Binomial Theorem · Question 1

Let R denote the set of all real numbers. Consider the polynomial function f: R R defined by f(x)= d10dx10 ((x2-1)10 ), for all x R. Here d10dx10 ((x2-1)10 ) is the 10th order derivative of the function (x2-1)10. Then which of the following statements is (are) TRUE ?

1. The coefficient of x8 in the polynomial f(x) is (-10) ( 18!8! )
2. The value of f(1)+f(-1) is equal to 10! 211
3. The degree of the polynomial f(x) is 10
4. The constant term of the polynomial f(x) is - ( 10!5! )

Answer

(1) The coefficient of x8 in the polynomial f(x) is (-10) ( 18!8! ), (2) The value of f(1)+f(-1) is equal to 10! 211, (3) The degree of the polynomial f(x) is 10

Solution

(x2-1)10= k=010 10k(-1)kx20-2k and d10dx10xm= m!(m-10)!xm-10 (zero if m<10).

So f(x)= k=05 10k(-1)k (20-2k)!(10-2k)!x10-2k; degree =10. (C) true.

Coefficient of x8 (k=1): -10 18!8!. (A) true.

Constant term (k=5): - 105 10!=-252 10! - 10!5!. (D) false.

Write (x2-1)10=(x-1)10(x+1)10; by Leibniz rule at x=1 only the term with all 10 derivatives on (x-1)10 survives: f(1)=10! 210. f is even, so f(-1)=f(1).

f(1)+f(-1)=10! 211. (B) true.

Answer: (A), (B), (C)