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 g\left(x\right) = \cos \left(2\pi sin x\right)x > 0 .X = \left\{x: f\left(x\right) = 0\right\}, Y = \left\{x: {f}^{'}\left(x\right) = 0\right\}Z = \left\{x: g\left(x\right) = 0\right\}, W = \left\{x: {g}^{'}\left(x\right) = 0\right\}IX, Y, ZW .II\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?
Which of the following is the only correct combination?
Options
- A
- B
- C
- D
Correct answer
C.
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 = m\pi, m \in I⇒ cos x = mcos x = 0, 1, - 1⇒ x = \dfrac{n\pi}{2}⇒ X = \left\{\dfrac{n\pi}{2}, n \in I\right\} ⇒ X = \left\{\pm \dfrac{\pi}{2}, \pm \pi, \pm \dfrac{3\pi}{2}, \pm 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) = 0⇒ 2\pi sin x = \left(2m + 1\right)\dfrac{\pi}{2}, m \in I⇒ 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\}, n \in I
\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(2m + 1\right)\dfrac{\pi}{2}, m \in I⇒ cos x = \dfrac{2m + 1}{2}⇒ cos x = - \dfrac{1}{2}, \dfrac{1}{2} ⇒ x = n\pi \pm \dfrac{\pi}{3}, n \in Isin x = 0 ⇒ x = n\piY = \left\{n\pi, n\pi \pm \dfrac{\pi}{3}\right\}, n \in IY = \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) = 0⇒ 2\pi sin x = \left(2m + 1\right)\dfrac{\pi}{2}, m \in I⇒ 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\}, n \in I
\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(2m + 1\right)\dfrac{\pi}{2}, m \in I⇒ cos x = \dfrac{2m + 1}{2}⇒ cos x = - \dfrac{1}{2}, \dfrac{1}{2} ⇒ x = n\pi \pm \dfrac{\pi}{3}, n \in Isin x = 0 ⇒ x = n\piY = \left\{n\pi, n\pi \pm \dfrac{\pi}{3}\right\}, n \in IY = \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