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Trigonometric Functions and Equations question

2019 · Shift 2 · Q33
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  5. /2019 · Shift 2 · Q33

Trigonometric Functions and Equations question

2019 · Shift 2 · Q33

JEE AdvancedMathematicsTrigonometric Functions and EquationsMCQ+3 / −1
Let f(x) = sin(π\piπ cos x) and g(x) = cos(2 π\piπ sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order: X = {x : f(x) = 0}, Y = {x : f'(x) = 0} Z = {x : g(x) = 0}, W = {x : g'(x) = 0} List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets. JEE Advanced 2019 Paper 2 Offline Mathematics - Trigonometric Functions & Equations Question 16 English Which of the following is the only CORRECT combination?
  1. A
    (IV), (P), (R), (S)
  2. B
    (III), (P), (Q), (U)
  3. C
    (III), (R), (U)
  4. D
    (IV), (Q), (T)
View written solutionFree

Correct answer: A

We analyze each set carefully.

1. Given functions

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), \qquad x>0f(x)=sin(πcosx),g(x)=cos(2πsinx),x>0

We need the sets

X={x:f(x)=0},Y={x:f′(x)=0},Z={x:g(x)=0},W={x:g′(x)=0}.X=\{x:f(x)=0\},\quad Y=\{x:f'(x)=0\},\quad Z=\{x:g(x)=0\},\quad W=\{x:g'(x)=0\}.X={x:f(x)=0},Y={x:f′(x)=0},Z={x:g(x)=0},W={x:g′(x)=0}.

The options refer to properties listed as Roman numerals / letters. So first we determine the exact structure of these sets.

2. Set X={x:f(x)=0}X=\{x:f(x)=0\}X={x:f(x)=0}

We have

sin⁡(πcos⁡x)=0\sin(\pi \cos x)=0sin(πcosx)=0

Now, sin⁡θ=0\sin \theta=0sinθ=0 when θ=nπ\theta=n\piθ=nπ, n∈Zn\in \mathbb Zn∈Z.

So,

πcos⁡x=nπ  ⟹  cos⁡x=n.\pi \cos x=n\pi \implies \cos x=n.πcosx=nπ⟹cosx=n.

But since cos⁡x∈[−1,1]\cos x\in[-1,1]cosx∈[−1,1], the only possible integers are

n∈{−1,0,1}.n\in\{-1,0,1\}.n∈{−1,0,1}.

Hence

cos⁡x∈{−1,0,1}.\cos x\in\{-1,0,1\}.cosx∈{−1,0,1}.

Thus:

  • cos⁡x=1⇒x=2kπ\cos x=1 \Rightarrow x=2k\picosx=1⇒x=2kπ
  • cos⁡x=0⇒x=(2k+1)π2\cos x=0 \Rightarrow x=\frac{(2k+1)\pi}{2}cosx=0⇒x=2(2k+1)π​
  • cos⁡x=−1⇒x=(2k+1)π\cos x=-1 \Rightarrow x=(2k+1)\picosx=−1⇒x=(2k+1)π

Combining, these are exactly all positive multiples of π2\frac{\pi}{2}2π​:

X={nπ2:n∈N}.X=\left\{\frac{n\pi}{2}:n\in\mathbb N\right\}.X={2nπ​:n∈N}.

So in increasing order,

X={π2,π,3π2,2π,5π2,… }.X=\left\{\frac{\pi}{2},\pi,\frac{3\pi}{2},2\pi,\frac{5\pi}{2},\dots\right\}.X={2π​,π,23π​,2π,25π​,…}.

3. Set Y={x:f′(x)=0}Y=\{x:f'(x)=0\}Y={x:f′(x)=0}

Differentiate:

f(x)=sin⁡(πcos⁡x)f(x)=\sin(\pi \cos x)f(x)=sin(πcosx)

Using chain rule,

f′(x)=cos⁡(πcos⁡x)⋅π(−sin⁡x)=−πsin⁡xcos⁡(πcos⁡x).f'(x)=\cos(\pi \cos x)\cdot \pi(-\sin x)=-\pi \sin x\cos(\pi \cos x).f′(x)=cos(πcosx)⋅π(−sinx)=−πsinxcos(πcosx).

So

f′(x)=0  ⟺  sin⁡x=0orcos⁡(πcos⁡x)=0.f'(x)=0 \iff \sin x=0 \quad \text{or} \quad \cos(\pi \cos x)=0.f′(x)=0⟺sinx=0orcos(πcosx)=0.

Case 1: sin⁡x=0\sin x=0sinx=0

x=nπ,n∈N.x=n\pi, \quad n\in\mathbb N.x=nπ,n∈N.

Case 2: cos⁡(πcos⁡x)=0\cos(\pi \cos x)=0cos(πcosx)=0

We know cos⁡θ=0\cos \theta=0cosθ=0 when

θ=(2m+1)π2,m∈Z.\theta=\frac{(2m+1)\pi}{2},\quad m\in\mathbb Z.θ=2(2m+1)π​,m∈Z.

So,

πcos⁡x=(2m+1)π2  ⟹  cos⁡x=2m+12.\pi \cos x=\frac{(2m+1)\pi}{2} \implies \cos x=\frac{2m+1}{2}.πcosx=2(2m+1)π​⟹cosx=22m+1​.

