \over 9}} $$</p>
<p>$$ = {5 \over 3}\pi {r^2}$$</p>
<p>$${{dA} \over {dt}} = {{10} \over 3}\pi r{{dr} \over {dt}}$$</p>
<p>$$ = {{10} \over 3}\pi \,.\,3\,.\,{1 \over {2\pi }}$$</p>
<p>$$ = 5$$</p> | integer | jee-main-2022-online-27th-july-evening-shift |
1lh2zk9lq | maths | application-of-derivatives | rate-of-change-of-quantity | <p> The number of points, where the curve $$y=x^{5}-20 x^{3}+50 x+2$$ crosses the $$\mathrm{x}$$-axis, is ____________.</p> | [] | null | 5 | Given equation of curve
<br><br>$$
\begin{aligned}
& y=x^5-20 x^3+50 x+2 \\\\
& \Rightarrow \frac{d y}{d x}=5 x^4-60 x^2+50
\end{aligned}
$$
<br><br>On putting $\frac{d y}{d x}=0$
<br><br>$$
\begin{array}{ll}
\Rightarrow & 5\left(x^4-12 x^2+10\right)=0 \\\\
\Rightarrow & x^2=\frac{12 \pm \sqrt{144-40}}{2}=6 \pm \sqrt{26} \\\\
\Rightarrow & x^2=6-\sqrt{26}, 6+\sqrt{26} \\\\
\Rightarrow & x^2=6-5.10,6+5.10 \\\\
\Rightarrow & x^2=09,11.1 \\\\
\Rightarrow & x= \pm \sqrt{0.9}, \pm \sqrt{11.1} \\\\
\Rightarrow & x=-0.95,0.95,-3.33,3.33
\end{array}
$$
<br><br>Now,
<br><br>$$
\begin{aligned}
y(0) & =2(+\mathrm{ve}) \Rightarrow y(1)=+\mathrm{ve} \\\\
y(2) & =-\mathrm{ve} \Rightarrow y(3.3)=-\mathrm{ve} \\\\
y(-1) & =-\mathrm{ve} \Rightarrow y(-2)=+\mathrm{ve} \\\\
y(-3.3) & =-\mathrm{ve}
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1locy9o68/7ed2dae1-7f5d-4feb-a54d-f2d90c2aabe3/3a30c2f0-772a-11ee-8319-49c6602bc172/file-6y3zli1locy9o69.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1locy9o68/7ed2dae1-7f5d-4feb-a54d-f2d90c2aabe3/3a30c2f0-772a-11ee-8319-49c6602bc172/file-6y3zli1locy9o69.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 - Application of Derivatives Question 29 English Explanation">
<br>$$
\because \text { Required number of points }=5
$$ | integer | jee-main-2023-online-6th-april-evening-shift |
jaoe38c1lse5ndcf | maths | application-of-derivatives | rate-of-change-of-quantity | <p>$$\text { If } f(x)=\left|\begin{array}{ccc}
x^3 & 2 x^2+1 & 1+3 x \\
3 x^2+2 & 2 x & x^3+6 \\
x^3-x & 4 & x^2-2
\end{array}\right| \text { for all } x \in \mathbb{R} \text {, then } 2 f(0)+f^{\prime}(0) \text { is equal to }$$</p> | [{"identifier": "A", "content": "24"}, {"identifier": "B", "content": "18"}, {"identifier": "C", "content": "42"}, {"identifier": "D", "content": "48"}] | ["C"] | null | <p>$$\begin{aligned}
& f(0)=\left|\begin{array}{ccc}
0 & 1 & 1 \\
2 & 0 & 6 \\
0 & 4 & -2
\end{array}\right|=12 \\
& f^{\prime}(x)=\left|\begin{array}{ccc}
3 x^2 & 4 x & 3 \\
3 x^2+2 & 2 x & x^3+6 \\
x^3-x & 4 & x^2-2
\end{array}\right|+
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \left|\begin{array}{ccc}
x^3 & 2 x^2+1 & 1+3 x \\
6 x & 2 & 3 x^2 \\
x^3-x & 4 & x^2-2
\end{array}\right|+ \\
& \left|\begin{array}{ccc}
x^3 & 2 x^2+1 & 1+3 x \\
3 x^2+2 & 2 x & x^3+6 \\
3 x^2-1 & 0 & 2 x
\end{array}\right| \\
& \therefore f^{\prime}(0)=\left|\begin{array}{ccc}
0 & 0 & 3 \\
2 & 0 & 6 \\
0 & 4 & -2
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 1 \\
0 & 2 & 0 \\
0 & 4 & -2
\end{array}\right|+\left|\begin{array}{ccc}
0 & 1 & 1 \\
2 & 0 & 6 \\
-1 & 0 & 0
\end{array}\right| \\
& =24-6=18 \\
& \therefore 2 f(0)+f^{\prime}(0)=42
\end{aligned}$$</p> | mcq | jee-main-2024-online-31st-january-morning-shift |
