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1l59jocof | maths | quadratic-equation-and-inequalities | modulus-function | <p>Let $$A = \{ x \in R:|x + 1| < 2\} $$ and $$B = \{ x \in R:|x - 1| \ge 2\} $$. Then which one of the following statements is NOT true?</p> | [{"identifier": "A", "content": "$$A - B = ( - 1,1)$$"}, {"identifier": "B", "content": "$$B - A = R - ( - 3,1)$$"}, {"identifier": "C", "content": "$$A \\cap B = ( - 3, - 1]$$"}, {"identifier": "D", "content": "$$A \\cup B = R - [1,3)$$"}] | ["B"] | null | <p>A = ($$-$$3, 1) and B = ($$-$$ $$\infty$$, $$-$$1] $$\cup$$ [3, $$\infty$$)</p>
<p>So, A $$-$$ B = ($$-$$1, 1)</p>
<p>B $$-$$ A = ($$-$$ $$\infty$$, $$-$$3] $$\cup$$ [3, $$\infty$$) = R $$-$$ ($$-$$3, 3)</p>
<p>A $$\cap$$ B = ($$-$$3, $$-$$1]</p>
<p>and A $$\cup$$ B = ($$-$$ $$\infty$$, 1) $$\cup$$ [3, $$\infty$$) = R $$-$$ [1, 3)</p> | mcq | jee-main-2022-online-25th-june-evening-shift |
1l6nl1la8 | maths | quadratic-equation-and-inequalities | modulus-function | <p>$$
\text { Let } S=\left\{x \in[-6,3]-\{-2,2\}: \frac{|x+3|-1}{|x|-2} \geq 0\right\} \text { and } $$<br/><br/>$$T=\left\{x \in \mathbb{Z}: x^{2}-7|x|+9 \leq 0\right\} \text {. }
$$</p>
<p>Then the number of elements in $$\mathrm{S} \cap \mathrm{T}$$ is :</p> | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}] | ["D"] | null | <p>$$|{x^2}| - 7|x| + 9 \le 0$$</p>
<p>$$ \Rightarrow |x| \in \left[ {{{7 - \sqrt {13} } \over 2},{{7 + \sqrt {13} } \over 2}} \right]$$</p>
<p>As $$x \in Z$$</p>
<p>So, x can be $$ \pm \,2, \pm \,3, \pm \,4, \pm \,5$$</p>
<p>Out of these values of x,</p>
<p>$$x = 3, - 4, - 5$$</p>
<p>satisfy S as well</p>
<p>$$n(S \cap T) = 3$$</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1ldybvfxu | maths | quadratic-equation-and-inequalities | modulus-function | <p>Let $$\lambda \in \mathbb{R}$$ and let the equation E be $$|x{|^2} - 2|x| + |\lambda - 3| = 0$$. Then the largest element in the set S = {$$x+\lambda:x$$ is an integer solution of E} is ______</p> | [] | null | 5 | $D \geq 0 \Rightarrow 4-4|\lambda-3| \geq 0$
<br/><br/>
$|\lambda-3| \leq 1$
<br/><br/>
$-1 \leq \lambda-3 \leq 1$
<br/><br/>
$2 \leq \lambda \leq 4$
<br/><br/>
$|x|=\frac{2 \pm \sqrt{4-4|\lambda-3|}}{2}$
<br/><br/>
$=1 \pm \sqrt{1-|\lambda-3|}$
<br/><br/>
$x_{\text {largest }}=1+1=2$, when $\lambda=3$
<br/><br/>
Largest element of $S=2+3=5$ | integer | jee-main-2023-online-24th-january-morning-shift |
lgnxyz9u | maths | quadratic-equation-and-inequalities | modulus-function | The number of real roots of the equation $x|x|-5|x+2|+6=0$, is : | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}] | ["B"] | null | $$
\begin{aligned}
& x|x|-5|x+2|+6=0 \\\\
& \text { Case-I : } \\\\
& \text { When } x<-2 \text { then } \\\\
& -x^2+5(x+2)+6=0 \\\\
& \Rightarrow x^2-5 x-16=0 \\\\
& \Rightarrow x=\frac{5 \pm \sqrt{25+64}}{2} \\\\
& \therefore x=\frac{5-\sqrt{89}}{2} \text { is accepted }
\end{aligned}
$$
<br/><br/><b>Case-II :</b>
<br/><br/>$$
\begin{aligned}
& \text { When } -2 \leq x < 0 \text { then } \\\\
& -x^2-5(x+2)+6=0 \\\\
\Rightarrow & x^2+5 x+4=0 \\\\
\Rightarrow & (x+1)(x+4)=0 \\\\
& x=-1 \text { is accepted }
\end{aligned}
$$
<br/><br/><b>Case-III :</b>
<br/><br/>$$
\begin{aligned}
& \text { When } x \geq 0 \text { then } \\\\
& x^2-5(x+2)+6=0 \\\\
\Rightarrow & x^2-5 x-4=0 \\\\
& x=\frac{5 \pm \sqrt{25+16}}{2} \\\\
& =\frac{5 \pm \sqrt{41}}{2} \\\\
& x=\frac{5 - \sqrt{41}}{2} \text { is accepted } \\\\
\therefore & 3 \text { real roots are possible. }
\end{aligned}
$$ | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgpxzk6k | maths | quadratic-equation-and-inequalities | modulus-function | <p>The set of all $$a \in \mathbb{R}$$ for which the equation $$x|x-1|+|x+2|+a=0$$ has exactly one real root, is :</p> | [{"identifier": "A", "content": "$$(-\\infty, \\infty)$$"}, {"identifier": "B", "content": "$$(-6, \\infty)$$"}, {"identifier": "C", "content": "$$(-\\infty,-3)$$"}, {"identifier": "D", "content": "$$(-6,-3)$$"}] | ["A"] | null | $$
x|x-1|+|x+2|+a=0
$$
<br/><br/>Case I : If $x<-2$ then
<br/><br/>$$
-x^2+x-x-2+a=0
$$
<br/><br/>$$
a=x^2+2
$$
<br/><br/>$y=x^2+2$ is decreasing $\forall x \in(-\infty,-2)$
<br/><br/>Case II : If $-2 \leq x<1$ then
<br/><br/>$$
\begin{aligned}
& -x^2+x+x+2+a=0 \\\\
& a=x^2-2 x-2
\end{aligned}
$$
<br/><br/>$y=x^2-2 x-2$ is decreasing $\forall x \in[-2,1)$.
<br/><br/>Case III: If $x \geq 1$ then
<br/><br/>$$
\begin{aligned}
& x^2-x+x+2+a=0 \\\\
& a=-\left(x^2+2\right)
\end{aligned}
$$
<br/><br/>$y=-\left(x^2+2\right)$ is decreasing $\forall x \in[1, \infty)$
<br/><br/>$\therefore$ Exactly one real root $\forall x \in R$ | mcq | jee-main-2023-online-13th-april-morning-shift |
1lsg52jkv | maths | quadratic-equation-and-inequalities | modulus-function | <p>The number of real solutions of the equation $$x\left(x^2+3|x|+5|x-1|+6|x-2|\right)=0$$ is _________.</p> | [] | null | 1 | <p>The given equation is $$x(x^2+3|x|+5|x-1|+6|x-2|)=0$$, which can be solved by analyzing it in parts. It can be broken down into: $$x=0$$ and $$x^2+3|x|+5|x-1|+6|x-2|=0$$. </p><p>For $$x=0$$, it's clear that it is a solution to the equation since it makes the entire expression equal to zero.</p><b>Case (I)</b>
<br/><br/>$$
x<0
$$
<br/><br/>$$
\begin{aligned}
& x^2-3 x-5(x-1)-6(x-2)=0 \\\\
& x^2-14 x+17=0
\end{aligned}
$$
<br/><br/>$\because$ All roots are positive $\Rightarrow$ no solution
<br/><br/><b>Case (II)</b>
<br/><br/>$$
\begin{aligned}
& 0 < x < 1 \\\\
& x^2+3 x-5(x-1)-6(x-2)=0 \\\\
& x^2-8 x+17=0 \\\\
& \because D < 0 \Rightarrow \text { no solution }
\end{aligned}
$$
<br/><br/><b>Case (III)</b>
<br/><br/>$$
1 < x < 2
$$
<br/><br/>$$
x^2+3 x+5(x-1)-6(x-2)=0
$$
<br/><br/>$$
x^2+2 x+7=0
$$
<br/><br/>$\Rightarrow$ no solution
<br/><br/><b>Case (IV)</b>
<br/><br/>$$
\begin{aligned}
& x > 2 \\\\
& x^2+3 x+5(x-1)+6(x-2)=0 \\\\
& x^2+14 x-19=0
\end{aligned}
$$
<br/><br/>All roots less than 2
<br/><br/>$\Rightarrow$ no solution
<br/><br/>Here $x=0$ is only solution. | integer | jee-main-2024-online-30th-january-evening-shift |
lv3ve4b4 | maths | quadratic-equation-and-inequalities | modulus-function | <p>The number of distinct real roots of the equation $$|x+1||x+3|-4|x+2|+5=0$$, is _______</p> | [] | null | 2 | <p>Let's analyze the equation $ |x+1||x+3|-4|x+2|+5=0 $ based on different intervals of $ x $.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw4nlbkg/60bbafa8-f162-4063-8d33-6f852bb91330/ef943d00-10fb-11ef-aaa0-17ca36a32505/file-1lw4nlbkh.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw4nlbkg/60bbafa8-f162-4063-8d33-6f852bb91330/ef943d00-10fb-11ef-aaa0-17ca36a32505/file-1lw4nlbkh.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 - Quadratic Equation and Inequalities Question 7 English Explanation"></p>
<p>(I) If $ x < -3 $ :</p>
<p>$ \begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 + 4x + 3 + 4x + 8 + 5 = 0 \\ & x^2 + 8x + 16 = 0 \\ & \Rightarrow x = -4 \quad (\text{one solution}) \end{aligned} $</p>
<p>(II) If $ -3 \leq x < -2 $ :</p>
<p>$ \begin{aligned} & |x+1| = -(x+1), \\ & |x+3| = -(x+3), \\ & |x+2| = -(x+2), \\ & -(x+1)*(-(x+3)) - 4(-(x+2)) + 5 = 0 \\ & x^2 - 10 = 0 \\ & \Rightarrow x = \pm \sqrt{10} \\ & \text{(Do not satisfy } -3 \leq x < -2) \end{aligned} $</p>
<p>(III) If $ -2 \leq x < -1 $ :</p>
<p>$ \begin{aligned} & |x+1|=-(x+1), \\ & |x+3|=x+3, \\ & |x+2|=-(x+2), \\ & -(x+1)(x+3)-4(-(x+2))+5=0 \\ & -x^2-4x-3-4x-8+5=0 \\ & -x^2-8x-6=0 \\ & \Rightarrow x^2+8x+6=0 \\ & x=\frac{-8 \pm 2\sqrt{10}}{2} = -4 \pm \sqrt{10} \end{aligned} $</p>
<p>(IV) If $ x \geq -1 $ :</p>
<p>$ \begin{aligned} & |x+1|=x+1, \\ & |x+3|=x+3, \\ & |x+2|=x+2, \\ & (x+1)(x+3)-4(x+2)+5=0 \\ & x^2+4x+3-4x-8+5=0 \\ & x^2=0 \\ & \Rightarrow x=0 \quad (\text{one solution}) \end{aligned} $</p>
<p>$\Rightarrow$ The number of distinct real roots are two : $ x = -4 $ and $ x = 0 $.</p> | integer | jee-main-2024-online-8th-april-evening-shift |
lv7v4o4e | maths | quadratic-equation-and-inequalities | modulus-function | <p>The number of distinct real roots of the equation $$|x||x+2|-5|x+1|-1=0$$ is __________.</p> | [] | null | 3 | <p>$$|x| \quad|x+2|-5|x+1|-1=0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/jaoe38c1lwgkotns/48dc3e18-87b2-4fd8-a3b9-f30ede35fcf3/fd635a80-1789-11ef-b185-b7bdfa18e39b/file-jaoe38c1lwgkotnt.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/jaoe38c1lwgkotns/48dc3e18-87b2-4fd8-a3b9-f30ede35fcf3/fd635a80-1789-11ef-b185-b7bdfa18e39b/file-jaoe38c1lwgkotnt.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 - Quadratic Equation and Inequalities Question 5 English Explanation"></p>
<p>$$\begin{aligned}
& \text { (I) if } x<-2 \\
& x^2+2 x+5 x+5-1=0 \\
& x^2+7 x+4=0 \Rightarrow \text { one root satisfying } x<-2
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { (II) if }-2 \leq x<-1 \\
& -x^2-2 x+5 x+5-1=0 \\
& x^2-3 x-4=0 \Rightarrow \text { not root satisfying }-2 \leq x<-1
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { (III) if }-1 \leq x<0 \\
& -x^2-2 x-5 x-5-1=0 \\
& x^2+7 x+6=0 \\
& x=-1 \text { is only root satisfying }-1 \leq x<0
\end{aligned}$$</p>
<p>(IV) if $$x \geq 0$$</p>
<p>$$\begin{aligned}
& x^2+2 x-5 x-5-1=0 \\
& x^2-3 x-6=0
\end{aligned}$$</p>
<p>one root satisfying $$x \geq 0$$</p>
<p>$$\Rightarrow$$ The number of distinct real roots are three.</p> | integer | jee-main-2024-online-5th-april-morning-shift |
lv9s20ns | maths | quadratic-equation-and-inequalities | modulus-function | <p>The number of real solutions of the equation $$x|x+5|+2|x+7|-2=0$$ is __________.</p> | [] | null | 3 | <p>$$x|x+5|+2|x+7|-2=0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwexk63n/0315e6c4-33b6-4400-9e04-b71afe90ed53/c1552330-16a2-11ef-afc3-d53e649859cd/file-1lwexk63o.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwexk63n/0315e6c4-33b6-4400-9e04-b71afe90ed53/c1552330-16a2-11ef-afc3-d53e649859cd/file-1lwexk63o.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 - Quadratic Equation and Inequalities Question 4 English Explanation"></p>
<p>$$\begin{aligned}
& \text { (i) } d x \geq-5 \Rightarrow x(x+5)+2(x+7)-2=0 \\
& x^2+7 x+12=0 \Rightarrow x=-3,-4
\end{aligned}$$</p>
<p>$$\begin{aligned}
\text{(ii)} \quad & x \in(-7,-5) \\
& x(-x-5)+2(x+7)-2=0 \\
& -x^2-3 x+12=0 \\
& \Rightarrow x^2+3 x-12=0 \\
& \Rightarrow x=\frac{-3-\sqrt{57}}{2} \text { satisfy }
\end{aligned}$$</p>
<p>$$\begin{aligned}
\text{(iii)} \quad & x \leq-7 \\
& \Rightarrow x(-x-5)+2(-x-7)-2=0 \\
& -x^2-7 x-16=0 \Rightarrow x^2+7 x+16=0
\end{aligned}$$</p>
<p>No solution</p> | integer | jee-main-2024-online-5th-april-evening-shift |
FDe9CsgnfSoRtuy8 | maths | quadratic-equation-and-inequalities | nature-of-roots | If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals | [{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$3$$ "}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$$1$$ "}] | ["D"] | null | <p>Let n and (n + 1) be the roots of x<sup>2</sup> $$-$$ bx + c = 0.</p>
<p>Then, n + (n + 1) = b and n(n + 1) = c</p>
<p>$$\therefore$$ b<sup>2</sup> $$-$$ 4c = (2n + 1)<sup>2</sup> $$-$$ 4n(n + 1)</p>
<p>= 4n<sup>2</sup> + 4n + 1 $$-$$ 4n<sup>2</sup> $$-$$ 4n = 1</p> | mcq | aieee-2005 |
qHu7OenJ4BgHkd9E | maths | quadratic-equation-and-inequalities | nature-of-roots | If the roots of the equation $${x^2} - bx + c = 0$$ be two consecutive integers, then $${b^2} - 4c$$ equals | [{"identifier": "A", "content": "$$-2$$ "}, {"identifier": "B", "content": "$$3$$ "}, {"identifier": "C", "content": "$$2$$"}, {"identifier": "D", "content": "$$1$$"}] | ["D"] | null | Let $$\alpha ,\,\,\alpha + 1\,\,$$ be roots
<br><br>Then $$\alpha + \alpha + 1 = b = $$ sum of -
<br><br>roots $$\alpha \left( {\alpha + 1} \right) = c$$
<br><br>$$=$$ product of roots
<br><br>$$\therefore$$ $${b^2} - 4c $$
<br><br>$$ = {\left( {2\alpha + 1} \right)^2} - 4\alpha \left( {\alpha + 1} \right)$$
<br><br>$$ = 1.$$ | mcq | aieee-2005 |
46LMmx62v4YiGyce | maths | quadratic-equation-and-inequalities | nature-of-roots | If the roots of the equation $$b{x^2} + cx + a = 0$$ imaginary, then for all real values of $$x$$, the expression $$3{b^2}{x^2} + 6bcx + 2{c^2}$$ is : | [{"identifier": "A", "content": "less than $$4ab$$ "}, {"identifier": "B", "content": "greater than $$-4ab$$"}, {"identifier": "C", "content": "less than $$-4ab$$"}, {"identifier": "D", "content": "greater than $$4ab$$"}] | ["B"] | null | Given that roots of the equation
<br><br>$$b{x^2} + cx + a = 0$$ are imaginary
<br><br>$$\therefore$$ $${c^2} - 4ab < 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>Let $$y = 3{b^2}{x^2} + 6bc\,x + 2{c^2}$$
<br><br>$$ \Rightarrow 3{b^2}{x^2} + 6bc\,x + 2{c^2} - y = 0$$
<br><br>As $$x$$ is real, $$D \ge 0$$
<br><br>$$ \Rightarrow 36{b^2}{c^2} - 12{b^2}\left( {2{c^2} - y} \right) \ge 0$$
<br><br>$$ \Rightarrow 12{b^2}\left( {3{c^2} - 2{c^2} + y} \right) \ge 0$$
<br><br>$$ \Rightarrow {c^2} + y \ge 0$$
<br><br>$$ \Rightarrow y \ge - {c^2}$$
<br><br>But from eqn. $$(i),$$ $${c^2} < 4ab$$
<br><br>or $$ - {c^2} > - 4ab$$
<br><br>$$\therefore$$ we get $$y \ge - {c^2} > - 4ab$$
<br><br>$$y > - 4ab$$ | mcq | aieee-2009 |
3ow2JWohjhrBtHTY | maths | quadratic-equation-and-inequalities | nature-of-roots | The equation $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$ has: | [{"identifier": "A", "content": "infinite number of real roots "}, {"identifier": "B", "content": "no real roots "}, {"identifier": "C", "content": "exactly one real root "}, {"identifier": "D", "content": "exactly four real roots "}] | ["B"] | null | Given equation is $${e^{\sin x}} - {e^{ - \sin x}} - 4 = 0$$
<br><br>Put $${e^{{\mathop{\rm sinx}\nolimits} \,}} = t$$ in the given equation,
<br><br>we get $${t^2} - 4t - 1 = 0$$
<br><br>$$ \Rightarrow t = {{4 \pm \sqrt {16 + 4} } \over 2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm \sqrt {20} } \over 2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = {{4 \pm 2\sqrt 5 } \over 2}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\, = 2 \pm \sqrt 5 $$
<br><br>$$ \Rightarrow {e^{\sin x}} = 2 \pm \sqrt 5 $$ $$\,\,\,\,\,$$ (as $$t = {e^{\sin x}}$$)
<br><br>$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 $$ and
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,$$ $${e^{\sin x}} = 2 + \sqrt 5 $$
<br><br>$$ \Rightarrow {e^{\sin x}} = 2 - \sqrt 5 < 0$$
<br><br>and $$\,\,\,\,\,\,\sin x = \ln \left( {2 + \sqrt 5 } \right) > 1$$ So, rejected
<br><br>Hence given equation has no solution.
<br><br>$$\therefore$$ The equation has no real roots. | mcq | aieee-2012 |
Fl2e9gnl2jKIkobMZcMni | maths | quadratic-equation-and-inequalities | nature-of-roots | The sum of all the real values of x satisfying the equation
<br/>2<sup>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50)</sup> = 1 is : | [{"identifier": "A", "content": "16"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "$$-$$ 5"}] | ["C"] | null | We know, 2<sup>x</sup> = 1 only when x = 0.
<br><br>Similarly, 2<sup>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50)</sup> = 1 when
<br><br>(x$$-$$1)(x<sup>2</sup> + 5x $$-$$ 50) = 0
<br><br>$$ \Rightarrow $$ (x - 1)(x + 10)(x - 5) = 0
<br><br>$$ \therefore $$ x = 1, -10, 5
<br><br>Sum of real values of x = 1 + (-10) + 5 = -4 | mcq | jee-main-2017-online-9th-april-morning-slot |
wQG21GrIK7fXq8Sa2uU4f | maths | quadratic-equation-and-inequalities | nature-of-roots | The number of all possible positive integral values of $$\alpha $$ for which the roots of the quadratic equation, 6x<sup>2</sup> $$-$$ 11x + $$\alpha $$ = 0 are rational numbers is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "5"}] | ["A"] | null | For rational D must be perfect square
<br><br>D = 121 $$-$$ 24$$\alpha $$
<br><br>for 121 $$-$$ 24$$\alpha $$ to be perfect square a must be 3, 4, 5
<br><br>So, ans $$\alpha $$ = 3 | mcq | jee-main-2019-online-9th-january-evening-slot |
wXK8RulvIwkGLmPr1c6RX | maths | quadratic-equation-and-inequalities | nature-of-roots | The number of integral values of m for which the
equation
<br/><br/>(1 + m<sup>2</sup>
)x<sup>2</sup>
– 2(1 + 3m)x + (1 + 8m) = 0
has no real root is : | [{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "infinitely many"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["B"] | null | (1 + m<sup>2</sup>
)x<sup>2</sup>
– 2(1 + 3m)x + (1 + 8m) = 0
<br><br>Given equation has no real solution,
<br><br>$$ \therefore $$ Discriminant (D) < 0
<br><br>$$ \Rightarrow $$ 4(1 + 3m)<sup>2</sup> - 4(1 + m<sup>2</sup>)(1 + 8m) < 0
<br><br>$$ \Rightarrow $$ 4[9m<sup>2</sup> + 6m + 1 - 8m - 1 - 8m<sup>3</sup> - m<sup>2</sup>] < 0
<br><br>$$ \Rightarrow $$ -8m<sup>3</sup> + 8m<sup>2</sup> - 2m < 0
<br><br>$$ \Rightarrow $$ -2m(4m<sup>2</sup> - 4m + 1) < 0
<br><br>$$ \Rightarrow $$ m(2m - 1)<sup>2</sup> > 0
<br><br>$$ \therefore $$ m > 0 and m $$ \ne $$ $${{1 \over 2}}$$
<br><br>So we can say number of integral values of m are infinitely
many. | mcq | jee-main-2019-online-8th-april-evening-slot |
r4cRgmuaQNaqnCyfRN18hoxe66ijvwvt235 | maths | quadratic-equation-and-inequalities | nature-of-roots | If m is chosen in the quadratic equation
<br/><br/>(m<sup>2</sup> + 1)
x<sup>2</sup> – 3x + (m<sup>2</sup> + 1)<sup>2</sup> = 0
<br/><br/>such that the sum of its
roots is greatest, then the absolute difference of
the cubes of its roots is :- | [{"identifier": "A", "content": "$$4\\sqrt 3 $$"}, {"identifier": "B", "content": "$$8\\sqrt 3 $$"}, {"identifier": "C", "content": "$$8\\sqrt 5 $$"}, {"identifier": "D", "content": "$$10\\sqrt 5 $$"}] | ["C"] | null | Given quadratic equation
<br><br>(m<sup>2</sup> + 1)
x<sup>2</sup> – 3x + (m<sup>2</sup> + 1)<sup>2</sup> = 0
<br><br>Let roots of the equation $$\alpha $$ and $$\beta $$.