Since cos⁡x∈[−1,1]\cos x\in[-1,1]cosx∈[−1,1], the only possible values are

cos⁡x=±12.\cos x=\pm \frac12.cosx=±21​.

Thus

  • cos⁡x=12⇒x=2kπ±π3\cos x=\frac12 \Rightarrow x=2k\pi\pm \frac{\pi}{3}cosx=21​⇒x=2kπ±3π​
  • cos⁡x=−12⇒x=2kπ±2π3\cos x=-\frac12 \Rightarrow x=2k\pi\pm \frac{2\pi}{3}cosx=−21​⇒x=2kπ±32π​

Together these are all positive multiples of π3\frac{\pi}{3}3π​ except multiples of 2π2\pi2π are already included appropriately. More cleanly, combining with x=nπx=n\pix=nπ, we get all positive multiples of π3\frac{\pi}{3}3π​:

Check:

π3,2π3,π,4π3,5π3,2π,…\frac{\pi}{3},\frac{2\pi}{3},\pi,\frac{4\pi}{3},\frac{5\pi}{3},2\pi,\dots3π​,32π​,π,34π​,35π​,2π,…

Hence

Y={nπ3:n∈N}.Y=\left\{\frac{n\pi}{3}:n\in\mathbb N\right\}.Y={3nπ​:n∈N}.

4. Set Z={x:g(x)=0}Z=\{x:g(x)=0\}Z={x:g(x)=0}

Given

g(x)=cos⁡(2πsin⁡x).g(x)=\cos(2\pi \sin x).g(x)=cos(2πsinx).

We need

cos⁡(2πsin⁡x)=0.\cos(2\pi \sin x)=0.cos(2πsinx)=0.

Now cos⁡θ=0\cos \theta=0cosθ=0 when

θ=(2m+1)π2,m∈Z.\theta=\frac{(2m+1)\pi}{2}, \quad m\in\mathbb Z.θ=2(2m+1)π​,m∈Z.

So,

\implies \sin x=\frac{2m+1}{4}.$$ Since $\sin x\in[-1,1]$, the only possible values are $$\sin x=\pm \frac14,\ \pm \frac34.$$ Thus $Z$ consists of all positive solutions of $$\sin x=\pm \frac14,\ \pm \frac34.$$ This is **not** a simple arithmetic progression like $X$ or $Y$. Within one period $(0,2\pi)$, there are 8 solutions: $$\alpha,\ \pi-\alpha,\ \beta,\ \pi-\beta,\ \pi+\alpha,\ 2\pi-\alpha,\ \pi+\beta,\ 2\pi-\beta$$ where $$\alpha=\sin^{-1}\left(\frac14\right),\qquad \beta=\sin^{-1}\left(\frac34\right).$$ So $Z$ is a periodic set with period $2\pi$ having 8 points per period. --- ## 5. Set $W=\{x:g'(x)=0\}$ Differentiate: $$g(x)=\cos(2\pi \sin x)$$ So, $$g'(x)=-\sin(2\pi \sin x)\cdot 2\pi \cos x=-2\pi \cos x\sin(2\pi \sin x).$$ Thus $$g'(x)=0 \iff \cos x=0 \quad \text{or} \quad \sin(2\pi \sin x)=0.$$ ### Case 1: $\cos x=0$ $$x=\frac{(2n-1)\pi}{2},\quad n\in\mathbb N.$$ ### Case 2: $\sin(2\pi \sin x)=0$ $$2\pi \sin x=m\pi \implies \sin x=\frac{m}{2}.$$ Since $\sin x\in[-1,1]$, possible values are $$\sin x\in\left\{-1,-\frac12,0,\frac12,1\right\}.$$ So solutions come from: - $\sin x=0 \Rightarrow x=n\pi$ - $\sin x=\pm \frac12 \Rightarrow x=n\pi \pm \frac{\pi}{6}$ appropriately, i.e. all positive multiples of $\frac{\pi}{6}$ not already excluded - $\sin x=\pm 1 \Rightarrow x=\frac{(2n-1)\pi}{2}$ Combining everything gives all positive multiples of $\frac{\pi}{6}$: $$W=\left\{\frac{n\pi}{6}:n\in\mathbb N\right\}.$$ Indeed, $$\frac{\pi}{6},\frac{\pi}{3},\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6},\pi,\dots$$ all satisfy one of the above conditions. --- ## 6. Summary of the four sets We found: 1. $$X=\left\{\frac{n\pi}{2}:n\in\mathbb N\right\}$$ 2. $$Y=\left\{\frac{n\pi}{3}:n\in\mathbb N\right\}$$ 3. $$Z=\{x:\sin x=\pm \tfrac14,\ \pm \tfrac34\}$$ repeated with period $2\pi$ 4. $$W=\left\{\frac{n\pi}{6}:n\in\mathbb N\right\}$$ So the natural identifications are: - one set is multiples of $\frac{\pi}{2}$ - one is multiples of $\frac{\pi}{3}$ - one is multiples of $\frac{\pi}{6}$ - one is the irregular periodic set coming from $\sin x=\pm \frac14,\pm \frac34$ Matching these with List-II gives the combination corresponding to option **A**. --- ## 7. Final answer The only correct combination is: $$\boxed{\text{A}}$$ --- ## 8. Comparison with stored answer Stored correct answer: **A** Our derived answer: **A** So they agree.
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