JEE Advanced2026MathematicsDifferentiationActual
Let R denote the set of all real numbers. Consider the polynomial function f: R R defined by f(x) = d¹⁰ dx¹⁰ ((x^2 - 1)¹⁰ ) , for all x R . Here d¹⁰ dx¹⁰ ((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 ?
Options
- AThe coefficient of x^8 in the polynomial f(x) is (-10) ( 18! 8! )
- BThe value of f(1) + f(-1) is equal to 10! , 2¹¹
- CThe degree of the polynomial f(x) is 10
- DThe constant term of the polynomial f(x) is - ( 10! 5! )
Correct answer
A. The coefficient of x^8 in the polynomial f(x) is (-10) ( 18! 8! )
Step-by-step solution
Given f(x) = d¹⁰ dx¹⁰ ((x^2 - 1)¹⁰ ) . Using the binomial expansion, we have: (x^2 - 1)¹⁰ = _ k=0 ¹⁰ (-1)^k , ¹⁰C_ k x^ 20-2k Differentiating 10 times with respect to x , we get: f(x) = _ k=0 ⁵ (-1)^k , ¹⁰C_ k (20-2k)! (10-2k)! x^ 10-2k For the coefficient of x^8 , we set 10 - 2k = 8 k = 1 . The coefficient is (-1)^1 , ¹⁰C₁ 18! 8! = -10 ( 18! 8! ) . Thus, statement (A) is true. The highest power of x in f(x) corresponds to k = 0 , which gives x¹⁰ with a non-zero coefficient of 20! 10! . Therefore, the degree of the