• # question_answer Directions: Questions No. 90 are Assertion - Reason type questions. Each of these questions contains two statements: Statement-1 (Assertion) and Statement-2 (Reason). Each of these questions also have four alternative choices, only one of which is the correct answer. You have to select the correct choice. Let$f(x)=x|x|$and$g(x)=sinx$ Statement 1: gof is differentiable at$x=0$and its derivative is continuous at that point Statement 2: gof is twice differentiable at$x=0$                                                             A) (a) Statement-1 is true, Statement-2 is true, Statement-2 is a correct explanation for statement-1B) (b) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for statement-1.C) (c) Statement-1 is true, statement-2 is false.D) (d) Statement-1 is false, Statement-2 is true

$g(f(x))=\sin (f(x))=\left\{ \begin{matrix} \sin {{x}^{2}}, & x\ge 0 \\ -2x\,\cos {{x}^{2}}, & x<0 \\ \end{matrix} \right.$ $(g(f(x)))'=\left\{ \begin{matrix} 2x\,\cos {{x}^{2}}, & x\ge 0 \\ -2x\,\cos {{x}^{2}}, & x<0 \\ \end{matrix} \right.$ R.H.D. of$(g(f(0)))'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{2h{{\cosh }^{2}}}{h}=2$ L.H.D. of$(g(f(0)))'=\underset{h\to {{0}^{+}}}{\mathop{\lim }}\,\frac{2h{{\cosh }^{2}}}{-h}=-2$ Clearly gof is twice differentiable at$x=0$hence it is differentiable at$x=0$and its derivative is continuous at$x=0$. You will be redirected in 3 sec 