Question

Let $f\left( x \right) = x\left| x \right|{\text{ and }}g\left( x \right) = \sin x$
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.
$\left( a \right)$ Statement – 1 is true, statement – 2 is true; statement – 2 is a correct explanation for statement – 1.
$\left( b \right)$ Statement – 1 is true, statement – 2 is true; statement – 2 is not a correct explanation for statement – 1.
$\left( c \right)$ Statement – 1 is true, statement – 2 is false.
$\left( d \right)$ Statement – 1 is false, statement – 2 is true.

Answer 1

In this particular question use the concept that gof is simply $g\left( {f\left( x \right)} \right)$ and use the concept that if left hand derivative (LHD) is equal to right hand derivative (RHD) at x = 0, then the function is differentiable at x = 0 otherwise not, and use the concept that if, LHD = RHD at x= 0 then the function is continuous at x = 0, otherwise not so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given data:
$f\left( x \right) = x\left| x \right|{\text{ and }}g\left( x \right) = \sin x$
$\Rightarrow gof = g\left( {f\left( x \right)} \right) = six\left( {x\left| x \right|} \right)$

\Rightarrow gof = \left\\{ {\sin {x^2},x \geqslant 0} \right. \\\ {\text{ }}\sin \left( { - {x^2}} \right),x < 0 \\\ \end{gathered} $$, $\left[ {\because \left| x \right| = x,x \geqslant 0{\text{ and }}\left| x \right| = - x,x < 0} \right]$ Now as we know that sin (-x) = -sin x so we have, $$\begin{gathered} \Rightarrow gof = \left\\{ {\sin {x^2},x \geqslant 0} \right. \\\ {\text{ }} - \sin \left( {{x^2}} \right),x < 0 \\\ \end{gathered} $$ Statement – 1: gof is differentiable at x = 0 and its derivative is continuous at that point. Now differentiate the above equation we have, $$\begin{gathered} \Rightarrow \dfrac{d}{{dx}}\left( {gof} \right) = \left\\{ {\dfrac{d}{{dx}}\sin {x^2},x \geqslant 0} \right. \\\ {\text{ }} - \dfrac{d}{{dx}}\sin \left( {{x^2}} \right),x < 0 \\\ \end{gathered} $$ Now as we know that $\dfrac{d}{{dx}}\sin {x^n} = \cos {x^n}\dfrac{d}{{dx}}{x^n} = \cos {x^n}\left( {n{x^{n - 1}}} \right)$ so use this property we have, $$\begin{gathered} \Rightarrow \dfrac{d}{{dx}}\left( {gof} \right) = \left\\{ {2x\cos {x^2},x \geqslant 0} \right. \\\ {\text{ }} - 2x\cos {x^2},x < 0 \\\ \end{gathered} $$.......................... (1) Now as we know that LHD = $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{d}{{dx}}\left( {gof} \right) = \mathop {\lim }\limits_{x \to {0^ - }} - 2x\cos {x^2} = 0$ And RHD = $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{d}{{dx}}\left( {gof} \right) = \mathop {\lim }\limits_{x \to {0^ + }} 2x\cos {x^2} = 0$ So as we see that at x = 0, $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{d}{{dx}}\left( {gof} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{d}{{dx}}\left( {gof} \right) = 0$ So the function is differentiable at x = 0. Hence the function is continuous at x = 0. Hence Statement – 1 is true Statement – 2: gof is twice differentiable at x = 0. Now again differentiate equation (1) we have, $$\begin{gathered} \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right) = \left\\{ {\dfrac{d}{{dx}}2x\cos {x^2},x \geqslant 0} \right. \\\ {\text{ }} - \dfrac{d}{{dx}}2x\cos {x^2},x < 0 \\\ \end{gathered} $$ Now as we know that $\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$ so use this property we have, $$\begin{gathered} \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right) = \left\\{ {2x\dfrac{d}{{dx}}\cos {x^2} + \cos {x^2}\dfrac{d}{{dx}}2x,x \geqslant 0} \right. \\\ {\text{ }} - 2x\dfrac{d}{{dx}}\cos {x^2} - \cos {x^2}\dfrac{d}{{dx}}2x,x < 0 \\\ \end{gathered} $$ $$\begin{gathered} \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right) = \left\\{ {2x\left( { - 2x\sin {x^2}} \right) + 2\cos {x^2},x \geqslant 0} \right. \\\ {\text{ }} - 2x\left( { - 2x\sin {x^2}} \right) - 2\cos {x^2},x < 0 \\\ \end{gathered} $$ $$\begin{gathered} \Rightarrow \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right) = \left\\{ { - 2x\left( {2x\sin {x^2}} \right) + 2\cos {x^2},x \geqslant 0} \right. \\\ {\text{ }}2x\left( {2x\sin {x^2}} \right) - 2\cos {x^2},x < 0 \\\ \end{gathered} $$ Now as we know that LHD = $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right) = \mathop {\lim }\limits_{x \to {0^ - }} \left( {2x\left( {2x\sin {x^2}} \right) - 2\cos {x^2}} \right) = - 2$ And RHD = $\mathop {\lim }\limits_{x \to {0^ + }} \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( { - 2x\left( {2x\sin {x^2}} \right) + 2\cos {x^2}} \right) = 2$ So as we see that at x = 0, $\mathop {\lim }\limits_{x \to {0^ - }} \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right) \ne \mathop {\lim }\limits_{x \to {0^ + }} \dfrac{d}{{dx}}\left( {\dfrac{d}{{dx}}\left( {gof} \right)} \right)$ $ \Rightarrow {\text{LHD}} \ne {\text{RHD}}$ So the function is not twice differentiable at x = 0. Hence statement – 2 is false. **Hence option (c) is the correct answer.** **Note:** Whenever we face such types of questions the key concept we have to remember is that always recall the basic differentiation property such as, $\dfrac{d}{{dx}}\sin {x^n} = \cos {x^n}\dfrac{d}{{dx}}{x^n} = \cos {x^n}\left( {n{x^{n - 1}}} \right)$, $\dfrac{d}{{dx}}mn = m\dfrac{d}{{dx}}n + n\dfrac{d}{{dx}}m$ so simply differentiate it as above and check whether LHD is equal to RHD at x = 0 if yes then it is differentiable as well as continuous otherwise not.