lv7v4o2e | maths | application-of-derivatives | rate-of-change-of-quantity | <p>Let $$f(x)=x^5+2 x^3+3 x+1, x \in \mathbf{R}$$, and $$g(x)$$ be a function such that $$g(f(x))=x$$ for all $$x \in \mathbf{R}$$. Then $$\frac{g(7)}{g^{\prime}(7)}$$ is equal to :</p> | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "14"}] | ["D"] | null | <p>$$\begin{aligned}
& f(x)=x^5+2 x^3+3 x+1 \\
& g(f(x))=x . \quad \Rightarrow g^{\prime}(f(x)) f^{\prime}(x)=1 \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
&\begin{aligned}
& \text { Now } \frac{g(7)}{g^{\prime}(7)} \\
& g(7) \Rightarrow f(x)=7 \\
& x^5+2 x^3+3 x+1=7 \\
& \Rightarrow x\left(x^4+2 x^2+3\right)=0 \\
& \Rightarrow x=1 \\
& \therefore g(7) \Rightarrow g(f(1))=1 \\
& \& ~g^{\prime}(f(x))=\frac{1}{f^{\prime}(x)} \\
& g^{\prime}(7) \\
& \Rightarrow f(x)=7 \Rightarrow x=1 \\
& \therefore g^{\prime}(7)=\frac{1}{f^{\prime}(1)} \\
& =\frac{1}{5 x^4+6 x^2+3} \\
& =\frac{1}{14} \\
& \therefore \frac{g(7)}{g^{\prime}(7)}=\frac{1}{\frac{1}{14}} \quad 14
\end{aligned}\\
\end{aligned}$$</p> | mcq | jee-main-2024-online-5th-april-morning-shift |
WzB2UMLEAvI0j8Sz | maths | application-of-derivatives | tangent-and-normal | A function $$y=f(x)$$ has a second order derivative $$f''\left( x \right) = 6\left( {x - 1} \right).$$ If its graph passes through the point $$(2, 1)$$ and at that point the tangent to the graph is $$y = 3x - 5$$, then the function is : | [{"identifier": "A", "content": "$${\\left( {x + 1} \\right)^2}$$ "}, {"identifier": "B", "content": "$${\\left( {x - 1} \\right)^3}$$"}, {"identifier": "C", "content": "$${\\left( {x + 1} \\right)^3}$$"}, {"identifier": "D", "content": "$${\\left( {x - 1} \\right)^2}$$"}] | ["B"] | null | $$f''\left( x \right) = 6\left( {x - 1} \right).$$ Inegrating,
<br><br>we get $$f'\left( x \right) = 3{x^2} - 6x + c$$
<br><br>Slope at $$\left( {2,1} \right) = f'\left( 2 \right) = c = 3$$
<br><br>$$\left[ {\,\,} \right.$$ As slope of tangent at $$(2, 1)$$ is $$3$$ $$\left. {\,\,} \right]$$
<br><br>$$\therefore$$ $$f'\left( x \right) = 3{x^2} - 6x + 3 = 3{\left( {x - 1} \right)^2}$$
<br><br>Inegrating again, we get
<br><br>$$f\left( x \right) = {\left( {x - 1} \right)^3} + D$$
<br><br>The curve passes through $$(2,1)$$
<br><br>$$ \Rightarrow 1 = {\left( {2 - 1} \right)^3} + D \Rightarrow D = 0$$
<br><br>$$\therefore$$ $$f\left( x \right) = {\left( {x - 1} \right)^3}$$ | mcq | aieee-2004 |
7xmtfPK7UFAAvV0N | maths | application-of-derivatives | tangent-and-normal | The normal to the curve x = a(1 + cos $$\theta $$), $$y = a\sin \theta $$ at $$'\theta '$$ always passes through the fixed point | [{"identifier": "A", "content": "$$(a, a)$$ "}, {"identifier": "B", "content": "$$(0, a)$$ "}, {"identifier": "C", "content": "$$(0, 0)$$ "}, {"identifier": "D", "content": "$$(a, 0)$$ "}] | ["D"] | null | $${{dx} \over {d\theta }} = - a\sin \theta $$ and $${{dy} \over {d\theta }} = a\cos \theta $$
<br><br>$$\therefore$$ $${{dy} \over {dx}} = - \cot \theta .$$
<br><br>$$\therefore$$ The slope of the normal at $$\theta $$ = $$ - {1 \over { - \cot \theta }}$$$$= \tan \theta $$
<br><br>$$\therefore$$ The equation of the normal at $$\theta $$ is
<br><br>$$y - a\sin \theta = \tan \theta \left( {x - a - a\cos \theta } \right)$$
<br><br>$$ \Rightarrow $$ $$y - a\sin \theta =$$ $${{\sin \theta } \over {\cos \theta }}$$ $$\left( {x - a - a\cos \theta } \right)$$
<br><br>$$ \Rightarrow y\cos \theta - a\sin \theta \cos \theta $$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,$$ $$ = x\sin \theta - a\sin \theta - a\sin \theta \cos \theta $$
