We need the values of $\lambda$ for which the quadratic

has exactly one root in $(0,1)$.

Let

Since $\lambda^2+1>0$ for all real $\lambda$, this is always a quadratic opening upward.

1. First check whether the equation has real roots

For real roots, discriminant must be nonnegative:

So real roots exist when

2. Use Vieta’s relations

If roots are $\alpha,\beta$, then

Since the product is positive, the two roots have the same sign.

We want exactly one root in $(0,1)$.

Because both roots have the same sign, they cannot be one positive and one negative. So the only possibility is that both are positive, with one lying in $(0,1)$ and the other at least $1$.

Thus we need positive roots, which requires

Combined with $|\lambda|\ge 1$, we get

3. Condition for exactly one root in $(0,1)$

For an upward opening parabola with two positive roots, having exactly one root in $(0,1)$ means that $x=0$ and $x=1$ lie on the same side or opposite side? Let's check carefully:

• $f(0)=2>0$ always.
• If exactly one root lies in $(0,1)$ and the other is $>1$, then $f(1)<0$ (since between the two roots the parabola is negative).
• If a root is exactly at $1$, then there is still exactly one root in $(0,1)$ provided the other root is in $(0,1)$.

So we need

with real roots and positivity already ensured.

Now

Thus

Let us verify endpoints:

At $\lambda=1$

Equation becomes

Root is $x=1$, which is not in $(0,1)$. So $\lambda=1$ is not allowed.

At $\lambda=3$

Equation becomes

Roots are

Exactly one root, $\frac15$, lies in $(0,1)$. So $\lambda=3$ is allowed.

Hence the required set is

4. Check options

• A: $(-3,-1)$ — no, roots would have negative sum, so not suitable.
• B: $(2,4]$ — includes values $>3$, where both roots are $<1$ or no longer exactly one in $(0,1)$.
• C: $(0,2)$ — includes $(0,1)$ where roots are not real, and excludes $3$.
• D: $(1,3]$ — correct.

Final Answer

So the correct option is D.