Question

Let $g\left(x\right)$ be a polynomial of degree one and $f\left(x\right)$ be defined by $f\left(x\right)=\{\begin{array}{cc}g\left(x\right),& x\le 0\\ |x{|}^{sinx},& x>0\end{array}$
If $f\left(x\right)$ is continuous satisfying $f^{\prime }\left(1\right)=f(-1)$, then $g\left(x\right)$ is

• A. $(1+\sin 1)x+1$
• B. $(1-\sin 1)x+1$
• C. $(1-\sin 1)x-1$
• D. $(1+\sin 1)x-1$

Answer 1

The correct option is B $(1-\sin 1)x+1$

$f\left(x\right)={\left|x\right|}^{sinx}forx>0f^{\prime }\left(x\right)={\left|x\right|}^{sinx}(sinx/x+cosx(lnx\left)\right)f^{\prime }\left(1\right)=1\ast sin1=sin1forx0f\left(x\right)=g\left(x\right)=ax+basgivenf(-1)=f^{\prime }\left(1\right)ascontinuousf\left(0^{+}\right)=1\left(using{L}^{\prime }Hospita{l}^{\prime }srule\right)=g\left(0\right)\Rightarrow b=1-a+1=sin1\Rightarrow a=1-sin1g\left(x\right)=(1-sin1)x+1$
Hence the answer is B