<br><br>$$ \Rightarrow x\sin \theta - y\cos \theta = a\sin \theta $$
<br><br>$$ \Rightarrow y = \left( {x - a} \right)\tan \theta $$
<br><br>which always passes through $$(a, 0)$$ | mcq | aieee-2004 |
xDpHAzVDO3wJz2Tq | maths | application-of-derivatives | tangent-and-normal | The normal to the curve
<br/>$$x = a\left( {\cos \theta + \theta \sin \theta } \right),y = a\left( {\sin \theta - \theta \cos \theta } \right)$$ at any point
<br/>$$\theta\, '$$ is such that
| [{"identifier": "A", "content": "it passes through the origin "}, {"identifier": "B", "content": "it makes an angle $${\\pi \\over 2} + \\theta $$ with the $$x$$-axis "}, {"identifier": "C", "content": "it passes through $$\\left( {a{\\pi \\over 2}, - a} \\right)$$ "}, {"identifier": "D", "content": "it is at a constant distance from the origin "}] | ["D"] | null | $$x = a\left( {\cos \theta + \theta \sin \theta } \right)$$
<br><br>$$ \Rightarrow {{dx} \over {d\theta }} = a\left( { - \sin \theta + \sin \theta + \theta \cos \theta } \right)$$
<br><br>$$ \Rightarrow {{dx} \over {d\theta }} = a\theta \cos \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>$$y = a\left( {\sin \theta - \theta \cos \theta } \right)$$
<br><br>$${{dy} \over {d\theta }} = a\left[ {\cos \theta - \cos \theta + \theta \sin \theta } \right]$$
<br><br>$$ \Rightarrow {{dy} \over {d\theta }} = a\theta \sin \theta \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>From equations $$(1)$$ and $$(2),$$ we get
<br><br>$${{dy} \over {dx}} = \tan \theta \Rightarrow $$ Slope of normal $$ = - \cot \,\theta $$
<br><br>Equation of normal at
<br><br>$$'\theta '$$ is $$y - a\left( {\sin \theta - \theta \cos \theta } \right)$$
<br><br>$$ = - \cot \theta \left( {x - a\left( {\cos \theta + \theta \sin \theta } \right)} \right.$$
<br><br>$$ \Rightarrow y\sin \theta - a{\sin ^2}\theta + a\,\,\theta \cos \theta \sin \theta $$
<br><br>$$ = - x\cos \theta + a{\cos ^2}\theta + a\,\theta \sin \theta \cos \theta $$
<br><br>$$ \Rightarrow x\cos \theta + y\sin \theta = a$$
<br><br>Clearly this is an equation of straight line -
<br><br>which is at a constant distance $$'a'$$ from origin. | mcq | aieee-2005 |
DLqakneawRUrbWd3 | maths | application-of-derivatives | tangent-and-normal | Angle between the tangents to the curve $$y = {x^2} - 5x + 6$$ at the points $$(2,0)$$ and $$(3,0)$$ is | [{"identifier": "A", "content": "$$\\pi $$ "}, {"identifier": "B", "content": "$${\\pi \\over 2}$$ "}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$${\\pi \\over 4}$$"}] | ["B"] | null | $${{dy} \over {dx}} = 2x - 5$$
<br><br>$$\therefore$$ $${m_1} = {\left( {2x - 5} \right)_{\left( {2,0} \right)}} = - 1,$$
<br><br> $${m_2} = {\left( {2x - 5} \right)_{\left( {3,0} \right)}} = 1 \Rightarrow {m_1}{m_2} = - 1$$
<br><br>i.e. the tangents are perpendicular to each other. | mcq | aieee-2006 |
TN45biXlr0c9O695 | maths | application-of-derivatives | tangent-and-normal | The equation of the tangent to the curve $$y = x + {4 \over {{x^2}}}$$, that
<br/>is parallel to the $$x$$-axis, is | [{"identifier": "A", "content": "$$y=1$$ "}, {"identifier": "B", "content": "$$y=2$$ "}, {"identifier": "C", "content": "$$y=3$$ "}, {"identifier": "D", "content": "$$y=0$$ "}] | ["C"] | null | Since tangent is parallel to $$x$$-axis,
<br><br>$$\therefore$$ $${{dy} \over {dx}} = 0 \Rightarrow 1 - {8 \over {{x^3}}} = 0 \Rightarrow x = 2 \Rightarrow y = 3$$
