Question

Let $f^{\prime }\left(\sin x\right)<0$ and $f^{\prime \prime }\left(\sin x\right)>0\forall x\in (0,\frac{\pi }{2})$ and $g\left(x\right)=f\left(\sin x\right)+f\left(\cos x\right)$ , then

• A. $g\left(x\right)$ increases if $x\in (0,\frac{\pi }{4})$
• B. $g\left(x\right)$ decreases if $x\in (0,\frac{\pi }{4})$
• C. $g\left(x\right)$ increases if $x\in (\frac{\pi }{4},\frac{\pi }{2})$
• D. $g\left(x\right)$ increases if $x\in (0,\frac{\pi }{2})$

Answer 1

Answer: (B), (C)

The correct options are
B $g\left(x\right)$ decreases if $x\in (0,\frac{\pi }{4})$
C $g\left(x\right)$ increases if $x\in (\frac{\pi }{4},\frac{\pi }{2})$

$g\left(x\right)=f\left(\sin x\right)+f\left(\cos x\right)$
$g^{\prime }\left(x\right)=f^{\prime }\left(\sin x\right)\cos x+f^{\prime }\left(\cos x\right)(-\sin x)$
$g^{\prime }\left(x\right)=\cos x{f}^{\prime }\left(\sin x\right)-\sin x{f}^{\prime }\left(\cos x\right)$
$g^{\prime }\left(x\right)=\cos x{f}^{\prime }\left(\sin x\right)-\sin x{f}^{\prime }\left(\sin \right(\frac{\pi }{2}-x\left)\right)$
$g^{\prime \prime }\left(x\right)=-\sin x{f}^{\prime }\left(\sin x\right)+{\cos }^{2}x{f}^{\prime \prime }\left(\sin x\right)-\cos x{f}^{\prime }\left(\cos x\right)+\sin x{f}^{\prime \prime }\left(\cos x\right)$
$f^{\prime }\left(\sin x\right)<0$ for all $x\in (0,\frac{\pi }{2})$
And $\sin x>0,\cos x>0\forall x\in (0,\frac{\pi }{2})$
So, g''(x) >0
$\Rightarrow$ g'(x) is increasing in $(0,\displaystyle \frac{\pi}{2} )$
Now , put g'(x)=0
$\Rightarrow x=\frac{\pi }{4}$
$f^{\prime }\left(\sin x\right)<0$ for all $x\in (0,\frac{\pi }{2})$
$\Rightarrow f^{\prime }\left(\sin x\right)<0$ for all $x\in (0,\frac{\pi }{4})$
Also, $\cos x>\sin x$ for $x\in (0,\frac{\pi }{4})$
Hence, g'(x)<0 ,for $x\in (0,\frac{\pi }{4})$

Also , $\sin x>\cos x$ for $x\in (\frac{\pi }{4},\frac{\pi }{2})$
Hence, g'(x)>0 for $x\in (\frac{\pi }{4},\frac{\pi }{2})$