<br><br>$$ \therefore $$ Sum of roots = $$\alpha $$ + $$\beta $$ = $${3 \over {{m^2} + 1}}$$
<br><br>Product of roots = $$\alpha $$$$\beta $$ = m<sup>2</sup> + 1
<br><br>$${3 \over {{m^2} + 1}}$$ is maximum when m = 0<br><br>
Hence equation becomes x<sup>2</sup> – 3x + 1 = 0<br><br>
$$\alpha + \beta = 3$$, $$\alpha \beta = 1$$ <br><br>$$\left| {\alpha - \beta } \right|$$ = $$\sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } $$ = $$\sqrt {{{\left( 3 \right)}^2} - 4.1} $$ = $$\sqrt 5 $$<br><br>
$$\left| {{\alpha ^3} - {\beta ^3}} \right| = \left| {(\alpha - \beta )({\alpha ^2} + {\beta ^2} + \alpha \beta )} \right| $$
<br><br>= $$\sqrt 5 \left| {{{\left( {\alpha + \beta } \right)}^2} - \alpha \beta } \right|$$
<br><br>$$= \sqrt 5 (9 - 1) = 8\sqrt 5 $$ | mcq | jee-main-2019-online-9th-april-evening-slot |
xt1r2MgSVHVAqiqUTS7k9k2k5h0294b | maths | quadratic-equation-and-inequalities | nature-of-roots | The least positive value of 'a' for which the
equation <br/><br/>2x<sup>2</sup> + (a – 10)x + $${{33} \over 2}$$
= 2a has real
roots is | [] | null | 8 | For real roots Discriminate $$ \ge $$ 0.
<br><br>(a – 10)<sup>2</sup>
– 4$$\left( {{{33} \over 2} - 2a} \right).2$$ $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ a<sup>2</sup>
+ 100 – 20a – 132 + 16a $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ a
<sup>2</sup>
– 4a – 32 $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ (a – 8) (a + 4) $$ \ge $$ 0
<br><br>$$ \Rightarrow $$ a $$ \le $$ -4 $$ \cup $$ a $$ \ge $$ 8
<br><br>$$ \therefore $$ least positive a = 8 | integer | jee-main-2020-online-8th-january-morning-slot |
V6ZrAEdn5TU3mCjJcDjgy2xukez5rybs | maths | quadratic-equation-and-inequalities | nature-of-roots | Let f(x) be a quadratic polynomial such that
<br/>f(–1) + f(2) = 0. If one of the roots of f(x) = 0
<br/>is 3, then its other root lies in : | [{"identifier": "A", "content": "(\u20133, \u20131)"}, {"identifier": "B", "content": "(1, 3)"}, {"identifier": "C", "content": "(\u20131, 0)"}, {"identifier": "D", "content": "(0, 1)"}] | ["C"] | null | Let the other root is $$\alpha $$.
<br><br>$$ \therefore $$ f(x)
=
a(x
–
3)
(x
–
$$\alpha $$)
<br><br>f(2) = a($$\alpha $$– 2)
<br><br>f(–1) = 4a(1 + $$\alpha $$)
<br><br>Given f(–1) + f(2) = 0
<br><br>$$ \Rightarrow $$a($$\alpha $$ – 2 + 4 + 4$$\alpha $$) = 0
<br><br>$$ \Rightarrow $$ 5$$\alpha $$ = -2 As a $$ \ne $$ 0
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $$-\frac{2}{5} $$ = - 0.4
<br><br>$$ \therefore $$ $$\alpha $$ $$ \in $$ (–1, 0) | mcq | jee-main-2020-online-2nd-september-evening-slot |
1ktbh3q6c | maths | quadratic-equation-and-inequalities | nature-of-roots | The sum of all integral values of k (k $$\ne$$ 0) for which the equation $${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$$ in x has no real roots, is ____________. | [] | null | 66 | $${2 \over {x - 1}} - {1 \over {x - 2}} = {2 \over k}$$<br><br>$$x \in R - \{ 1,2\} $$<br><br>$$ \Rightarrow k(2x - 4 - x + 1) = 2({x^2} - 3x + 2)$$<br><br>$$ \Rightarrow k(x - 3) = 2({x^2} - 3x + 2)$$<br><br>for x $$\ne$$ 3, $$k = 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right)$$<br><br>$$x - 3 + {2 \over {x - 3}} \ge 2\sqrt 2 ,\forall x > 3$$<br><br>& $$x - 3 + {2 \over {x - 3}} \le - 2\sqrt 2 ,\forall x < - 3$$<br><br>$$ \Rightarrow 2\left( {x - 3 + {2 \over {x - 3}} + 3} \right) \in \left( { - \infty ,6 - 4\sqrt 2 } \right] \cup \left[ {6 + 4\sqrt 2 ,\infty } \right)$$<br><br>for no real roots<br><br>$$k \in (6 - 4\sqrt 2 ,6 + 4\sqrt 2 ) - \{ 0\} $$<br><br>Integral k$$\in$${1, 2 ..... 11}<br><br>Sum of k = 66 | integer | jee-main-2021-online-26th-august-morning-shift |
1ktg0otbe | maths | quadratic-equation-and-inequalities | nature-of-roots | The set of all values of K > $$-$$1, for which the equation $${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$ has real roots, is : | [{"identifier": "A", "content": "$$\\left( {1,{5 \\over 2}} \\right]$$"}, {"identifier": "B", "content": "[2, 3)"}, {"identifier": "C", "content": "$$\\left[ { - {1 \\over 2},1} \\right)$$"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},{3 \\over 2}} \\right] - \\{ 1\\} $$"}] | ["A"] | null | $${(3{x^2} + 4x + 3)^2} - (k + 1)(3{x^2} + 4x + 3)(3{x^2} + 4x + 2) + k{(3{x^2} + 4x + 2)^2} = 0$$<br><br>Let $$3{x^2} + 4x + 3 = a$$<br><br>and $$3{x^2} + 4x + 2 = b \Rightarrow b = a - 1$$<br><br>Given equation becomes<br><br>$$ \Rightarrow {a^2} - (k + 1)ab + k{b^2} = 0$$<br><br>$$ \Rightarrow a(a - kb) - b(a - kb) = 0$$<br><br>$$ \Rightarrow (a - kb)(a - b) = 0 \Rightarrow a = kb$$ or a = b (reject) $$\because$$ a = kb<br><br>$$ \Rightarrow 3{x^2} + 4x + 3 = k(3{x^2} + 4x + 2)$$<br><br>$$ \Rightarrow 3(k - 1){x^2} + 4(k - 1)x + (2k - 3) = 0$$ for real roots<br><br>D $$\ge$$ 0<br><br>$$ \Rightarrow 16{(k - 1)^2} - 4(3(k - 1))(2k - 3) \ge 0$$<br><br>$$ \Rightarrow 4(k - 1)\{ 4(k - 1) - 3(2k - 3)\} \ge 0$$<br><br>$$ \Rightarrow 4(k - 1)\{ - 2k + 5\} \ge 0$$<br><br>$$ \Rightarrow - 4(k - 1)\{ 2k - 5\} \ge 0$$<br><br>$$ \Rightarrow (k - 1)(2k - 5) \le 0$$<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265611/exam_images/toc4l28ppbnq2m2dc3jk.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266450/exam_images/rhcaoo76lkifjqgvybn7.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265538/exam_images/ms3gmbxipmmuyc7ugdvd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th August Evening Shift Mathematics - Quadratic Equation and Inequalities Question 70 English Explanation"></picture> <br><br>$$\therefore$$ $$k \in \left[ {1,{5 \over 2}} \right]$$<br><br>$$\therefore$$ $$k \ne 1$$<br><br>$$\therefore$$ $$k \in \left( {1,{5 \over 2}} \right]$$ | mcq | jee-main-2021-online-27th-august-evening-shift |
1kto6yf27 | maths | quadratic-equation-and-inequalities | nature-of-roots | The numbers of pairs (a, b) of real numbers, such that whenever $$\alpha$$ is a root of the equation x<sup>2</sup> + ax + b = 0, $$\alpha$$<sup>2</sup> $$-$$ 2 is also a root of this equation, is : | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "8"}] | ["A"] | null | Consider the equation x<sup>2</sup> + ax + b = 0<br><br>If has two roots (not necessarily real $$\alpha$$ & $$\beta$$)<br><br>Either $$\alpha$$ = $$\beta$$ or $$\alpha$$ $$\ne$$ $$\beta$$<br><br>Case (1) If $$\alpha$$ = $$\beta$$, then it is repeated root. Given that $$\alpha$$<sup>2</sup> $$-$$ 2 is also a root<br><br>So, $$\alpha$$ = $$\alpha$$<sup>2</sup> $$-$$ 2 $$\Rightarrow$$ ($$\alpha$$ + 1)($$\alpha$$ $$-$$ 2) = 0<br><br>$$\Rightarrow$$ $$\alpha$$ = $$-$$1 or $$\alpha$$ = 2<br><br>When $$\alpha$$ = $$-$$1 then (a, b) = (2, 1)<br><br>$$\alpha$$ = 2 then (a, b) = ($$-$$4, 4)<br><br>Case (2) If $$\alpha$$ $$\ne$$ $$\beta$$<br><br>Then<br><br>(I) $$\alpha$$ = $$\alpha$$<sup>2</sup> $$-$$ 2 and $$\beta$$ = $$\beta$$<sup>2</sup> $$-$$ 2<br><br>Hence, (a, b) = ($$-$$($$\alpha$$ + $$\beta$$), $$\alpha$$$$\beta$$)<br><br>($$-$$1, $$-$$2)<br><br>(II) $$\alpha$$ = $$\beta$$<sup>2</sup> $$-$$ 2 and $$\beta$$ = $$\alpha$$<sup>2</sup> $$-$$ 2<br><br>Then $$\alpha$$ $$-$$ $$\beta$$ = $$\beta$$<sup>2</sup> $$-$$ $$\alpha$$<sup>2</sup> = ($$\beta$$ $$-$$ $$\alpha$$) ($$\beta$$ + $$\alpha$$)<br><br>Since $$\alpha$$ $$\ne$$ $$\beta$$ we get $$\alpha$$ + $$\beta$$ = $$\beta$$<sup>2</sup> + $$\alpha$$<sup>2</sup> $$-$$ 4<br><br>$$\alpha$$ + $$\beta$$ = ($$\alpha$$ + $$\beta$$)<sup>2</sup> $$-$$ 2$$\alpha$$$$\beta$$ $$-$$ 4<br><br>Thus $$-$$1 = 1 $$-$$2 $$\alpha$$$$\beta$$ $$-$$ 4 which implies<br><br>$$\alpha$$$$\beta$$ = $$-$$1 Therefore (a, b) = ($$-$$($$\alpha$$ + $$\beta$$), $$\alpha$$$$\beta$$)<br><br>= (1, $$-$$1)<br><br>(III) $$\alpha$$ = $$\alpha$$<sup>2</sup> $$-$$ 2 = $$\beta$$<sup>2</sup> $$-$$ 2 and $$\alpha$$ $$\ne$$ $$\beta$$<br><br>$$\Rightarrow$$ $$\alpha$$ = $$-$$ $$\beta$$<br><br>Thus $$\alpha$$ = 2, $$\beta$$ = $$-$$2<br><br>$$\alpha$$ = $$-$$1, $$\beta$$ = 1<br><br>Therefore (a, b) = (0, $$-$$4) & (0, 1)<br><br>(IV) $$\beta$$ = $$\alpha$$<sup>2</sup> $$-$$ 2 = $$\beta$$<sup>2</sup> $$-$$ 2 and $$\alpha$$ $$\ne$$ $$\beta$$ is same as (III) Therefore we get 6 pairs of (a, b)<br><br>Which are (2, 1), ($$-$$4, 4), ($$-$$1, $$-$$2), (1, $$-$$1), (0, $$-$$4)<br><br>Option (a) | mcq | jee-main-2021-online-1st-september-evening-shift |
1l567jtfa | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of real solutions of the equation $${e^{4x}} + 4{e^{3x}} - 58{e^{2x}} + 4{e^x} + 1 = 0$$ is ___________.</p> | [] | null | 2 | <p>Dividing by e<sup>2x</sup></p>
<p>$${e^{2x}} + 4{e^x} - 58 + 4{e^{ - x}} + {e^{ - 2x}} = 0$$</p>
<p>$$ \Rightarrow {({e^x} + {e^{ - x}})^2} + 4({e^x} + {e^{ - x}}) - 60 = 0$$</p>
<p>Let $${e^x} + {e^{ - x}} = t \in [2,\infty )$$</p>
<p>$$ \Rightarrow {t^2} + 4t - 60 = 0$$</p>
<p>$$ \Rightarrow t = 6$$ is only possible solution</p>
<p>$${e^x} + {e^{ - x}} = 6 \Rightarrow {e^{2x}} - 6{e^x} + 1 = 0$$</p>
<p>Let $${e^x} = p$$,</p>
<p>$${p^2} - 6p + 1 = 0$$</p>
<p>$$ \Rightarrow p = {{3 + \sqrt 5 } \over 2}$$ or $${{3 - \sqrt 5 } \over 2}$$</p>
<p>So $$x = \ln \left( {{{3 + \sqrt 5 } \over 2}} \right)$$ or $$\ln \left( {{{3 - \sqrt 5 } \over 2}} \right)$$</p> | integer | jee-main-2022-online-28th-june-morning-shift |
1l57o4z2c | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of distinct real roots of x<sup>4</sup> $$-$$ 4x + 1 = 0 is :</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}] | ["B"] | null | <p>$$f(x) = {x^4} - 4x + 1 = 0$$</p>
<p>$$f'(x) = 4{x^3} - 4$$</p>
<p>$$ = 4(x - 1)({x^2} + 1 + x)$$</p>
<p> <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5q9sp3c/20ce78f3-fabd-4839-b609-84bc56bb070f/63f06880-0655-11ed-903e-c9687588b3f3/file-1l5q9sp3d.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5q9sp3c/20ce78f3-fabd-4839-b609-84bc56bb070f/63f06880-0655-11ed-903e-c9687588b3f3/file-1l5q9sp3d.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 27th June Morning Shift Mathematics - Quadratic Equation and Inequalities Question 60 English Explanation"> </p>
<p>$$\Rightarrow$$ Two solution</p> | mcq | jee-main-2022-online-27th-june-morning-shift |
1l58gxgm4 | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>Let p and q be two real numbers such that p + q = 3 and p<sup>4</sup> + q<sup>4</sup> = 369. Then $${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}}$$ is equal to _________.</p> | [] | null | 4 | <p>$$\because$$ $$p + q = 3$$ ...... (i)</p>
<p>and $${p^4} + {q^4} = 369$$ ...... (ii)</p>
<p>$${\{ {(p + q)^2} - 2pq\} ^2} - 2{p^2}{q^2} = 369$$</p>
<p>or $${(9 - 2pq)^2} - 2{(pq)^2} = 369$$</p>
<p>or $${(pq)^2} - 18pq - 144 = 0$$</p>
<p>$$\therefore$$ $$pq = - 6$$ or 24</p>
<p>But $$pq = 24$$ is not possible</p>
<p>$$\therefore$$ $$pq = - 6$$</p>
<p>Hence, $${\left( {{1 \over p} + {1 \over q}} \right)^{ - 2}} = {\left( {{{pq} \over {p + q}}} \right)^2} = {( - 2)^2} = 4$$</p> | integer | jee-main-2022-online-26th-june-evening-shift |
1l5b7s8ay | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The sum of all the real roots of the equation <br/><br/>$$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$ is</p> | [{"identifier": "A", "content": "$${\\log _e}3$$"}, {"identifier": "B", "content": "$$ - {\\log _e}3$$"}, {"identifier": "C", "content": "$${\\log _e}6$$"}, {"identifier": "D", "content": "$$ - {\\log _e}6$$"}] | ["B"] | null | <p>$$({e^{2x}} - 4)(6{e^{2x}} - 5{e^x} + 1) = 0$$</p>
<p>Let $${e^x} = t$$</p>
<p>$$\therefore$$ $$({t^2} - 4)(6{t^2} - 5t + 1) = 0$$</p>
<p>$$ \Rightarrow ({t^2} - 4)(2t - 1)(3t - 1) = 0$$</p>
<p>$$\therefore$$ t = 2, $$-$$2, $${1 \over 2}$$, $${1 \over 3}$$</p>
<p>$$\therefore$$ $${e^x} = 2 \Rightarrow x = \ln 2$$</p>
<p>$${e^x} = - 2$$ (not possible)</p>
<p>$${e^x} = {1 \over 2} \Rightarrow x = - \ln 2$$</p>
<p>$${e^x} = {1 \over 3} \Rightarrow x = - \ln 3$$</p>
<p>$$\therefore$$ Sum of all real roots</p>
<p>$$ = \ln 2 - \ln 2 - \ln 3$$</p>
<p>$$ = - \ln 3$$</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5b8b8oc | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of distinct real roots of the equation <br/><br/>x<sup>7</sup> $$-$$ 7x $$-$$ 2 = 0 is</p> | [{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "3"}] | ["D"] | null | <p>Given equation $${x^7} - 7x - 2 = 0$$</p>
<p>Let $$f(x) = {x^7} - 7x - 2$$</p>
<p>$$f'(x) = 7{x^6} - 7 = 7({x^6} - 1)$$</p>
<p>and $$f'(x) = 0 \Rightarrow x = \, + \,1$$</p>
<p>and $$f( - 1) = - 1 + 7 - 2 = 5 > 0$$</p>
<p>$$f(1) = 1 - 7 - 2 = - 8 < 0$$</p>
<p>So, roughly sketch of f(x) will be</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5v5wwhx/6e6ad712-3c5b-4e48-ba35-59fbeae28bd0/1fa05b50-0906-11ed-a790-b11fa70c8a36/file-1l5v5wwhy.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5v5wwhx/6e6ad712-3c5b-4e48-ba35-59fbeae28bd0/1fa05b50-0906-11ed-a790-b11fa70c8a36/file-1l5v5wwhy.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 24th June Evening Shift Mathematics - Quadratic Equation and Inequalities Question 55 English Explanation"></p>
<p>So, number of real roots of f(x) = 0 and 3</p> | mcq | jee-main-2022-online-24th-june-evening-shift |
1l5vz59za | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>Let S be the set of all integral values of $$\alpha$$ for which the sum of squares of two real roots of the quadratic equation $$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$ is minimum. Then S :</p> | [{"identifier": "A", "content": "is an empty set"}, {"identifier": "B", "content": "is a singleton"}, {"identifier": "C", "content": "contains exactly two elements"}, {"identifier": "D", "content": "contains more than two elements"}] | ["A"] | null | <p>Given quadratic equation,</p>
<p>$$3{x^2} + (\alpha - 6)x + (\alpha + 3) = 0$$</p>
<p>Let, a and b are the roots of the equation,</p>
<p>$$\therefore$$ $$a + b = - {{\alpha - 6} \over 3}$$</p>
<p>and $$ab = {{\alpha + 3} \over 3}$$</p>