<br><br>Equation of tangent is $$y - 3 = 0\left( {x - 2} \right) \Rightarrow y = 3$$ | mcq | aieee-2010 |
da6RXvhPYMZWxEIr | maths | application-of-derivatives | tangent-and-normal | The shortest distance between line $$y-x=1$$ and curve $$x = {y^2}$$ is | [{"identifier": "A", "content": "$${{3\\sqrt 2 } \\over 8}$$ "}, {"identifier": "B", "content": "$${8 \\over {3\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "$${4 \\over {\\sqrt 3 }}$$ "}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4}$$ "}] | ["A"] | null | Shortest distance between two curve occurred along -
<br><br>the common normal
<br><br>Slope of normal to $${y^2} = x$$ at point
<br><br>$$P\left( {{t^2},t} \right)$$ is $$-2t$$ and
<br><br>slope of line $$y - x = 1$$ is $$1.$$
<br><br>As they are perpendicular to each other
<br><br>$$\therefore$$ $$\left( { - 2t} \right) = - 1 \Rightarrow t = {1 \over 2}$$
<br><br>$$\therefore$$ $$P\left( {{1 \over 4},{1 \over 2}} \right)$$
<br><br>and shortest distance $$ = \left| {{{{1 \over 2} - {1 \over 4} - 1} \over {\sqrt 2 }}} \right|$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266469/exam_images/f698mokq1rh01m1s0toe.webp" loading="lazy" alt="AIEEE 2011 Mathematics - Application of Derivatives Question 182 English Explanation">
<br><br>So shortest distance between them is $${{3\sqrt 2 } \over 8}$$ | mcq | aieee-2011 |
i0kX6KoVoB3amGZz | maths | application-of-derivatives | tangent-and-normal | The intercepts on $$x$$-axis made by tangents to the curve,
<br/>$$y = \int\limits_0^x {\left| t \right|dt,x \in R,} $$ which are parallel to the line $$y=2x$$, are equal to : | [{"identifier": "A", "content": "$$ \\pm 1$$ "}, {"identifier": "B", "content": "$$ \\pm 2$$"}, {"identifier": "C", "content": "$$ \\pm 3$$"}, {"identifier": "D", "content": "$$ \\pm 4$$"}] | ["A"] | null | Since, $$y = \int\limits_0^x {\left| t \right|} dt,x \in R$$
<br><br>therefore $${{dy} \over {dx}} = \left| x \right|$$
<br><br>But from $$y = 2x,{{dy} \over {dx}} = 2$$
<br><br>$$ \Rightarrow \left| x \right| = 2 \Rightarrow x = \pm 2$$
<br><br>Points $$y = \int\limits_0^{ \pm 2} {\left| t \right|dt} = \pm 2$$
<br><br>$$\therefore$$ equation of tangent is
<br><br>$$y - 2 = 2\left( {x - 2} \right)$$ or $$y + 2 = 2\left( {x + 2} \right)$$
<br><br>$$ \Rightarrow $$ $$x$$-intercept $$ = \pm 1.$$ | mcq | jee-main-2013-offline |
nbnpebQ2nYcrhYcK | maths | application-of-derivatives | tangent-and-normal | The normal to the curve, $${x^2} + 2xy - 3{y^2} = 0$$, at $$(1,1)$$ | [{"identifier": "A", "content": "meets the curve again in the third quadrant. "}, {"identifier": "B", "content": "meets the curve again in the fourth quadrant. "}, {"identifier": "C", "content": "does not meet the curve again."}, {"identifier": "D", "content": "meets the curve again in the second quadrant."}] | ["B"] | null | Given curve is
<br><br>$${x^2} + 2xy - 3{y^2} = 0$$
<br><br>Difference $$w.r.t.x,$$
<br><br>$$2x + 2x{{dy} \over {dx}} + 2y - 6y{{dy} \over {dx}} = 0$$
<br><br>$${\left( {{{dy} \over {dx}}} \right)_{\left( {1,1} \right)}} = 1$$
<br><br>Equation of normal at $$(1,1)$$ is
<br><br>$$y=2-x$$
<br><br>Solving eq. $$(1)$$ and $$(2),$$ we get $$x=1,3$$
<br><br>Point of intersection $$\left( {1,1} \right),\left( {3, - 1} \right)$$
<br><br>Normal cuts the curve again in $$4$$th quadrant. | mcq | jee-main-2015-offline |
