JEE Advanced 2019Paper 2
- Exam
- JEE Advanced
- Year
- 2019
- Paper
- Paper 2
- Subject
- Mathematics
- Chapter
- Trigonometric Equations
JEE Advanced 2019 · Paper 2 · Mathematics · Trigonometric Equations
Let f(x) = (π cos x) and g(x) = (2π sin x) be two functions defined...
Let and be two functions defined for Define the following sets whose elements are written in the increasing order:
List- contains the sets and List- contains some information regarding these sets.
Which of the following is the only correct combination?
List- contains the sets and List- contains some information regarding these sets.
| List- | List- | ||
|---|---|---|---|
| \left(I\right)X | |||
| \left(II\right)Y | an arithmetic progression | ||
| \left(III\right)Z | NOT an arithmetic progression | ||
| \left(IV\right)W | |||
| \left(T\right)⊇ \left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi\right\} |
Which of the following is the only correct combination?
Options
- A
- B
- C
- D
Correct answer
B.
Step-by-step solution
f\left(x\right) = \sin \left(\pi cos x\right)X = \left\{x: f\left(x\right) = 0\right\}f\left(x\right) = 0\sin \left(\pi cos x\right) = 0\pi cos x = n\picos x = ncos x = 0, 1, - 1⇒ x = \dfrac{n\pi}{2}X = \left\{\dfrac{n\pi}{2}, n \in N\right\} ⇒ X = \left\{\dfrac{\pi}{2}, \pi, \dfrac{3\pi}{2}, 2\pi \ldots \ldots\right\}
\left(II\right)g\left(x\right) = \cos \left(2\pi sin x\right)Z = \left\{x: g\left(x\right) = 0\right\}\cos \left(2\pi sin x\right) = 02\pi sin x = \left(2n + 1\right)\dfrac{\pi}{2}sin x = \dfrac{2n + 1}{4}sin x = - \dfrac{1}{4}, \dfrac{1}{4}, \dfrac{- 3}{4}, \dfrac{3}{4}Z = \left\{n\pi \pm \sin^{- 1} \left(\dfrac{1}{4}\right), n\pi \pm \sin^{- 1} \left(\dfrac{3}{4}\right)\right\}
\left(III\right)f\left(x\right) = \sin \left(\pi cos x\right)Y = \left(x: {f}^{'}\left(x\right) = 0\right){f}^{'}\left(x\right) = \cos \left(\pi cos x\right) . \left(- \pi sin x\right) = 0\cos \left(\pi cos x\right) = 0 ⇒ \pi cos x = \left(2n + 1\right)\dfrac{\pi}{2}cos x = \dfrac{2n + 1}{2}cos x = - \dfrac{1}{2}, \dfrac{1}{2} ⇒ x = n\pi \pm \dfrac{\pi}{3}sin x = 0 ⇒ x = n\piY = \left\{n\pi, n\pi \pm \dfrac{\pi}{3}\right\}Y = \left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}, 2\pi \ldots\right\}
\left(IV\right) and
now
or
\left(IV\right) - P, R, S
\left(II\right)g\left(x\right) = \cos \left(2\pi sin x\right)Z = \left\{x: g\left(x\right) = 0\right\}\cos \left(2\pi sin x\right) = 02\pi sin x = \left(2n + 1\right)\dfrac{\pi}{2}sin x = \dfrac{2n + 1}{4}sin x = - \dfrac{1}{4}, \dfrac{1}{4}, \dfrac{- 3}{4}, \dfrac{3}{4}Z = \left\{n\pi \pm \sin^{- 1} \left(\dfrac{1}{4}\right), n\pi \pm \sin^{- 1} \left(\dfrac{3}{4}\right)\right\}
\left(III\right)f\left(x\right) = \sin \left(\pi cos x\right)Y = \left(x: {f}^{'}\left(x\right) = 0\right){f}^{'}\left(x\right) = \cos \left(\pi cos x\right) . \left(- \pi sin x\right) = 0\cos \left(\pi cos x\right) = 0 ⇒ \pi cos x = \left(2n + 1\right)\dfrac{\pi}{2}cos x = \dfrac{2n + 1}{2}cos x = - \dfrac{1}{2}, \dfrac{1}{2} ⇒ x = n\pi \pm \dfrac{\pi}{3}sin x = 0 ⇒ x = n\piY = \left\{n\pi, n\pi \pm \dfrac{\pi}{3}\right\}Y = \left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}, 2\pi \ldots\right\}
\left(IV\right) and
now
or
\left(IV\right) - P, R, S