Question

$f\left(x\right)={\left({\sin }^{-1}x\right)}^2.\cos \left(1/x\right)$ if $x\ne 0,f\left(0\right)=0,f\left(x\right)$ is:

• A. continuous no where in $-1\le x\le 1$
• B. continuous every where in $-1\le x\le 1$
• C. differentiable no where in $-1\le x\le 1$
• D. differentiable everywhere $-1<x<1$

Answer 1

Answer: (B), (D)

continuous every where in $-1\le x\le 1$
differentiable everywhere $-1<x<1$

Since ${\sin }^{-1}x$ and $\cos \frac{1}{x}$ are continuous an ddifferentiable in $x\in [-1,1]-\left\{0\right\}$
Now at $x=0$
$f^{\prime }\left(0^{-}\right)=\underset{h\to 0}{lim}\frac{{\left({\sin }^{-1}(0-h)\right)}^2\cos \left(\frac{-1}{h}\right)-0}{-h}=0$
$f^{\prime }\left(0^{+}\right)=\underset{h\to 0}{lim}\frac{{\left({\sin }^{-1}h\right)}^2\cos \left(\frac{1}{h}\right)-0}{h}=0$
Hence LHD$=$RHD
so $f\left(x\right)$ is continuous and differentiable everywhere in $-1\le x\le 1$