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]