JEE Main 2025 — 24 January, Shift 1. Previous Year Question.
Problem
If $I(m,n) = \displaystyle\int_0^1 x^{m-1}(1-x)^{n-1}\,dx$, where $m,n$ are positive, then $I(9,14)+I(10,13)$ is equal to:
Key insight. $I(m,n)$ is the Beta function in disguise, but knowing that isn’t necessary here. Since the question only asks to identify which option the sum matches, substituting the given values directly and adding the two integrands algebraically is faster than evaluating any integral.
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Approach
Rather than trying to compute $I(9,14)$ and $I(10,13)$ as actual numbers — which would mean evaluating two nontrivial definite integrals — it’s far quicker to substitute the given values directly into the general form $x^{m-1}(1-x)^{n-1}$, add the two resulting integrands, and simplify. If the answer choices are themselves written in $I(m,n)$ form, the simplified sum will match one of them exactly, without ever computing a numerical value.
Solution
Step 1 — Write out both integrals explicitly
$$I(9,14) = \int_0^1 x^{8}(1-x)^{13}\,dx, \qquad I(10,13) = \int_0^1 x^{9}(1-x)^{12}\,dx$$
Step 2 — Add the two integrands
$$I(9,14)+I(10,13) = \int_0^1 \Big[x^8(1-x)^{13} + x^9(1-x)^{12}\Big]\,dx$$
Step 3 — Factor out the common terms
Both terms share $x^8(1-x)^{12}$:
$$x^8(1-x)^{13} + x^9(1-x)^{12} = x^8(1-x)^{12}\Big[(1-x) + x\Big] = x^8(1-x)^{12}\cdot 1 = x^8(1-x)^{12}$$
Step 4 — Match against the definition of $I(m,n)$
$$\int_0^1 x^8(1-x)^{12}\,dx$$
Comparing with $I(m,n) = \int_0^1 x^{m-1}(1-x)^{n-1}\,dx$: matching $m-1=8$ gives $m=9$, and matching $n-1=12$ gives $n=13$.
Answer
$$I(9,14)+I(10,13) = I(9,13)$$
Common mistakes
- Trying to evaluate the Beta function formula $I(m,n) = \dfrac{(m-1)!(n-1)!}{(m+n-1)!}$ for each term separately and comparing decimal values against the options — this works in principle but takes far longer than the direct algebraic simplification.
- Forgetting that $(1-x)+x=1$ is the whole trick — without spotting this cancellation, the two terms look unrelated rather than combinable.
Practise next
- Simplify $I(5,8) + I(6,7)$ the same way, by writing both integrands explicitly and factoring out their common power.
Show answer
$I(5,7)=\dfrac{1}{2310}$. Writing both integrands over $[0,1]$ and factoring out the common power,
$$I(m+1,n)+I(m,n+1)=\int_0^1 x^{m-1}(1-x)^{n-1}\bigl[x+(1-x)\bigr]dx=I(m,n).$$
With $m=5,\ n=7$ the left side is exactly $I(6,7)+I(5,8)$, so the sum collapses to $I(5,7)=\dfrac{4!\,6!}{11!}=\dfrac{1}{2310}$.
Doubts are answered by Shiwam, usually within a day. Ask about this question specifically — a general question about the chapter is better asked in class.