Question

If largets subset of $(0,\pi )$ at each point of which the function $f\left(x\right)=3{\cos }^{4}x+10{\cos }^{3}x+6{\cos }^{2}x-3$ is decreasing is $(0,\frac{\pi }{p})\cup (\frac{2\pi }{r},\pi )$, then find the value of $(p+r)$

• A. $5$
• B. $4$
• C. $3$
• D. $7$

Answer 1

Answer: (A)

$5$

$f\left(x\right)=3{\cos }^{4}x+10{\cos }^{3}x+6{\cos }^{2}x-3$

$f^{\prime }\left(x\right)=12{\cos }^{3}x(-\sin x)+30{\cos }^{2}x(-\sin x)+12\cos x(-\sin x)$
$=-3\sin 2x(2{\cos }^{2}x+5\cos x+2)$
$=-3\sin 2x(2\cos x+1)(\cos x+2)$
When $f^{\prime }\left(x\right)=0\Rightarrow \sin 2x=0\Rightarrow x=0,\frac{\pi }{2},\pi$
and $2\cos x+1=0\Rightarrow x=\frac{2\pi }{3}$

As $\cos x+2\ne 0\Rightarrow \cos x\ne -2$
Sign scheme for $f^{\prime }\left(x\right)$ in $[0,\pi ]$ is as below (see image)
$f\left(x\right)$ decreases on $(0,\frac{\pi }{2})\cup (\frac{2\pi }{3},\pi )$ and increases on $(\frac{\pi }{2},\frac{2\pi }{3})$