<p>For real roots,</p>
<p>$$D \ge 0$$</p>
<p>$$ \Rightarrow {(\alpha - 6)^2} - 4\,.\,3\,.\,(\alpha + 9) \ge 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} - 12\alpha + 36 - 12\alpha - 36 \ge 0$$</p>
<p>$$ \Rightarrow {\alpha ^2} - 24\alpha \ge 0$$</p>
<p>$$ \Rightarrow \alpha (\alpha - 24) \ge 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l5ykeybm/9c9e29ec-41a0-441a-a6fa-621f2f2c715d/2f5b3b10-0ae5-11ed-a51c-73986e88f75f/file-1l5ykeybn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l5ykeybm/9c9e29ec-41a0-441a-a6fa-621f2f2c715d/2f5b3b10-0ae5-11ed-a51c-73986e88f75f/file-1l5ykeybn.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 30th June Morning Shift Mathematics - Quadratic Equation and Inequalities Question 51 English Explanation"></p>
<p>$$\therefore$$ $$\alpha > 24$$ or $$\alpha < 0$$</p>
<p>$$\therefore$$ Real roots of the equation possible for $$\alpha > 24$$ or $$\alpha < 0$$.</p>
<p>Now, sum of square of roots</p>
<p>$$ = {a^2} + {b^2}$$</p>
<p>$$ = {(a + b)^2} - 2ab$$</p>
<p>$$ = {{{{(\alpha - 6)}^2}} \over 9} - 2\,.\,{{(\alpha + 3)} \over 3}$$</p>
<p>$$ = {{{\alpha ^2} - 12\alpha + 36 - 6\alpha - 18} \over 9}$$</p>
<p>$$ = {{{\alpha ^2} - 18\alpha + 18} \over 9} = f(x)$$</p>
<p>$$\therefore$$ Sum of square of roots are minimum when $${a^2} + {b^2} = $$ minimum.</p>
<p>$$\therefore$$ $$f{(\alpha )_{\min }} = {{{\alpha ^2} - 18\alpha + 18} \over 9}$$</p>
<p>Value of quadratic equation $${\alpha ^2} - 18\alpha + 18$$ is minimum at $$\alpha = - {b \over {2a}} = - {{( - 18)} \over {2\,.\,1}} = 9$$</p>
<p>But for real roots $$\alpha$$ should be less than 0 or greater than 24.</p>
<p>So, there is no value of $$\alpha$$ in the range $$\alpha > 24 \cup \alpha < 0$$ where sum of squares of two real roots is minimum.</p>
<p>$$\therefore$$ S is an empty set.</p> | mcq | jee-main-2022-online-30th-june-morning-shift |
1l6gje0t4 | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of distinct real roots of the equation $$x^{5}\left(x^{3}-x^{2}-x+1\right)+x\left(3 x^{3}-4 x^{2}-2 x+4\right)-1=0$$ is ______________.</p> | [] | null | 3 | <p>$${x^8} - {x^7} - {x^6} + {x^5} + 3{x^4} - 4{x^3} - 2{x^2} + 4x - 1 = 0$$</p>
<p>$$ \Rightarrow {x^7}(x - 1) - {x^5}(x - 1) + 3{x^3}(x - 1) - x({x^2} - 1) + 2x(1 - x) + (x - 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - x(x + 1) - 2x + 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^7} - {x^5} + 3{x^3} - {x^2} - 3x + 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^5}({x^2} - 1) + 3x({x^2} - 1) - 1({x^2} - 1)) = 0$$</p>
<p>$$ \Rightarrow (x - 1)({x^2} - 1)({x^5} + 3x - 1) = 0$$</p>
<p>$$\therefore$$ $$x = \, \pm \,1$$ are roots of above equation and $${x^5} + 3x - 1$$ is a monotonic term hence vanishs at exactly one value of x other than 1 or $$-$$1.</p>
<p>$$\therefore$$ 3 real roots.</p> | integer | jee-main-2022-online-26th-july-morning-shift |
1l6m705hn | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The sum of all real values of $$x$$ for which $$\frac{3 x^{2}-9 x+17}{x^{2}+3 x+10}=\frac{5 x^{2}-7 x+19}{3 x^{2}+5 x+12}$$ is equal to __________.</p> | [] | null | 6 | <p>$${{3{x^2} - 9x + 17} \over {{x^2} + 3x + 10}} = {{5{x^2} - 7x + 19} \over {3{x^2} + 5x + 12}}$$</p>
<p>$$ \Rightarrow {{3{x^2} - 9x + 17} \over {5{x^2} - 7x + 19}} = {{{x^2} + 3x + 10} \over {3{x^2} + 5x + 12}}$$</p>
<p>$${{ - 2{x^2} - 2x - 2} \over {5{x^2} - 7x + 19}} = {{ - 2{x^2} - 2x - 2} \over {3{x^2} + 5x + 12}}$$</p>
<p>Either $${x^2} + x + 1 = 0$$ or No real roots</p>
<p>$$ \Rightarrow 5{x^2} - 7x + 19 = 3{x^2} + 5x + 12$$</p>
<p>$$2{x^2} - 12x + 7 = 0$$</p>
<p>sum of roots = 6</p> | integer | jee-main-2022-online-28th-july-morning-shift |
ldo7nrak | maths | quadratic-equation-and-inequalities | nature-of-roots | The equation $\mathrm{e}^{4 x}+8 \mathrm{e}^{3 x}+13 \mathrm{e}^{2 x}-8 \mathrm{e}^{x}+1=0, x \in \mathbb{R}$ has : | [{"identifier": "A", "content": "two solutions and both are negative"}, {"identifier": "B", "content": "two solutions and only one of them is negative"}, {"identifier": "C", "content": "four solutions two of which are negative"}, {"identifier": "D", "content": "no solution"}] | ["A"] | null | $e^{4 x}+8 e^{3 x}+13 e^{2 x}-8 e^{x}+1=0$
<br/><br/>Let $\mathrm{e}^{\mathrm{x}}=\mathrm{t}$
<br/><br/>Now, $\mathrm{t}^{4}+8 \mathrm{t}^{3}+13 \mathrm{t}^{2}-8 \mathrm{t}+1=0$
<br/><br/>Dividing equation by $\mathrm{t}^{2}$
<br/><br/>$$
\begin{aligned}
& t^{2}+8 t+13-\frac{8}{t}+\frac{1}{t^{2}}=0 \\\\
& t^{2}+\frac{1}{t^{2}}+8\left(t-\frac{1}{t}\right)+13=0 \\\\
& \left(t-\frac{1}{t}\right)^{2}+2+8\left(t-\frac{1}{t}\right)+13=0
\end{aligned}
$$
<br/><br/>Let $\mathrm{t}-\frac{1}{\mathrm{t}}=\mathrm{z}$
<br/><br/>$$
z^{2}+8 z+15=0
$$
<br/><br/>$$
\begin{aligned}
& (z+3)(z+5)=0
\end{aligned}
$$
<br/><br/>$$
z=-3 \text { or } z=-5
$$
<br/><br/>So, $\mathrm{t}-\frac{1}{\mathrm{t}}=-3$ or $\mathrm{t}-\frac{1}{\mathrm{t}}=-5$
<br/><br/>$t^{2}+3 t-1=0$ or $t^{2}+5 t-1=0$
<br/><br/>$\mathrm{t}=\frac{-3 \pm \sqrt{13}}{2}$ or $\mathrm{t}=\frac{-5 \pm \sqrt{29}}{2}$
<br/><br/>as $t=e^{x}$ so $t$ must be positive,
<br/><br/>$$
t=\frac{\sqrt{13}-3}{2} \text { or } \frac{\sqrt{29}-5}{2}
$$
<br/><br/>So, $x=\ln \left(\frac{\sqrt{13}-3}{2}\right)$ or $x=\ln \left(\frac{\sqrt{29}-5}{2}\right)$
<br/><br/>Hence two solutions and both are negative. | mcq | jee-main-2023-online-31st-january-evening-shift |
1ldpre7c6 | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of real roots of the equation $$\sqrt{x^{2}-4 x+3}+\sqrt{x^{2}-9}=\sqrt{4 x^{2}-14 x+6}$$, is :</p> | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}] | ["B"] | null | $\sqrt{(x-1)(x-3)}+\sqrt{(x-3)(x+3)}$
<br/><br/>$=\sqrt{4\left(x-\frac{12}{4}\right)\left(x-\frac{2}{4}\right)}$
<br/><br/>$\Rightarrow \sqrt{\mathrm{x}-3}=0 \Rightarrow \mathrm{x}=3$ which is in domain
<br/><br/>or
<br/><br/>$\sqrt{\mathrm{x}-1}+\sqrt{\mathrm{x}+3}=\sqrt{4 \mathrm{x}-2}$
<br/><br/>$2 \sqrt{(x-1)(x+3)}=2 x-4$
<br/><br/>$x^{2}+2 x-3=x^{2}-4 x+4$
<br/><br/>$6 \mathrm{x}=7$
<br/><br/>$\mathrm{x}=7 / 6$ (rejected) | mcq | jee-main-2023-online-31st-january-morning-shift |
1ldww04sg | maths | quadratic-equation-and-inequalities | nature-of-roots | <p>The number of real solutions of the equation $$3\left( {{x^2} + {1 \over {{x^2}}}} \right) - 2\left( {x + {1 \over x}} \right) + 5 = 0$$, is</p> | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $3\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)+5=0$
<br/><br/>
$3\left[\left(x+\frac{1}{x}\right)^{2}-2\right]-2\left(x+\frac{1}{x}\right)+5=0$
<br/><br/>
Put $x+\frac{1}{x}=t \Rightarrow t \in(-\infty,-2] \cup[2, \infty)$
<br/><br/>
$$
\begin{aligned}
& 3 t^{2}-2 t-1=0 \\\\
& 3 t^{2}-3 t+t-1=0 \\\\
& \Rightarrow 3 t(t-1)+1(t-1)=0 \Rightarrow t=1,=-\frac{1}{3} \\\\
& \Rightarrow \quad t=1,-\frac{1}{3} \\\\
& \because t \in(-\infty,-2] \cup[2, \infty)
\end{aligned}
$$
<br/><br/>
No real value of $t \Rightarrow$ no real value of $x$. | mcq | jee-main-2023-online-24th-january-evening-shift |
1lgyonh22 | maths | quadratic-equation-and-inequalities | nature-of-roots | <p> Let m and $$\mathrm{n}$$ be the numbers of real roots of the quadratic equations $$x^{2}-12 x+[x]+31=0$$ and $$x^{2}-5|x+2|-4=0$$ respectively, where $$[x]$$ denotes the greatest integer $$\leq x$$. Then $$\mathrm{m}^{2}+\mathrm{mn}+\mathrm{n}^{2}$$ is equal to __________.</p> | [] | null | 9 | The givne eqn is : $x^2-12 x+[x]+31=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow\{x\}-x=x^2-12 x+31 \\\\
& \Rightarrow\{x\}=x^2-11 x+31
\end{aligned}
$$
<br/><br/>So, $0 \leq x^2-11 x+31<1$
<br/><br/>$$
\begin{aligned}
& \Rightarrow x^2-11 x+30 \leq 0 \\\\
& \Rightarrow(x-5)(x-6)<0 \\\\
& \Rightarrow x \in(5,6) \\\\
& \therefore[x]=5
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore x^2-12 x+5+31=0 \\\\
& \Rightarrow x^2-12 x+36=0 \\\\
& \Rightarrow(x-6)^2=0 \Rightarrow x=6
\end{aligned}
$$
<br/><br/>Hence, $x \in \phi$
<br/><br/>$$
(\because x \in(5,6))
$$
<br/><br/>$$
\therefore m=0
$$
<br/><br/>Another equation is $x^2-5[x+2]-4=0$
<br/><br/><b>Case I :</b> $x \geq-2$
<br/><br/>$$
x^2-5 x-14=0 \Rightarrow x=7,-2
$$
<br/><br/><b>Case II :</b> $x<-2$
<br/><br/>$$
\begin{aligned}
& x^2+5 x+6=0 \Rightarrow x=-3-2 \\\\
& \therefore x \in\{-3,-2,7\}
\end{aligned}
$$
<br/><br/>$$
\therefore n=3
$$
<br/><br/>Hence, $m^2+m x+n^2=0+0+9=9$ | integer | jee-main-2023-online-8th-april-evening-shift |
p18BDSZMK3wfynvD | maths | quadratic-equation-and-inequalities | range-of-quadratic-expression | If $$x$$ is real, the maximum value of $${{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$ is | [{"identifier": "A", "content": "$${1 \\over 4}$$ "}, {"identifier": "B", "content": "$$41$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$${17 \\over 7}$$ "}] | ["B"] | null | $$y = {{3{x^2} + 9x + 17} \over {3{x^2} + 9x + 7}}$$
<br><br>$$3{x^2}\left( {y - 1} \right) + 9x\left( {y - 1} \right) + 7y - 17 = 0$$
<br><br>$$D \ge 0$$ as $$x$$ is real
<br><br>$$81{\left( {y - 1} \right)^2} - 4 \times 3\left( {y - 1} \right)\left( {7y - 17} \right) \ge 0$$
<br><br>$$ \Rightarrow \left( {y - 1} \right)\left( {y - 41} \right) \le 0$$
<br><br>$$ \Rightarrow 1 \le y \le 41$$
<br><br>$$\therefore$$ Max value of $$y$$ is $$41$$ | mcq | aieee-2006 |
cX7V3ZAuOS6Qua2u | maths | quadratic-equation-and-inequalities | range-of-quadratic-expression | The sum of all real values of $$x$$ satisfying the equation $${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}}\, = 1$$ is : | [{"identifier": "A", "content": "$$6$$"}, {"identifier": "B", "content": "$$5$$ "}, {"identifier": "C", "content": "$$3$$"}, {"identifier": "D", "content": "$$-4$$ "}] | ["C"] | null | Given equation,
$${\left( {{x^2} - 5x + 5} \right)^{{x^2} + 4x - 60}} = 1$$
<br><br><b>Case 1 : </b>When x<sup>2</sup> - 5x + 5 = 1 and x<sup>2</sup> + 4x - 60 is any real no then this equation satisfy.
<br><b>Note :</b> When we put any real number as a power of 1 the value stays always 1 (1<sup> any real no</sup> = 1).
<br> x<sup>2</sup> - 5x + 5 = 1
<br>(x - 1)(x - 4) = 0
<br>$$\therefore$$ x = 1, 4
<br><br><b>Case 2 : </b>When x<sup>2</sup> - 5x + 5 is a real no and x<sup>2</sup> + 4x - 60 = 0 then the given equation satisfy. As we know if power of any real no is zero then it will become 1((any real number)<sup>0</sup> = 1).
<br>For, x<sup>2</sup> + 4x - 60 = 0
<br>(x - 6)(x + 10) = 0
<br>$$\therefore$$ x = 6, -10
<br><br><b>Case 3 :</b> When x<sup>2</sup> - 5x + 5 = -1 and x<sup>2</sup> + 4x - 60 is even this equation satisfy. As we know (-1)<sup>even</sup> = 1.
<br>For, x<sup>2</sup> - 5x + 5 = -1
<br>(x - 2)(x - 3) = 0
<br>$$\therefore$$ x = 2, 3
<br>But x can't be 3 because when x = 3 the value of x<sup>2</sup> + 4x - 60 becomes 3<sup>2</sup> + 4.3 - 60 = - 39 which is an odd number, then (-1)<sup>-39</sup> = -1. So for x = 3 equation does not satisfy.
<p>$$\therefore$$ The sum of all the real values = 1 + 4 + 6 + (-10) + 2 = 3</p> | mcq | jee-main-2016-offline |
o19HmX9Z6V1iojIJW4fqh | maths | quadratic-equation-and-inequalities | range-of-quadratic-expression | The number of integral values of m for which the quadratic expression, (1 + 2m)x<sup>2</sup> – 2(1 + 3m)x + 4(1 + m), x $$ \in $$ R, is always positive, is : | [{"identifier": "A", "content": "7"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "6"}] | ["A"] | null | Expression is always positive it
<br><br>2m + 1 > 0 $$ \Rightarrow $$ m > $$-$$ $${1 \over 2}$$ &
<br><br>D < 0 $$ \Rightarrow $$ m<sup>2</sup> $$-$$ 6m $$-$$ 3 < 0
<br><br> 3 $$-$$ $$\sqrt {12} $$ < m < 3 + $$\sqrt {12} $$ . . . . (iii)
<br><br>$$ \therefore $$ Common interval is
<br><br>3 $$-$$ $$\sqrt {12} $$ < m < 3 + $$\sqrt {12} $$
<br><br>$$ \therefore $$ Intgral value of m {0, 1, 2, 3, 4, 5, 6} | mcq | jee-main-2019-online-12th-january-evening-slot |
jaoe38c1lsd52291 | maths | quadratic-equation-and-inequalities | range-of-quadratic-expression | <p>Let $$a, b, c$$ be the lengths of three sides of a triangle satistying the condition $$\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$$. If the set of all possible values of $$x$$ is the interval $$(\alpha, \beta)$$, then $$12\left(\alpha^2+\beta^2\right)$$ is equal to __________.</p> | [] | null | 36 | <p>$$\left(a^2+b^2\right) x^2-2 b(a+c) x+b^2+c^2=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow a^2 x^2-2 a b x+b^2+b^2 x^2-2 b c x+c^2=0 \\
& \Rightarrow(a x-b)^2+(b x-c)^2=0 \\
& \Rightarrow a x-b=0, \quad b x-c=0 \\
& \Rightarrow a+b>c \quad b+c>a \quad c+a>b
\end{aligned}$$</p>
<p>$$\begin{array}{l|l|l}
a+a x>b x & a x+b x>a & a x^2+a>a x \\
a+a x>a x^2 & a x+a x^2>a & x^2-x+1>0 \\
x^2-x-1<0 & x^2+x-1>0 & \text { always true }
\end{array}$$</p>
<p>$$\begin{aligned}
& \frac{1-\sqrt{5}}{2}< x<\frac{1+\sqrt{5}}{2} \\
& x< \frac{-1-\sqrt{5}}{2}, \text { or } x >\frac{-1+\sqrt{5}}{2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{\sqrt{5}-1}{2}< x<\frac{\sqrt{5}+1}{2} \\
& \Rightarrow \alpha=\frac{\sqrt{5}-1}{2}, \beta=\frac{\sqrt{5}+1}{2} \\
& 12\left(\alpha^2+\beta^2\right)=12\left(\frac{(\sqrt{5}-1)^2+(\sqrt{5}+1)^2}{4}\right)=36
\end{aligned}$$</p> | integer | jee-main-2024-online-31st-january-evening-shift |
3YsoBigo1uuZD4hJ | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Difference between the corresponding roots of $${x^2} + ax + b = 0$$ and $${x^2} + bx + a = 0$$ is same and $$a \ne b,$$ then | [{"identifier": "A", "content": "$$a + b + 4 = 0$$ "}, {"identifier": "B", "content": "$$a + b - 4 = 0$$ "}, {"identifier": "C", "content": "$$a - b - 4 = 0$$ "}, {"identifier": "D", "content": "$$a - b + 4 = 0$$ "}] | ["A"] | null | Let $$\alpha ,\beta $$ and $$\gamma ,\delta $$ be the roots of the equations $${x^2} + ax + b = 0$$
<br><br>and $${x^2} + bx + a = 0$$ respectively.
