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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 f(x)=sin⁡(πcosx)f\left(x\right) = \sin \left(\pi cos x\right) and g(x)=cos⁡(2πsinx)g\left(x\right) = \cos \left(2\pi sin x\right) be two functions defined for x>0.x > 0 . Define the following sets whose elements are written in the increasing order:
X={x:f(x)=0},Y={x:f′(x)=0}X = \left\{x: f\left(x\right) = 0\right\}, Y = \left\{x: {f}^{'}\left(x\right) = 0\right\}
Z={x:g(x)=0},W={x:g′(x)=0}Z = \left\{x: g\left(x\right) = 0\right\}, W = \left\{x: {g}^{'}\left(x\right) = 0\right\}
List- II contains the sets X,Y,ZX, Y, Z and W.W . List- IIII contains some information regarding these sets.
List- II List- IIII
\left(I\right)</td><td></td> <td>X</td><td>(P)</td> <td>\left(P\right) ⊇{π2,3π2,4π,7π}⊇ \left\{\dfrac{\pi}{2}, \dfrac{3\pi}{2}, 4\pi, 7\pi\right\}
\left(II\right)</td><td></td> <td>Y</td><td>(Q)</td> <td>\left(Q\right) an arithmetic progression
\left(III\right)</td><td></td> <td>Z</td><td>(R)</td> <td>\left(R\right) NOT an arithmetic progression
\left(IV\right)</td><td></td> <td>W</td><td>(S)</td> <td>\left(S\right) ⊇{π6,7π6,13π6}⊇ \left\{\dfrac{\pi}{6}, \dfrac{7\pi}{6}, \dfrac{13\pi}{6}\right\}
\left(T\right)</td><td></td> <td>⊇ \left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi\right\}</td></tr><tr><td></td><td></td><td>(U)</td> </tr> <tr> <td> </td> <td> </td> <td>\left(U\right) ⊇{π6,3π4}⊇ \left\{\dfrac{\pi}{6}, \dfrac{3\pi}{4}\right\}

Which of the following is the only correct combination?

Options

  1. A(IV)−(Q),(T)\left(IV\right) - \left(Q\right), \left(T\right)
  2. B(IV)−(P),(R),(S)\left(IV\right) - \left(P\right), \left(R\right), \left(S\right)
  3. C(III)−(R),(U)\left(III\right) - \left(R\right), \left(U\right)
  4. D(III)−(P),(Q),(U)\left(III\right) - \left(P\right), \left(Q\right), \left(U\right)

Correct answer

B. (IV)−(P),(R),(S)\left(IV\right) - \left(P\right), \left(R\right), \left(S\right)

Step-by-step solution

(I)\left(I\right)f\left(x\right) = \sin \left(\pi cos x\right)andandX = \left\{x: f\left(x\right) = 0\right\}<br/><br />f\left(x\right) = 0<br/><br />\sin \left(\pi cos x\right) = 0<br/><br />\pi cos x = n\pi<br/><br />cos x = n<br/><br />cos x = 0, 1, - 1<br/><br />⇒ x = \dfrac{n\pi}{2}<br/><br />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\}<br/>(I)−P,Q<br />\left(I\right) - P, Q
\left(II\right)g\left(x\right) = \cos \left(2\pi sin x\right)andandZ = \left\{x: g\left(x\right) = 0\right\}<br/><br />\cos \left(2\pi sin x\right) = 0<br/><br />2\pi sin x = \left(2n + 1\right)\dfrac{\pi}{2}<br/><br />sin x = \dfrac{2n + 1}{4}<br/><br />sin x = - \dfrac{1}{4}, \dfrac{1}{4}, \dfrac{- 3}{4}, \dfrac{3}{4}<br/><br />Z = \left\{n\pi \pm \sin^{- 1} \left(\dfrac{1}{4}\right), n\pi \pm \sin^{- 1} \left(\dfrac{3}{4}\right)\right\}<br/>(II)−Q,T<br />\left(II\right) - Q, T
\left(III\right)f\left(x\right) = \sin \left(\pi cos x\right)<em>and</em><em>and</em>Y = \left(x: {f}^{'}\left(x\right) = 0\right)<br/><br />{f}^{'}\left(x\right) = \cos \left(\pi cos x\right) . \left(- \pi sin x\right) = 0<br/>now<br />now\cos \left(\pi cos x\right) = 0 ⇒ \pi cos x = \left(2n + 1\right)\dfrac{\pi}{2}<br/><br />cos x = \dfrac{2n + 1}{2}<br/><br />cos x = - \dfrac{1}{2}, \dfrac{1}{2} ⇒ x = n\pi \pm \dfrac{\pi}{3}<br/>or<br />orsin x = 0 ⇒ x = n\pi<br/>hence<br />henceY = \left\{n\pi, n\pi \pm \dfrac{\pi}{3}\right\}<br/><br />Y = \left\{\dfrac{\pi}{3}, \dfrac{2\pi}{3}, \pi, \dfrac{4\pi}{3}, \dfrac{5\pi}{3}, 2\pi \ldots\right\}<br/>(III)→R<br />\left(III\right) \to R
\left(IV\right)g(x)=cos⁡(2πsinx)g\left(x\right) = \cos \left(2\pi sin x\right) and W={x:g′(x)=0}W = \left\{x: {g}^{'}\left(x\right) = 0\right\}
g′(x)=−sin⁡(2πsinx)(2πcosx)=0{g}^{'}\left(x\right) = - \sin \left(2\pi sin x\right)\left(2\pi cos x\right) = 0
now cosx=0⇒x=(2n+1)π2cos x = 0 ⇒ x = \left(2n + 1\right)\dfrac{\pi}{2}
or sin⁡(2πsinx)=0\sin \left(2\pi sin x\right) = 0
2πsinx=nπ2\pi sin x = n\pi
sinx=n2sin x = \dfrac{n}{2}
sinx=−1,−12,0,12,1sin x = - 1, - \dfrac{1}{2}, 0, \dfrac{1}{2}, 1
⇒x=nπ2,nπ±π6⇒ x = \dfrac{n\pi}{2}, n\pi \pm \dfrac{\pi}{6}
W={nπ2,nπ±π6}W = \left\{\dfrac{n\pi}{2}, n\pi \pm \dfrac{\pi}{6}\right\}
W={π6,π2,5π6,π,7π6,3π2….}W = \left\{\dfrac{\pi}{6}, \dfrac{\pi}{2}, \dfrac{5\pi}{6}, \pi, \dfrac{7\pi}{6}, \dfrac{3\pi}{2} \ldots .\right\}
\left(IV\right) - P, R, S
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