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.

Let $$f\left( x \right) = {x^7} + 14{x^5} + 16{x^3} + 30x - 560$$

$$ \Rightarrow f'\left( x \right) = 7{x^6} + 70{x^4} + 48{x^2} + 30 > 0,\,\forall x \in R$$

$$ \Rightarrow f$$ is an increasing function on $$R$$

Also $$\mathop {\lim }\limits_{x \to \infty } \,\,f\left( x \right) = \infty $$ and $$\mathop {\lim }\limits_{x \to - \infty } \,\,f\left( x \right) = - \infty $$

$$ \Rightarrow $$ The curve $$y = f\left( x \right)$$ crosses $$x$$-axis only once.

$$\therefore$$ $$f\left( x \right) = 0$$ has exactly one real root.

Let $$y = {x^3} - px + q$$

$$ \Rightarrow {{dy} \over {dx}} = 3{x^2} - p$$

For $${{dy} \over {dx}} = 0 \Rightarrow 3{x^2} - p = 0$$

$$ \Rightarrow x = \pm \sqrt {{p \over 3}} $$

$${{{d^2}y} \over {d{x^2}}} = 6x$$

$${\left. {{{{d^2}y} \over {d{x^2}}}} \right|_{x = \sqrt {{p \over 3}} }} = + ve\,\,\,\,$$ and

$$\,\,\,\,\,\,\,\,\,\,$$ $${\left. {\,\,\,{{{d^2}y} \over {d{x^2}}}} \right|_{x = - \sqrt {{p \over 3}} }} = - ve$$

$$\therefore$$ $$y$$ has ninima at $$x = \sqrt {{p \over 3}} $$

and maxima at $$x = - \sqrt {{p \over 3}} $$

Given that $${p^2} + {q^2} = 1$$

$$\therefore$$ $$p = \cos \theta $$ and $$q = \sin \theta $$

Then $$p+q$$ $$ = \cos \theta + \sin \theta $$

We know that

$$ - \sqrt {{a^2} + {b^2}} \le a\cos \theta + b\sin \theta \le \sqrt {{a^2} + {b^2}} $$

$$\therefore$$ $$ - \sqrt 2 \le \cos \theta + \sin \theta \le \sqrt 2 $$

Hence max. value of $$p + q$$ is $$\sqrt 2 $$