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}X = \{x : f(x) = 0\}, Y={x:f′(x)=0}Y = \{x : f'(x) = 0\}, Z={x:g(x)=0}Z = \{x : g(x) = 0\}, W={x:g′(x)=0}W = \{x : g'(x) = 0\}. Column-I contains the sets XX, YY, ZZ and WW. Column-II contains some information regarding these sets. Column-I: (I) XX, (II) YY, (III) ZZ, (IV) WW. Column-II: (p) ⊇{π/2,3π/2,4π,7π}\supseteq \{\pi/2, 3\pi/2, 4\pi, 7\pi\}; (q) an arithmetic progression; (r) NOT an arithmetic progression; (s) ⊇{π/6,7π/6,13π/6}\supseteq \{\pi/6, 7\pi/6, 13\pi/6\}; (t) ⊇{π/3,2π/3,π}\supseteq \{\pi/3, 2\pi/3, \pi\}; (u) ⊇{π/6,3π/4}\supseteq \{\pi/6, 3\pi/4\}.

Answer: A. (IV),(p),(r),(s)(IV), (p), (r), (s)

Watch this solved on the board The full walkthrough, written out a line at a time with narration explaining each step as it appears. Pause anywhere and ask why a line is there. Free to watch

Or read the worked solution below. Jump to the steps ↓

Written solution

Given: f(x)=sin⁡(πcos⁡x),g(x)=cos⁡(2πsin⁡x)\text{Given: } f(x) = \sin(\pi \cos x), \quad g(x) = \cos(2\pi \sin x)

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\}

To find: the matching of X,Y,Z,W\text{To find: the matching of } X, Y, Z, W

with the correct statements from Column II (p)–(u)\text{with the correct statements from Column II } (p)\text{--}(u)

1. find zeros of f, f′, g, g′1.\ \text{find zeros of } f,\ f',\ g,\ g'

2. write each set and check AP2.\ \text{write each set and check AP}

3. check membership and match options3.\ \text{check membership and match options}

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;cos⁡x∈[−1,1]⇒n=−1,0,1\Rightarrow \cos x = n,\ n\in\mathbb{Z};\quad \cos x\in[-1,1]\Rightarrow n=-1,0,1

X={x:cos⁡x∈{−1,0,1}}={mπ2:m≥1}X = \left\{x:\cos x\in\{-1,0,1\}\right\} = \left\{\frac{m\pi}{2}: m\ge 1\right\}

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

successive gaps=π2⇒X is an AP⇒(q)\text{successive gaps} = \frac{\pi}{2} \Rightarrow X \text{ is an AP} \Rightarrow (q)

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

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

f′(x)=0⇒sin⁡x=0 or cos⁡(πcos⁡x)=0f'(x)=0 \Rightarrow \sin x = 0 \ \text{or}\ \cos(\pi\cos x)=0

sin⁡x=0⇒x=nπ\sin x=0 \Rightarrow x = n\pi

cos⁡(πcos⁡x)=0⇒πcos⁡x=π2+nπ\cos(\pi\cos x)=0 \Rightarrow \pi\cos x = \frac{\pi}{2}+n\pi

⇒cos⁡x=12+n⇒cos⁡x=±12\Rightarrow \cos x = \frac{1}{2}+n \Rightarrow \cos x = \pm\frac{1}{2}

⇒x=±π3+2kπ, ±2π3+2kπ\Rightarrow x = \pm\frac{\pi}{3}+2k\pi,\ \pm\frac{2\pi}{3}+2k\pi

Y={mπ3:m≥1}={π3,2π3,π,4π3,5π3,…}Y = \left\{\frac{m\pi}{3}: m\ge1\right\} = \left\{\frac{\pi}{3}, \frac{2\pi}{3}, \pi, \frac{4\pi}{3}, \frac{5\pi}{3}, \ldots\right\}

common difference=π3⇒Y is an AP⇒(q)\text{common difference} = \frac{\pi}{3} \Rightarrow Y \text{ is an AP} \Rightarrow (q)

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

g(x)=0⇒cos⁡(2πsin⁡x)=0⇒2πsin⁡x=π2+nπg(x)=0 \Rightarrow \cos(2\pi\sin x)=0 \Rightarrow 2\pi\sin x = \frac{\pi}{2}+n\pi

⇒sin⁡x=14+n2=±14, ±34\Rightarrow \sin x = \frac{1}{4}+\frac{n}{2} = \pm\frac{1}{4},\ \pm\frac{3}{4}

Z={x:sin⁡x∈{±14,±34}}Z = \left\{x:\sin x\in\left\{\pm\frac{1}{4},\pm\frac{3}{4}\right\}\right\}

gaps between consecutive Z-elements are not equal\text{gaps between consecutive } Z\text{-elements are not equal}

⇒Z is NOT an AP⇒(r)\Rightarrow Z \text{ is NOT an AP} \Rightarrow (r)

sin⁡π2=1⇒π2∉Z\sin\frac{\pi}{2}=1 \Rightarrow \frac{\pi}{2}\notin Z

sin⁡π6=12⇒π6∉Z\sin\frac{\pi}{6}=\frac{1}{2} \Rightarrow \frac{\pi}{6}\notin Z

Z matches only (r)Z \text{ matches only } (r)

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

g′(x)=0⇒cos⁡x=0 or sin⁡(2πsin⁡x)=0g'(x)=0 \Rightarrow \cos x = 0 \ \text{or}\ \sin(2\pi\sin x)=0

cos⁡x=0⇒x=(2k+1)π2\cos x = 0 \Rightarrow x = \frac{(2k+1)\pi}{2}

sin⁡(2πsin⁡x)=0⇒sin⁡x=n2=0, ±12, ±1\sin(2\pi\sin x)=0 \Rightarrow \sin x = \frac{n}{2} = 0,\ \pm\frac{1}{2},\ \pm1

⇒x=nπ, π6+2kπ, 5π6+2kπ, …\Rightarrow x = n\pi,\ \frac{\pi}{6}+2k\pi,\ \frac{5\pi}{6}+2k\pi,\ \ldots

W={mπ2}∪{π6+2kπ, 5π6+2kπ, 7π6+2kπ, 11π6+2kπ}W = \left\{\frac{m\pi}{2}\right\} \cup \left\{\frac{\pi}{6}+2k\pi,\ \frac{5\pi}{6}+2k\pi,\ \frac{7\pi}{6}+2k\pi,\ \frac{11\pi}{6}+2k\pi\right\}

gaps of W alternate π3,π6⇒NOT an AP⇒(r)\text{gaps of } W \text{ alternate } \frac{\pi}{3}, \frac{\pi}{6} \Rightarrow \text{NOT an AP} \Rightarrow (r)

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

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

X→(p),(q);Y→(q),(t)X \to (p),(q);\quad Y \to (q),(t)

Z→(r);W→(p),(r),(s)Z \to (r);\quad W \to (p),(r),(s)

(1) (IV)=W: (p),(r),(s) ✓(1)\ (IV)=W:\ (p),(r),(s)\ \checkmark

(2), (3) claim Z is AP / contains (u) ×(2),\ (3)\ \text{claim } Z \text{ is AP / contains } (u)\ \times

(4) claims W is AP / contains (t) ×(4)\ \text{claims } W \text{ is AP / contains } (t)\ \times

X→(p,q); Y→(q,t); Z→(r); W→(p,r,s)\boxed{X\to(p,q);\ Y\to(q,t);\ Z\to(r);\ W\to(p,r,s)}

correct option (1)\boxed{\text{correct option } (1)}

More from Trigonometric Functions

All 28 Trigonometric Functions questions →