The set of all real values of $\lambda $ for which thequadratic equations,
($\lambda $² + 1)x² – 4$\lambda $x + 2 = 0 always have exactly one root in the interval (0, 1) is :

Given quadratic:
$ (\lambda^2 + 1)x^2 - 4\lambda x + 2 = 0 $

Let $ f(x) = (\lambda^2 + 1)x^2 - 4\lambda x + 2 $

We need exactly one root in $(0,1)$

Condition:

For exactly one root in $(0,1)$:

$ f(0)\cdot f(1) < 0 $

Step 1: Compute values

$ f(0) = 2 > 0 $

$ f(1) = (\lambda^2 + 1) - 4\lambda + 2 = \lambda^2 - 4\lambda + 3 $

$ = (\lambda - 1)(\lambda - 3) $

Step 2: Sign condition

$ f(0)\cdot f(1) < 0 \Rightarrow 2 \cdot (\lambda - 1)(\lambda - 3) < 0 $

$ (\lambda - 1)(\lambda - 3) < 0 $

Step 3: Solve

$ 1 < \lambda < 3 $

Step 4: Check endpoints

At $ \lambda = 1 $:
$ f(1) = 0 \Rightarrow x = 1 $ (not in open interval)

At $ \lambda = 3 $:
$ f(1) = 0 \Rightarrow x = 1 $ (not in open interval)

So endpoints excluded.