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 $$\sqrt 2 $$

Using Lagrange's Mean Value Theorem

Let $$f(x)$$ be a function defined on $$\left[ {a,b} \right]$$

then, $$f'\left( c \right) = {{f\left( b \right) - f\left( a \right)} \over {b - a}}\,\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$$

$$c\,\, \in \left[ {a,b} \right]$$

$$\therefore$$ Given $$f\left( x \right) = {\log _e}x$$

$$\therefore$$ $$f'\left( x \right) = {1 \over x}$$

$$\therefore$$ equation $$(i)$$ become $${1 \over c} = {{f\left( 3 \right) - f\left( 1 \right)} \over {3 - 1}}$$

$$ \Rightarrow {1 \over c} = {{{{\log }_e}3 - {{\log }_e}1} \over 2} = {{{{\log }_e}3} \over 2}$$

$$ \Rightarrow c = {2 \over {{{\log }_e}3}} \Rightarrow c = 2\,{\log _3}e$$