Question

Let ∫ sin⁹ 𝑥 cos(11𝑥) 𝑑𝑥 = cos(10𝑥) 𝑓(𝑥) + 𝑐 , where 𝑐 is a constant. If 𝑓 ′′ (\(\frac{\pi}{4}\)) − 𝑘𝑓 ′(\(\frac{ \pi}{4}\)) = 0, then 𝑘 is equal to ______ (in integer)

Answer 1

Correct Answer: 8

Given:
\[ \int \sin^{9}x\cos(11x)\,dx=\cos(10x)\,f(x)+c. \] Differentiate both sides to relate \(f\) to the integrand: \[ \frac{d} {dx}\big(\cos(10x)f(x)\big)=-10\sin(10x)f(x)+\cos(10x)f'(x) =\sin^{9}x\cos(11x). \] Use \(\cos(11x)=\cos(10x+x)=\cos10x\cos x-\sin10x\sin x\). Then \[ \sin^{9}x\cos(11x)=\cos10x\big(\sin^{9}x\cos x\big)-\sin10x\big(\sin^{10}x\big). \] Equate coefficients of \(\cos10x\) and \(\sin10x\) on both sides: \[ \begin{cases} \cos10x:\quad f'(x)=\sin^{9}x\cos x,\\[6pt] \sin10x:\quad -10\,f(x)=-\sin^{10}x. \end{cases} \] From the second equation \[ f(x)=\frac{1}{10}\sin^{10}x, \] and this indeed satisfies the first (check: \(f'=(1/10)\cdot10\sin^{9}x\cos x=\sin^{9}x\cos x\)). Now compute \(f'\) and \(f''\): \[ f'(x)=\sin^{9}x\cos x, \] \[ f''(x)=\frac{d}{dx}\big(\sin^{9}x\cos x\big) =9\sin^{8}x\cos^{2}x-\sin^{10}x. \] Evaluate at \(x=\dfrac{\pi}{4}\). Here \(\sin x=\cos x=\dfrac{\sqrt2}{2}\), so \[ \sin^{2}x=\cos^{2}x=\tfrac12,\qquad \sin^{8}x=(\tfrac12)^{4}=\tfrac{1}{16},\qquad \sin^{10}x=\tfrac{1}{32}. \] Thus \[ f'\!\left(\tfrac{\pi}{4}\right)=\sin^{9}x\cos x =\sin^{8}x(\sin x\cos x)=\tfrac{1}{16}\cdot\tfrac12=\tfrac{1}{32}, \] \[ f''\!\left(\tfrac{\pi}{4}\right)=9\cdot\tfrac{1}{16}\cdot\tfrac12-\tfrac{1}{32} =9\cdot\tfrac{1}{32}-\tfrac{1}{32}=\tfrac{8}{32}=\tfrac{1}{4}. \] We are given \[ f''\!\left(\tfrac{\pi}{4}\right)-k\,f'\!\left(\tfrac{\pi}{4}\right)=0. \] Substitute the values: \[ \tfrac{1}{4}-k\cdot\tfrac{1}{32}=0 \quad\Rightarrow\quad k=\frac{1/4}{1/32}=8. \] Answer: \(\boxed{8}\)