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Sets and Relations question

2016 · 10 Apr · Shift 1 · Q27
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  5. /2016 · 10 Apr · Shift 1 · Q27

Sets and Relations question

2016 · 10 Apr · Shift 1 · Q27

JEE MainMathematicsSets and RelationsMCQ+4 / −1
Let P = {θ\thetaθ : sin θ−\theta -θ− cos θ\thetaθ=2 cos⁡θ\sqrt 2 \,\cos \theta2​cosθ} and Q = {θ\thetaθ : sin θ\thetaθ + cos θ\thetaθ=2 sin⁡θ\sqrt 2 \,\sin \theta2​sinθ} be two sets. Then
  1. A
    P ⊂\subset⊂ Q and Q −-− P eϕe \phieϕ
  2. B
    Q ot⊂ot\subsetot⊂ P
  3. C
    P ot⊂ot\subsetot⊂ Q
  4. D
    P = Q
View written solutionFree

Correct answer: D

  1. Solve for set PPP

Given

sin⁡θ−cos⁡θ=2 cos⁡θ\sin\theta - \cos\theta = \sqrt{2}\,\cos\thetasinθ−cosθ=2​cosθ

Rearrange:

sin⁡θ=(2+1)cos⁡θ\sin\theta = (\sqrt{2}+1)\cos\thetasinθ=(2​+1)cosθ

If cos⁡θ=0\cos\theta=0cosθ=0, then LHS =sin⁡θ=\sin\theta=sinθ and RHS =0=0=0, so equation is not satisfied. Hence we may divide by cos⁡θ\cos\thetacosθ:

tan⁡θ=2+1\tan\theta = \sqrt{2}+1tanθ=2​+1

Now use the standard identity

tan⁡3π8=tan⁡67.5∘=2+1\tan\frac{3\pi}{8}=\tan 67.5^\circ=\sqrt{2}+1tan83π​=tan67.5∘=2​+1

So,

P={θ:θ=nπ+3π8,  n∈Z}P=\left\{\theta:\theta=n\pi+\frac{3\pi}{8},\; n\in\mathbb Z\right\}P={θ:θ=nπ+83π​,n∈Z}
  1. Solve for set QQQ

Given

sin⁡θ+cos⁡θ=2 sin⁡θ\sin\theta + \cos\theta = \sqrt{2}\,\sin\thetasinθ+cosθ=2​sinθ

Rearrange:

cos⁡θ=(2−1)sin⁡θ\cos\theta = (\sqrt{2}-1)\sin\thetacosθ=(2​−1)sinθ

If sin⁡θ=0\sin\theta=0sinθ=0, then LHS =cos⁡θ=\cos\theta=cosθ and RHS =0=0=0, so equation is not satisfied. Hence we may divide by sin⁡θ\sin\thetasinθ:

cot⁡θ=2−1\cot\theta = \sqrt{2}-1cotθ=2​−1

Thus

tan⁡θ=12−1=2+1\tan\theta = \frac{1}{\sqrt{2}-1}=\sqrt{2}+1tanθ=2​−11​=2​+1

Therefore,

Q={θ:θ=nπ+3π8,  n∈Z}Q=\left\{\theta:\theta=n\pi+\frac{3\pi}{8},\; n\in\mathbb Z\right\}Q={θ:θ=nπ+83π​,n∈Z}
  1. Compare PPP and QQQ

We obtained

P=Q={θ:tan⁡θ=2+1}P=Q=\left\{\theta:\tan\theta=\sqrt{2}+1\right\}P=Q={θ:tanθ=2​+1}

Hence,

P=QP=QP=Q
  1. Check options
  • A: P⊂QP\subset QP⊂Q and Q−P≠ϕQ-P\ne\phiQ−P=ϕ — false, because if P=QP=QP=Q, then Q−P=ϕQ-P=\phiQ−P=ϕ.
  • B: Q⊄PQ\not\subset PQ⊂P — false, since Q=PQ=PQ=P.
  • C: P⊄QP\not\subset QP⊂Q — false, since P=QP=QP=Q.
  • D: P=QP=QP=Q — true.

Therefore, the correct option is

D\boxed{D}D​
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