For a real number λ, consider the improper integrals Iλ = ∫₀¹ dx/(1 − x)^λ and Kλ = ∫₁^∞ dx/x^λ. Which of the following statements are true?

A, B, D. Iλ converges iff λ 1. So they never converge together, and at λ = 1 neither does.

P integral thresholds — CSIR NET December 2023 Part C, solved

P integral thresholds

For a real number $\lambda$ , consider the improper integrals $I_\lambda = \int_{0}^{1}$ dx $/(1 - x)^\lambda$ and $K_\lambda = \int_{1}^\infty$ dx $/x^\lambda$ . Which of the following statements are true?

A.There exists $\lambda$ such that $I_\lambda$ converges, but $K_\lambda$ does not converge.✓ Correct
B.There exists $\lambda$ such that $K_\lambda$ converges, but $I_\lambda$ does not converge.✓ Correct
C.There exists $\lambda$ such that $I_\lambda, K_\lambda$ both converge.
D.There exists $\lambda$ such that neither $I_\lambda$ nor $K_\lambda$ converges.✓ Correct

The idea, in plain words

Both integrals stretch to infinity in some sense, but at opposite ends: one has trouble where the function blows up at a point, the other where the region runs off forever. Raising the power helps one and hurts the other, so they can never both behave at once.

Solution

$I_\lambda$ converges iff $\lambda < 1; K_\lambda$ converges iff $\lambda > 1$ .

So they never converge together, and at $\lambda = 1$ neither does.

Why each option is right or wrong

Checked against 5 symbolic computations

A.Correct
True.

$I_\lambda$ has its singularity at the endpoint x = 1 and converges exactly when $\lambda < 1; K_\lambda$ runs to infinity and converges exactly when $\lambda > 1$ .

Take $\lambda = 1/2$ : SymPy evaluates $I_{1/2}$ = 2

while $K_{1/2}$ diverges.

The two thresholds point in opposite directions — that is the whole content of the question.
B.Correct
True.

Take $\lambda = 2$ : SymPy gives K_2 = 1, and I_2 diverges (verified as $\infty)$ .

Away from the origin a large exponent helps convergence; at a singularity it hurts.
C.WrongBoundary and endpoint
False, and it is the option the question is built to catch.

Both converging would need $\lambda < 1$ and $\lambda > 1$ at once.

The two conditions are not merely different, they are complementary — there is no overlap window, and the single point $\lambda = 1$ they share is where both diverge.
D.Correct
True, at $\lambda = 1$ exactly.

$K_1 = \int_{1}^\infty$ dx/x diverges (verified as $\infty)$ and $I_1 = \int_{0}^{1}$ dx/(1 − x) diverges too — both are the logarithm, from the two ends.

The borderline exponent fails on both sides.

1 of these 1 wrong options correspond to a named reasoning error — the same errors recur across subjects.

The trap it tests

Boundary and endpoint

The statement turns at the edge of the interval, the domain, or the parameter range.

Drill statements like this

Last revised 12 Sept 2026. Found a mistake? Tell us — corrections are the fastest thing we act on.

P integral thresholds

More on this topic