Question

Let $f\left(x\right)=\{\begin{array}{cc}-2\sin x,& x\le -m\\ A\sin x+B,& -m<x<m\\ \cos x,& x\ge m\end{array}$ be a continuous function, where $m$ is the principal solution of the equation ${\sin }^{3}x+\sin x\cos x+{\cos }^{3}x=1$. If $m>0$, then

• A. $A=-1,B=1$
• B. $m=\pi$
• C. $m=\frac{\pi }{2}$
• D. $A=1,B=-1$

Answer 1

Answer: (A), (C)

The correct options are
A $A=-1,B=1$
C $m=\frac{\pi }{2}$

${\sin }^{3}x+\sin x\cos x+{\cos }^{3}x=1$
$\Rightarrow ({\sin }^{3}x+{\cos }^{3}x)-(1-\sin x\cos x)=0$
$\Rightarrow (\sin x+\cos x)(1-\sin x\cos x)-(1-\sin x\cos x)=0$
$\Rightarrow (1-\sin x\cos x)(\sin x+\cos x-1)=0\Rightarrow \sin x\cos x=1\text{or}\sin x+\cos x=1\Rightarrow \sin 2x=2\text{or}\sin x+\cos x=1$
$\sin 2x=2$ is not possible.

$\sin x+\cos x=1$
$\Rightarrow \sin x\sin \frac{\pi }{4}+\cos x\cos \frac{\pi }{4}=\frac{1}{\sqrt{2}}$
$\Rightarrow \cos (x-\frac{\pi }{4})=\cos \frac{\pi }{4}$
$\Rightarrow x=0,\frac{\pi }{2},2\pi$
But $m$ is a principal solution with $m>0$. So, $m=\frac{\pi }{2}$

Now, $f\left(x\right)=\{\begin{array}{cc}-2\sin x,& x\le -\frac{\pi }{2}\\ A\sin x+B,& -\frac{\pi }{2}<x<\frac{\pi }{2}\\ \cos x,& x\ge \frac{\pi }{2}\end{array}$

Continuity at $x=-\frac{\pi }{2}$
$\underset{x\to -\pi /2^{-}}{lim}f\left(x\right)=\underset{x\to -\pi /2^{+}}{lim}f\left(x\right)$
$\Rightarrow 2=-A+B\cdots \left(1\right)$

Continuity at $x=\frac{\pi }{2}$
$\underset{x\to \pi /2^{-}}{lim}f\left(x\right)=\underset{x\to \pi /2^{+}}{lim}f\left(x\right)$
$\Rightarrow A+B=0\cdots \left(2\right)$

Solving equation $\left(1\right)$ and $\left(2\right)$, we get
$B=1,A=-1$