<br><br>$$\therefore$$ $$\alpha + \beta = - a,\alpha \beta = b$$
<br><br>and $$\gamma + \delta = - b,\gamma \delta = a.$$
<br><br>Given $$\left| {\alpha - \beta } \right| = \left| {\gamma - \delta } \right|$$
<br><br>$$ \Rightarrow {\left( {\alpha - \beta } \right)^2} = {\left( {\gamma - \delta } \right)^2}$$
<br><br>$$ \Rightarrow {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta = {\left( {\gamma + \delta } \right)^2} - 4\gamma \delta $$
<br><br>$$ \Rightarrow {a^2} - 4b = {b^2} - 4a$$
<br><br>$$ \Rightarrow \left( {{a^2} - {b^2}} \right) + 4\left( {a - b} \right) = 0$$
<br><br>$$ \Rightarrow a + b + 4 = 0$$
<br><br>( as $$a \ne b$$ ) | mcq | aieee-2002 |
WCMl8h2haO23S7FT | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$p$$ and $$q$$ are the roots of the equation $${x^2} + px + q = 0,$$ then | [{"identifier": "A", "content": "$$p = 1,\\,\\,q = - 2$$ "}, {"identifier": "B", "content": "$$p = 0,\\,\\,q = 1$$ "}, {"identifier": "C", "content": "$$p = - 2,\\,\\,q = 0$$ "}, {"identifier": "D", "content": "$$p = - 2,\\,\\,q = 1$$ "}] | ["A"] | null | $$p + q = - p$$ and $$pq = q \Rightarrow q\left( {p - 1} \right) = 0$$
<br><br>$$ \Rightarrow q = 0$$ or $$p=1.$$
<br><br>If $$q = 0,$$ then $$p=0.$$ i.e.$$p=q$$
<br><br>$$\therefore$$ $$p=1$$ and $$q=-2.$$ | mcq | aieee-2002 |
6tmOFTAoGjVrDmnk | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha \ne \beta $$ but $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3$$ then the equation having $$\alpha /\beta $$ and $$\beta /\alpha \,\,$$ as its roots is | [{"identifier": "A", "content": "$$3{x^2} - 19x + 3 = 0$$"}, {"identifier": "B", "content": "$$3{x^2} + 19x - 3 = 0$$ "}, {"identifier": "C", "content": "$$3{x^2} - 19x - 3 = 0$$ "}, {"identifier": "D", "content": "$${x^2} - 5x + 3 = 0$$ "}] | ["A"] | null | We have $${\alpha ^2} = 5\alpha - 3$$ and $${\beta ^2} = 5\beta - 3;$$
<br><br>$$ \Rightarrow \alpha \,\,\& \,\,\beta $$ are roots of
<br><br>equation, $${x^2} = 5x - 3$$ or $${x^2} - 5x + 3 = 0$$
<br><br>$$\therefore$$ $$\alpha + \beta = 5$$ and $$\alpha \beta = 3$$
<br><br>Thus, the equation having $${\alpha \over \beta }\,\,\& \,\,{\beta \over \alpha }$$ as its roots is
<br><br>$${x^2} - x\left( {{\alpha \over \beta } + {\beta \over \alpha }} \right) + {{\alpha \beta } \over {\alpha \beta }} = 0$$
<br><br>$$ \Rightarrow {x^2} - x\left( {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right) + 1 = 0$$
<br><br>or $$3{x^2} - 19x + 3 = 0$$ | mcq | aieee-2002 |
nDip6hROBe7bOvoK | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If the sum of the roots of the quadratic equation $$a{x^2} + bx + c = 0$$ is equal to the sum of the squares of their reciprocals, then $${a \over c},\,{b \over a}$$ and $${c \over b}$$ are in | [{"identifier": "A", "content": "Arithmetic - Geometric Progression "}, {"identifier": "B", "content": "Arithmetic Progression "}, {"identifier": "C", "content": "Geometric Progression "}, {"identifier": "D", "content": "Harmonic Progression "}] | ["D"] | null | $$a{x^2} + bx + c = 0,$$ $$\alpha + \beta = {{ - b} \over a},\alpha \beta = {c \over a}$$
<br><br>As for given condition, $$\alpha + \beta = {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}}$$
<br><br>$$\alpha + \beta = {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} - {b \over a}$$
<br><br>$$ = {{{{{b^2}} \over {{a^2}}} - {{2c} \over a}} \over {{{{c^2}} \over {{a^2}}}}}$$
<br><br>On simplification $$2{a^2}c = a{b^2} + b{c^2}$$
<br><br>$$ \Rightarrow {{2a} \over b} = {c \over a} + {b \over c}$$
<br><br>$$ \Rightarrow {c \over a},{a \over b},{b \over c}$$ are in $$A.P.$$
<br><br>$$\therefore$$ $${a \over c},{b \over a},\,\,\& \,\,$$ are in $$H.P.$$ | mcq | aieee-2003 |
QaXT2TPJqyZ9tKIE | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of '$$a$$' for which one root of the quadratic equation
$$$\left( {{a^2} - 5a + 3} \right){x^2} + \left( {3a - 1} \right)x + 2 = 0$$$
<br/>is twice as large as the other is | [{"identifier": "A", "content": "$$ - {1 \\over 3}$$ "}, {"identifier": "B", "content": "$$ {2 \\over 3}$$"}, {"identifier": "C", "content": "$$ - {2 \\over 3}$$"}, {"identifier": "D", "content": "$$ {1 \\over 3}$$ "}] | ["B"] | null | Let the roots of given equation be $$\alpha $$ and $$2$$$$\alpha $$ then
<br><br>$$\alpha + 2\alpha = 3\alpha = {{1 - 3a} \over {{a^2} - 5a + 3}}$$
<br><br>and $$\alpha .2\alpha = 2{\alpha ^2} = {2 \over {{a^2} - 5a + 3}}$$
<br><br>$$ \Rightarrow \alpha = {{1 - 3a} \over {3\left( {{a^2} - 5a + 3} \right)}}$$
<br><br>$$\therefore$$ $$2\left[ {{1 \over 9}{{{{\left( {1 - 3a} \right)}^2}} \over {{{\left( {{a^2} - 5a + 3} \right)}^2}}}} \right]$$
<br><br>$$ = {2 \over {{a^2} - 5a + 3}}$$
<br><br>$${{{{\left( {1 - 3a} \right)}^2}} \over {\left( {{a^2} - 5a + 3} \right)}} = 9$$
<br><br>or $$9{a^2} - 6a + 1$$
<br><br>$$ = 9{a^2} - 45a + 27$$
<br><br>or $$39a = 26$$ or $$a = {2 \over 3}$$ | mcq | aieee-2003 |
yQOuJuXRcD24ppfE | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation | [{"identifier": "A", "content": "$${x^2} - 18x - 16 = 0$$ "}, {"identifier": "B", "content": "$${x^2} - 18x + 16 = 0$$"}, {"identifier": "C", "content": "$${x^2} + 18x - 16 = 0$$"}, {"identifier": "D", "content": "$${x^2} + 18x + 16 = 0$$"}] | ["B"] | null | Let two numbers be a and b then $${{a + b} \over 2} = 9$$
<br><br>and $$\sqrt {ab} = 4$$
<br><br>$$\therefore$$ Equation with roots $$a$$ and $$b$$ is
<br><br>$${x^2} - \left( {a + b} \right)x + ab = 0$$
<br><br>$$ \Rightarrow {x^2} - 18x + 16 = 0$$ | mcq | aieee-2004 |
eS6pxDZE0UywetZu | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\left( {1 - p} \right)$$ is a root of quadratic equation $${x^2} + px + \left( {1 - p} \right) = 0$$ then its root are | [{"identifier": "A", "content": "$$ - 1,2$$ "}, {"identifier": "B", "content": "$$ - 1,1$$"}, {"identifier": "C", "content": "$$ 0,-1$$"}, {"identifier": "D", "content": "$$0,1$$ "}] | ["C"] | null | Let the second root be $$\alpha .$$
<br><br>Then $$\alpha + \left( {1 - p} \right) = - p \Rightarrow \alpha = - 1$$
<br><br>Also $$\alpha .\left( {1 - p} \right) = 1 - p$$
<br><br>$$ \Rightarrow \left( {\alpha - 1} \right)\left( {1 - p} \right) = 0$$
<br><br>$$ \Rightarrow p = 1$$ [as $$\alpha = - 1$$]
<br><br>$$\therefore$$ Roots are $$\alpha = - 1$$ and $$p-1=0$$ | mcq | aieee-2004 |
vg7MAc99xWYqkfya | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | In a triangle $$PQR,\;\;\angle R = {\pi \over 2}.\,\,If\,\,\tan \,\left( {{P \over 2}} \right)$$ and $$ \tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0,\,\,a \ne 0$$ then | [{"identifier": "A", "content": "$$a = b + c$$ "}, {"identifier": "B", "content": "$$c = a + b$$ "}, {"identifier": "C", "content": "$$b = c$$ "}, {"identifier": "D", "content": "$$b = a + c$$ "}] | ["B"] | null | $$\angle $$R = 90<sup>o</sup> $$ \therefore $$ $$\angle $$P + $$\angle $$Q = 90<sup>o</sup>
<br><br>$$ \Rightarrow $$ $${P \over 2} + {Q \over 2} = {{90} \over 2} = 45$$<sup>o</sup>
<br><br>$$\tan \left( {{P \over 2}} \right),\tan \left( {{Q \over 2}} \right)$$ are the roots of $$a{x^2} + bx + c = 0$$
<br><br>$$ \therefore $$ $$\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right) = - {b \over a},\,\,$$
<br><br>and $$\tan \left( {{P \over 2}} \right).\tan \left( {{Q \over 2}} \right) = {c \over a}$$
<br><br>$${{\tan \left( {{P \over 2}} \right) + \tan \left( {{Q \over 2}} \right)} \over {1 - \tan \left( {{P \over 2}} \right)\tan \left( {{Q \over 2}} \right)}}$$
<br><br>$$ = \tan \left( {{P \over 2} + {Q \over 2}} \right) $$= tan 45<sup>o</sup> = 1
<br><br>$$ \Rightarrow {{ - {b \over a}} \over {1 - {c \over a}}} = 1$$
<br><br>$$ \Rightarrow - {b \over a} = {a \over a} - {c \over a}$$
<br><br>$$ \Rightarrow - b = a - c$$ or $$c = a + b.$$ | mcq | aieee-2005 |
j6vQurOyjyj7GX1T | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$a$$ for which the sum of the squares of the roots of the equation $${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
assume the least value is : | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["A"] | null | $${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
<br><br>$$ \Rightarrow \alpha + \beta = a - 2;\,\,\alpha \beta = - \left( {a + 1} \right)$$
<br><br>$${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$
<br><br>$$ = {a^2} - 2a + 6 = {\left( {a - 1} \right)^2} + 5$$
<br><br>For min. value of $${\alpha ^2} + {\beta ^2}$$ where $$\alpha $$ is an integer
<br><br>$$ \Rightarrow \,\,a = 1.$$ | mcq | aieee-2005 |
fSshcquXjrbG2JCO | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$a$$ for which the sum of the squares of the roots of the equation <br/>$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$ assume the least value is | [{"identifier": "A", "content": "$$1$$ "}, {"identifier": "B", "content": "$$0$$ "}, {"identifier": "C", "content": "$$3$$ "}, {"identifier": "D", "content": "$$2$$ "}] | ["A"] | null | Given quadratic equation,
<br><br>$${x^2} - \left( {a - 2} \right)x - a - 1 = 0$$
<br><br>Let $$\alpha $$ and $$\beta $$ are the roots of the equation.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$a - 2$$
<br><br>and $$\alpha $$$$\beta $$ = $$ - a - 1$$
<br><br>Now $${\alpha ^2} + {\beta ^2} = {\left( {\alpha + \beta } \right)^2} - 2\alpha \beta $$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 2} \right)^2} + 2\left( {a + 1} \right)$$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {a^2} - 2a + 6$$
<br><br>$$ \Rightarrow $$ $${\alpha ^2} + {\beta ^2} = {\left( {a - 1} \right)^2} + 5$$
<br><br>$$ \Rightarrow $$ The value of $${\alpha ^2} + {\beta ^2}$$ will be minimum, when $${a - 1}$$ = 0
<br><br>$$ \Rightarrow $$ $${a = 1}$$ | mcq | aieee-2005 |
JPyMtvr5Uy9w4a1n | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If the roots of the quadratic equation $${x^2} + px + q = 0$$ are $$\tan {30^ \circ }$$ and $$\tan {15^ \circ }$$, respectively, then the value of $$2 + q - p$$ is | [{"identifier": "A", "content": "2 "}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "0 "}, {"identifier": "D", "content": "1 "}] | ["B"] | null | $${x^2} + px + q = 0$$
<br><br>Sum of roots $$ = \tan {30^ \circ } + \tan {15^ \circ } = - p$$
<br><br>Products of roots $$ = \tan {30^ \circ }.\tan {15^ \circ } = q$$
<br><br>$$\tan {45^ \circ } = {{\tan {{30}^ \circ } + \tan {{15}^ \circ }} \over {1 - \tan {{30}^ \circ }.\tan {{15}^ \circ }}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {{ - p} \over {1 - q}} = 1$$
<br><br>$$ \Rightarrow - p = 1 - q \Rightarrow q - p = 1$$
<br><br>$$\therefore$$ $$2 + q - p = 3$$ | mcq | aieee-2006 |
TpLwVDBYA4fxcyBw | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If the difference between the roots of the equation $${x^2} + ax + 1 = 0$$ is less than $$\sqrt 5 ,$$ then the set of possible values of $$a$$ is | [{"identifier": "A", "content": "$$\\left( {3,\\infty } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\infty , - 3} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - 3,3} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( { - 3,\\infty } \\right)$$ "}] | ["C"] | null | Let $$\alpha $$ and $$\beta $$ are roots of the equation $${x^2} + ax + 1 = 0$$
<br><br>So, $$\alpha + \beta = - a$$ and $$\alpha \beta = 1$$
<br><br>given $$\left| {\alpha - \beta } \right| < \sqrt 5 $$
<br><br>$$ \Rightarrow \sqrt {{{\left( {\alpha - \beta } \right)}^2} - 4\alpha \beta } < \sqrt 5 $$
<br><br>(as $${\left( {\alpha - \beta } \right)^2} = {\left( {\alpha + \beta } \right)^2} - 4\alpha \beta $$ )
<br><br>$$ \Rightarrow \sqrt {{a^2} - 4} < \sqrt 5 $$
<br><br>$$ \Rightarrow {a^2} - 4 < 5$$
<br><br>$$ \Rightarrow {a^2} - 9 < 0 \Rightarrow {a^2} < 9$$
<br><br>$$ \Rightarrow - 3 < a < 3$$
<br><br>$$ \Rightarrow a \in \left( { - 3,3} \right)$$ | mcq | aieee-2007 |
DOpL5LxqjKLhacNf | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the equation $${x^2} - x + 1 = 0,$$ then $${\alpha ^{2009}} + {\beta ^{2009}} = $$ | [{"identifier": "A", "content": "$$\\, - 1$$ "}, {"identifier": "B", "content": "$$\\, 1$$"}, {"identifier": "C", "content": "$$\\, 2$$ "}, {"identifier": "D", "content": "$$\\, - 2$$ "}] | ["B"] | null | $${x^2} - x + 1 = 0$$
<br><br>$$ \Rightarrow x = {{1 \pm \sqrt {1 - 4} } \over 2}$$
<br><br>$$x = {{1 \pm \sqrt 3 i} \over 2}$$
<br><br>$$\alpha = {1 \over 2} + i{{\sqrt 3 } \over 2} = - {\omega ^2}$$
<br><br>$$\beta = {1 \over 2} - {{i\sqrt 3 } \over 2} = - \omega $$
<br><br>$${\alpha ^{2009}} + {\beta ^{2009}} = {\left( { - {\omega ^2}} \right)^{2009}} + {\left( { - \omega } \right)^{2009}}$$
<br><br>$$ = - {\omega ^2} - \omega = 1$$ | mcq | aieee-2010 |
GyCCDYFP4RhqNHz9 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of equation $$p{x^2} + qx + r = 0,$$ $$p \ne 0.$$ If $$p,\,q,\,r$$ in A.P. and $${1 \over \alpha } + {1 \over \beta } = 4,$$ then the value of $$\left| {\alpha - \beta } \right|$$ is : | [{"identifier": "A", "content": "$${{\\sqrt {34} } \\over 9}$$ "}, {"identifier": "B", "content": "$${{2\\sqrt 13 } \\over 9}$$ "}, {"identifier": "C", "content": "$${{\\sqrt {61} } \\over 9}$$"}, {"identifier": "D", "content": "$${{2\\sqrt 17 } \\over 9}$$"}] | ["B"] | null | Let $$p,q,r$$ are in $$AP$$
<br><br>$$ \Rightarrow 2q = p + r\,\,\,\,\,\,\,\,\,\,\,\,...\left( i \right)$$
<br><br>Given $${1 \over \alpha } + {1 \over \beta } = 4 \Rightarrow {{\alpha + \beta } \over {\alpha \beta }} = 4$$
<br><br>We have $$\alpha + \beta = - q/p$$ and $$\alpha \beta = {r \over p}$$
<br><br>$$ \Rightarrow {{ - {q \over p}} \over {{r \over p}}} = 4 \Rightarrow q = - 4r\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
<br><br>From $$(i),$$ we have
<br><br>$$2\left( { - 4r} \right) = p + r \Rightarrow p = - 9r$$
<br><br>$$q = - 4r$$
<br><br>Now $$\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } $$
<br><br>$$ = \sqrt {{{\left( {{{ - q} \over p}} \right)}^2} - {{4r} \over p}} $$
<br><br>$$ = {{\sqrt {{q^2} - 4pr} } \over {\left| p \right|}}$$
<br><br>$$ = {{\sqrt {16{r^2} + 36{r^2}} } \over {\left| { - 9r} \right|}}$$
<br><br>$$ = {{2\sqrt {13} } \over 9}$$ | mcq | jee-main-2014-offline |
RL48R3LIbXTYoq2f | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of equation $${x^2} - 6x - 2 = 0$$. If $${a_n} = {\alpha ^n} - {\beta ^n},$$ for $$n \ge 1,$$ then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is equal to : | [{"identifier": "A", "content": "$$3$$"}, {"identifier": "B", "content": "$$ - 3$$ "}, {"identifier": "C", "content": "$$6$$ "}, {"identifier": "D", "content": "$$ - 6$$ "}] | ["A"] | null | Given equation, x<sup>2</sup> - 6x - 2 = 0
<br><br> Roots are $$\alpha $$ and $$\beta $$.
<br><br>So, $$\alpha + \beta = 6$$ and $$\alpha \beta = - 2$$
<br><br>In the question given, $${a_n} = {\alpha ^n} - {\beta ^n}$$
<br><br>$$\therefore$$ $${a_8} = {\alpha ^8} - {\beta ^8}$$
<br><br>and $${a_9} = {\alpha ^9} - {\beta ^9}$$
<br><br>and $${a_{{10}}} = {\alpha ^{{10}}} - {\beta ^{10}}$$
<br><br>Now, the given equation
<br><br>$${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$
<br><br>= $${{{\alpha ^{10}} - {\beta ^{10}} - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$
<br><br>=$${{{\alpha ^{10}} - {\beta ^{10}} + \alpha \beta \left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$ (as $$\alpha \beta = - 2$$)
<br><br>=$${{{\alpha ^{10}} - {\beta ^{10}} + {\alpha ^9}\beta - \alpha {\beta ^9}} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$
<br><br>= $${{{\alpha ^9}\left( {\alpha + \beta } \right) - {\beta ^9}\left( {\alpha + \beta } \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$
<br><br>= $${{\left( {\alpha + \beta } \right)\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$
<br><br>= $${{\left( {\alpha + \beta } \right)} \over 2}$$
<br><br>= $${6 \over 2}$$ (as $${ {\alpha + \beta } }$$ = 6)
<br><br>= 3 | mcq | jee-main-2015-offline |
0unmlbvFhbWLTnKlsl7ae | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If x is a solution of the equation, $$\sqrt {2x + 1} $$ $$ - \sqrt {2x - 1} = 1,$$ $$\,\,\left( {x \ge {1 \over 2}} \right),$$ then $$\sqrt {4{x^2} - 1} $$ is equal to : | [{"identifier": "A", "content": "$${3 \\over 4}$$"}, {"identifier": "B", "content": "$${1 \\over 2}$$"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "$$2\\sqrt 2 $$"}] | ["A"] | null | Given,
<br><br>$$\sqrt {2x + 1} - \sqrt {2x - 1} = 1$$
<br><br>$$ \Rightarrow $$ $$\sqrt {2x + 1} = 1 + \sqrt {2x - 1} $$
<br><br>Squaring both sides, we get
<br><br>2x + 1 $$=$$ 1 + 2x $$-$$ 1 + 2$$\sqrt {2x - 1} $$
<br><br>$$ \Rightarrow $$ 1 $$=$$ 2$$\sqrt {2x - 1} $$
<br><br>$$ \Rightarrow $$ 1 $$=$$ 4(2x $$-$$ 1)
<br><br>$$ \Rightarrow $$ 8x $$-$$ 4 $$=$$ 1
<br><br>$$ \Rightarrow $$ x $$=$$ $${5 \over 8}$$
<br><br>So, $$\sqrt {4{x^2} - 1} $$
<br><br>$$ = \sqrt {4\left( {{{25} \over {64}}} \right) - 1} $$
<br><br>$$ = \sqrt {{{36} \over {64}}} $$
<br><br>$$ = {6 \over 8}$$
<br><br>$$ = {3 \over 4}$$ | mcq | jee-main-2016-online-10th-april-morning-slot |
7IrvwEVArDgbLPCv | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If for a positive integer n, the quadratic equation
<br/><br/>$$x\left( {x + 1} \right) + \left( {x + 1} \right)\left( {x + 2} \right)$$$$ + .... + \left( {x + \overline {n - 1} } \right)\left( {x + n} \right)$$$$ = 10n$$
<br/><br/>has two consecutive integral solutions, then n is equal to : | [{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "11"}, {"identifier": "D", "content": "12"}] | ["C"] | null | $$\sum\limits_{r = 1}^n {\left( {x + r - 1} \right)\left( {x + r} \right)} = 10n$$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + xr + \left( {r - 1} \right)x + {r^2} - r} \right)} = 10n$$
<br><br>$$ \Rightarrow $$ $$\sum\limits_{r = 1}^n {\left( {{x^2} + \left( {2r - 1} \right)x + r\left( {r - 1} \right)} \right)} = 10n$$
<br><br>$$ \Rightarrow $$ $$n{x^2} + \left\{ {1 + 3 + 5 + .... + \left( {2n - 1} \right)} \right\}x$$
<br><br>$$ + \left\{ {1.2 + 2.3 + ... + \left( {n - 1} \right)n} \right\}$$ = 10n
<br><br>$$ \Rightarrow $$ $$n{x^2} + {n^2}x + {{n\left( {{n^2} - 1} \right)} \over 3} = 10n$$
<br><br>$$ \Rightarrow $$ $${x^2} + nx + {{\left( {{n^2} - 31} \right)} \over 3} = 0$$
<br><br>Let $$\alpha $$ and $$\alpha $$ + 1 be its two solutions
<br><br>$$ \therefore $$ $$\alpha $$ + ($$\alpha $$ + 1) = -n
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $${{ - n - 1} \over 2}$$ ....(1)
<br><br>Also $$\alpha $$($$\alpha $$ + 1) = $${{\left( {{n^2} - 31} \right)} \over 3}$$ ......(2)
<br><br>Putting value of (1) in (2), we get
<br><br>$$ - \left( {{{n + 1} \over 2}} \right)\left( {{{1 - n} \over 2}} \right) = {{\left( {{n^2} - 31} \right)} \over 3}$$
<br><br>$$ \Rightarrow $$ n<sup>2</sup> = 121
<br><br>$$ \Rightarrow $$ n = 11 | mcq | jee-main-2017-offline |
0pxENP4lfSOmWKuYrR5Mm | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let p(x) be a quadratic polynomial such that p(0)=1. If p(x) leaves remainder 4 when divided by x$$-$$ 1 and it leaves remainder 6 when divided by x + 1; then : | [{"identifier": "A", "content": "p(2) = 11"}, {"identifier": "B", "content": "p(2) = 19"}, {"identifier": "C", "content": "p($$-$$ 2) = 19"}, {"identifier": "D", "content": "p($$-$$ 2) = 11"}] | ["C"] | null | Let, P(x) = ax<sup>2</sup> + bx + c
<br><br>As, P(0) = 1,
<br><br>$$\therefore\,\,\,$$ a(0)<sup>2</sup> + b(0) + c = 1
<br><br>$$ \Rightarrow $$$$\,\,\,$$ c = 1
<br><br>$$\therefore\,\,\,$$ P(x) = ax<sup>2</sup> + bx + 1
<br><br>If P(x) is divided by x $$-$$ 1, remainder = 4
<br><br>$$ \Rightarrow $$$$\,\,\,$$ P$$\left( 1 \right) = 4$$
<br><br>$$\therefore\,\,\,$$ a + b + 1 = 4 . . . . . (1)
<br><br>If P(x) is divided by x + 1, remainder = 6
<br><br>$$ \Rightarrow $$$$\,\,\,$$ P($$-$$ 1) = 6
<br><br>$$\therefore\,\,\,$$ a $$-$$ b + 1 = 6 . . . .(2)
<br><br>By solving (1) and (2) we get,
<br><br>a = 4, and b = $$-$$1
<br><br>$$\therefore\,\,\,$$ P(x) = 4x<sup>2</sup> $$-$$ x + 1
<br><br>P(2) = 4(2)<sup>2</sup> $$-$$ 2 + 1 = 15
<br><br>P($$-$$ 2) = 4 ($$-$$2)<sup>2</sup> $$-$$ ($$-$$ 2) + 1 = 19 | mcq | jee-main-2017-online-8th-april-morning-slot |
XAqhFZbkmjsfKVvXYMvuA | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If tanA and tanB are the roots of the quadratic equation, 3x<sup>2</sup> $$-$$ 10x $$-$$ 25 = 0, then the value of 3 sin<sup>2</sup>(A + B) $$-$$ 10 sin(A + B).cos(A + B) $$-$$ 25 cos<sup>2</sup>(A + B) is : | [{"identifier": "A", "content": "$$-$$ 10"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "$$-$$ 25"}, {"identifier": "D", "content": "25"}] | ["C"] | null | As tan A and tan B are the roots of 3x<sup>2</sup> $$-$$ 10x $$-$$ 25 = 0,
<br><br>So, tan(A + B) = $${{\tan A + \tan B} \over {1 - \tan A\tan B}}$$
<br><br>= $${{{{10} \over 3}} \over {1 + {{25} \over 3}}}$$ = $${{10/3} \over {28/3}}$$ = $${5 \over {14}}$$
<br><br>Now, cos<sup>2</sup> (A + B) = $$-$$ 1 + 2 cos<sup>2</sup> (A + B)
<br><br>= $${{1 - {{\tan }^2}(A + B)} \over {1 + {{\tan }^2}(A + B)}}\,$$ $$ \Rightarrow $$ cos<sup>2</sup>(A + B) = $${{196} \over {221}}$$
<br><br>$$\therefore\,\,\,$$ 3sin<sup>2</sup>(A + B) $$-$$ 10sin(A + B)cos(A + B) $$-$$ 25 cos<sup>2</sup>(A + B)
<br><br>= cos<sup>2</sup>(A + B) [ 3tan<sup>2</sup>(A + B) $$-$$ 10tan(A + B) $$-$$ 25]
<br><br>= $${{75 - 700 - 4900} \over {196}} \times {{196} \over {221}}$$
<br><br>= $$-$$ $${{5525} \over {196}}$$ $$ \times $$ $${{196} \over {221}}$$ = $$-$$ 25 | mcq | jee-main-2018-online-15th-april-morning-slot |
DCEEzNPGePYIxhKKzY7yv | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If an angle A of a $$\Delta $$ABC satiesfies 5 cosA + 3 = 0, then the roots of the quadratic equation, 9x<sup>2</sup> + 27x + 20 = 0 are : | [{"identifier": "A", "content": "secA, cotA"}, {"identifier": "B", "content": "sinA, secA"}, {"identifier": "C", "content": "secA, tanA"}, {"identifier": "D", "content": "tanA, cosA"}] | ["C"] | null | Here, 9x<sup>2</sup> + 27x + 20 = 0
<br><br>$$\therefore\,\,\,$$ x = $${{ - b \pm \sqrt {{b^2} - 4ac} } \over {2a}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $${{ - 27 \pm \sqrt {{{27}^2} - 4 \times 9 \times 20} } \over {2 \times 9}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $$-$$ $${4 \over 3}$$, $$-$$ $${5 \over 3}$$
<br><br>Given, cosA = $$-$$ $${3 \over 5}$$
<br><br>$$\therefore\,\,\,$$ sec A = $${1 \over {\cos A}}$$ = $$-$$ $${5 \over 3}$$
<br><br>Here, A is an obtuse angle.
