Question

If $f\left(x\right)\{\begin{array}{cc}\frac{x\log \cos x}{\log (1+x^2)},& x\ne 0\\ 0,& x=0\end{array}$ then (more than one option can be correct)

• A. f(x) is not continuous at x $=$ 0
• B. f(x) is continuous at x $=$ 0
• C. f(x) is continuous at x $=$ 0 but not differentiable at x $=$ 0
• D. f(x) is differentiable at x $=$ 0

Answer 1

Answer: (B), (D)

The correct options are
B f(x) is continuous at x $=$ 0
D f(x) is differentiable at x $=$ 0

$f\left(x\right)\{\begin{array}{cc}\frac{xlogcosx}{log(1+x^2)}& x\ne 0\\ 0& x=0\end{array}$
$f\left(o^{-}\right)=f\left(o^{+}\right){lim}_{x\to 0}=\frac{xlog(1+1-cosx)}{log(1+x^2)}=\frac{x(1-cosx)}{x^2}=\frac{x}{2}=0$
$\therefore f\left(o\right)=f\left(o^{+}\right)=f\left(o^{-}\right)=0$
$\therefore$ f is continuous
$\frac{xlogcosx}{log(1+x^2)}$ all three functions are differentiable