JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 15)
Let f(x) = sin($$\pi $$ cos x) and g(x) = cos(2$$\pi $$ sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order:
X = {x : f(x) = 0}, Y = {x : f'(x) = 0}
Z = {x : g(x) = 0}, W = {x : g'(x) = 0}
List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets.
Which of the following is the only CORRECT combination?
X = {x : f(x) = 0}, Y = {x : f'(x) = 0}
Z = {x : g(x) = 0}, W = {x : g'(x) = 0}
List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets.
Which of the following is the only CORRECT combination?
(IV), (P), (R), (S)
(III), (P), (Q), (U)
(III), (R), (U)
(IV), (Q), (T)
Explanation
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.
$$ \because $$ 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.
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