k0fZVE4svRgWgQQCzYBRu | maths | application-of-derivatives | tangent-and-normal | If the tangent at a point P, with parameter t, on the curve x = 4t<sup>2</sup> + 3, y = 8t<sup>3</sup>β1, <i>t</i> $$ \in $$ <b>R</b>, meets the curve again at a point Q, then the coordinates of Q are : | [{"identifier": "A", "content": "(t<sup>2</sup> + 3, \u2212 t<sup>3</sup> \u22121)\n"}, {"identifier": "B", "content": "(4t<sup>2</sup> + 3, \u2212 8t<sup>3</sup> \u22121)\n"}, {"identifier": "C", "content": "(t<sup>2</sup> + 3, t<sup>3</sup> \u22121)\n"}, {"identifier": "D", "content": "(16t<sup>2</sup> + 3, \u2212 64t<sup>3</sup> \u22121)"}] | ["A"] | null | Given, x = 4t<sup>2</sup> + 3 and y = 8t<sup>3</sup> $$-$$ 1
<br><br>$$ \therefore $$ P $$ \equiv $$ (4t<sup>2</sup> + 3, 8t<sup>3</sup> $$-$$ 1)
<br><br>$${{dx} \over {dt}} = 8t$$ and $${{dy} \over {dt}}$$ $$=$$ 24t<sup>2</sup>
<br><br>Slope of tangent at
<br><br>P $$=$$ $${{dy} \over {dx}}$$ $$=$$ $${{dy/dt} \over {dx/dt}}$$ $$=$$ 3t
<br><br>Assume Q $$=$$ (4$$\lambda $$<sup>2</sup> + 3, 8$$\lambda $$<sup>3</sup> $$-$$ 1)
<br><br>$$ \therefore $$ Slope of PQ $$=$$ 3t
<br><br>$$ \Rightarrow $$ $${{8{\lambda ^3} - 8{t^3}} \over {4{\lambda ^2} - 4{t^2}}} = 3t$$
<br><br>$$ \Rightarrow $$ $${{8\left( {\lambda - t} \right)\left( {{\lambda ^2} + {t^2} + \lambda t} \right)} \over {4\left( {\lambda - t} \right)\left( {\lambda + t} \right)}} = 3t$$
<br><br>$$ \Rightarrow $$ 2$$\lambda $$<sup>2</sup> + 2t<sup>2</sup> + 2$$\lambda $$t $$=$$ 3t (t + $$\lambda $$)
<br><br>$$ \Rightarrow $$ t<sup>2</sup> + $$\lambda $$t $$-$$ 2$$\lambda $$<sup>2</sup> = 0
<br><br>$$ \Rightarrow $$ (t $$-$$ $$\lambda $$) (t + 2$$\lambda $$) $$=$$ 0
<br><br>$$ \therefore $$ $$\lambda $$ $$=$$ t or $$\lambda $$ $$=$$ $$-$$ $${t \over 2}$$
<br><br>$$ \therefore $$ Q $$=$$ (4t<sup>2</sup> + 3, 8t<sup>3</sup> $$-$$ 1) Or
<br><br>Q $$=$$ $$\left( {4 \times {{\left( { - {t \over 2}} \right)}^2} + 3, - 8 \times {{{t^3}} \over 8} - 1} \right)$$
<br><br> $$=$$ (t<sup>2</sup> + 3, $$-$$ t<sup>3</sup> $$-$$ 1) | mcq | jee-main-2016-online-9th-april-morning-slot |
idGZjiTMQyZ5SsmMHXupk | maths | application-of-derivatives | tangent-and-normal | Let C be a curve given by y(x) = 1 + $$\sqrt {4x - 3} ,x > {3 \over 4}.$$ If P is a point
on C, such that the tangent at P has slope $${2 \over 3}$$, then a point through which the normal at P passes, is : | [{"identifier": "A", "content": "(2, 3)"}, {"identifier": "B", "content": "(4, $$-$$3)"}, {"identifier": "C", "content": "(1, 7) "}, {"identifier": "D", "content": "(3, $$-$$ 4), "}] | ["C"] | null | Given,
<br><br>y = 1 + $$\sqrt {4x - 3} $$
<br><br>$$ \therefore $$ $${{dy} \over {dx}}$$ = $${1 \over {2\sqrt {4x - 3} }} \times 4 = {2 \over 3}$$
<br><br>$$ \Rightarrow $$ 4x $$-$$ 3 = 9
<br><br>$$ \Rightarrow $$ x = 3
<br><br>$$ \therefore $$ y = 1 + $$\sqrt {12 - 3} $$ = 4
<br><br>$$ \therefore $$ Equation of normal at point P(3,4)
<br><br>y $$-$$ 4 = $$-$$ $${3 \over 2}$$ (x $$-$$ 3)
<br><br>$$ \Rightarrow $$ 2y $$-$$ 8 = $$-$$ 3x + 9
<br><br>$$ \Rightarrow $$ 3x + 2y $$-$$ 17 = 0 | mcq | jee-main-2016-online-10th-april-morning-slot |
v0O67CCyo8mfgAHJ | maths | application-of-derivatives | tangent-and-normal | Consider :
<br/>f $$\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin x} \over {1 - \sin x}}} } \right),x \in \left( {0,{\pi \over 2}} \right).$$