<br><br>$$\therefore\,\,\,$$ tan A = $$-$$ $$\sqrt {{{\sec }^2}A - 1} = - {4 \over 3}.$$
<br><br>Hence, roots of the equation are sec A and tan A. | mcq | jee-main-2018-online-16th-april-morning-slot |
Vi5B6NQIKaeP2PUAHr7rQ | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let p, q and r be real numbers (p $$ \ne $$ q, r $$ \ne $$ 0), such that the roots of the equation $${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$ are equal in magnitude but opposite in sign, then the sum of squares of these roots is equal to : | [{"identifier": "A", "content": "$${{{p^2} + {q^2}} \\over 2}$$"}, {"identifier": "B", "content": "p<sup>2</sup> + q<sup>2</sup> "}, {"identifier": "C", "content": "2(p<sup>2</sup> + q<sup>2</sup>)"}, {"identifier": "D", "content": "p<sup>2</sup> + q<sup>2</sup> + r<sup>2</sup>"}] | ["B"] | null | Given,
<br><br>$${1 \over {x + p}} + {1 \over {x + q}} = {1 \over r}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $${{x + p + x + q} \over {\left( {x + p} \right)\left( {x + q} \right)}} = {1 \over r}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ (2x + p + q) r = x<sup>2</sup> + px + qx + pq
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x<sup>2</sup> + (p + q $$-$$ 2r) x + pq $$-$$ pr $$-$$ qr = 0
<br><br>Let $$\alpha $$ and $$\beta $$ are the roots,
<br><br>$$\therefore\,\,\,$$ $$\alpha $$ + $$\beta $$ = $$-$$ (p + q $$-$$ 2r)
<br><br>and $$\alpha $$ $$\beta $$ = pq $$-$$ pr $$-$$ qr
<br><br>Given that, $$\alpha $$ = $$-$$ $$\beta $$ $$ \Rightarrow $$ $$\alpha $$
+ $$\beta $$ = 0
<br><br>$$\therefore\,\,\,$$ $$-$$ (p + q $$-$$ 2r) = 0
<br><br>Now, $$\alpha $$<sup>2</sup> + $$\beta $$<sup>2</sup>
<br><br>= ($$\alpha $$ + $$\beta $$)<sup>2</sup> $$-$$ 2$$\alpha $$ $$\beta $$
<br><br>= ($$-$$ (p + q $$-$$ 2r))<sup>2</sup> $$-$$ 2 (pq $$-$$ pr $$-$$ qr)
<br><br>= p<sup>2</sup> +q<sup>2</sup> + 4r<sup>2</sup> + 2pq $$-$$ 4pr $$-$$ 4qr $$-$$ 2pq + 2pr + 2qr
<br><br>= p<sup>2</sup> + q<sup>2</sup> + 4r<sup>2</sup> $$-$$ 2pr $$-$$ 2qr
<br><br>= p<sup>2</sup> + q<sup>2</sup> $$-$$ 2r (p + q $$-$$ 2r)
<br><br>= p<sup>2</sup> + q<sup>2</sup> $$-$$ 2r (0)
<br><br>= p<sup>2</sup> + q<sup>2</sup> | mcq | jee-main-2018-online-16th-april-morning-slot |
PVuqvBx83lZ1nES9eOeNX | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If f(x) is a quadratic expression such that f (1) + f (2) = 0, and $$-$$ 1 is a root of f (x) = 0, then the other root of f(x) = 0 is : | [{"identifier": "A", "content": "$$-$$ $${5 \\over 8}$$"}, {"identifier": "B", "content": "$$-$$ $${8 \\over 5}$$"}, {"identifier": "C", "content": "$${5 \\over 8}$$"}, {"identifier": "D", "content": "$${8 \\over 5}$$"}] | ["D"] | null | Let $$\alpha $$ and $$\beta $$ = - 1 are the roots of the polynomial, then we get
<br><br>f(x) = x<sup>2</sup> + (1 - $$\alpha $$)x - $$\alpha $$
<br><br>$$ \therefore $$ f(1) = 2 - 2$$\alpha $$
<br><br>and f(2) = 6 - 3$$\alpha $$
<br><br>Also given,
<br><br> f (1) + f (2) = 0
<br><br>$$ \therefore $$ 2 - 2$$\alpha $$ + 6 - 3$$\alpha $$ = 0
<br><br>$$ \Rightarrow $$ $$\alpha $$ = $${8 \over 5}$$ | mcq | jee-main-2018-online-15th-april-evening-slot |
nmugPDXgqKsFPiI0tLcu4 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\lambda $$ $$ \in $$ <b>R</b> is such that the sum of the cubes of the roots of the equation,
<br/>x<sup>2</sup> + (2 $$-$$ $$\lambda $$) x + (10 $$-$$ $$\lambda $$) = 0 is minimum, then the magnitude of the difference of the roots of this equation is : | [{"identifier": "A", "content": "$$4\\sqrt 2 $$"}, {"identifier": "B", "content": "$$2\\sqrt 5 $$"}, {"identifier": "C", "content": "$$2\\sqrt 7 $$"}, {"identifier": "D", "content": "20"}] | ["B"] | null | Let $$\alpha $$, $$\beta $$ are the roots of the equation,
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 2 and $$\alpha $$$$\beta $$ = 10 $$-$$ $$\lambda $$
<br><br>$${\alpha ^3} + {\beta ^3}$$ = ($$\alpha $$ + $$\beta $$)<sup>3</sup> $$-$$ 3$$\alpha $$$$\beta $$ ($$\alpha $$ + $$\beta $$)
<br><br>= ($$\lambda $$ $$-$$ 2)<sup>3</sup> $$-$$ 3(10 $$-$$ $$\lambda $$)($$\lambda $$ $$-$$ 2)
<br><br>= $$\lambda ^3$$ $$-$$ 3$$\lambda ^2$$ $$-$$ 24$$\lambda $$ + 52
<br><br>Let $$f(\lambda $$) = $$\lambda ^3$$ $$-$$ 3$$\lambda ^2$$ $$-$$ 24$$\lambda $$ + 52
<br><br>$$ \therefore $$ $${{df(\lambda )} \over {d\lambda }}$$ = 3$$\lambda ^2$$ $$-$$ 6$$\lambda $$ $$-$$ 24
<br><br>$$ \therefore $$ at maximum of minimum $${{df(\lambda )} \over {d\lambda }}$$ = 0
<br><br>$$ \therefore $$ $$\lambda ^2$$ $$-$$ 2$$\lambda $$ $$-$$ 8 = 0
<br><br>$$ \Rightarrow $$ ($$\lambda $$ + 2) ($$\lambda $$ $$-$$ 4) = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $$-$$2, 4
<br><br>$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 2$$\lambda $$ $$-$$ 2
<br><br>When $$\lambda $$ = $$-$$2<br><br>
$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = $$-$$ 6 < 0
<br><br>$$ \therefore $$ at $$\lambda $$ = $$-$$2, f($$\lambda $$) has maximum value.<br><br>
When $$\lambda $$ = 4
<br><br>$${{{d^2}f(\lambda )} \over {d{\lambda ^2}}}$$ = 6 > 0
<br><br>$$ \therefore $$ at $$\lambda $$ = 4, f($$\lambda $$) has minimum value.
<br><br>$$ \therefore $$ When $$\lambda $$ = 4 equation is,
<br><br>x<sup>2</sup> $$-$$ 2x + 6 = 0
<br><br>$$ \therefore $$ ($$\alpha $$ $$-$$ $$\beta $$)<sup>2</sup> = ($$\alpha $$ + $$\beta $$)<sup>2</sup> $$-$$ 4$$\alpha \beta$$
<br><br>$$ \Rightarrow $$ x<sup>2</sup> $$-$$ 4 $$ \times $$ 6
<br><br>= $$-$$ 20<br><br>
$$ \Rightarrow $$ ($$\alpha $$ $$-$$ $$\beta $$) = $$2\sqrt 5 i$$
<br><br>$$ \Rightarrow $$ $$\left| {\alpha - \beta } \right|$$ = $$2\sqrt 5$$ (ans) | mcq | jee-main-2018-online-15th-april-morning-slot |
Z0kE4u5jpksipfnZxwdCg | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$\lambda $$ such that sum of the squares of the roots of the quadratic equation, x<sup>2</sup> + (3 – $$\lambda $$)x + 2 = $$\lambda $$ has the least value is - | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${{15} \\over 8}$$"}, {"identifier": "D", "content": "$${4 \\over 9}$$"}] | ["B"] | null | $$\alpha $$ + $$\beta $$ = $$\lambda $$ $$-$$ 3
<br><br>$$\alpha $$$$\beta $$ = 2 $$-$$ $$\lambda $$
<br><br>$$\alpha $$<sup>2</sup> + $$\beta $$<sup>2</sup> = ($$\alpha $$ + $$\beta $$)<sup>2</sup> $$-$$ 2$$\alpha $$$$\beta $$ = ($$\lambda $$ $$-$$ 3)<sup>2</sup> $$-$$ 2$$\left( {2 - \lambda } \right)$$
<br><br>= $$\lambda $$<sup>2</sup> + 9 $$-$$ 6$$\lambda $$ $$-$$ 4 + 2$$\lambda $$
<br><br>= $$\lambda $$<sup>2</sup> $$-$$ 4$$\lambda $$ + 5
<br><br>= ($$\lambda $$ $$-$$ 2)<sup>2</sup> + 1
<br><br>$$ \therefore $$ $$\lambda $$ = 2 | mcq | jee-main-2019-online-10th-january-evening-slot |
cSCA6jlUnOtHa6FTlPLZK | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If one real root of the quadratic equation 81x<sup>2</sup> + kx + 256 = 0 is cube of the other root, then a value of k is | [{"identifier": "A", "content": "$$-$$ 81"}, {"identifier": "B", "content": "$$-$$ 300"}, {"identifier": "C", "content": "100"}, {"identifier": "D", "content": "144"}] | ["B"] | null | 81x<sup>2</sup> + kx + 256 = 0 ; x = $$\alpha $$, $$\alpha $$<sup>3</sup>
<br><br>$$ \Rightarrow $$ $$\alpha $$<sup>4</sup> = $${{256} \over {81}}$$ $$ \Rightarrow $$ $$\alpha $$ = $$ \pm $$ $${{4} \over {3}}$$
<br><br>Now $$-$$ $${k \over {81}}$$ = $$\alpha $$ + $$\alpha $$<sup>3</sup> = $$ \pm $$ $${{100} \over {27}}$$
<br><br>$$ \Rightarrow $$ k = $$ \pm $$300 | mcq | jee-main-2019-online-11th-january-morning-slot |
UrxNTlitqqL3D9jwUZty7 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of the quadratic equation x<sup>2</sup>
sin $$\theta $$ – x(sin $$\theta $$ cos $$\theta $$ + 1) + cos $$\theta $$ = 0 (0 < $$\theta $$ < 45<sup>o</sup>), and $$\alpha $$ < $$\beta $$. Then $$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{{{\left( { - 1} \right)}^n}} \over {{\beta ^n}}}} \right)} $$ is equal to :
| [{"identifier": "A", "content": "$${1 \\over {1 + \\cos \\theta }} + {1 \\over {1 - \\sin \\theta }}$$"}, {"identifier": "B", "content": "$${1 \\over {1 - \\cos \\theta }} + {1 \\over {1 + \\sin \\theta }}$$"}, {"identifier": "C", "content": "$${1 \\over {1 - \\cos \\theta }} - {1 \\over {1 + \\sin \\theta }}$$"}, {"identifier": "D", "content": "$${1 \\over {1 + \\cos \\theta }} - {1 \\over {1 - \\sin \\theta }}$$"}] | ["B"] | null | D = (1 + sin$$\theta $$ cos$$\theta $$)<sup>2</sup> $$-$$ 4sin$$\theta $$cos$$\theta $$ = (1 $$-$$ sin$$\theta $$ cos$$\theta $$)<sup>2</sup>
<br><br>$$ \Rightarrow $$ roots are $$\beta $$ = cosec$$\theta $$ and $$\alpha $$ = cos$$\theta $$
<br><br>$$\sum\limits_{n = 0}^\infty {\left( {{\alpha ^n} + {{\left( { - {1 \over \beta }} \right)}^n}} \right)} = \sum\limits_{n = 0}^\infty {{{\left( {\cos \theta } \right)}^n}} + \sum\limits_{n = 0}^n {{{\left( { - \sin \theta } \right)}^n}} $$
<br><br>$$ = {1 \over {1 - \cos \theta }} + {1 \over {1 + \sin \theta }}$$ | mcq | jee-main-2019-online-11th-january-evening-slot |
YIRrhh126EC0R8efYDcTv | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\lambda $$ be the ratio of the roots of the quadratic equation in x, 3m<sup>2</sup>x<sup>2</sup> + m(m – 4)x + 2 = 0, then the least value of m for which $$\lambda + {1 \over \lambda } = 1,$$ is | [{"identifier": "A", "content": "$$ - 2 + \\sqrt 2 $$"}, {"identifier": "B", "content": "4$$-$$3$$\\sqrt 2 $$"}, {"identifier": "C", "content": "2 $$-$$ $$\\sqrt 3 $$"}, {"identifier": "D", "content": "4 $$-$$ 2$$\\sqrt 3 $$"}] | ["B"] | null | 3m<sup>2</sup>x<sup>2</sup> + m(m $$-$$ 4) x + 2 = 0
<br><br>$$\lambda + {1 \over \lambda } = 1,{\alpha \over \beta } + {\beta \over \alpha } = 1,{\alpha ^2} + {\beta ^2} = \alpha \beta $$
<br><br>($$\alpha $$ + $$\beta $$)<sup>2</sup> = 3$$\alpha $$$$\beta $$
<br><br>$${\left( { - {{m\left( {m - 4} \right)} \over {3{m^2}}}} \right)^2} = {{3\left( 2 \right)} \over {3{m^2}}},{{{{\left( {m - 4} \right)}^2}} \over {9{m^2}}} = {6 \over {3m}}$$
<br><br>$${\left( {m - 4} \right)^2} = 18,m = 4 \pm \sqrt {18,} \,\,4 \pm 3\sqrt 2 $$ | mcq | jee-main-2019-online-12th-january-morning-slot |
ASH34YkiuFpNO2zeGRoYa | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let p, q $$ \in $$ R. If 2 - $$\sqrt 3$$ is a root of the quadratic
equation, x<sup>2</sup> + px + q = 0, then : | [{"identifier": "A", "content": "p<sup>2</sup> \u2013 4q \u2013 12 = 0"}, {"identifier": "B", "content": "q<sup>2</sup> \u2013 4p \u2013 16 = 0"}, {"identifier": "C", "content": "q<sup>2</sup> + 4p + 14 = 0"}, {"identifier": "D", "content": "p<sup>2</sup> \u2013 4q + 12 = 0"}] | ["A"] | null | If a quadratic equation with rational coefficient has one irrational root then other root will be the conjugate of the irrational root.
<br><br>Here x<sup>2</sup> + px + q = 0 has one root 2 - $$\sqrt 3$$.
<br><br>$$ \therefore $$ Other root will be 2 + $$\sqrt 3$$.
<br><br>Sum of the roots = -p = 4
<br><br>and product of the roots = q = 1
<br><br>You can see p<sup>2</sup> – 4q – 12 = 0 satisfy value of p = -4 and q = 1. | mcq | jee-main-2019-online-9th-april-morning-slot |
W9EBjBBLvv4KN4kY2B3rsa0w2w9jwxdqfnm | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the quadratic equation,
<br/>x<sup>2</sup> + x sin $$\theta $$ - 2 sin $$\theta $$ = 0, $$\theta \in \left( {0,{\pi \over 2}} \right)$$, then
<br/>$${{{\alpha ^{12}} + {\beta ^{12}}} \over {\left( {{\alpha ^{ - 12}} + {\beta ^{ - 12}}} \right).{{\left( {\alpha - \beta } \right)}^{24}}}}$$ is equal to : | [{"identifier": "A", "content": "$${{{2^{12}}} \\over {{{\\left( {\\sin \\theta - 8} \\right)}^6}}}$$"}, {"identifier": "B", "content": "$${{{2^6}} \\over {{{\\left( {\\sin \\theta + 4} \\right)}^{12}}}}$$"}, {"identifier": "C", "content": "$${{{2^{12}}} \\over {{{\\left( {\\sin \\theta + 8} \\right)}^{12}}}}$$"}, {"identifier": "D", "content": "$${{{2^{12}}} \\over {{{\\left( {\\sin \\theta - 4} \\right)}^{12}}}}$$"}] | ["C"] | null | Given $$\alpha + \beta = - \sin \theta $$ and$$\alpha \beta = - 2\sin \theta $$<br><br>
$${{\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){\alpha ^{12}}{\beta ^{12}}} \over {\left( {{\alpha ^{12}} + {\beta ^{12}}} \right){{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}}$$<br><br>
$$\left| {\alpha - \beta } \right| = \sqrt {{{\left( {\alpha + \beta } \right)}^2} - 4\alpha \beta } = \sqrt {{{\sin }^2}\theta + 8\sin \theta } $$<br><br>
Hence required quantity<br><br>
$${{{{\left( {\alpha \beta } \right)}^{12}}} \over {{{\left( {\alpha - \beta } \right)}^{24}}}} = {{{{\left( {2\sin \theta } \right)}^{12}}} \over {{{\sin }^{12}}\theta {{\left( {\sin \theta + 8} \right)}^{12}}}} = {{{2^{12}}} \over {{{\left( {\sin \theta + 8} \right)}^{12}}}}$$ | mcq | jee-main-2019-online-10th-april-morning-slot |
fpfAhxHaFIjNWVXwdijgy2xukfw0t2zw | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ be two roots of the equation<br/> x<sup>2</sup> – 64x + 256 = 0. Then the value of
<br/>$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$ is : | [{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}] | ["C"] | null | x<sup>2</sup> – 64x + 256 = 0
<br><br>$$\alpha $$ + $$\beta $$ = 64, $$\alpha $$$$\beta $$ = 256
<br><br>$${\left( {{{{\alpha ^3}} \over {{\beta ^5}}}} \right)^{1/8}} + {\left( {{{{\beta ^3}} \over {{\alpha ^5}}}} \right)^{1/8}}$$
<br><br>= $${{{\alpha ^{{3 \over 8}}}} \over {{\beta ^{{5 \over 8}}}}} + {{{\beta ^{{3 \over 8}}}} \over {{\alpha ^{{5 \over 8}}}}}$$
<br><br>= $${{\alpha + \beta } \over {{{\left( {\alpha \beta } \right)}^{{5 \over 8}}}}}$$
<br><br>= $${{64} \over {{{\left( {256} \right)}^{{5 \over 8}}}}}$$
<br><br>= 2 | mcq | jee-main-2020-online-6th-september-morning-slot |
2VBumVZIDcyB2i8HAGjgy2xukg0cjhh6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the equation
<br/>2x(2x + 1) = 1, then $$\beta $$ is equal to :
| [{"identifier": "A", "content": "$$ - 2\\alpha \\left( {\\alpha + 1} \\right)$$"}, {"identifier": "B", "content": "$$ 2\\alpha \\left( {\\alpha + 1} \\right)$$"}, {"identifier": "C", "content": "$$2{\\alpha ^2}$$"}, {"identifier": "D", "content": "$$ 2\\alpha \\left( {\\alpha - 1} \\right)$$"}] | ["A"] | null | $$\alpha $$ and
$$\beta $$ are the roots of the equation
<br>4x<sup>2</sup> + 2x – 1 = 0.
