JEE Advanced 2026 Paper 2, Mathematics Section 2 Q1: Introduction to Binomial Expansion

JEE Advanced2026Paper 2Mathematics Section 2

Let $R$ denote the set of all real numbers. Consider the polynomial function $f : R \to R$ defined by

$f (x) = \frac{d ^{10}}{d x ^{10}} ((x^{2} - 1)^{10})$ , for all $x \in R$ .

Here $\frac{d ^{10}}{d x ^{10}} ((x^{2} - 1)^{10})$ is the 10th order derivative of the function $(x^{2} - 1)^{10}$ .

Then which of the following statements is (are) TRUE?

(A) The coefficient of $x^{8}$ in the polynomial $f (x)$ is $(- 10) (\frac{18 !}{8 !})$

(B) The value of $f (1) + f (- 1)$ is equal to $10! 2^{11}$

(D) The constant term of the polynomial $f (x)$ is $- (\frac{10 !}{5 !})$

A
The coefficient of $x^{8}$ in the polynomial $f (x)$ is $(- 10) (\frac{18 !}{8 !})$
B
The value of $f (1) + f (- 1)$ is equal to $10! 2^{11}$
C
The degree of the polynomial $f (x)$ is $10$
D
The constant term of the polynomial $f (x)$ is $- (\frac{10 !}{5 !})$

Solution

Let $P (x) = (x^{2} - 1)^{10}$ . Its binomial expansion is $P (x) = r = 0 \sum 10 (r 10) (- 1)^{r} x^{20 - 2 r}$ , which is a polynomial of degree $20$ .

$f (x) = P^{(10)} (x)$ , the 10th derivative, so $de g f = 20 - 10 = 10$ . Option (C) is correct.

Coefficient of $x^{8}$ in $f$ : we need the term where $20 - 2 r - 10 = 8$ , i.e. $r = 1$ . The contribution from $(1 10) (- 1)^{1} x^{18}$ after ten differentiations is $- (1 10) \cdot 18 \cdot 17 \cdot 16 \dots 9 \cdot x^{8} = - 10 \cdot \frac{18 !}{8 !} x^{8}$ . Option (A) is correct.

Constant term of $f$ : we need $20 - 2 r - 10 = 0$ , i.e. $r = 5$ . The contribution is $(5 10) (- 1)^{5} \cdot 10 \cdot 9 \cdot 8 \dots 1 = - (5 10) \cdot 10!$ , which is not $- \frac{10 !}{5 !}$ . Option (D) is incorrect.

For (B): write $P (x) = (x - 1)^{10} (x + 1)^{10}$ . By Leibniz's rule, $P^{(10)} (1) = k = 0 \sum 10 (k 10) [(x - 1)^{10}]^{(k)} [(x + 1)^{10}]^{(10 - k)}_{x = 1}$ . Only $k = 10$ contributes at $x = 1$ , giving $10! \cdot 2^{10}$ . Similarly $P^{(10)} (- 1) = 10! \cdot 2^{10}$ . Hence $f (1) + f (- 1) = 2 \cdot 10! \cdot 2^{10} = 10! \cdot 2^{11}$ . Option (B) is correct.

$∴$ Correct options: (A), (B), (C).

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JEE Advanced 2026 Paper 2, Mathematics Section 2 Q1: Introduction to Binomial Expansion

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