JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 16)
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 combinations is correct?
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 combinations is correct?
(II), (Q), (T)
(II), (R), (S)
(I), (P), (R)
(I), (Q), (U)
Explanation
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.
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.
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