Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
For the second part $J$, substitute $u = \tan\theta$ and use integration by parts to get $J = \frac{1}{2\ln 2}[u\ln(1+u^2) - 2u + 2\arctan(u)]_0^{\sqrt{3}} = \sqrt{3} - \frac{\sqrt{3}}{\ln 2} + \frac{\pi}{3\ln 2}$
If $\alpha = 1$ and $\beta = 1 + i\sqrt{2}$, where $i = \sqrt{-1}$ are two roots of the equation
$x^3 + ax^2 + bx + c = 0$, $a, b, c \in \mathbb{R}$, then $\int_{-1}^{1} (x^3 + ax^2 + bx + c)dx$ is equal to:
(1) $-2$
(2) $-4$
(3) $-8$
(4) $-10$
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Answer: Option (3)
Key idea: Complex roots and Vieta's formulas for cubic polynomials with real coefficients
Since the polynomial has real coefficients and $\beta = 1 + i\sqrt{2}$ is a root, the complex conjugate $\gamma = 1 - i\sqrt{2}$ must also be a root.
Using Vieta's formulas: Sum of roots = $1 + (1+i\sqrt{2}) + (1-i\sqrt{2}) = 3 = -a$, so $a = -3$
Sum of products of pairs: $\alpha\beta + \beta\gamma + \gamma\alpha = (1+i\sqrt{2}) + 3 + (1-i\sqrt{2}) = 5 = b$
Product of roots: $\alpha\beta\gamma = 1 \cdot (1+i\sqrt{2})(1-i\sqrt{2}) = 1 \cdot 3 = 3 = -c$, so $c = -3$
Key idea: Inverse trigonometric functions, Integration by parts, Trigonometric identities
Step 1: Use the identity $\cot^{-1}(y) = \tan^{-1}(1/y)$ to write $\cot^{-1}(1+x+x^2) = \tan^{-1}\left(\frac{1}{1+x+x^2}\right)$
Step 2: Apply the formula $\tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a-b}{1+ab}\right)$ with $a=1+x, b=x$ to get $\cot^{-1}(1+x+x^2) = \tan^{-1}(1+x) - \tan^{-1}(x)$
Let $(2^{1-a} + 2^{1+a})$, $f(a)$, $(3^a + 3^{-a})$ be in A.P. and $\alpha$ be the minimum value of $f(a)$. Then the value of the integral $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{(e^{2x} - e^{-2x})}$ is :
(1) $\frac{1}{2} \log_e\left(\frac{4}{3}\right)$
(2) $\frac{1}{4} \log_e\left(\frac{4}{3}\right)$
(3) $\frac{1}{2} \log_e\left(\frac{8}{5}\right)$
(4) $\frac{1}{4} \log_e\left(\frac{8}{5}\right)$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Let $f: \mathbf{R} \to \mathbf{R}$ be such that $f(xy) = f(x)f(y)$, for all $x, y \in \mathbf{R}$ and $f(0) \neq 0$. Let $g: [1, \infty) \to \mathbf{R}$ be a differentiable function such that
$x^2 g(x) = \int\limits_1^x (t^2 f(t) - tg(t)) dt$.
Then $g(2)$ is equal to:
(1) $\frac{13}{8}$
(2) $\frac{11}{16}$
(3) $\frac{15}{32}$
(4) $\frac{17}{64}$
Show answer and solution
Answer: Option (3)
Key idea: Functional equations and first-order linear differential equations
Step 1: From $f(xy) = f(x)f(y)$ and $f(0) \neq 0$, set $x=y=0$ to get $f(0) = [f(0)]^2$, so $f(0) = 1$. Then $f(0) = f(x \cdot 0) = f(x)f(0)$ gives $f(x) = 1$ for all $x$.
Step 2: Substitute $f(t) = 1$ into the equation: $x^2 g(x) = \int_1^x (t^2 - tg(t)) dt$
Step 3: Differentiate both sides using Leibniz rule: $2xg(x) + x^2g'(x) = x^2 - xg(x)$
Step 4: Simplify to get the linear ODE: $g'(x) + \frac{3}{x}g(x) = 1$
Step 5: Solve using integrating factor $\mu(x) = x^3$: $\frac{d}{dx}[x^3g(x)] = x^3$, so $g(x) = \frac{x}{4} + \frac{C}{x^3}$
Step 6: Use initial condition $g(1) = 0$ to find $C = -\frac{1}{4}$, giving $g(x) = \frac{x}{4} - \frac{1}{4x^3}$
Let $f:(1,\infty)\to\mathbb{R}$ be a function defined as $f(x)=\frac{x-1}{x+1}$. Let $f^{i+1}(x)=f(f^i(x))$, $i=1, 2,...,25$, where $f^1(x)=f(x)$. If $g(x)+f^{26}(x)=0$, $x\in(1,\infty)$, then the area of the region bounded by the curves $y=g(x)$, $2y=2x-3$, $y=0$ and $x=4$ is:
(1) $\frac{1}{8}+\log_e 2$
(2) $\frac{1}{4}+\log_e 2$
(3) $\frac{5}{6}+3\log_e 2$
(4) $\frac{5}{6}+\log_e 2$
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
If $\int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \left[\cot\left(x - \frac{\pi}{3}\right)\cot\left(x + \frac{\pi}{3}\right) + 1\right]dx = \alpha \log_e(\sqrt{3}-1)$, then $9\alpha^2$ is equal to __________.
Show answer and solution
Answer: 12
Key idea: Definite Integration with Trigonometric Identities and Substitution
Step 1: Simplify $\cot(x - \frac{\pi}{3})\cot(x + \frac{\pi}{3}) + 1$ using product formulas to get $\frac{2}{2\cos(2x) + 1}$
Step 2: Use $\cos(2x) = 1 - 2\sin^2(x)$ to rewrite as $\frac{2\sec^2(x)}{3 - \tan^2(x)}$
Step 3: Substitute $u = \tan(x)$, so $du = \sec^2(x)dx$. Limits: $u = \frac{1}{\sqrt{3}}$ to $u = 1$
Step 4: Use partial fractions: $\frac{2}{3-u^2} = \frac{1}{\sqrt{3}}\left[\frac{1}{\sqrt{3}-u} + \frac{1}{\sqrt{3}+u}\right]$
Step 5: Integrate to get $\frac{1}{\sqrt{3}}\ln\left|\frac{\sqrt{3}+u}{\sqrt{3}-u}\right|$ and evaluate from $\frac{1}{\sqrt{3}}$ to $1$
Step 6: Simplify using $\sqrt{3}+1 = \frac{2}{\sqrt{3}-1}$ to get $I = -\frac{2}{\sqrt{3}}\ln(\sqrt{3}-1) = -\frac{2\sqrt{3}}{3}\ln(\sqrt{3}-1)$
Step 7: Compare with $\alpha\log_e(\sqrt{3}-1)$ to find $\alpha = -\frac{2\sqrt{3}}{3}$, so $9\alpha^2 = 9 \cdot \frac{4 \cdot 3}{9} = 12$
FAQs
Frequently asked questions
How many questions come from Definite Integration in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 14 questions on it, about 1.6 per shift.
Where can I practise Definite Integration PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
