For, X = {x : f(x) = 0}, x > 0

Now, f(x) = 0

$$ \Rightarrow $$ sin($$\pi $$ cos x) = 0, x > 0

$$ \Rightarrow $$ $$\pi $$ cos x = n$$\pi $$, n $$ \in $$ Integer.

$$ \Rightarrow $$ cos x = n

$$ \Rightarrow $$ cos x = $$-$$1, 0, 1

{$$ \because $$ cos x $$ \in $$[$$-$$1, 1]}

When cos x = ± 1$$ \Rightarrow $$ x = n$$\pi $$

When cos x = 0 $$ \Rightarrow $$ x = (2n + 1)$${{\pi \over 2}}$$

Hence, (i) $$ \to $$ (P), (Q)

For, Y = {x : f'(x) = 0}, x > 0

Now, f'(x) = 0

$$ \Rightarrow $$ $$-$$$$\pi $$ sin x cos($$\pi $$ cos x) = 0

$$ \Rightarrow $$ either sin x = 0 $$ \Rightarrow $$ x = n$$\pi $$, n is an integer,

or cos($$\pi $$ cos x) = 0

$$ \Rightarrow $$ $$\pi $$ cos x = (2n + 1)$${{\pi \over 2}}$$, n is an integer

$$ \Rightarrow $$ cos x = $${{{2n + 1} \over 2}}$$

$$ \Rightarrow $$ $$\cos x = \pm {1 \over 2}$$ {$$ \because $$ cos x $$ \in $$[$$-$$1, 1]}

$$ \Rightarrow $$ x = $$2n\pi \pm {\pi \over 3}$$ or $$2n\pi \pm {{2\pi } \over 3}$$, n is an integer.

So, (ii) $$ \to $$ (Q), (T)

Hence, option (a) is correct.

For Z = {x : g(x) = 0}, x > 0

$$ \because $$ g(x) = cos(2$$\pi $$ sin x) = 0

$$ \Rightarrow $$ $$2\pi \sin x = (2n + 1){\pi \over 2},\,n \in $$ Integer

$$ \Rightarrow $$ $$\sin x = - {3 \over 4}, - {1 \over 4},{1 \over 4},{3 \over 4}$$ [$$ \because $$ sin x $$ \in $$ [$$-$$1, 1]]

here values of sin x, $$ - {3 \over 4}, - {1 \over 4},{1 \over 4},{3 \over 4}$$ are in an A.P. but corresponding values of x are not in an AP so, (iii) $$ \to $$ R.

For W = {x : g'(x) = 0}, x > 0

So, g'(x) = $$-$$2 $$\pi $$ cos x sin(2$$\pi $$ sin x) = 0

$$ \Rightarrow $$ either cos x = 0 or sin(2$$\pi $$ sin x) = 0

$$ \Rightarrow $$ either $$x = (2n + 1){\pi \over 2}$$ or 2$$\pi $$ sin x = n$$\pi $$, n$$ \in $$ Integers.

$$ \because $$ $$2\pi \sin x = nx$$

$$ \Rightarrow $$ $$\sin x = {n \over 2} = - 1, - {1 \over 2},0,{1 \over 2},1$$ {$$ \because $$ sin x $$ \in $$[$$-$$1, 1)}

$$ \because $$ $$x = n\pi ,\,(2n + 1){\pi \over 2}$$ or $$x = n\pi + {( - 1)^n}\left( { \pm {\pi \over 6}} \right)$$

$$ \Rightarrow $$ (iv) $$ \to $$ P, R, S

Hence, option (a) is correct.

It is given that,

$$\sum\limits_{k = 1}^{13} {{1 \over {\sin \left( {{\pi \over 4} + {{(k - 1)\pi } \over 6}} \right)\sin \left( {{\pi \over 4} + {{k\pi } \over 6}} \right)}}} $$

Let $$\alpha = {\pi \over 4}$$ and $$\beta = {\pi \over 6}$$. Therefore,

$$\sum\limits_{k = 1}^{13} {{1 \over {\sin (\alpha + k\beta )sin(\alpha + (k - 1)\beta )}}} $$

$$ = {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {{{\sin ((\alpha + k\beta ) - (\alpha + (k - 1)\beta ))} \over {\sin (\alpha + k\beta )\sin (\alpha + (k - 1)\beta )}}} $$

$$ = {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {(\cot (\alpha + (k - 1)\beta ) - \cot (\alpha + k\beta ))} $$

$$ = {1 \over {\sin \beta }}\{ [\cot (\alpha ) - \cot (\alpha + \beta )] + [\cot (\alpha + \beta ) - \cot (\alpha + 2\beta )] + ...... + [\cot (\alpha + 12\beta ) - \cot (\alpha + 13\beta )]\} $$

$$ = {1 \over {\sin \beta }}(\cot \alpha - \cot (\alpha + 13\beta ))$$

$$ = {1 \over {\sin (\pi /6)}}\left( {\cot {\pi \over 4} - \cot \left( {{\pi \over 4} + {{13\pi } \over 6}} \right)} \right)$$

$$ = 2(1 - 2 + \sqrt 3 ) = 2(\sqrt 3 - 1)$$