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $$ - {1 \over 2}$$
<br><br>$$ \Rightarrow $$ -1 = 2$$\alpha $$ + 2$$\beta $$
<br><br>and 4$$\alpha $$<sup>2</sup> + 2$$\alpha $$ - 1 = 0
<br><br>$$ \Rightarrow $$ 4$$\alpha $$<sup>2</sup> + 2$$\alpha $$ + 2$$\alpha $$ + 2$$\beta $$ = 0
<br><br>$$ \Rightarrow $$ $$\beta $$ = $$ - 2\alpha \left( {\alpha + 1} \right)$$ | mcq | jee-main-2020-online-6th-september-evening-slot |
ADtDSoqLJjuj5nPt5Zjgy2xukfqcls6x | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the equation,
<br/>7x<sup>2</sup> – 3x – 2 = 0, then the value of
<br/>$${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$$ is equal to : | [{"identifier": "A", "content": "$${1 \\over {24}}$$"}, {"identifier": "B", "content": "$${{27} \\over {32}}$$"}, {"identifier": "C", "content": "$${{27} \\over {16}}$$"}, {"identifier": "D", "content": "$${3 \\over 8}$$"}] | ["C"] | null | Given, 7x<sup>2</sup> – 3x – 2 = 0
<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = $${3 \over 7}$$
<br><br>$$\alpha $$$$\beta $$ = - $${2 \over 7}$$
<br><br>$${\alpha \over {1 - {\alpha ^2}}} + {\beta \over {1 - {\beta ^2}}}$$
<br><br>= $${{\alpha + \beta - \alpha \beta \left( {\alpha + \beta } \right)} \over {1 - {\alpha ^2} - {\beta ^2} + {\alpha ^2}{\beta ^2}}}$$
<br><br>= $${{{3 \over 7} + {2 \over 7}\left( {{3 \over 7}} \right)} \over {1 - {{\left( {\alpha + \beta } \right)}^2} + 2\alpha \beta + {{\left( { - {2 \over 7}} \right)}^2}}}$$
<br><br>= $${{{3 \over 7} + {2 \over 7}\left( {{3 \over 7}} \right)} \over {1 - {{\left( {{3 \over 7}} \right)}^2} + 2\left( { - {2 \over 7}} \right) + {{\left( { - {2 \over 7}} \right)}^2}}}$$
<br><br>= $${{27} \over {16}}$$ | mcq | jee-main-2020-online-5th-september-evening-slot |
PAZAL0qE1kEzLNFYmvjgy2xukfak9s9b | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\lambda \ne 0$$ be in R. If $$\alpha $$ and $$\beta $$ are the roots of the <br/>equation, x<sup>2</sup> - x + 2$$\lambda $$ = 0 and $$\alpha $$ and $$\gamma $$ are the roots of <br/>the equation, $$3{x^2} - 10x + 27\lambda = 0$$, then $${{\beta \gamma } \over \lambda }$$ is equal to:
| [{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "27"}, {"identifier": "D", "content": "18"}] | ["D"] | null | $$\alpha $$ and $$\beta $$ are the roots of the <br>equation x<sup>2</sup> - x + 2$$\lambda $$ = 0 .....(1)
<br><br>$$ \therefore $$ $$\alpha + \beta = 1,\,\alpha \beta = 2\lambda $$
<br><br>$$\alpha $$ and $$\gamma $$ are the roots of <br>the equation, $$3{x^2} - 10x + 27\lambda = 0$$ ......(2)
<br><br>$$ \therefore $$ $$\alpha + \gamma $$ = $${{10} \over 3}$$, $$\alpha \gamma $$ = $${{27\lambda } \over 3}$$ = 9$$\lambda $$
<br><br>Multiplying equation (1) by 3 and subtracting form equation (2) we get
<br><br>-7x + 21$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ x = 3$$\lambda $$
<br><br>$$ \therefore $$ $$\alpha $$ = 3$$\lambda $$
<br><br>As $$\alpha $$ is root of equation (1) so
<br><br>$$\alpha $$<sup>2</sup> - $$\alpha $$ + 2$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ 9$$\lambda $$<sup>2</sup> - 3$$\lambda $$ + 2$$\lambda $$ = 0
<br><br>$$ \Rightarrow $$ $$\lambda $$ = $${1 \over 9}$$
<br><br>$$ \Rightarrow $$ $$\alpha $$ = 3$$ \times $$ $${1 \over 9}$$ = $${1 \over 3}$$
<br><br>Also $$\alpha \beta = 2\lambda $$ = $${2 \over 9}$$
<br><br>$$ \Rightarrow $$ $$\beta $$ = $${2 \over 3}$$
<br><br>Also $$\alpha \gamma $$ = 9$$\lambda $$ = 9$$ \times $$ $${1 \over 9}$$ = 1
<br><br>$$ \Rightarrow $$ $$\gamma $$ = 3
<br><br>$$ \therefore $$ $${{\beta \gamma } \over \lambda }$$ = $${{{2 \over 3}.3} \over {{1 \over 9}}}$$ = 18 | mcq | jee-main-2020-online-4th-september-evening-slot |
0EWeqQcicU7tlKL9gijgy2xukf8z77mb | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of x<sup>2</sup> - 3x + p=0 and $$\gamma $$ and $$\delta $$ be the roots of x<sup>2</sup> - 6x + q = 0. If $$\alpha, \beta, \gamma, \delta $$
form a geometric progression.Then ratio (2q + p) : (2q - p) is: | [{"identifier": "A", "content": "9 : 7"}, {"identifier": "B", "content": "5 : 3\n"}, {"identifier": "C", "content": "3 : 1 "}, {"identifier": "D", "content": "33 :31 "}] | ["A"] | null | $$\alpha $$ and $$\beta $$ are the roots of x<sup>2</sup> $$-$$ 3x + p = 0<br><br>$$ \therefore $$ $$\alpha $$ + $$\beta $$ = 3 and $$\alpha \beta $$ = p<br><br>$$\gamma $$ and $$\delta $$ are the roots of x<sup>2</sup> $$-$$ 6x + q = 0<br><br>$$ \therefore $$ $$\gamma $$ + $$\delta $$ = 6 and $$\gamma \delta $$ = q<br><br>Given that, $$\alpha $$, $$\beta $$, $$\gamma $$, $$\delta $$ are in G.P.<br><br>$$ \therefore $$ Let $$\alpha $$ = a, $$\beta $$ = ar, $$\gamma $$ = ar<sup>2</sup> and $$\delta $$ = ar<sup>3</sup><br><br>As $$\alpha $$ + $$\beta $$ = 3<br><br>$$ \Rightarrow $$ a + ar = 3<br><br>$$ \Rightarrow $$ a(1 + r) = 3 ...(1)<br><br>Also $$\gamma $$ + $$\delta $$ = 6<br><br>$$ \Rightarrow $$ ar<sup>2</sup> + ar<sup>3</sup> = 6<br><br>$$ \Rightarrow $$ ar<sup>2</sup> (1 + r) = 6 ... (2)<br><br>Dividing (2) by (1),<br><br>r<sup>2</sup> = 2<br><br>$$ \Rightarrow r = \sqrt 2 $$<br><br>$$ \therefore $$ $$a = {3 \over {1 + \sqrt 2 }}$$<br><br>So, $$\alpha = {3 \over {1 + \sqrt 2 }}$$, $$\beta = {{3\sqrt 2 } \over {1 + \sqrt 2 }}$$, $$\gamma = {{3 \times 2} \over {1 + \sqrt 2 }}$$, $$\delta = {{3(2\sqrt {2)} } \over {1 + \sqrt 2 }}$$<br><br>$$ \therefore $$ $$p = \alpha \beta = {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}$$<br><br>and $$q = \gamma \delta = {{36\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}$$<br><br>Now, $${{2q + p} \over {2q - p}}$$<br><br>$$ = {{{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} + {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}} \over {{{72\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}} - {{9\sqrt 2 } \over {{{\left( {1 + \sqrt 2 } \right)}^2}}}}}$$<br><br>$$ = {{72\sqrt 2 + 9\sqrt 2 } \over {72\sqrt 2 - 9\sqrt 2 }}$$<br><br>$$ = {{81} \over {63}}$$<br><br>$$ = {9 \over 7}$$ | mcq | jee-main-2020-online-4th-september-morning-slot |
vQfn0SPLSnaGH8uZOkjgy2xukewro3a6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let
$$\alpha $$ and
$$\beta $$ be the roots of the equation
<br/>5x<sup>2</sup> + 6x – 2 = 0. If S<sub>n</sub>
=
$$\alpha $$<sup>n</sup> +
$$\beta $$<sup>n</sup>, n = 1, 2, 3....,
then : | [{"identifier": "A", "content": "5S<sub>6</sub>\n + 6S<sub>5</sub>\n = 2S<sub>4</sub>"}, {"identifier": "B", "content": "5S<sub>6</sub>\n + 6S<sub>5</sub>\n + 2S<sub>4</sub> = 0"}, {"identifier": "C", "content": "6S<sub>6</sub>\n + 5S<sub>5</sub>\n + 2S<sub>4</sub> = 0"}, {"identifier": "D", "content": "6S<sub>6</sub>\n + 5S<sub>5</sub>\n = 2S<sub>4</sub>"}] | ["A"] | null | $$\alpha $$ and
$$\beta $$ be the roots of the equation
<br>5x<sup>2</sup> + 6x – 2 = 0.
<br><br>$$ \Rightarrow $$ 5$$\alpha $$<sup>2</sup> + 6$$\alpha $$ - 2 = 0
<br><br>$$ \Rightarrow $$ 5$$\alpha $$<sup>n + 2</sup> + 6$$\alpha $$<sup>n + 2</sup> - 2$$\alpha $$<sup>n</sup> = 0 ......(1)
<br><br>(By multiplying $$\alpha $$<sup>n</sup>)
<br><br>Similarly 5$$\beta $$<sup>n + 2</sup> + 6$$\beta $$<sup>n + 2</sup> - 2$$\beta $$<sup>n</sup> = 0 ......(2)
<br><br>By adding (1) & (2)
<br><br>5S<sub>n+2</sub> + 6S<sub>n+1</sub> – 2S<sub>n</sub> = 0
<br><br>For n = 4
<br><br>5S<sub>6</sub>
+ 6S<sub>5</sub>
= 2S<sub>4</sub> | mcq | jee-main-2020-online-2nd-september-morning-slot |
8XBpeSuAbU3N3i9RH5jgy2xukf0x1fao | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha $$ and $$\beta $$ are the roots of the equation
<br/>x<sup>2</sup>
+ px + 2 = 0 and $${1 \over \alpha }$$ and $${1 \over \beta }$$ are the<br/> roots of
the equation 2x<sup>2</sup>
+ 2qx + 1 = 0, then
<br/>$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right)$$ is equal to :
| [{"identifier": "A", "content": "$${9 \\over 4}\\left( {9 - {q^2}} \\right)$$"}, {"identifier": "B", "content": "$${9 \\over 4}\\left( {9 + {q^2}} \\right)$$"}, {"identifier": "C", "content": "$${9 \\over 4}\\left( {9 - {p^2}} \\right)$$"}, {"identifier": "D", "content": "$${9 \\over 4}\\left( {9 + {p^2}} \\right)$$"}] | ["C"] | null | $$\alpha $$ and $$\beta $$ are the roots of the <br><br>equation
x<sup>2</sup>
+ px + 2 = 0
<br><br>$$ \therefore $$ $$\alpha + \beta = - p,\,\alpha \beta = 2$$
<br><br>$${1 \over \alpha }$$ and $${1 \over \beta }$$ are the roots of
the <br><br>equation 2x<sup>2</sup>
+ 2qx + 1 = 0
<br><br>$$ \therefore $$ $${1 \over \alpha } + {1 \over \beta } = - q,\,{1 \over {\alpha \beta }} = {1 \over 2}$$
<br><br>$$ \Rightarrow $$ $${{\alpha + \beta } \over {\alpha \beta }} = - q \Rightarrow {{ - p} \over 2} = - q$$<br><br>$$ \Rightarrow p = 2q$$<br><br>$$\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \alpha \beta + {1 \over {\alpha \beta }} + 2$$ $$ = 2 + {1 \over 2} + 2 = {9 \over 2}$$<br><br>$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right) = \alpha \beta + {1 \over {\alpha \beta }} - {\alpha \over \beta } - {\beta \over \alpha }$$<br><br>$$ = 2 + {1 \over 2} - \left[ {{{{\alpha ^2} + {\beta ^2}} \over {\alpha \beta }}} \right]$$$$<br><br>= {5 \over 2} - \left[ {{{{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \over {\alpha \beta }}} \right]$$<br><br>$$ = {5 \over 2} - \left[ {{{{p^2} - 4} \over 2}} \right]$$<br><br>$$ = {{9 - {p^2}} \over 2}$$<br><br>$$\left( {\alpha - {1 \over \alpha }} \right)\left( {\beta - {1 \over \beta }} \right)\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = \left( {{{9 - {p^2}} \over 2}} \right)\left( {{9 \over 2}} \right)$$<br><br>$$ = {9 \over 4}(9 - {p^2})$$ | mcq | jee-main-2020-online-3rd-september-morning-slot |
B9WR7JawEEY279wmzL7k9k2k5e2h7g2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be two real roots of the equation <br/>(k + 1)tan<sup>2</sup>x - $$\sqrt 2 $$ . $$\lambda $$tanx = (1 - k), where k($$ \ne $$ - 1)
and $$\lambda $$ are real numbers. if tan<sup>2</sup> ($$\alpha $$ + $$\beta $$) = 50, then a value of $$\lambda $$ is:
| [{"identifier": "A", "content": "5$$\\sqrt 2 $$"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "10$$\\sqrt 2 $$"}] | ["B"] | null | Let tan$$\alpha $$ and tan$$\beta $$ are the roots of
<br><br>(k + 1)tan<sup>2</sup>x - $$\sqrt 2 $$ . $$\lambda $$tanx - (1 - k) = 0
<br><br>$$ \therefore $$ tan$$\alpha $$ + tan$$\beta $$ = $${{\sqrt 2 \lambda } \over {k + 1}}$$
<br><br>and an$$\alpha $$.tan$$\beta $$ = $${{k - 1} \over {k + 1}}$$
<br><br>Now tan($$\alpha $$ + $$\beta $$)
<br><br>= $${{\tan \alpha + \tan \beta } \over {1 - \tan \alpha \tan \beta }}$$
<br><br>= $${{{{\lambda \sqrt 2 } \over {k + 1}}} \over {1 - {{k - 1} \over {k + 1}}}}$$
<br><br>= $${{{\lambda \sqrt 2 } \over 2}}$$ = $${{\lambda \over {\sqrt 2 }}}$$
<br><br>Given $${{{{\lambda ^2}} \over 2}}$$ = 50
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 10 | mcq | jee-main-2020-online-7th-january-morning-slot |
oXLFdyeUzYjCQopsqF7k9k2k5hk5f09 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$. <br/>If $$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$ and<br/> $$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} $$, then a and b are the roots of the quadratic equation : | [{"identifier": "A", "content": "x<sup>2</sup> + 101x + 100 = 0"}, {"identifier": "B", "content": "x<sup>2</sup> + 102x + 101 = 0"}, {"identifier": "C", "content": "x<sup>2</sup> \u2013 102x + 101 = 0"}, {"identifier": "D", "content": "x<sup>2</sup> \u2013 101x + 100 = 0"}] | ["C"] | null | $$\alpha = {{ - 1 + i\sqrt 3 } \over 2}$$ = $$\omega $$
<br><br>$$a = \left( {1 + \alpha } \right)\sum\limits_{k = 0}^{100} {{\alpha ^{2k}}} $$
<br><br>= $$\left( {1 + \omega } \right)\sum\limits_{k = 0}^{100} {{\omega ^{2k}}} $$
<br><br>= $$\left( {1 + \omega } \right){{1\left( {1 - {{\left( {{\omega ^2}} \right)}^{101}}} \right)} \over {1 - {\omega ^2}}}$$
<br><br>= $${{1 - {\omega ^{202}}} \over {1 - \omega }}$$
<br><br>= $${{1 - \omega } \over {1 - \omega }}$$ = 1
<br><br>$$b = \sum\limits_{k = 0}^{100} {{\alpha ^{3k}}} $$
<br><br>= $$\sum\limits_{k = 0}^{100} {{\omega ^{3k}}} $$
<br><br>= 101 [as $${\omega ^3}$$ = 1]
<br><br>$$ \therefore $$ roots are 101 and 1
<br><br>Then equation is = x<sup>2</sup>
– 102x + 101 = 0 | mcq | jee-main-2020-online-8th-january-evening-slot |
jDnLrPSQv7zEMajD7o7k9k2k5flw7xw | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha $$ and $$\beta $$ be the roots of the equation x<sup>2</sup>
- x - 1 = 0. <br/>If p<sub>k</sub>
= $${\left( \alpha \right)^k} + {\left( \beta \right)^k}$$
, k $$ \ge $$ 1, then which one
of the following statements is not true? | [{"identifier": "A", "content": "(p<sub>1</sub> + p<sub>2</sub> + p<sub>3</sub> + p<sub>4</sub> + p<sub>5</sub>) = 26"}, {"identifier": "B", "content": "p<sub>5</sub> = 11"}, {"identifier": "C", "content": "p<sub>3</sub> = p<sub>5</sub> \u2013 p<sub>4</sub>"}, {"identifier": "D", "content": "p<sub>5</sub> = p<sub>2</sub> \u00b7 p<sub>3</sub>"}] | ["D"] | null | x<sup>2</sup>
- x - 1 = 0
<br><br>$$ \therefore $$ $$\alpha $$<sup>2</sup>
- $$\alpha $$ - 1 = 0
<br><br>$$ \Rightarrow $$ $$\alpha $$<sup>2</sup> = $$\alpha $$ + 1
<br><br>$$ \therefore $$ $$\alpha $$<sup>3</sup> = $$\alpha $$<sup>2</sup> + $$\alpha $$
<br><br>= $$\alpha $$ + 1 + $$\alpha $$
<br><br>= 2$$\alpha $$ + 1
<br><br>Now $$\alpha $$<sup>4</sup> = 2$$\alpha $$<sup>2</sup> + $$\alpha $$
<br><br>= 2($$\alpha $$ + 1) + $$\alpha $$
<br><br>= 3$$\alpha $$ + 2
<br><br>Now $$\alpha $$<sup>5</sup> = 3$$\alpha $$<sup>2</sup> + 2$$\alpha $$
<br><br>= 3($$\alpha $$ + 1) + 2$$\alpha $$
<br><br>= 5$$\alpha $$ + 3
<br><br>Given p<sub>k</sub>
= $${\left( \alpha \right)^k} + {\left( \beta \right)^k}$$
<br><br>$$ \therefore $$ p<sub>5</sub> = $${\left( \alpha \right)^5} + {\left( \beta \right)^5}$$
<br><br>= 5$$\alpha $$ + 3 + 5$$\beta $$ + 3
<br><br>= 5($$\alpha $$ + $$\beta $$) + 6
<br><br>= 5(1) + 6 [As $$\alpha $$ + $$\beta $$ = 1]
<br><br>= 11
<br><br>Now p<sub>2</sub> · p<sub>3</sub>
<br><br>= ($$\alpha $$<sup>2</sup> + $$\beta $$<sup>2</sup>).($$\alpha $$<sup>3</sup> + $$\beta $$<sup>3</sup>)
<br><br>= ( $$\alpha $$ + 1 + $$\beta $$ + 1)(2$$\alpha $$ + 1 + 2$$\beta $$ + 1)
<br><br>= ( $$\alpha $$ + $$\beta $$ + 2)(2($$\alpha $$ + $$\beta $$) + 2)
<br><br>= (1 + 2)(2 + 2) = 12
<br><br>$$ \therefore $$ p<sub>5</sub> $$ \ne $$ p<sub>2</sub> ·p<sub>3</sub>
<br><br>So option (D) is wrong. | mcq | jee-main-2020-online-7th-january-evening-slot |
qlRl2Oq6OSlxdJ5RT61klt7pumo | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha$$ and $$\beta$$ be the roots of x<sup>2</sup> $$-$$ 6x $$-$$ 2 = 0. If a<sub>n</sub> = $$\alpha$$<sup>n</sup> $$-$$ $$\beta$$<sup>n</sup> for n $$ \ge $$ 1, then the value of $${{{a_{10}} - 2{a_8}} \over {3{a_9}}}$$ is : | [{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}] | ["B"] | null | Given, $$\alpha$$ and $$\beta$$ be the roots of $${x^2} - 6x - 2 = 0$$<br><br>$$\matrix{
{\alpha + \beta = 6} \cr
{\alpha \beta = - 2} \cr
} $$<br><br>and $${\alpha ^2} - 6\alpha - 2 = 0 \Rightarrow {\alpha ^2} - 2 = 6\alpha $$<br><br>$${\beta ^2} - 6\beta - 2 = 0 \Rightarrow {\beta ^2} - 2 = 6\beta $$<br><br>$${{{a_{10}} - 2{a_8}} \over {3{a_9}}} = {{\left( {{\alpha ^{10}} - {\beta ^{10}}} \right) - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$<br><br>$$ = {{\left( {{\alpha ^{10}} - 2{\alpha ^8}} \right) - \left( {{\beta ^{10}} - 2{\beta ^8}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$<br><br>Now, $$ = {{{\alpha ^8}\left( {{\alpha ^2} - 2} \right) - {\beta ^8}\left( {{\beta ^2} - 2} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}}$$<br><br>$$ = {{{\alpha ^8}(6\alpha ) - {\beta ^8}(6\beta )} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {{6\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {3\left( {{\alpha ^9} - {\beta ^9}} \right)}} = {6 \over 3} = 2$$ | mcq | jee-main-2021-online-25th-february-evening-slot |
