QJEEM20Mae6aa: JEE Main 2020 Mathematics - Class 11 JEEM20Mae6aa Solution: The set of all real values of for which the quadratic…

Given quadratic equation,

$({\lambda ^2} + 1){x^2} - 4\lambda x + 2 = 0$

Here coefficient of x² is ( $\lambda$ ² + 1) which is always positive. So quadratic equation is upward parabola.



So, either f(0) < 0 and f(1) > 1

or f(0) > 0 and f(1) < 0

$\therefore$ In both those cases,

f(0) f(1) $\le$ 0

$\Rightarrow$ 2( $\lambda$ ² $-$ 4 $\lambda$ + 3) $\le$ 0

$\Rightarrow$ $\lambda$ $\in$ [1, 3]

At $\lambda$ = 1 :

Quadratic equation becomes

2x² $-$ 4x + 2 = 0

$\Rightarrow$ (x $-$ 1)² = 0

$\Rightarrow$ x = 1, 1

As both roots can't lie between (0, 1)

So, $\lambda$ = 1 can't be possible.

At $\lambda$ = 3 :

10x² $-$ 12x + 2 = 0

$\Rightarrow$ 5x² $-$ 6x + 1 = 0

$\Rightarrow$ (5x $-$ 1) (x $-$ 1) = 0

$\Rightarrow$ x = 1, ${1 \over 5}$

In the interval (0, 1) exactly one root ${1 \over 5}$ present.

$\therefore$ $\lambda$ $\in$ (1, 3]