Question

The arithmetic mean of the roots of the equation $4{\cos }^{3}x-4{\cos }^{2}x-\cos (\pi +x)-1=0$ in the interval $(0,315)$ is equal to

• A. $49\pi$
• B. $50\pi$
• C. $51\pi$
• D. $100\pi$

Answer 1

The correct option is C $51\pi$

$4{\cos }^3\left(x\right)-4{\cos }^{2}x+\cos \left(x\right)-1=0$
Or
$\cos \left(x\right)[4{\cos }^{2}x+1]-1(4{\cos }^{2}x+1)=0$
Or
$(\cos x-1)(4{\cos }^2+1)=0$
Or
$\cos x=1$ and ${\cos }^2\left(x\right)=\frac{-1}{4}$ ...(not possible).
Hence
$\cos \left(x\right)=1$ implies $x=2k\pi$ where $kϵN$.
Now
$100\pi <315<101\pi$
Hence
$x=0,2\pi ,4\pi ,...100\pi$.
Thus
$Sum$
$=2\pi +4\pi +6\pi +..100\pi$
$=2\pi [1+2+\mathrm{3..50}]$
$=2\pi .\frac{50.51}{2}$
Hence
$A.M$$=\frac{Sum}{50}=2\pi .\frac{50.51}{2.50}=51\pi$.