APggZAnowmhcpBCqH71kluy92o5 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha$$ and $$\beta$$ be two real numbers such that $$\alpha$$ + $$\beta$$ = 1 and $$\alpha$$$$\beta$$ = $$-$$1. Let p<sub>n</sub> = ($$\alpha$$)<sup>n</sup> + ($$\beta$$)<sup>n</sup>, p<sub>n$$-$$1</sub> = 11 and p<sub>n+1</sub> = 29 for some integer n $$ \ge $$ 1. Then, the value of p$$_n^2$$ is ___________. | [] | null | 324 | Given, $$\alpha$$ + $$\beta$$ = 1, $$\alpha$$$$\beta$$ = $$-$$ 1<br><br>$$ \therefore $$ Quadratic equation with roots $$\alpha$$, $$\beta$$ is x<sup>2</sup> $$-$$ x $$-$$ 1 = 0<br><br>$$ \Rightarrow $$ $$\alpha$$<sup>2</sup> = $$\alpha$$ + 1<br><br>Multiplying both sides by $$\alpha$$<sup>n$$-$$1</sup><br><br>$$\alpha$$<sup>n$$+$$1</sup> = $$\alpha$$<sup>n</sup> + $$\alpha$$<sup>n$$-$$1</sup> ......(1)<br><br>Similarly,<br><br>$$\beta$$<sup>n + 1</sup> = $$\beta$$<sup>n</sup> + $$\beta$$<sup>n + 1</sup> ..... (2)<br><br>Adding (1) & (2)<br><br>$${\alpha ^{n + 1}} + {\beta ^{n + 1}} = ({\alpha ^n} + {\beta ^n}) + ({\alpha ^{n - 1}} + {\beta ^{n - 1}})$$<br><br>$$ \Rightarrow $$ P<sub>n+1</sub> = P<sub>n</sub> + P<sub>n$$-$$1</sub><br><br>$$ \Rightarrow $$ 29 = P<sub>n</sub> + 11 (Given, P<sub>n + 1</sub> = 29, P<sub>n $$-$$ 1</sub> = 11)<br><br>$$ \Rightarrow $$ P<sub>n</sub> = 18<br><br>$$ \therefore $$ $$P_n^2$$ = 18<sup>2</sup> = 324 | integer | jee-main-2021-online-26th-february-evening-slot |
5gwO0y8VqkWaKc4sGK1kmjavirx | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$4 + {1 \over {5 + {1 \over {4 + {1 \over {5 + {1 \over {4 + ......\infty }}}}}}}}$$ is : | [{"identifier": "A", "content": "2 + $${2 \\over 5}\\sqrt {30} $$"}, {"identifier": "B", "content": "2 + $${4 \\over {\\sqrt 5 }}\\sqrt {30} $$"}, {"identifier": "C", "content": "5 + $${2 \\over 5}\\sqrt {30} $$"}, {"identifier": "D", "content": "4 + $${4 \\over {\\sqrt 5 }}\\sqrt {30} $$"}] | ["A"] | null | $$y = 4 + {1 \over {5 + {1 \over y}}}$$<br><br>$$ \Rightarrow y = 4 + {y \over {5y + 1}}$$<br><br>$$ \Rightarrow 5{y^2} - 20y - 4 = 0$$<br><br>$$ \Rightarrow y = {{20 \pm \sqrt {400 + 80} } \over {10}}$$<br><br>$$ \Rightarrow y = {{20 \pm 4\sqrt {30} } \over {10}},y > 0$$<br><br>$$ \therefore $$ $$y = {{10 + 2\sqrt {30} } \over 5}$$ | mcq | jee-main-2021-online-17th-march-morning-shift |
Y7qXRAQT5BMwMqf4Iw1kmlj8n1j | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | The value of $$3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$$ is equal to | [{"identifier": "A", "content": "1.5 + $$\\sqrt 3 $$"}, {"identifier": "B", "content": "2 + $$\\sqrt 3 $$"}, {"identifier": "C", "content": "3 + 2$$\\sqrt 3 $$"}, {"identifier": "D", "content": "4 + $$\\sqrt 3 $$"}] | ["A"] | null | Let $$x = 3 + {1 \over {4 + {1 \over {3 + {1 \over {4 + {1 \over {3 + ....\infty }}}}}}}}$$<br><br>So, $$x = 3 + {1 \over {4 + {1 \over x}}} = 3 + {1 \over {{{4x + 1} \over x}}}$$<br><br>$$ \Rightarrow (x - 3) = {x \over {(4x + 1)}}$$<br><br>$$ \Rightarrow (4x + 1)(x - 3) = x$$<br><br>$$ \Rightarrow 4{x^2} - 12x + x - 3 = x$$<br><br>$$ \Rightarrow 4{x^2} - 12x - 3 = 0$$<br><br>$$x = {{12 \pm \sqrt {{{(12)}^2} + 12 \times 4} } \over {2 \times 4}} = {{12 \pm \sqrt {12(16)} } \over 8}$$<br><br>$$ = {{12 \pm 4 \times 2\sqrt 3 } \over 8} = {{3 \pm 2\sqrt 3 } \over 2}$$<br><br>$$x = {3 \over 2} \pm \sqrt 3 = 1.5 \pm \sqrt 3 $$.<br><br>But only positive value is accepted<br><br>So, x = $$1.5 + \sqrt 3 $$ | mcq | jee-main-2021-online-18th-march-morning-shift |
1krpth9pg | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha$$ and $$\beta$$ are the distinct roots of the equation $${x^2} + {(3)^{1/4}}x + {3^{1/2}} = 0$$, then the value of $${\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1)$$ is equal to : | [{"identifier": "A", "content": "56 $$\\times$$ 3<sup>25</sup>"}, {"identifier": "B", "content": "56 $$\\times$$ 3<sup>24</sup>"}, {"identifier": "C", "content": "52 $$\\times$$ 3<sup>24</sup>"}, {"identifier": "D", "content": "28 $$\\times$$ 3<sup>25</sup>"}] | ["C"] | null | As, $$({\alpha ^2} + \sqrt 3 ) = - {(3)^{1/4}}.\alpha $$<br><br>$$ \Rightarrow ({\alpha ^4} + 2\sqrt 3 {\alpha ^2} + 3) = \sqrt 3 {\alpha ^2}$$ (On squaring)<br><br>$$\therefore$$ $$({\alpha ^4} + 3) = ( - )\sqrt 3 {\alpha ^2}$$<br><br>$$ \Rightarrow {\alpha ^8} + 6{\alpha ^4} + 9 = 3{\alpha ^4}$$ (Again squaring)<br><br>$$\therefore$$ $${\alpha ^8} + 3{\alpha ^4} + 9 = 0$$<br><br>$$ \Rightarrow {\alpha ^8} = - 9 - 3{\alpha ^4}$$<br><br>(Multiply by $$\alpha$$<sup>4</sup>)<br><br>So, $${\alpha ^{12}} = - 9{\alpha ^4} - 3{\alpha ^8}$$<br><br>$$\therefore$$ $${\alpha ^{12}} = - 9{\alpha ^4} - 3( - 9 - 3{\alpha ^4})$$<br><br>$$ \Rightarrow {\alpha ^{12}} = - 9{\alpha ^4} + 27 + 9{\alpha ^4}$$<br><br>Hence, $${\alpha ^{12}} = {(27)^2}$$<br><br>$$ \Rightarrow {({\alpha ^{12}})^8} = {(27)^8}$$<br><br>$$ \Rightarrow {\alpha ^{96}} = {(3)^{24}}$$<br><br>Similarly $${\beta ^{96}} = {(3)^{24}}$$<br><br>$$\therefore$$ $${\alpha ^{96}}({\alpha ^{12}} - 1) + {\beta ^{96}}({\beta ^{12}} - 1) = {(3)^{24}} \times 52$$<br><br>$$\Rightarrow$$ Option (3) is correct. | mcq | jee-main-2021-online-20th-july-morning-shift |
1krw3bqey | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | If $$\alpha$$, $$\beta$$ are roots of the equation $${x^2} + 5(\sqrt 2 )x + 10 = 0$$, $$\alpha$$ > $$\beta$$ and $${P_n} = {\alpha ^n} - {\beta ^n}$$ for each positive integer n, then the value of $$\left( {{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}} \right)$$ is equal to _________. | [] | null | 1 | $${x^2} + 5\sqrt 2 x + 10 = 0$$<br><br>& $${P_n} = {\alpha ^n} - {\beta ^n}$$ (Given)<br><br>Now, $${{{{P_{17}}{P_{20}} + 5\sqrt 2 {P_{17}}{P_{19}}} \over {{P_{18}}{P_{19}} + 5\sqrt 2 P_{18}^2}}}$$ = $${{{{P_{17}}({P_{20}} + 5\sqrt 2 {P_{19}})} \over {{P_{18}}({P_{19}} + 5\sqrt 2 P_{18}^{})}}}$$<br><br>$${{{P_{17}}({\alpha ^{20}} - {\beta ^{20}} + 5\sqrt 2 ({\alpha ^{19}} - {\beta ^{19}}))} \over {{P_{18}}({\alpha ^{19}} - {\beta ^{19}} + 5\sqrt 2 ({\alpha ^{18}} - {\beta ^{18}}))}}$$<br><br>$${{{P_{17}}({\alpha ^{19}}(\alpha + 5\sqrt 2 ) - {\beta ^{19}}(\beta + 5\sqrt 2 ))} \over {{P_{18}}({\alpha ^{18}}(\alpha + 5\sqrt 2 ) - {\beta ^{18}}(\beta + 5\sqrt 2 ))}}$$<br><br>Since, $$\alpha + 5\sqrt 2 = - 10/\alpha $$ ..... (1)<br><br>& $$\beta + 5\sqrt 2 = - 10/\beta $$ ....... (2)<br><br>Now, put there values in above expression $$ = - {{10{P_{17}}{P_{18}}} \over { - 10{P_{18}}{P_{17}}}} = 1$$ | integer | jee-main-2021-online-25th-july-morning-shift |
1krxkxou6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | Let $$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$ and $$\beta = \mathop {\min }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$. If $$8{x^2} + bx + c = 0$$ is a quadratic equation whose roots are $$\alpha$$<sup>1/5</sup> and $$\beta$$<sup>1/5</sup>, then the value of c $$-$$ b is equal to : | [{"identifier": "A", "content": "42"}, {"identifier": "B", "content": "47"}, {"identifier": "C", "content": "43"}, {"identifier": "D", "content": "50"}] | ["A"] | null | $$\alpha = \mathop {\max }\limits_{x \in R} \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} $$<br><br>$$ = \max \{ {2^{6\sin 3x}}{.2^{8\cos 3x}}\} $$<br><br>$$ = max\{ {2^{6\sin 3x + 8\cos 3x}}\} $$<br><br>and $$\beta = \min \{ {8^{2\sin 3x}}{.4^{4\cos 3x}}\} = \min \{ {2^{6\sin 3x + 8\cos 3x}}\} $$<br><br>Now range of $$6sin3x + 8cos3x$$<br><br>$$ = \left[ { - \sqrt {{6^2} + {8^2}} , + \sqrt {{6^2} + {8^2}} } \right] = [ - 10,10]$$<br><br>$$\alpha$$ = 2<sup>10</sup> & $$\beta$$ = 2<sup>$$-$$10</sup><br><br>So, $$\alpha$$<sup>1/5</sup> = 2<sup>2</sup> = 4<br><br>$$\Rightarrow$$ $$\beta$$<sup>1/5</sup> = 2<sup>$$-$$2</sup> = 1/4<br><br>quadratic 8x<sup>2</sup> + bx + c = 0
<br><br>$$ - {b \over 8} = {{17} \over 4}$$ $$ \Rightarrow $$ b = -34
<br><br>$${c \over 8} = 1$$ $$ \Rightarrow $$ c = 8
<br><br>$$ \therefore $$ c – b = 8 + 34 = 42 | mcq | jee-main-2021-online-27th-july-evening-shift |
1ktiq58td | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | cosec18$$^\circ$$ is a root of the equation : | [{"identifier": "A", "content": "x<sup>2</sup> + 2x $$-$$ 4 = 0"}, {"identifier": "B", "content": "4x<sup>2</sup> + 2x $$-$$ 1 = 0"}, {"identifier": "C", "content": "x<sup>2</sup> $$-$$ 2x + 4 = 0"}, {"identifier": "D", "content": "x<sup>2</sup> $$-$$ 2x $$-$$ 4 = 0"}] | ["D"] | null | $$\cos ec18^\circ = {1 \over {\sin 18^\circ }} = {4 \over {\sqrt 5 - 1}} = \sqrt 5 + 1$$<br><br>Let $$\cos ec18^\circ = x = \sqrt 5 + 1$$<br><br>$$ \Rightarrow x - 1 = \sqrt 5 $$<br><br>Squaring both sides, we get<br><br>$${x^2} - 2x + 1 = 5$$<br><br>$$ \Rightarrow {x^2} - 2x - 4 = 0$$ | mcq | jee-main-2021-online-31st-august-morning-shift |
1l55h39ss | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let f(x) be a quadratic polynomial such that f($$-$$2) + f(3) = 0. If one of the roots of f(x) = 0 is $$-$$1, then the sum of the roots of f(x) = 0 is equal to :</p> | [{"identifier": "A", "content": "$${{11} \\over 3}$$"}, {"identifier": "B", "content": "$${{7} \\over 3}$$"}, {"identifier": "C", "content": "$${{13} \\over 3}$$"}, {"identifier": "D", "content": "$${{14} \\over 3}$$"}] | ["A"] | null | <p>$$\because$$ x = $$-$$1 be the roots of f(x) = 0</p>
<p>$$\therefore$$ Let $$f(x) = A(x + 1)(x - 1)$$ ...... (i)</p>
<p>Now, $$f( - 2) + f(3) = 0$$</p>
<p>$$ \Rightarrow A[ - 1( - 2 - b) + 4(3 - b)] = 0$$</p>
<p>$$b = {{14} \over 3}$$</p>
<p>$$\therefore$$ Second root of f(x) = 0 will be $${{14} \over 3}$$</p>
<p>$$\therefore$$ Sum of roots $$ = {{14} \over 3} - 1 = {{11} \over 3}$$</p> | mcq | jee-main-2022-online-28th-june-evening-shift |
1l56rw4ul | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha$$, $$\beta$$ be the roots of the equation $${x^2} - 4\lambda x + 5 = 0$$ and $$\alpha$$, $$\gamma$$ be the roots of the equation $${x^2} - \left( {3\sqrt 2 + 2\sqrt 3 } \right)x + 7 + 3\lambda \sqrt 3 = 0$$, $$\lambda$$ > 0. If $$\beta + \gamma = 3\sqrt 2 $$, then $${(\alpha + 2\beta + \gamma )^2}$$ is equal to __________.</p> | [] | null | 98 | <p>$$\because$$ $$\alpha$$, $$\beta$$ are roots of x<sup>2</sup> $$-$$ 4$$\lambda$$x + 5 = 0</p>
<p>$$\therefore$$ $$\alpha$$ + $$\beta$$ = 4$$\lambda$$ and $$\alpha$$$$\beta$$ = 5</p>
<p>Also, $$\alpha$$, $$\gamma$$ are roots of</p>
<p>$${x^2} - (3\sqrt 2 + 2\sqrt 3 )x + 7 + 3\sqrt 3 \lambda = 0,\,\lambda > 0$$</p>
<p>$$\therefore$$ $$\alpha + \gamma = 3\sqrt 2 + 2\sqrt 3 $$, $$\alpha \gamma = 7 + 3\sqrt 3 \lambda $$</p>
<p>$$\because$$ $$\alpha$$ is common root</p>
<p>$$\therefore$$ $${\alpha ^2} - 4\lambda \,\,\alpha + 5 = 0$$ ....... (i)</p>
<p>and $${\alpha ^2} - (3\sqrt 2 + 2\sqrt 3 )\alpha + 7 + 3\sqrt 3 \lambda = 0$$ ...... (ii)</p>
<p>From (i) - (ii) : we get $$\alpha = {{2 + 3\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$$</p>
<p>$$\because$$ $$\beta + \gamma = 3\sqrt 2 $$</p>
<p>$$\therefore$$ $$4\lambda + 3\sqrt 2 + 2\sqrt 3 - 2\alpha = 3\sqrt 2 $$</p>
<p>$$ \Rightarrow 3\sqrt 2 = 4\lambda + 3\sqrt 2 + 2\sqrt 3 - {{4 + 6\sqrt 3 \lambda } \over {3\sqrt 2 + 2\sqrt 3 - 4\lambda }}$$</p>
<p>$$ \Rightarrow 8{\lambda ^2} + 3(\sqrt 3 + 2\sqrt 2 )\lambda - 4 - 3\sqrt 6 = 0$$</p>
<p>$$\therefore$$ $$\lambda = {{6\sqrt 2 - 3\sqrt 2 \pm \sqrt {9(11 - 4\sqrt 6 ) + 32(4 + 3\sqrt 6 )} } \over {16}}$$</p>
<p>$$\therefore$$ $$\lambda = \sqrt 2 $$</p>
<p>$$\therefore$$ $${(\alpha + 2\beta + \gamma )^2} = {(\alpha + \beta + \beta + \gamma )^2}$$</p>
<p>$$ = {(4\sqrt 2 + 3\sqrt 2 )^2}$$</p>
<p>$$ = {(7\sqrt 2 )^2} = 98$$</p> | integer | jee-main-2022-online-27th-june-evening-shift |
1l58advm7 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The sum of the cubes of all the roots of the equation <br/><br/>$${x^4} - 3{x^3} - 2{x^2} + 3x + 1 = 0$$ is _________.</p> | [] | null | 36 | <p>$${x^4} - 3{x^3} - {x^2} - {x^2} + 3x + 1 = 0$$</p>
<p>$$({x^2} - 1)({x^2} - 3x - 1) = 0$$</p>
<p>Let the root of $${x^2} - 3x - 1 = 0$$ be $$\alpha$$ and $$\beta$$ and other two roots of given equation are 1 and $$-$$1</p>
<p>So sum of cubes of roots</p>
<p>$$ = {1^3} + {( - 1)^3} + {\alpha ^3} + {\beta ^3}$$</p>
<p>$$ = {(\alpha + \beta )^3} - 3\alpha \beta (\alpha + \beta )$$</p>
<p>$$ = {(3)^3} - 3( - 1)(3)$$</p>
<p>$$ = 36$$</p> | integer | jee-main-2022-online-26th-june-morning-shift |
1l5c0zzpy | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If the sum of the squares of the reciprocals of the roots $$\alpha$$ and $$\beta$$ of <br/><br/>the equation 3x<sup>2</sup> + $$\lambda$$x $$-$$ 1 = 0 is 15, then 6($$\alpha$$<sup>3</sup> + $$\beta$$<sup>3</sup>)<sup>2</sup> is equal to :</p> | [{"identifier": "A", "content": "18"}, {"identifier": "B", "content": "24"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "96"}] | ["B"] | null | <p>$$3{x^2} + \lambda x - 1 = 0$$</p>
<p>Given, two roots are $$\alpha$$ and $$\beta$$.</p>
<p>$$\therefore$$ Sum of roots $$ = \alpha + \beta = {-\lambda \over 3}$$</p>
<p>And product of roots $$ = \alpha \beta = {-1 \over 3}$$</p>
<p>Given that,</p>
<p>Sum of square of reciprocal of roots $$\alpha$$ and $$\beta$$ is 15.</p>
<p>$$\therefore$$ $${1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} = 15$$</p>
<p>$$ \Rightarrow {{{\alpha ^2} + {\beta ^2}} \over {{\alpha ^2}{\beta ^2}}} = 15$$</p>
<p>$$ \Rightarrow {{{{(\alpha + \beta )}^2} - 2\alpha \beta } \over {{{(\alpha \beta )}^2}}} = 15$$</p>
<p>$$ \Rightarrow {{{{{\lambda ^2}} \over 9} + 2 \times {1 \over 3}} \over {{1 \over 9}}} = 15$$</p>
<p>$$ \Rightarrow {{{{{\lambda ^2} + 6} \over 9}} \over {{1 \over 9}}} = 15$$</p>
<p>$$ \Rightarrow {\lambda ^2} + 6 = 15$$</p>
<p>$$ \Rightarrow {\lambda ^2} = 9$$</p>
<p>Now, $$6{({\alpha ^3} + {\beta ^3})^2}$$</p>
<p>$$ = 6{\{ (\alpha + \beta )({\alpha ^2} + {\beta ^2} - \alpha \beta )\} ^2}$$</p>
<p>$$ = 6{(\alpha + \beta )^2}{\left[ {{{(\alpha + \beta )}^2} - 2\alpha \beta - \alpha \beta } \right]^2}$$</p>
<p>$$ = 6{\left( {{-\lambda \over 3}} \right)^2}{\left[ {{{\left( {{-\lambda \over 3}} \right)}^2} - 3\,.\,{-1 \over 3}} \right]^2}$$</p>
<p>$$ = 6 \times {{{\lambda ^2}} \over 9} \times \left[ {{{{\lambda ^2}} \over 9} + 1} \right]$$</p>
<p>$$ = 6 \times {{9} \over 9} \times {\left[ {{{9} \over 9} + 1} \right]^2}$$</p>
<p>$$ = 6 \times {\left( {{2}} \right)^2}$$</p>
<p>$$ = {{6 \times 4}} = 24$$</p> | mcq | jee-main-2022-online-24th-june-morning-shift |
1l6hxh4o8 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>The minimum value of the sum of the squares of the roots of $$x^{2}+(3-a) x+1=2 a$$ is:</p> | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "8"}] | ["C"] | null | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7nny14a/71f56c77-eb10-4874-b074-072a3d7a554d/cce478a0-2c7e-11ed-a18d-5933e4fde865/file-1l7nny14b.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7nny14a/71f56c77-eb10-4874-b074-072a3d7a554d/cce478a0-2c7e-11ed-a18d-5933e4fde865/file-1l7nny14b.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 26th July Evening Shift Mathematics - Quadratic Equation and Inequalities Question 47 English Explanation"></p>
<p>$$\alpha + \beta = a - 3,\,\alpha \beta = 1 - 2a$$</p>
<p>$$ \Rightarrow {\alpha ^2} + {\beta ^2} = {(a - 3)^2} - 2(1 - 2a)$$</p>
<p>$$ = {a^2} - 6a + 9 - 2 + 4a$$</p>
<p>$$ = {a^2} - 2a + 7$$</p>
<p>$$ = {(a - 1)^2} + 6$$</p>
<p>So, $${\alpha ^2} + {\beta ^2} \ge 6$$</p> | mcq | jee-main-2022-online-26th-july-evening-shift |
1l6kihzs6 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If $$\alpha, \beta$$ are the roots of the equation</p>
<p>$$
x^{2}-\left(5+3^{\sqrt{\log _{3} 5}}-5^{\sqrt{\log _{5} 3}}\right)x+3\left(3^{\left(\log _{3} 5\right)^{\frac{1}{3}}}-5^{\left(\log _{5} 3\right)^{\frac{2}{3}}}-1\right)=0
$$,</p>
<p>then the equation, whose roots are $$\alpha+\frac{1}{\beta}$$ and $$\beta+\frac{1}{\alpha}$$, is :</p> | [{"identifier": "A", "content": "$$3 x^{2}-20 x-12=0$$"}, {"identifier": "B", "content": "$$3 x^{2}-10 x-4=0$$"}, {"identifier": "C", "content": "$$3 x^{2}-10 x+2=0$$"}, {"identifier": "D", "content": "$$3 x^{2}-20 x+16=0$$"}] | ["B"] | null | <p>$${3^{\sqrt {{{\log }_3}5} }} - {5^{\sqrt {{{\log }_5}3} }} = {3^{\sqrt {{{\log }_3}5} }} - {\left( {{3^{{{\log }_3}5}}} \right)^{\sqrt {{{\log }_5}3} }}$$</p>
<p>$${3^{{{\left( {{{\log }_3}5} \right)}^{{1 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} - {5^{{{\left( {{{\log }_5}3} \right)}^{{2 \over 3}}}}} = 0$$</p>
