If the sum of the square of the roots of the equation x^2-( α-2) x-(1+ α)=0 is least, then α is equal to

Correct answer: D. (π)/(2) Given equation is x²-( α- 2) x-(1+ α)= 0 Let x_1 and x_2 be two roots of quadratic equation. aligned & x_1+x_2= α-2 and x_1 x_2=-(1+ α) & (x_1+x_2)^2=( α-2)^2= ^2 α+4-4 α aligned aligned & x_1^2+x_2^2= ^2 α+4-4 α-2 x_1 x_2 & = ^2 α+4-4 α+2(1+ α) & = ^2 α-2 α+6 aligned Now, By putting α= (π)/(6) , α= (π)/(4) , α= (π)/(3) and α= (π)/(2) in (A) one by one We get least value ofx_1^2+x_2^2 at (π)/(2) Hence, α= (π)/(2)

JEE Main 201212 MayOnline

Exam: JEE Main
Year: 2012
Date: 12 May
Mode: Online
Subject: Mathematics
Chapter: Quadratic Equation

JEE Main 2012 · 12 May · Online · Mathematics · Quadratic Equation

If the sum of the square of the roots of the equation x^2-( α-2)...

If the sum of the square of the roots of the equation

x^2-(\sin \alpha-2) x-(1+\sin \alpha)=0

is least, then

\alpha

is equal to

Options

A $\frac{\pi}{6}$
B $\frac{\pi}{4}$
C $\frac{\pi}{3}$
D $\frac{\pi}{2}$

Correct answer

D. $\frac{\pi}{2}$

Step-by-step solution

Given equation is x²-(\sin

\alpha-

2) x-(1+\sin

\alpha)=

0 Let

x_1 and x_2

be two roots of quadratic equation.

\begin{aligned}& \therefore x_1+x_2=\sin \alpha-2 \text{\text{ and } }x_1 x_2=-(1+\sin\alpha)\\ & \left(x_1+x_2\right)^2=(\sin \alpha-2)^2=\sin ^2 \alpha+4-4 \sin \alpha \end{aligned}

\begin{aligned}& \Rightarrow \quad x_1^2+x_2^2=\sin ^2 \alpha+4-4 \sin \alpha-2 x_1 x_2 \\ & =\sin ^2 \alpha+4-4 \sin \alpha+2(1+\sin \alpha)\\ & =\sin ^2 \alpha-2 \sin \alpha+6 \end{aligned}

Now, By putting

\alpha=

\frac{\pi}{6}

, \alpha=

\frac{\pi}{4}

, \alpha=

\frac{\pi}{3} and

\alpha=

\frac{\pi}{2}

in (A) one by one We get least value ofx_1^2+x_2^2

at \frac{\pi}{2} Hence,

\alpha=

\frac{\pi}{2}$

Practice Quadratic Equation on Quantrex Academy →

More from Quadratic Equation