Question

If $f^{\prime \prime }\left(x\right)>0$ and $f^{\prime }\left(1\right)=0$ such that $g\left(x\right)=f({\cot }^{2}x+2\cot x+2)$ where $0<x<\pi$ then the interval in which $g\left(x\right)$ is decreasing is

• A. $(0,\pi )$
• B. $(\frac{\pi }{2},\pi )$
• C. $(\frac{3\pi }{4},\pi )$
• D. $(0,\frac{3\pi }{4})$

Answer 1

The correct option is D $(0,\frac{3\pi }{4})$

$g^{\prime }\left(x\right)$
$=-f^{\prime }\left(\right(1+\cot \left(x\right){)}^2+1).(2(1+\cot (x\left)\right)).{cosec}^{2}x$
Hence
$g^{\prime }\left(x\right)<0$ for g(x) to be a decreasing function.
$-f^{\prime }\left(\right(1+\cot \left(x\right){)}^2+1).(2(1+\cot (x\left)\right)).{cosec}^{2}x<0$
Or
$f^{\prime }\left(\right(1+\cot \left(x\right){)}^2+1).(2(1+\cot (x\left)\right)).{cosec}^{2}x>0$
Now
${cosec}^{2}x$ is always greater than $0.$
Hence
$f^{\prime }\left(\right(1+\cot \left(x\right){)}^2+1).(2(1+\cot (x\left)\right))>0$
Or
$\left(2\right(1+\cot \left(x\right)\left)\right)>0$ ...(i) and $f^{\prime }\left(\right(1+\cot \left(x\right){)}^2+1)>0$ ...(ii)
Or
$2(1+\cot (x\left)\right)>0$
$\cot \left(x\right)<-1$
$x<\pi -\frac{\pi }{4}$
Or
$x<\frac{3\pi }{4}$ ...(a)
Now consider ii,
$f^{\prime }\left(\right(1+\cot x{)}^2+1)$
Let $x=\frac{3\pi }{4}$
$\cot \left(x\right)=-1$
Hence
$f^{\prime }(1{)}_{x=\frac{3\pi }{4}}$
$=0$
Hence
$x=\frac{3\pi }{4}$ is a critical point for $f\left(x\right)$.
Thus
$g\left(x\right)$ is decreasing on the interval $(0,\frac{3\pi }{4})$... from a.