<p>Note : In the given equation 'x' is missing.</p>
<p>So <img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7q9vxzm/482edcf8-ee96-45ef-83c7-507c2a3fb4dc/2cff2b20-2dee-11ed-a744-1fb8f3709cfa/file-1l7q9vxzn.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7q9vxzm/482edcf8-ee96-45ef-83c7-507c2a3fb4dc/2cff2b20-2dee-11ed-a744-1fb8f3709cfa/file-1l7q9vxzn.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 - Quadratic Equation and Inequalities Question 46 English Explanation"></p>
<p>$$\alpha + \beta + {1 \over \alpha } + {1 \over \beta } = (\alpha + \beta ) + {{\alpha + \beta } \over {\alpha \beta }}$$</p>
<p>$$ = 5 - {5 \over 3} = {{10} \over 3}$$</p>
<p>$$\left( {\alpha + {1 \over \beta }} \right)\left( {\beta + {1 \over \alpha }} \right) = 2 + \alpha \beta + {1 \over {\alpha \beta }} = 2 - 3 - {1 \over 3} = {{ - 4} \over 3}$$</p>
<p>So Equation must be option (B).</p> | mcq | jee-main-2022-online-27th-july-evening-shift |
1l6nl5cn4 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha$$, $$\beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+\sqrt{6}=0$$ and $$\frac{1}{\alpha^{2}}+1, \frac{1}{\beta^{2}}+1$$ be the roots of the equation $$x^{2}+a x+b=0$$. Then the roots of the equation $$x^{2}-(a+b-2) x+(a+b+2)=0$$ are :</p> | [{"identifier": "A", "content": "non-real complex numbers"}, {"identifier": "B", "content": "real and both negative"}, {"identifier": "C", "content": "real and both positive"}, {"identifier": "D", "content": "real and exactly one of them is positive"}] | ["B"] | null | <p>$$\alpha + \beta = \sqrt 2 $$, $$\alpha \beta = \sqrt 6 $$</p>
<p>$${1 \over {{\alpha ^2}}} + 1 + {1 \over {{\beta ^2}}} + 1 = 2 + {{{\alpha ^2} + {\beta ^2}} \over 6}$$</p>
<p>$$ = 2 + {{2 - 2\sqrt 6 } \over 6} = - a$$</p>
<p>$$\left( {{1 \over {{\alpha ^2}}} + 1} \right)\left( {{1 \over {{\beta ^2}}} + 1} \right) = 1 + {1 \over {{\alpha ^2}}} + {1 \over {{\beta ^2}}} + {1 \over {{\alpha ^2}{\beta ^2}}}$$</p>
<p>$$ = {7 \over 6} + {{2 - 2\sqrt 6 } \over 6} = b$$</p>
<p>$$ \Rightarrow a + b = {{ - 5} \over 6}$$</p>
<p>So, equation is $${x^2} + {{17x} \over 6} + {7 \over 6} = 0$$</p>
<p>OR $$6{x^2} + 17x + 7 = 0$$</p>
<p>Both roots of equation are $$-$$ve and distinct</p> | mcq | jee-main-2022-online-28th-july-evening-shift |
1l6p0or7p | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>If $$\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots+\frac{1}{(180-a)(200-a)}=\frac{1}{256}$$, then the maximum value of $$\mathrm{a}$$ is :</p> | [{"identifier": "A", "content": "198"}, {"identifier": "B", "content": "202"}, {"identifier": "C", "content": "212"}, {"identifier": "D", "content": "218"}] | ["C"] | null | <p>$${1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {40 - a}} + {1 \over {40 - a}} - {1 \over {60 - a}}\, + \,....\, + \,{1 \over {180 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$</p>
<p>$$ \Rightarrow {1 \over {20}}\left( {{1 \over {20 - a}} - {1 \over {200 - a}}} \right) = {1 \over {256}}$$</p>
<p>$$ \Rightarrow {1 \over {20}}\left( {{{180} \over {(20 - a)(200 - a)}}} \right) = {1 \over {256}}$$</p>
<p>$$ \Rightarrow (20 - a)(200 - a) = 9.256$$</p>
<p>OR $${a^2} - 220a + 1696 = 0$$</p>
<p>$$ \Rightarrow a = 212,\,8$$</p> | mcq | jee-main-2022-online-29th-july-morning-shift |
1l6rfi0r3 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta(\alpha>\beta)$$ be the roots of the quadratic equation $$x^{2}-x-4=0 .$$ If $$P_{n}=\alpha^{n}-\beta^{n}$$, $$n \in \mathrm{N}$$, then $$\frac{P_{15} P_{16}-P_{14} P_{16}-P_{15}^{2}+P_{14} P_{15}}{P_{13} P_{14}}$$ is equal to __________.</p> | [] | null | 16 | <p>$$\alpha$$ and $$\beta$$ are the roots of the quadratic equation $${x^2} - x - 4 = 0$$.</p>
<p>$$\therefore$$ $$\alpha$$ and $$\beta$$ are satisfy the given equation.</p>
<p>$${\alpha ^2} - \alpha - 4 = 0$$</p>
<p>$$ \Rightarrow {\alpha ^{n + 1}} - {\alpha ^n} - 4{\alpha ^{n - 1}} = 0$$ ...... (1)</p>
<p>and $${\beta ^2} - \beta - 4 = 0$$</p>
<p>$$ \Rightarrow {\beta ^{n + 1}} - {\beta ^n} - 4{\beta ^{n - 1}} = 0$$ ...... (2)</p>
<p> Subtracting (2) from (1), we get,</p>
<p>$$({\alpha ^{n + 1}} - {\beta ^{n + 1}}) - ({\alpha ^n} - {\beta ^n}) - 4({\alpha ^{n - 1}} - {\beta ^{n - 1}}) = 0$$</p>
<p>$$ \Rightarrow {P_{n + 1}} - {P_n} - 4{P_{n - 1}} = 0$$</p>
<p>$$ \Rightarrow {P_{n + 1}} = {P_n} + 4{P_{n - 1}}$$</p>
<p>$$ \Rightarrow {P_{n + 1}} - {P_n} = 4{P_{n - 1}}$$</p>
<p>For $$n = 14$$, $${P_{15}} - {P_{14}} = 4{P_{13}}$$</p>
<p>For $$n = 15$$, $${P_{16}} - {P_{15}} = 4{P_{14}}$$</p>
<p>Now, $${{{P_{15}}{P_{16}} - {P_{14}}{P_{16}} - P_{15}^2 + {P_{14}}{P_{15}}} \over {{P_{13}}{P_{14}}}}$$</p>
<p>$$ = {{{P_{16}}({P_{15}} - {P_{14}}) - {P_{15}}({P_{15}} - {P_{14}})} \over {{P_{13}}{P_{14}}}}$$</p>
<p>$$ = {{({P_{15}} - {P_{14}})({P_{16}} - {P_{15}})} \over {{P_{13}}{P_{14}}}}$$</p>
<p>$$ = {{(4{P_{13}})(4{P_{14}})} \over {{P_{13}}{P_{14}}}}$$</p>
<p>$$ = 16$$</p> | integer | jee-main-2022-online-29th-july-evening-shift |
1ldomtn4p | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$S = \left\{ {x:x \in \mathbb{R}\,\mathrm{and}\,{{(\sqrt 3 + \sqrt 2 )}^{{x^2} - 4}} + {{(\sqrt 3 - \sqrt 2 )}^{{x^2} - 4}} = 10} \right\}$$. Then $$n(S)$$ is equal to</p> | [{"identifier": "A", "content": "6"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "2"}] | ["B"] | null | Let $(\sqrt{3}+\sqrt{2})^{x^{2}-4}=t$
<br/><br/>$$
\begin{aligned}
& t+\frac{1}{t}=10 \\\\
\Rightarrow & t^{2}-10 t+1=0 \\\\
\Rightarrow & t=\frac{10 \pm \sqrt{100-4}}{2}=5 \pm 2 \sqrt{6}
\end{aligned}
$$
<br/><br/><b>Case-I</b>
<br/><br/>$$
\begin{aligned}
& t=5+2 \sqrt{6} = (\sqrt{3}+\sqrt{2})^{2} \\\\
\Rightarrow & (\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}+\sqrt{2})^{2} \\\\
\Rightarrow & x^{2}-4=2 \Rightarrow x^{2}=6 \Rightarrow x=\pm \sqrt{6}
\end{aligned}
$$
<br/><br/><b>Case-II :</b>
<br/><br/>$t=5-2 \sqrt{6}$ = $(\sqrt{3}-\sqrt{2})^{2}$
<br/><br/>$(\sqrt{3}+\sqrt{2})^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}$
<br/><br/>$\Rightarrow\left((\sqrt{3}-\sqrt{2})^{-1}\right)^{x^{2}-4}=(\sqrt{3}-\sqrt{2})^{2}$
<br/><br/>$\Rightarrow 4-x^{2}=2$
<br/><br/>$\Rightarrow x^{2}=2$
<br/><br/>$\Rightarrow x=\pm \sqrt{2}$ | mcq | jee-main-2023-online-1st-february-morning-shift |
1ldsgawzn | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha_1,\alpha_2,....,\alpha_7$$ be the roots of the equation $${x^7} + 3{x^5} - 13{x^3} - 15x = 0$$ and $$|{\alpha _1}| \ge |{\alpha _2}| \ge \,...\, \ge \,|{\alpha _7}|$$. Then $$\alpha_1\alpha_2-\alpha_3\alpha_4+\alpha_5\alpha_6$$ is equal to _________.</p> | [] | null | 9 | <p>$${x^7} + 3{x^5} - 13{x^3} - 15x = 0$$</p>
<p>$$x({x^6} + 3{x^4} - 13{x^2} - 15) = 0$$</p>
<p>$$x = 0 = {\alpha _7}$$</p>
<p>Let $${x^2} = t$$</p>
<p>$${t^3} + 3{t^2} - 13t - 15 = 0$$</p>
<p>$$(t + 1)(t + 5)(t - 3) = 0$$</p>
<p>$$t = {x^2} = - 1, - 5,3$$</p>
<p>$$x\, = \, \pm \,i, \pm \,\sqrt 5 i, \pm \,\sqrt 3 $$</p>
<p>$${\alpha _1},{\alpha _2} = \pm \,\sqrt 5 i,{\alpha _3},{\alpha _4} = \pm \,\sqrt 3 ,{\alpha _5},{\alpha _6} = \pm \,i$$</p>
<p>$${\alpha _1}{\alpha _2} - {\alpha _3}{\alpha _4} + {\alpha _5}{\alpha _6} = 5 + 3 + 1 = 9$$</p> | integer | jee-main-2023-online-29th-january-evening-shift |
1ldsu3cx2 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\lambda \ne 0$$ be a real number. Let $$\alpha,\beta$$ be the roots of the equation $$14{x^2} - 31x + 3\lambda = 0$$ and $$\alpha,\gamma$$ be the roots of the equation $$35{x^2} - 53x + 4\lambda = 0$$. Then $${{3\alpha } \over \beta }$$ and $${{4\alpha } \over \gamma }$$ are the roots of the equation</p> | [{"identifier": "A", "content": "$$7{x^2} - 245x + 250 = 0$$"}, {"identifier": "B", "content": "$$49{x^2} - 245x + 250 = 0$$"}, {"identifier": "C", "content": "$$49{x^2} + 245x + 250 = 0$$"}, {"identifier": "D", "content": "$$7{x^2} + 245x - 250 = 0$$"}] | ["B"] | null | $14 x^{2}-31 x+3 \lambda=0$
<br/><br/>
$$
\begin{aligned}
& \alpha+\beta=\frac{31}{14} \ldots .(1) \text { and } \alpha \beta=\frac{3 \lambda}{14}\quad...(2) \\\\
& 35 x^{2}-53 x+4 \lambda=0 \\\\
& \alpha+\gamma=\frac{53}{35} \ldots(3) \text { and } \alpha \gamma=\frac{4 \lambda}{35} \quad\ldots(4) \\\\
& \frac{(2)}{(4)} \Rightarrow \frac{\beta}{\gamma}=\frac{3 \times 35}{4 \times 14}=\frac{15}{8} \Rightarrow \beta=\frac{15}{8} \gamma
\end{aligned}
$$
<br/><br/>
(1) $-(3) \Rightarrow \beta-\gamma=\frac{31}{14}-\frac{53}{35}=\frac{155-106}{70}=\frac{7}{10}$
<br/><br/>
$\frac{15}{8} \gamma-\gamma=\frac{7}{10} \Rightarrow \gamma=\frac{4}{5}$
<br/><br/>
$\Rightarrow \beta=\frac{15}{8} \times \frac{4}{5}=\frac{3}{2}$
<br/><br/>
$\Rightarrow \alpha=\frac{31}{14}-\beta=\frac{31}{14}-\frac{3}{2}=\frac{5}{7}$
<br/><br/>
$\Rightarrow \lambda=\frac{14}{3} \alpha \beta=\frac{14}{3} \times \frac{5}{7} \times \frac{3}{2}=5$
<br/><br/>
so, sum of roots $\frac{3 \alpha}{\beta}+\frac{4 \alpha}{\gamma}=\left(\frac{3 \alpha \gamma+4 \alpha \beta}{\beta \gamma}\right)$<br/><br/>$$
\begin{aligned}
& =\frac{\left(3 \times \frac{4 \lambda}{35}+4 \times \frac{3 \lambda}{14}\right)}{\beta \gamma}=\frac{12 \lambda(14+35)}{14 \times 35 \beta \gamma} \\\\
& =\frac{49 \times 12 \times 5}{490 \times \frac{3}{2} \times \frac{4}{5}}=5
\end{aligned}
$$<br/><br/>
Product of roots
<br/><br/>
$$
=\frac{3 \alpha}{\beta} \times \frac{4 \alpha}{\gamma}=\frac{12 \alpha^{2}}{\beta \gamma}=\frac{12 \times \frac{25}{49}}{\frac{3}{2} \times \frac{4}{5}}=\frac{250}{49}
$$
<br/><br/>
So, required equation is $x^{2}-5 x+\frac{250}{49}=0$
<br/><br/>
$\Rightarrow 49 \mathrm{x}^{2}-245 \mathrm{x}+250=0$ | mcq | jee-main-2023-online-29th-january-morning-shift |
1ldu6c9z7 | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha \in\mathbb{R}$$ and let $$\alpha,\beta$$ be the roots of the equation $${x^2} + {60^{{1 \over 4}}}x + a = 0$$. If $${\alpha ^4} + {\beta ^4} = - 30$$, then the product of all possible values of $$a$$ is ____________.</p> | [] | null | 45 | $x^{2}+60^{\frac{1}{4}} x+a=0$
<br/><br/>
$$
\therefore \alpha+\beta=-60^{\frac{1}{4}}, \alpha \beta=a
$$
<br/><br/>
Now $\alpha^{4}+\beta^{4}=-30$
<br/><br/>
$\Rightarrow\left(\alpha^{2}+\beta^{2}\right)^{2}-2 a^{2}=-30$
<br/><br/>
$\Rightarrow\left[(\alpha+\beta)^{2}-2 a\right]^{2}-2 a^{2}=-30$
<br/><br/>
$\Rightarrow\left(60^{\frac{1}{2}}-2 a\right)^{2}-2 a^{2}=-30$
<br/><br/>
$\Rightarrow 60+4 a^{2}-4 \cdot 60^{\frac{1}{2}} a-2 a^{2}+30=0$
<br/><br/>
$\Rightarrow 2 a^{2}-8 \sqrt{15} a+90=0$
<br/><br/>
Product of value of $a=45$ | integer | jee-main-2023-online-25th-january-evening-shift |
1lgowfari | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be the roots of the equation $$x^{2}-\sqrt{2} x+2=0$$. Then $$\alpha^{14}+\beta^{14}$$ is equal to</p> | [{"identifier": "A", "content": "$$-64$$"}, {"identifier": "B", "content": "$$-64 \\sqrt{2}$$"}, {"identifier": "C", "content": "$$-128 \\sqrt{2}$$"}, {"identifier": "D", "content": "$$-128$$"}] | ["D"] | null |
<ol>
<li><p><strong>Find the roots of the quadratic equation:</strong></p>
<p>The quadratic formula is $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. For the quadratic equation $$x^2 - \sqrt{2}x + 2 = 0$$, we have $$a = 1, b = -\sqrt{2}, c = 2$$. Plugging these into the quadratic formula gives:
<br/><br/>$$x = \frac{\sqrt{2} \pm \sqrt{(-\sqrt{2})^2 - 4(1)(2)}}{2(1)} = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{6}i}{2}.$$
<br/><br/>So, we have two roots, $\alpha$ and $\beta$, which are:
<br/><br/>$$\alpha = \frac{\sqrt{2} + \sqrt{6}i}{2},$$
<br/><br/>$$\beta = \frac{\sqrt{2} - \sqrt{6}i}{2}.$$</p>
</li>
<li><p><strong>Express the roots in exponential form:</strong></p>
<p>We can express complex numbers in the form $$re^{i\theta}$$. For $\alpha$ and $\beta$, we find the magnitude $$r = \sqrt{2}$$ and the arguments $$\theta = \frac{\pi}{3}, -\frac{\pi}{3}$$ respectively. So, we have:
<br/><br/>$$\alpha = \sqrt{2}e^{i\frac{\pi}{3}},$$
<br/><br/>$$\beta = \sqrt{2}e^{-i\frac{\pi}{3}}.$$</p>
</li>
<li><p><strong>Calculate the 14th power of the roots:</strong></p>
<p>To find $$\alpha^{14}$$ and $$\beta^{14}$$, we use the property of exponents which says that $$(a^m)^n = a^{mn}$$. So, we have:
<br/><br/>$$\alpha^{14} = (\sqrt{2}e^{i\frac{\pi}{3}})^{14} = 2^7e^{i\frac{14\pi}{3}} = 128e^{i\frac{2\pi}{3}},$$
<br/><br/>$$\beta^{14} = (\sqrt{2}e^{-i\frac{\pi}{3}})^{14} = 2^7e^{-i\frac{14\pi}{3}} = 128e^{-i\frac{2\pi}{3}}.$$</p>
</li>
<li><p><strong>Add the 14th powers of the roots:</strong></p>
<p>We want to find the real part of $$\alpha^{14} + \beta^{14}$$. To do this, we use the property that $$e^{ix} = \cos(x) + i\sin(x)$$. We have:
<br/><br/>$$\alpha^{14} + \beta^{14} = 128e^{i\frac{2\pi}{3}} + 128e^{-i\frac{2\pi}{3}} = 128(2)\cos\left(\frac{2\pi}{3}\right) = -128.$$</p>
</li>
</ol>
| mcq | jee-main-2023-online-13th-april-evening-shift |
1lgrg43qz | maths | quadratic-equation-and-inequalities | relation-between-roots-and-coefficients | <p>Let $$\alpha, \beta$$ be the roots of the quadratic equation $$x^{2}+\sqrt{6} x+3=0$$. Then $$\frac{\alpha^{23}+\beta^{23}+\alpha^{14}+\beta^{14}}{\alpha^{15}+\beta^{15}+\alpha^{10}+\beta^{10}}$$ is equal to :</p> | [{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "729"}, {"identifier": "D", "content": "81"}] | ["D"] | null | Given quadratic equation: $$x^{2}+\sqrt{6} x+3=0$$
<br/><br/>We can find the roots using the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
<br/><br/>Here, $$a = 1$$, $$b = \sqrt{6}$$, and $$c = 3$$.
<br/><br/>Substituting these values into the quadratic formula, we have:
<br/><br/>$$x = \frac{-\sqrt{6} \pm \sqrt{(\sqrt{6})^2 - 4(1)(3)}}{2(1)}$$
<br/><br/>Simplifying further :
<br/><br/>$$x = \frac{-\sqrt{6} \pm \sqrt{6 - 12}}{2}$$
<br/><br/>$$x = \frac{-\sqrt{6} \pm \sqrt{-6}}{2}$$
<br/><br/>Since the discriminant is negative, the roots are complex numbers. We can express them using the imaginary unit $$i$$ :
<br/><br/>$$x = \frac{-\sqrt{6} \pm \sqrt{6}i}{2}$$
<br/><br/>$$ \Rightarrow $$ $$x = \frac{-1}{2}\sqrt{6} \pm \frac{1}{2}\sqrt{6}i$$
<br/><br/>$$ \therefore $$ $$\alpha, \beta=\sqrt{3} \mathrm{e}^{ \pm \frac{3 \pi \mathrm{i}}{4}}$$.
<br/><br/>The required expression can be rewritten in terms of the argument of the exponential form of the roots, which simplifies the calculation:
<br/><br/>$$
\begin{aligned}
& =\frac{(\sqrt{3})^{23}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{14}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{15}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{10}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\
\end{aligned}
$$
<br/><br/>Here, the exponential power of $$\sqrt{3}$$ in the numerator is larger by 8 compared to the denominator, so we can divide the numerator and denominator by $$(\sqrt{3})^{8} = 81$$ to simplify:
<br/><br/>$$
\begin{aligned}
& =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{69 \pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{42 \pi}{4}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{45 \pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{30 \pi}{4}\right)} \\\\
\end{aligned}
$$
<br/><br/>Since the cosine function has a period of $$2\pi$$, we can reduce the arguments of the cosine function in the numerator and denominator.
<br/><br/>We have $$69\pi/4 = \pi/4 + 17\pi = \pi/4$$,
<br/><br/>$$42\pi/4 = 2\pi/4 + 10\pi = \pi/2$$,
<br/><br/>$$45\pi/4 = \pi/4 + 11\pi = \pi/4$$, and
<br/><br/>$$30\pi/4 = 2\pi/4 + 7\pi = \pi/2$$.
<br/><br/>Therefore, the required expression simplifies to :
<br/><br/>$$
\begin{aligned}
& =\frac{(\sqrt{3})^{15}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{6}\left(2 \cos \frac{\pi}{2}\right)}{(\sqrt{3})^{7}\left(2 \cos \frac{\pi}{4}\right)+(\sqrt{3})^{2}\left(2 \cos \frac{\pi}{2}\right)} \\\\
& =\frac{(\sqrt{3})^{15}\sqrt{2}+(\sqrt{3})^{6}\cdot 0}{(\sqrt{3})^{7}\sqrt{2}+(\sqrt{3})^{2}\cdot 0} \\\\
& =\frac{(\sqrt{3})^{15}\sqrt{2}}{(\sqrt{3})^{7}\sqrt{2}} \\\\
& = (\sqrt{3})^{15-7} \\\\
& = (\sqrt{3})^{8} \\\\
& = 81.
\end{aligned}
$$ | mcq | jee-main-2023-online-12th-april-morning-shift |
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