<p>A normal to $$y = $$ f$$\left( x \right)$$ at $$x = {\pi \over 6}$$ also passes through the point:</p> | [{"identifier": "A", "content": "$$\\left( {{\\pi \\over 6},0} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{\\pi \\over 4},0} \\right)$$"}, {"identifier": "C", "content": "$$(0,0)$$ "}, {"identifier": "D", "content": "$$\\left( {0,{{2\\pi } \\over 3}} \\right)$$ "}] | ["D"] | null | $$f\left( x \right) = {\tan ^{ - 1}}\left( {\sqrt {{{1 + \sin \,x} \over {1 - \sin x}}} } \right)$$
<br><br>$$ = {\tan ^{ - 1}}\left( {\sqrt {{{{{\left( {\sin {x \over 2} + \cos {x \over 2}} \right)}^2}} \over {{{\left( {\sin {x \over x} - \cos {x \over 2}} \right)}^2}}}} } \right)$$
<br><br>$$ = {\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)$$
<br><br>$$ = {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)$$
<br><br>$$ \Rightarrow y = {\pi \over 4} + {x \over 2}$$
<br><br>$$ \Rightarrow {{dy} \over {dx}} = {1 \over 2}$$
<br><br>Slope of normal $$ = {{ - 1} \over {\left( {{{dy} \over {dx}}} \right)}} = - 2$$
<br><br>At $$\left( {{\pi \over 6},{\pi \over 4} + {\pi \over {12}}} \right)$$
<br><br>$$y - \left( {{\pi \over 4} + {\pi \over {12}}} \right) = - 2\left( {x - {\pi \over 6}} \right)$$
<br><br>$$y - {{4\pi } \over {12}} = - 2x + {{2\pi } \over 6}$$
<br><br>$$y - {\pi \over 3} = - 2x + {\pi \over 3}$$
<br><br>$$y = - 2x + {{2\pi } \over 3}$$
<br><br>This equation is satisfied only by the point
<br><br>$$\left( {0,{{2\pi } \over 3}} \right)$$ | mcq | jee-main-2016-offline |
ApbhaGnJQ9LIEHm8 | maths | application-of-derivatives | tangent-and-normal | The normal to the curve y(x β 2)(x β 3) = x + 6 at the point where the curve intersects the y-axis passes
through the point : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 2},{1 \\over 2}} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over 2}, - {1 \\over 3}} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{1 \\over 2},{1 \\over 3}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {1 \\over 2}, - {1 \\over 3}} \\right)$$"}] | ["A"] | null | Given $$y = {{x + 6} \over {\left( {x - 2} \right)\left( {x - 2} \right)}}$$
<br><br>At y-axis, x = 0 $$ \Rightarrow $$ y = 1
<br><br>On differentiating, we get
<br><br>$${{dy} \over {dx}} = {{\left( {{x^2} - 5x + 6} \right)\left( 1 \right) - \left( {x + 6} \right)\left( {2x - 5} \right)} \over {{{\left( {{x^2} - 5x + 6} \right)}^2}}}$$
<br><br>$${{dy} \over {dx}} = 1$$ at point (0, 1)
<br><br>$$ \therefore $$ Slope of normal = β 1
<br><br>Now equation of normal is y β 1 = β1 (x β 0)
<br><br>$$ \Rightarrow $$ y β 1 = β x
<br><br>x + y = 1 ......(1)
<br><br>By checking each option you can see point $$\left( {{1 \over 2},{1 \over 2}} \right)$$ satisfy equation (1). | mcq | jee-main-2017-offline |
2Vz6ubbB5NN9kiDT6g4bK | maths | application-of-derivatives | tangent-and-normal | The tangent at the point (2, $$-$$2) to the curve, x<sup>2</sup>y<sup>2</sup> $$-$$ 2x = 4(1 $$-$$ y) <b>does not</b> pass through the point : | [{"identifier": "A", "content": "$$\\left( {4,{1 \\over 3}} \\right)$$"}, {"identifier": "B", "content": "(8, 5)"}, {"identifier": "C", "content": "($$-$$4, $$-$$9)"}, {"identifier": "D", "content": "($$-$$2, $$-$$7)"}] | ["D"] | null | As, $${{dy} \over {dx}}$$ = $$-$$ $$\left[ {{{{{\delta f} \over {\delta x}}} \over {{{\delta f} \over {\delta y}}}}} \right]$$
<br><br>$${{{\delta f} \over {\delta x}}}$$ = y<sup>2</sup> $$ \times $$2x $$-$$ 2
<br><br>$${{{\delta f} \over {\delta y}}}$$ = x<sup>2</sup> $$ \times $$ 2y + 4
