The set of all real values of \[\lambda \] for which the
quadratic equations,
(\[\lambda \]2
+ 1)x2
-- 4\[\lambda \]x + 2 = 0
always have exactly one root in the interval
(0, 1) is :
Solutions · 1
Official solution
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⇒(x−1)2 = 0 \Rightarrow \[ x = 1, 1
As both roots can't lie between (0, 1)
So, \]\lambda = 1 can't be possible.
Atλ= 3 :
10x2−12x + 2 = 0⇒5x2−6x + 1 = 0⇒(5x−1) (x−1) = 0⇒x = 1,{1 \over 5}\[
In the interval (0, 1) exactly one root \]{1 \over 5}present.∴λ∈