JEE Advanced 2026 · Paper 2 · Mathematics Q5

← JEE Advanced 2026 · Paper 2

MathematicsMultiple correct+4 / -1Question 5 of 54

Let $\mathbb R$ denote the set of all real numbers. Consider the polynomial function $f:\mathbb R\to\mathbb R$ defined by
$f(x)=\frac{d^{10}}{dx^{10}}\left(\left(x^2-1\right)^{10}\right),\qquad\text{for all }x\in\mathbb R.$
Here $\dfrac{d^{10}}{dx^{10}}\left(\left(x^2-1\right)^{10}\right)$ is the $10^{\text{th}}$ order derivative of the function $\left(x^2-1\right)^{10}$ .

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

The coefficient of

x^8

in the polynomial

f(x)

(-10)\left(\dfrac{18!}{8!}\right)

The value of

f(1)+f(-1)

is equal to

10!\,2^{11}

The degree of the polynomial

f(x)

is 10

The constant term of the polynomial

f(x)

-\left(\dfrac{10!}{5!}\right)

Answer and worked solution

Official answer: A, B, C

Expand. $(x^2-1)^{10}=\sum_{k=0}^{10}\binom{10}k(-1)^{10-k}x^{2k}$ . The 10th derivative keeps the terms with $2k\ge10$ :
$f(x)=\sum_{k=5}^{10}\binom{10}k(-1)^k\frac{(2k)!}{(2k-10)!}\,x^{2k-10}.$

(C) $k=10$ gives $\frac{20!}{10!}x^{10}\ne0$ : degree 10. True.
(A) $x^8$ comes from $k=9$ : $\binom{10}9(-1)^9\frac{18!}{8!}=(-10)\frac{18!}{8!}$ . True.
(D) The constant term ( $k=5$ ) is $\binom{10}5(-1)^5\,10!=-252\cdot10!$ , not $-\frac{10!}{5!}$ . False.
(B) By Rodrigues' formula $f=2^{10}\,10!\,P_{10}(x)$ , and $P_{10}(\pm1)=1$ . So $f(1)+f(-1)=2\cdot2^{10}\,10!=10!\,2^{11}$ . True.

Topics:Higher-order derivatives Binomial expansion

← Previous question Next question →

JEE Advanced 2026 · Paper 2, Mathematics question 5