<br><br>$$\therefore\,\,\,$$ $${{dy} \over {dx}}$$ = $$-$$ $$\left( {{{2x{y^2} - 2} \over {2{x^2}y + 4}}} \right)$$
<br><br>$${\left[ {{{dy} \over {dx}}} \right]_{(2, - 2)}}$$ = $$-$$ $$\left( {{{2 \times 2 \times 4 - 2} \over {2 \times 4 \times ( - 2) + 4}}} \right)$$ = $$-$$ $$\left( {{{14} \over { - 12}}} \right)$$ = $${7 \over 6}$$
<br><br>$$\therefore\,\,\,$$ Slope of tangent to the curve = $${7 \over 6}$$
<br><br>Equation of tangent passes through (2, $$-$$ 2) is
<br><br>y + 2 = $${7 \over 6}$$ (x $$-$$ 2)
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 7x $$-$$ 6y = 26 . . . . .(1)
<br><br>Now put each option in equation (1) and see which one does not satisfy the equation. <br><br>By verifying each points you can see ($$-$$ 2, $$-$$ 7) does not satisfy the equation. | mcq | jee-main-2017-online-8th-april-morning-slot |
2Cp4wkEasxlGmlzJJmg0m | maths | application-of-derivatives | tangent-and-normal | A tangent to the curve, y = f(x) at P(x, y) meets x-axis at A and y-axis at B. If AP : BP = 1 : 3 and f(1) = 1, then the curve also passes through the point : | [{"identifier": "A", "content": "$$\\left( {{1 \\over 3},24} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( {{1 \\over 2},4} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( {2,{1 \\over 8}} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {3,{1 \\over 28}} \\right)$$"}] | ["C"] | null | <p>We have</p>
<p>$${{(y - {y_2})} \over {(x - {x_1})}} = f'({x_1})$$</p>
<p>$$ \Rightarrow y - {y_1} = f'({x_1})(x - {x_1})$$</p>
<p>$$\bullet$$ When y = 0: $${{ - {y_1}} \over {f'({x_1})}} = x - {x_1}$$</p>
<p>$$ \Rightarrow x = {x_1} - {{{y_1}} \over {f'({x_1})}}$$</p>
<p>Therefore, point A is $$A\left( {{x_1} - {{{y_1}} \over {f'({x_1})}},0} \right)$$</p>
<p>$$\bullet$$ When x = 0: $$y - {y_1} = f'(x)\,.\,( - {x_1})$$</p>
<p>$$ \Rightarrow y = {y_1} - {x_1}f'({x_1})$$</p>
<p>Therefore, point B is $$B(0,{y_1} - {x_1}f'({x_1}))$$</p>
<p>Point P divides AB in the ratio 1 : 3.</p>
<p>$${x_1} = {{\left[ {3\left( {{x_1} - {{{y_1}} \over {f'({x_1})}}} \right)} \right]} \over 4}$$</p>
<p>$${y_1} = {{{y_1} - {x_1}f'({x_1})} \over 4}$$</p>
<p>Therefore, $$4{y_1} = {y_1} - {x_1}f'({x_1})$$</p>
<p>$$ \Rightarrow f'({x_1}) = {{ - 3{y_1}} \over {{x_1}}} \Rightarrow f'(x) = {{ - 3y} \over x}$$</p>
<p>Now, $${{dy} \over {dx}} = {{ - 3y} \over x} \Rightarrow {{dy} \over y} = {{ - 3dx} \over x}$$</p>
<p>On integrating, we get</p>
<p>$$\ln y = - 3\ln x + C \Rightarrow y = k{x^{ - 3}}$$</p>
<p>$$y(1) = 1 \Rightarrow k = 1$$</p>
<p>$$y = {1 \over {{x^3}}}$$</p>
<p>When we substitute the values from the given options, only option (c) satisfies the above equation.</p> | mcq | jee-main-2017-online-9th-april-morning-slot |
9tt20aC5xbn0orfm | maths | application-of-derivatives | tangent-and-normal | If the curves y<sup>2</sup> = 6x, 9x<sup>2</sup> + by<sup>2</sup> = 16 intersect each other at right angles, then the value of b is : | [{"identifier": "A", "content": "$${9 \\over 2}$$"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "$${7 \\over 2}$$"}, {"identifier": "D", "content": "4"}] | ["A"] | null | When two curves intersect each other at right angle, then at the point of intersection the product of tangent of slopes = $$-1$$.
<br><br>Let m<sub>1</sub>, and m<sub>2</sub> are the tangent of the slope of the two curves respectively
<br><br>$$\therefore\,\,\,$$ m<sub>1</sub> m<sub>2</sub> = $$-$$ 1.
