Let f x = sin ⁡ π cos ⁡ x and g x = cos ⁡ 2 π sin ⁡ x be two functions…

Let fx=sin⁡πcos⁡x and gx=cos⁡2πsin⁡x be two functions defined for x>0. Define the following sets whose elements are written in the increasing order:
X=x:fx=0, Y=x:f'x=0
Z=x:gx=0, W=x:g'x=0
List- I contains the sets X, Y, Z and W. List- II contains some information regarding these sets.
 
  List-  I   List-  II
I X P ⊇π2,3π2, 4π, 7π
II Y Q an arithmetic progression
III Z R NOT an arithmetic progression
IV W S ⊇π6,7π6,13π6
    T ⊇π3,2π3, π
    U ⊇π6,3π4

Which of the following is the only correct combination?
  1. II-R, S
  2. I-P, R
  3. II-Q, T
  4. I-Q, U

Solution

I fx=sin⁡πcos⁡x  and X=x:fx=0
fx=0
⇒sin⁡πcos⁡x=0
⇒πcos⁡x=mπ, m∈I
⇒cos⁡x=m
cos⁡x=0, 1, -1
⇒x=nπ2
⇒X=nπ2, n∈I⇒X=±π2, ±π,±3π2,± 2π……
I-P, Q
II gx=cos⁡2πsin⁡x and Z=x:gx=0
cos⁡2πsin⁡x=0
⇒2πsin⁡x=2m+1π2,m∈I
⇒sin⁡x=2n+14
⇒sin⁡x=-14,14,-34,34
⇒Z=nπ±sin-1⁡14, nπ±sin-1⁡34,n∈I
II-Q, T
III fx=sin⁡πcos⁡x and Y=x:f'x=0
f'x=cos⁡πcos⁡x.-πsin⁡x=0
Now, cos⁡πcos⁡x=0⇒πcos⁡x=2m+1π2,m∈I
⇒cos⁡x=2m+12
⇒cos⁡x=-12,12⇒x=nπ±π3,n∈I
or sin⁡x=0⇒x=nπ
Hence, Y=nπ, nπ±π3,n∈I
Y=π3,2π3, π,4π3,5π3, 2π…
III→R
IV gx=cos⁡2πsin⁡x and W=x:g'x=0
g'x=-sin⁡2πsin⁡x2πcos⁡x=0
Now, cos⁡x=0⇒x=2m+1π2,m∈I
or sin⁡2πsin⁡x=0
2πsin⁡x=mπ,m∈I
⇒sin⁡x=m2
⇒sin⁡x=-1, -12, 0,12, 1
⇒x=nπ2, nπ±π6,n∈I
W=nπ2, nπ±π6
W=π6,π2,5π6,π, 7π6,3π2….
IV-P, R, S

Asked in: JEE Advanced 2019 (Paper 2)

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