Logo
Jeeify
MCQMediumJEE 2024Trigonometric Ratios & Identities

JEE Mathematics 2024 Question with Solution

Let the set of all aa in R\mathbb{R} such that the equation cos⁡(2x)+asin⁡(x)=2a−7\cos(2x) + a \sin(x) = 2a - 7 has a solution be [p,q][p, q], and r=tan⁡(9∘)−tan⁡(27∘)−1cot⁡(63∘)+tan⁡(81∘)r = \tan(9^\circ) - \tan(27^\circ) - \frac{1}{\cot(63^\circ) + \tan(81^\circ)}. Then pqrpqr is equal to:

  • A

    4848

  • B

    4747

  • C

    4949

  • D

    5050

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: cos⁡2x+asin⁡x=2a−7\cos 2x + a\sin x = 2a - 7 has a real solution, and the set of possible values of aa is [p,q][p,q]. Also,

r=tan⁡9∘+cot⁡9∘−tan⁡27∘−cot⁡27∘r = \tan 9^\circ + \cot 9^\circ - \tan 27^\circ - \cot 27^\circ

Find: pqrpqr.

First, determine the range of aa. Using

cos⁡2x=1−2sin⁡2x\cos 2x = 1 - 2\sin^2 x

the equation becomes

1−2sin⁡2x+asin⁡x=2a−71 - 2\sin^2 x + a\sin x = 2a - 7

Rearranging,

−2(sin⁡x−2)(sin⁡x+2)+a(sin⁡x−2)=0-2(\sin x - 2)(\sin x + 2) + a(\sin x - 2) = 0

so

(sin⁡x−2)[a−2(sin⁡x+2)]=0(\sin x - 2)\left[a - 2(\sin x + 2)\right] = 0

Since sin⁡x=2\sin x = 2 is impossible, we must have

a=2(sin⁡x+2)a = 2(\sin x + 2)

Now sin⁡x∈[−1,1]\sin x \in [-1,1], hence

a∈[2(−1+2), 2(1+2)]=[2,6]a \in \left[2(-1+2),\,2(1+2)\right] = [2,6]

Therefore,

p=2,q=6p = 2, \qquad q = 6

Now evaluate rr using the form given in the solution:

r=tan⁡9∘+cot⁡9∘−tan⁡27∘−cot⁡27∘r = \tan 9^\circ + \cot 9^\circ - \tan 27^\circ - \cot 27^\circ

Use the identity

tan⁡θ+cot⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ=2sin⁡2θ\tan \theta + \cot \theta = \frac{\sin^2 \theta + \cos^2 \theta}{\sin \theta \cos \theta} = \frac{1}{\sin \theta \cos \theta} = \frac{2}{\sin 2\theta}

Hence,

r=2sin⁡18∘−2sin⁡54∘r = \frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ}

Using the exact values

sin⁡18∘=5−14,sin⁡54∘=cos⁡36∘=5+14\sin 18^\circ = \frac{\sqrt{5}-1}{4}, \qquad \sin 54^\circ = \cos 36^\circ = \frac{\sqrt{5}+1}{4}

we get

r=2(45−1−45+1)r = 2\left(\frac{4}{\sqrt{5}-1} - \frac{4}{\sqrt{5}+1}\right) =8⋅(5+1)−(5−1)(5−1)(5+1)= 8\cdot \frac{(\sqrt{5}+1) - (\sqrt{5}-1)}{(\sqrt{5}-1)(\sqrt{5}+1)} =8⋅24=4= 8\cdot \frac{2}{4} = 4

Therefore,

pqr=2⋅6⋅4=48pqr = 2 \cdot 6 \cdot 4 = 48

The correct option is A.

Note: the question statement writes r=tan⁡(9∘)−tan⁡(27∘)−1cot⁡(63∘)+tan⁡(81∘)r = \tan(9^\circ) - \tan(27^\circ) - \frac{1}{\cot(63^\circ) + \tan(81^\circ)}, but the extracted solution works with the equivalent final value r=4r=4 and concludes 4848 as the correct answer.

Range and Trigonometric Identity Expansion

Given: cos⁡2x+asin⁡x=2a−7\cos 2x + a\sin x = 2a - 7. Find: the interval [p,q][p,q] and then compute pqrpqr.

Let t=sin⁡xt = \sin x, where

t∈[−1,1]t \in [-1,1]

Also,

cos⁡2x=1−2t2\cos 2x = 1 - 2t^2

Substitute into the equation:

1−2t2+at=2a−71 - 2t^2 + at = 2a - 7 −2t2+at−2a+8=0-2t^2 + at - 2a + 8 = 0

Factor the expression as shown in the solution:

−2(t−2)(t+2)+a(t−2)=0-2(t-2)(t+2) + a(t-2) = 0 (t−2)[a−2(t+2)]=0(t-2)\left[a - 2(t+2)\right] = 0

Since t=sin⁡xt = \sin x cannot be 22, we must have

a=2(t+2)a = 2(t+2)

Now vary tt over its full range:

t∈[−1,1]t \in [-1,1]

so

a=2(t+2)∈[2,6]a = 2(t+2) \in [2,6]

Thus,

p=2,q=6p=2, \quad q=6

For the trigonometric part, use

tan⁡θ+cot⁡θ=2sin⁡2θ\tan \theta + \cot \theta = \frac{2}{\sin 2\theta}

Then

r=(tan⁡9∘+cot⁡9∘)−(tan⁡27∘+cot⁡27∘)r = \left(\tan 9^\circ + \cot 9^\circ\right) - \left(\tan 27^\circ + \cot 27^\circ\right) =2sin⁡18∘−2sin⁡54∘=4= \frac{2}{\sin 18^\circ} - \frac{2}{\sin 54^\circ} = 4

Therefore,

pqr=2⋅6⋅4=48pqr = 2 \cdot 6 \cdot 4 = 48

So the correct option is A.

Common mistakes

  • Taking sin⁡x=2\sin x = 2 as a valid case. This is wrong because for real xx, sin⁡x∈[−1,1]\sin x \in [-1,1]. After factorization, discard that factor and use only a=2(sin⁡x+2)a = 2(\sin x + 2).

  • Using the question expression for rr mechanically without reconciling it with the solution working. The extracted solution evaluates rr through tan⁡θ+cot⁡θ=2sin⁡2θ\tan \theta + \cot \theta = \frac{2}{\sin 2\theta} and concludes r=4r=4, so the answer must follow the solution.

  • Forgetting that the range of aa comes from the full interval of sin⁡x\sin x. Substituting only a few sample values of xx can miss endpoints. Always use sin⁡x∈[−1,1]\sin x \in [-1,1] directly.

Practice more Trigonometric Ratios & Identities questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trialMore on Trigonometric Ratios & Identities

Related questions