A) h(x) is not differentiable at x = 0.
B) h(x) is differentiable at x = 0, but h¢(x) is not continuous at x = 0
C) h?(x) is continuous at x = 0 but it is not differentiable at x = 0
D) h?(x) is differentiable at x = 0
Correct Answer: C
Solution :
Let \[f(x)=x|x|,g(x)=sinx\] and h \[h(x)=gof(x)=g[f(x)]\] \[\therefore \]\[h(x)=\left\{ \begin{matrix} \sin {{x}^{2}}, & x\ge 0 \\ -\sin {{x}^{2}}, & x<0 \\ \end{matrix} \right.\] Now, \[h'(x)=\left\{ \begin{matrix} 2x\cos {{x}^{2}}, & x\ge 0 \\ -2x\cos {{x}^{2}}, & x<0 \\ \end{matrix} \right.\] Since, L.H.L and R.H.L at x = 0 of h?? (x) is equal to 0 therefore h? (x) is continuous at x = 0 Now, suppose h? (x) is differentiable Since, L.H.L and R.H.L at x = 0 of h¢¢ (x) are different therefore h?? (x) is not continuous. \[\Rightarrow \]h??(x) is not differentiable \[\Rightarrow \] our assumption is wrong Hence h?(x) is not differentiable at x = 0.
You need to login to perform this action.
You will be redirected in
3 sec
