Question

The arithmetic mean of roots of the equation $4{\cos }^{3}x-4{\cos }^{2}x-\cos (315\pi +x)=1$ is

• A. $50\pi$
• B. $51\pi$
• C. $100\pi$
• D. $315\pi$

Answer 1

Answer: (B)

$\because$ $\cos (315\pi +x)={(-1)}^{315}.\cos x=-\cos x$ $\therefore$ $4{\cos }^{3}x-4{\cos }^{2}x-\cos (315\pi x+x)=1$ $\Rightarrow$ $4{\cos }^{3}x-4{\cos }^{2}x+\cos x-1=0$ $\Rightarrow$ $(4{\cos }^{2}x+1)(\cos x-1)=0$ $\therefore$ $cox=1,4{\cos }^{2}x+1\ne 0$ $\Rightarrow$ $\cos x=\cos 0{}^{\circ }$ $\therefore$ $x=2n\pi ,$ $n\in I$ $\therefore$ $x=2\pi ,4\pi ,6\pi ,8\pi .....,100\pi$ $(\because 0<x<315)$ $\therefore$ Required arithmetic mean $=\frac{2\pi +4\pi +6\pi +8\pi +...+100\pi }{50}$ $=\frac{2\pi (1+2+3+4+...+50)}{50}$ $=2\pi .\frac{\frac{50}{2}.51}{50}=51\pi$