arthham

JEE Advanced 2019 · Trigonometric Functions

Let f(x)=sin⁡(πcos⁡x)f(x) = \sin(\pi \cos x) and g(x)=cos⁡(2πsin⁡x)g(x) = \cos(2\pi \sin x) be two functions defined for x>0x > 0. Define the following sets whose elements are written in the increasing order. X={x:f(x)=0},Y={x:f′(x)=0}X = \{x : f(x) = 0\}, Y = \{x : f'(x) = 0\} Z={x:g(x)=0},W={x:g′(x)=0}Z = \{x : g(x) = 0\}, W = \{x : g'(x) = 0\} Column - I contains the sets X, Y, Z and W. Column - II contains some information regarding these sets. | Column I | Column II | | --- | --- | | (I) X | (p) ⊇{π2,3π2,4π,7π}\supseteq \left\{\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi\right\} | | (II) Y | (q) an arithmetic progression | | (III) Z | (r) NOT an arithmetic progression | | (IV) W | (s) ⊇{π6,7π6,13π6}\supseteq \left\{\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6}\right\} | | | (t) ⊇{π3,2π3,π}\supseteq \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \pi\right\} | | | (u) ⊇{π6,3π4}\supseteq \left\{\frac{\pi}{6}, \frac{3\pi}{4}\right\} | Which of the following is the only CORRECT combination?

Answer: D. (III),(q),(t)(III), (q), (t)

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Written solution

f(x)=sin⁡(πcos⁡x),g(x)=cos⁡(2πsin⁡x),x>0f(x) = \sin(\pi \cos x), \qquad g(x) = \cos(2\pi \sin x), \quad x > 0

X={x:f(x)=0},Y={x:f′(x)=0}X = \{x : f(x) = 0\}, \qquad Y = \{x : f'(x) = 0\}

Z={x:g(x)=0},W={x:g′(x)=0}Z = \{x : g(x) = 0\}, \qquad W = \{x : g'(x) = 0\}

Match X,Y,Z,W with the correct statement (p)–(u).\text{Match } X, Y, Z, W \text{ with the correct statement } (\text{p})\text{--}(\text{u}).

Choose the only CORRECT combination.\text{Choose the only CORRECT combination.}

f(x)=0⇒sin⁡(πcos⁡x)=0⇒πcos⁡x=nπf(x) = 0 \Rightarrow \sin(\pi\cos x) = 0 \Rightarrow \pi\cos x = n\pi

⇒cos⁡x=n,n∈Z\Rightarrow \cos x = n, \quad n \in \mathbb{Z}

cos⁡x∈{−1,0,1}⇒x=nπ2,n=1,2,3,…\cos x \in \{-1, 0, 1\} \Rightarrow x = \frac{n\pi}{2}, \quad n = 1, 2, 3, \dots

X={π2,π,3π2,2π,5π2,… }X = \left\{\frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi, \frac{5\pi}{2}, \dots\right\}

common difference=π2 ⇒ X is an A.P. (q)\text{common difference} = \frac{\pi}{2} \ \Rightarrow \ X \text{ is an A.P. } (q)

π2,3π2,4π,7π ∈X ⇒(p)\frac{\pi}{2}, \frac{3\pi}{2}, 4\pi, 7\pi \ \in X \ \Rightarrow (p)

X⟶(p), (q)X \longrightarrow (p), \ (q)

f′(x)=cos⁡(πcos⁡x)⋅(−πsin⁡x)f'(x) = \cos(\pi\cos x)\cdot(-\pi\sin x)

f′(x)=−πsin⁡x cos⁡(πcos⁡x)f'(x) = -\pi\sin x\,\cos(\pi\cos x)

−πsin⁡x cos⁡(πcos⁡x)=0-\pi\sin x\,\cos(\pi\cos x) = 0

sin⁡x=0  or  cos⁡(πcos⁡x)=0\sin x = 0 \ \text{ or } \ \cos(\pi\cos x) = 0

x=nπ  or  cos⁡x=2k+12=±12x = n\pi \ \text{ or } \ \cos x = \frac{2k+1}{2} = \pm\frac{1}{2}

Y={π3,2π3,π,4π3,5π3,2π,… }Y = \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, 2\pi, \dots\right\}

={nπ3}, n=1,2,3,… ⇒ A.P. (q)= \left\{\frac{n\pi}{3}\right\}, \ n = 1, 2, 3, \dots \ \Rightarrow \ \text{A.P. } (q)

π3,2π3,π ∈Y ⇒(t)\frac{\pi}{3}, \frac{2\pi}{3}, \pi \ \in Y \ \Rightarrow (t)

Y⟶(q), (t)Y \longrightarrow (q), \ (t)

g(x)=0⇒cos⁡(2πsin⁡x)=0g(x) = 0 \Rightarrow \cos(2\pi\sin x) = 0

2πsin⁡x=(2k+1)π2⇒sin⁡x=±14, ±342\pi\sin x = \frac{(2k+1)\pi}{2} \Rightarrow \sin x = \pm\frac{1}{4}, \ \pm\frac{3}{4}

gaps are not constant⇒Z is NOT an A.P. (r)\text{gaps are not constant} \Rightarrow Z \text{ is NOT an A.P. } (r)

g′(x)=−sin⁡(2πsin⁡x)⋅(2πcos⁡x)g'(x) = -\sin(2\pi\sin x)\cdot(2\pi\cos x)

g′(x)=−2πcos⁡x sin⁡(2πsin⁡x)g'(x) = -2\pi\cos x\,\sin(2\pi\sin x)

−2πcos⁡x sin⁡(2πsin⁡x)=0-2\pi\cos x\,\sin(2\pi\sin x) = 0

cos⁡x=0  or  sin⁡(2πsin⁡x)=0\cos x = 0 \ \text{ or } \ \sin(2\pi\sin x) = 0

sin⁡x=n2⇒sin⁡x∈{0,±12,±1}\sin x = \frac{n}{2} \Rightarrow \sin x \in \left\{0, \pm\frac{1}{2}, \pm 1\right\}

W={x:sin⁡x∈{0,±12,±1}}W = \left\{x : \sin x \in \left\{0, \pm\tfrac{1}{2}, \pm 1\right\}\right\}

residues (units of π6)={0,1,3,5} ⇒NOT an A.P. (r)\text{residues (units of } \tfrac{\pi}{6}\text{)} = \{0,1,3,5\} \ \Rightarrow \text{NOT an A.P. } (r)

π6,7π6,13π6 ∈W ⇒(s)\frac{\pi}{6}, \frac{7\pi}{6}, \frac{13\pi}{6} \ \in W \ \Rightarrow (s)

SetColumn IIX(p), (q)Y(q), (t)Z(r)W(r), (s)\begin{array}{c|c} \text{Set} & \text{Column II} \\ X & (p),\ (q) \\ Y & (q),\ (t) \\ Z & (r) \\ W & (r),\ (s) \end{array}

The only correct combination is option (2)\text{The only correct combination is option } (2)

boxed(I) X → (p), (III) Z → (r)

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