Let f\left( x \right) = x\left| x \right| and g\left( x \right) = \sin x. 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.

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Answer: B. Statement-1 is true, Statement-2 is false Given that f\left( x ight) = x\left| x ight|\,\, and \,\,g\left( x ight) = \sin x So that go f\left( x ight) = g\left( {f\left( x ight)} ight) = g\left( {x\left| x ight|} ight) = \sin x\left| x ight| = \left\{ {\begin{matrix} {\sin \left( { - {x^2}} ight),} & {if\,\,\,x < 0} \ {\sin \left( {{x^2}} ight),} & {if\,\,\,x \ge 0} \ \end{matrix} } ight. = \left\{ {\begin{matrix} { - \sin \,{x^2},} & {if\,\,\,x < 0} \ {\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \ \end{matrix} } ight. 	herefore \left( {go\,f} ight)'\,\,\left( x ight) = \left\{ {\begin{matrix} { - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \ {2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \ \end{matrix} } ight. Here we observe L\left( {gof} ight)'\left( 0 ight) = 0 = R\left( {gof} ight)'\left( 0 ight) \Rightarrow go f is differentiable at x=0 and \left( {go\,f} ight)' is continuous at x=0 Now \left( {go\,f} ight)''\left( x ight) = \left\{ {\begin{matrix} { - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \ {2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \ \end{matrix} } ight. Here L\left( {gof} ight)''\left( 0 ight) = - 2 and R\left( {go\,f} ight)''\left( 0 ight) = 2 As L{\left( {go\,f} ight)^{''}}\left( 0 ight) 
e R\left( {go\,f} ight)''\,\,\left( 0 ight) \Rightarrow go\,f\left( x ight) is not twice differentiable at x=0. 	herefore Statement - 1 is true but statement -2 is false.

Let $$f\left( x \right) = x\left| x \right|$$ and $$g\left( x \right) = \sin x.$$ Statement-1: gof is differentiable at $$x=0$$ and its derivative is continuous at that point. Statement-2: gof is twice differentiable at $$x=0$$.

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