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Step-by-Step Solution
Step 1: Write down the given quadratic equation
The quadratic equation is
$$(\lambda^2 + 1)x^2 - 4\lambda x + 2 = 0.$$
Step 2: Identify the nature of the quadratic parabola
The coefficient of $x^2$ is $\lambda^2 + 1$, which is always positive for any real $\lambda$.
Therefore, the parabola always opens upwards.
Step 3: Interpret the condition “exactly one root in (0, 1)”
For an upward-opening parabola to have exactly one root between 0 and 1, the function
$$f(x) = (\lambda^2 + 1)x^2 - 4\lambda x + 2$$
at $x=0$ and $x=1$ must be of opposite signs or one of them be zero while ensuring that only one intersection point lies strictly in (0, 1).
Step 4: Evaluate $f(0)$ and $f(1)$
$f(0) = 2.$
$f(1) = (\lambda^2 + 1)\cdot 1^2 - 4\lambda \cdot 1 + 2
= \lambda^2 + 1 - 4\lambda + 2
= \lambda^2 - 4\lambda + 3.$
Step 5: Sign change condition for exactly one root in (0, 1)
For the parabola to intersect (0, 1) exactly once, one basic requirement is
$$f(0) \cdot f(1) \le 0.$$
We already have $f(0) = 2$. So,
$$2 \cdot (\lambda^2 - 4\lambda + 3) \le 0 \quad \implies \quad \lambda^2 - 4\lambda + 3 \le 0.$$
Factorizing:
$$(\lambda - 1)(\lambda - 3) \le 0 \quad \implies \quad \lambda \in [1, 3].$$
Step 6: Check the boundary points $\lambda = 1$ and $\lambda = 3$
At $\lambda = 1$:
The equation becomes
$$ (\,1^2 + 1\,)x^2 - 4 \cdot 1 \cdot x + 2 = 0 \quad \implies \quad 2x^2 - 4x + 2 = 0.$$
This simplifies to
$$ 2(x^2 - 2x + 1) = 2(x - 1)^2 = 0.$$
The only root is $x = 1$, and it is a repeated root.
However, $x=1$ does not lie strictly within (0,1).
Thus, $\lambda = 1$ is not included.
At $\lambda = 3$:
The equation becomes
$$ (\,3^2 + 1\,)x^2 - 4 \cdot 3 \cdot x + 2 = 0 \quad \implies \quad 10x^2 - 12x + 2 = 0.$$
Factor out 2:
$$ 2(5x^2 - 6x + 1) = 0.$$
Factorizing the inner quadratic:
$$ 5x^2 - 6x + 1 = (5x - 1)(x - 1) = 0,$$
so the roots are
$$ x = \frac{1}{5} \quad\text{and}\quad x = 1.$$
Notice $\frac{1}{5}$ lies in the interval (0, 1), while 1 is at the boundary. Therefore, exactly one root lies in (0,1).
Step 7: Conclusion
From the above, the suitable interval for $\lambda$ that ensures exactly one root in (0, 1) is
$$(1, 3].$$
Hence, the correct answer is
$$\lambda \in (1, 3].$$
