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 $g\left(x\right)$ is decreasing in

• A. $(\frac{\pi }{4},\frac{\pi }{2})$
• B. $(0,\frac{\pi }{4})$
• C. $(0,\frac{\pi }{2})$
• D. $(\frac{\pi }{6},\frac{\pi }{2})$

Answer 1

Answer: (B)

$(0,\frac{\pi }{4})$

Given $g\left(x\right)=f\left(\sin x\right)+f\left(\cos x\right)$

Differentiating w.r.t $x$, we get

$g^{\prime }\left(x\right)=f^{\prime }\left(\sin x\right).\cos x-f^{\prime }\left(\cos x\right).\sin x$

Again differentiating w.r.t $x$, we get

$g^{\prime \prime }\left(x\right)=-f^{\prime }\left(\sin x\right).\sin x+f^{\prime \prime }\left(\sin x\right).{\cos }^{2}x-f^{\prime }\left(\cos x\right).\cos x+f^{\prime \prime }\left(\cos x\right).{\sin }^{2}x$ ...(2)

Now, given $f^{\prime }\left(\sin x\right)<0\Rightarrow f^{\prime }\left(\sin (\frac{\pi }{2}-x)\right)<0(\because x\in (0,\frac{\pi }{2})\Rightarrow \frac{\pi }{2}-x\in (0,\frac{\pi }{2}))$

$\Rightarrow f^{\prime }\left(\cos x\right)<0$ ...(2)

Similarly $f^{\prime \prime }\left(\sin x\right)>0\Rightarrow f^{\prime \prime }\left(\sin (\frac{\pi }{2}-x)\right)>0\Rightarrow f^{\prime \prime }\left(\cos x\right)>0$ ...(3)

$\therefore g^{\prime \prime }\left(x\right)>0$ $($ Using (1),(2) and (3) $)$

$\Rightarrow g^{\prime }\left(x\right)$ is increasing in $(0,\frac{\pi }{2})$.

Also $g^{\prime }\left(\frac{\pi }{4}\right)=0$

$\Rightarrow g^{\prime }\left(x\right)>0\forall x\in (\frac{\pi }{4},\frac{\pi }{2})$ and $g^{\prime }\left(x\right)<0\forall x\in (0,\frac{\pi }{4})$

Thus $g\left(x\right)$ is decreasing in $(0,\frac{\pi }{4})$