Here, $$\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$$

$$ \Rightarrow \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = {1 \over {14}}$$ [using L' Hospital's rule] ....... (i)

As $$F(x) = \int_{ - 1}^x {f(t)dt} $$

$$ \Rightarrow F'(x) = f(x)$$ ...... (ii)

and $$G(x) = \int_{ - 1}^x {t|f\{ f(t)\} |dt} $$

$$ \Rightarrow G'(x) = x|f\{ f(x)\} |$$ ...... (iii)

$$\therefore$$ $$\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = \mathop {\lim }\limits_{x \to 1} {{F'(x)} \over {G'(x)}} = \mathop {\lim }\limits_{x \to 1} {{f(x)} \over {x|f\{ f(x)\} |}}$$

$$ = {{f(1)} \over {1|f\{ f(1)\} |}} = {{1/2} \over {|f(1/2)|}}$$ ....... (iv)

Given, $$\mathop {\lim }\limits_{x \to 1} {{F(x)} \over {G(x)}} = {1 \over {14}}$$

$$\therefore$$ $${{{1 \over 2}} \over {\left| {f\left( {{1 \over 2}} \right)} \right|}} = {1 \over {14}} \Rightarrow \left| {f\left( {{1 \over 2}} \right)} \right| = 7$$

$$\alpha = \int\limits_0^1 {{e^{(9x + 3{{\tan }^{ - 1}}x)}}\left( {{{12 + 9{x^2}} \over {1 + {x^2}}}} \right)dx} $$

Set $$9x + 3{\tan ^{ - 1}}x = t$$

so that $${{dt} \over {dx}} = 9 + {3 \over {1 + {x^2}}} = {{12 + 9{x^2}} \over {1 + {x^2}}}$$

We have, $$\alpha = \int\limits_0^{9 + {{3\pi } \over 4}} {{e^t}dt = {e^{9 + {{3\pi } \over 4}}} - 1} $$

$$\therefore$$ $$\ln \left| {\alpha + 1} \right| = 9 + {{3\pi } \over 4}$$

Thus $$\ln \left| {\alpha + 1} \right| - {{3\pi } \over 4} = 9$$