<br><br>Now let they intersect at point (x<sub>1</sub>, y<sub>1</sub>)
<br><br>$$\therefore\,\,\,$$ $$y_1^2 = 6x,$$ and $$9x_1^2 + b\,y_1^2 = 16$$
<br><br>y<sup>2</sup> = 6x
<br><br>$$ \Rightarrow \,\,\,\,2y{{dy} \over {dx}} = 6$$
<br><br>$$ \Rightarrow \,\,\,\,{{dy} \over {dx}} = {3 \over y}$$
<br><br>$$\therefore\,\,\,$$ $${\left( {{{dy} \over {dx}}} \right)_{\left( {{x_1},{y_1}} \right)}} = {3 \over {{y_1}}} = {m_1}$$
<br><br>9x<sup>2</sup> + by<sup>2</sup> = 16
<br><br>$$ = 18x + 2by{{dy} \over {dx}} = O$$
<br><br>$$ \Rightarrow \,\,\,\,{\left( {{{dy} \over {dx}}} \right)_{({x_1},{y_1})}} = - {{9{x_1}} \over {b{y_1}}} = {m_2}$$
<br><br>As m<sub>1</sub> m<sub>2</sub> = $$-$$1
<br><br>$$\therefore\,\,\,$$ $${3 \over {y{}_1}} \times - {{9{x_1}} \over {b{y_1}}} = - 1$$
<br><br>$$ \Rightarrow \,\,\,\,27{x_1} = by_1^2$$
<br><br>$$ \Rightarrow \,\,\,\,\,27{x_1} = b.6{x_1}$$ $$\,\,\,$$ [as $$y_1^2 = 6{x_1}\,]$$
<br><br>$$ \Rightarrow \,\,\,\,b = {{27} \over 6} = {9 \over 2}$$ | mcq | jee-main-2018-offline |
boajRKD8w1QSY5sgpYNZA | maths | application-of-derivatives | tangent-and-normal | If $$\beta $$ is one of the angles between the normals to the ellipse, x<sup>2</sup> + 3y<sup>2</sup> = 9 at the points (3 cos $$\theta $$, $$\sqrt 3 \sin \theta $$) and ($$-$$ 3 sin $$\theta $$, $$\sqrt 3 \,\cos \theta $$); $$\theta \in \left( {0,{\pi \over 2}} \right);$$ then $${{2\,\cot \beta } \over {\sin 2\theta }}$$ is equal to : | [{"identifier": "A", "content": "$${2 \\over {\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 3 }}$$"}, {"identifier": "C", "content": "$$\\sqrt 2 $$"}, {"identifier": "D", "content": "$${{\\sqrt 3 } \\over 4}$$ "}] | ["A"] | null | Since, x<sup>2</sup> + 3y<sup>2</sup> = 9
<br><br>$$ \Rightarrow $$ 2x + 6y $${{dy} \over {dx}}$$ = 0
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ = $${{ - x} \over {3y}}$$
<br><br>Slope of normal is $$-$$ $${{dx} \over {dy}}$$ = $${{3y} \over x}$$
<br><br>$$ \Rightarrow $$ $${\left( { - {{dx} \over {dy}}} \right)_{\left( {3\cos \theta ,\sqrt 3 \sin \theta } \right)}}$$
<br><br> = $${{3\sqrt 3 \sin \theta } \over {3\cos \theta }}$$ = $$\sqrt 3 \tan \theta $$ = m<sub>1</sub>
<br><br>& $${\left( { - {{dx} \over {dy}}} \right)_{\left( { - 3\sin \theta ,\sqrt 3 \cos \theta } \right)}}$$
<br><br> = $${{3\sqrt 3 \cos \theta } \over { - 3\sin \theta }}$$ = $$ - \sqrt 3 \cot \theta $$ = m<sub>2</sub>
<br><br>As, $$\beta $$ is the angle between the normals to the given ellipse then
<br><br>tan$$\beta $$ = $$\left| {{{{m_1} - {m_2}} \over {1 + {m_1}{m_2}}}} \right|$$
<br><br>= $$\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3\tan \theta \cot \theta }}} \right|$$ = $$\left| {{{\sqrt 3 \tan \theta + \sqrt 3 \cot \theta } \over {1 - 3}}} \right|$$
<br><br>So, tan $$\beta $$ = $${{\sqrt 3 } \over 2}$$ $$\left| {\tan \theta + \cot \theta } \right|$$
<br><br>$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}\left| {{{\sin \theta } \over {\cos \theta }} + {{\cos \theta } \over {\sin \theta }}} \right|$$
<br><br>$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over 2}$$ $$\left| {{1 \over {\sin \theta \cos \theta }}} \right|$$
<br><br>$$ \Rightarrow $$ $${1 \over {\cot \beta }} = {{\sqrt 3 } \over {\sin 2\theta }}$$
<br><br>$$ \Rightarrow $$ $${{2\cot \beta } \over {\sin 2\theta }}$$ = $${2 \over {\sqrt 3 }}$$ | mcq | jee-main-2018-online-15th-april-morning-slot |
8ffCZxoD0TRKsnKUmshBK | maths | application-of-derivatives | tangent-and-normal | The tangent to the curve y = x<sup>2</sup> β 5x + 5, parallel to the line 2y = 4x + 1, also passes through the point : | [{"identifier": "A", "content": "$$\\left\\{ {{1 \\over 4},{7 \\over 2}} \\right\\}$$"}, {"identifier": "B", "content": "$$\\left( { - {1 \\over 8},7} \\right)$$"}, {"identifier": "C", "content": "$$\\left( {{7 \\over 2},{1 \\over 4}} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 8}, - 7} \\right)$$"}] | ["D"] | null | y = x<sup>2</sup> $$-$$ 5x + 5
<br><br>$${{dy} \over {dx}} = 2x - 5 = 2 \Rightarrow x = {7 \over 2}$$
<br><br>at x = $${7 \over 2}$$, y = $${{ - 1} \over 4}$$
<br><br>Equation of tangent at
<br><br>$$\left( {{7 \over 2},{{ - 1} \over 4}} \right)$$ is 2x $$-$$ y $$-$$ $${{29} \over 4}$$ = 0
<br><br>Now check options
<br><br>x = $${1 \over 8}$$, y = $$-$$7 | mcq | jee-main-2019-online-12th-january-evening-slot |
Subsets and Splits