Let R denote the set of all real numbers. Consider the polynomial function f: R R defined by f(x) = d^ 10 dx^ 10 ((x^2 - 1)^ 10 ), for all x R . Here d^ 10 dx^ 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 ?
C. The degree of the polynomial f(x) is 10
Given f(x) = d^ 10 dx^ 10 ((x^2 - 1)^ 10 ). Using the binomial expansion, we have: (x^2 - 1)^ 10 = _ k=0 ^ 10 (-1)^k \, ^ 10 C_ k x^ 20-2k Differentiating 10 times with respect to x, we get: f(x) = _ k=0 ^ 5 (-1)^k \, ^ 10 C_ 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 \, ^ 10 C_ 1 18! 8! = -10 ( 18! 8! ). Thus, statement (A) is true. The highest power of x in f(x) corresponds to k = 0, which gives x^ 10 with a non-zero coefficient of 20! 10! . Therefore, the degree of the polynomial f(x) is 10. Thus, statement (C) is true. For the constant term, we set 10 - 2k = 0 k = 5. The constant term is (-1)^5 \, ^ 10 C_ 5 10! 0! = - 10! 5!5! 10! = - ( 10! 5! )^2. Thus, statement (D) is false. To find f(1) and f(-1), we use the Leibniz rule for the n-th derivative of a product: f(x) = d^ 10 dx^ 10 ((x-1)^ 10 (x+1)^ 10 ) = _ r=0 ^ 10 \, ^ 10 C_ r d^ 10-r dx^ 10-r ((x-1)^ 10 ) d^r dx^r ((x+1)^ 10 ) Evaluating at x = 1, all terms in the sum are zero except when r = 0 (since d^ 10-r dx^ 10-r ((x-1)^ 10 ) contains a factor of (x-1) for r > 0). f(1) = \, ^ 10 C_ 0 (10!) (1+1)^ 10 = 10! \, 2^ 10 Evaluating at x = -1, all terms are zero except when r = 10. f(-1) = \, ^ 10 C_ 10 (-1-1)^ 10 (10!) = 10! \, 2^ 10 Adding these values gives: f(1) + f(-1) = 10! \, 2^ 10 + 10! \, 2^ 10 = 10! \, 2^ 11 Thus, statement (B) is true. Answer: The coefficient of x^8 in the polynomial f(x) is (-10) ( 18! 8! ); The value of f(1) + f(-1) is equal to 10!\, 2^ 11 ; The degree of the polynomial f(x) is 10
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