JEE MAIN - Mathematics (2009 - No. 12)
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 $$x=0$$.
Statement-1: gof is differentiable at $$x=0$$ and its derivative is continuous at that point.
Statement-2: gof is twice differentiable at $$x=0$$.
Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
Statement-1 is true, Statement-2 is false
Statement-1 is false, Statement-2 is true
Statement-1 is true, Statement-2 is true Statement-2 is a correct explanation for Statement-1
Explanation
Given that $$f\left( x \right) = x\left| x \right|\,\,$$ and $$\,\,g\left( x \right) = \sin x$$
So that go
$$f\left( x \right) = g\left( {f\left( x \right)} \right)$$
$$ = g\left( {x\left| x \right|} \right) = \sin x\left| x \right|$$
$$ = \left\{ {\matrix{ {\sin \left( { - {x^2}} \right),} & {if\,\,\,x < 0} \cr {\sin \left( {{x^2}} \right),} & {if\,\,\,x \ge 0} \cr } } \right.$$
$$ = \left\{ {\matrix{ { - \sin \,{x^2},} & {if\,\,\,x < 0} \cr {\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \cr } } \right.$$
$$\therefore$$ $$\left( {go\,f} \right)'\,\,\left( x \right) = \left\{ {\matrix{ { - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \cr {2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \cr } } \right.$$
Here we observe
$$L\left( {gof} \right)'\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right)$$
$$ \Rightarrow $$ go $$f$$ is differentiable at $$x=0$$
and $$\left( {go\,f} \right)'$$ is continuous at $$x=0$$
Now $$\left( {go\,f} \right)''\left( x \right) = \left\{ {\matrix{ { - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \cr {2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \cr } } \right.$$
Here $$L\left( {gof} \right)''\left( 0 \right) = - 2$$ and $$R\left( {go\,f} \right)''\left( 0 \right) = 2$$
As $$L{\left( {go\,f} \right)^{''}}\left( 0 \right) \ne R\left( {go\,f} \right)''\,\,\left( 0 \right)$$
$$ \Rightarrow go\,f\left( x \right)$$ is not twice differentiable at $$x=0.$$
$$\therefore$$ Statement - $$1$$ is true but statement $$-2$$ is false.
So that go
$$f\left( x \right) = g\left( {f\left( x \right)} \right)$$
$$ = g\left( {x\left| x \right|} \right) = \sin x\left| x \right|$$
$$ = \left\{ {\matrix{ {\sin \left( { - {x^2}} \right),} & {if\,\,\,x < 0} \cr {\sin \left( {{x^2}} \right),} & {if\,\,\,x \ge 0} \cr } } \right.$$
$$ = \left\{ {\matrix{ { - \sin \,{x^2},} & {if\,\,\,x < 0} \cr {\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \cr } } \right.$$
$$\therefore$$ $$\left( {go\,f} \right)'\,\,\left( x \right) = \left\{ {\matrix{ { - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \cr {2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \cr } } \right.$$
Here we observe
$$L\left( {gof} \right)'\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right)$$
$$ \Rightarrow $$ go $$f$$ is differentiable at $$x=0$$
and $$\left( {go\,f} \right)'$$ is continuous at $$x=0$$
Now $$\left( {go\,f} \right)''\left( x \right) = \left\{ {\matrix{ { - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \cr {2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \cr } } \right.$$
Here $$L\left( {gof} \right)''\left( 0 \right) = - 2$$ and $$R\left( {go\,f} \right)''\left( 0 \right) = 2$$
As $$L{\left( {go\,f} \right)^{''}}\left( 0 \right) \ne R\left( {go\,f} \right)''\,\,\left( 0 \right)$$
$$ \Rightarrow go\,f\left( x \right)$$ is not twice differentiable at $$x=0.$$
$$\therefore$$ Statement - $$1$$ is true but statement $$-2$$ is false.
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