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Trigonometric Ratio and Identites question

2023 · 29 Jan · Shift 2 · Q31
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  5. /2023 · 29 Jan · Shift 2 · Q31

Trigonometric Ratio and Identites question

2023 · 29 Jan · Shift 2 · Q31

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
The set of all values of λ\lambdaλ for which the equation cos⁡22x−2sin⁡4x−2cos⁡2x=λ{\cos ^2}2x - 2{\sin ^4}x - 2{\cos ^2}x = \lambdacos22x−2sin4x−2cos2x=λ has a real solution xxx, is :
  1. A
    [−2,−1]\left[ { - 2, - 1} \right][−2,−1]
  2. B
    [−32,−1]\left[ { - {3 \over 2}, - 1} \right][−23​,−1]
  3. C
    [−2,−32]\left[ { - 2, - {3 \over 2}} \right][−2,−23​]
  4. D
    [−1,−12]\left[ { - 1, - {1 \over 2}} \right][−1,−21​]
View written solutionFree

Correct answer: B

  1. Given expression

We need the set of all values of λ\lambdaλ for which

cos⁡22x−2sin⁡4x−2cos⁡2x=λ\cos^2 2x-2\sin^4 x-2\cos^2 x=\lambdacos22x−2sin4x−2cos2x=λ

has a real solution.

So we need the range of

f(x)=cos⁡22x−2sin⁡4x−2cos⁡2x.f(x)=\cos^2 2x-2\sin^4 x-2\cos^2 x.f(x)=cos22x−2sin4x−2cos2x.
  1. Rewrite everything in terms of cos⁡2x\cos 2xcos2x

Use the identities:

sin⁡2x=1−cos⁡2x2,cos⁡2x=1+cos⁡2x2.\sin^2 x=\frac{1-\cos 2x}{2}, \qquad \cos^2 x=\frac{1+\cos 2x}{2}.sin2x=21−cos2x​,cos2x=21+cos2x​.

Hence,

sin⁡4x=(1−cos⁡2x2)2.\sin^4 x=\left(\frac{1-\cos 2x}{2}\right)^2.sin4x=(21−cos2x​)2.

Let

t=cos⁡2x,−1≤t≤1.t=\cos 2x, \qquad -1\le t\le 1.t=cos2x,−1≤t≤1.

Then

cos⁡22x=t2,\cos^2 2x=t^2,cos22x=t2, 2sin⁡4x=2(1−t2)2=(1−t)22,2\sin^4 x=2\left(\frac{1-t}{2}\right)^2=\frac{(1-t)^2}{2},2sin4x=2(21−t​)2=2(1−t)2​, 2cos⁡2x=2⋅1+t2=1+t.2\cos^2 x=2\cdot \frac{1+t}{2}=1+t.2cos2x=2⋅21+t​=1+t.

Therefore,

λ=t2−(1−t)22−(1+t).\lambda=t^2-\frac{(1-t)^2}{2}-(1+t).λ=t2−2(1−t)2​−(1+t).
  1. Simplify

Expand:

(1−t)2=1−2t+t2.(1-t)^2=1-2t+t^2.(1−t)2=1−2t+t2.

So,

λ=t2−1−2t+t22−1−t.\lambda=t^2-\frac{1-2t+t^2}{2}-1-t.λ=t2−21−2t+t2​−1−t.

Now simplify:

λ=t2−12+t−t22−1−t.\lambda=t^2-\frac12+t-\frac{t^2}{2}-1-t.λ=t2−21​+t−2t2​−1−t.

The ttt terms cancel, giving

λ=t22−32.\lambda=\frac{t^2}{2}-\frac32.λ=2t2​−23​.

Thus,

λ=12t2−32,−1≤t≤1.\lambda=\frac12 t^2-\frac32, \qquad -1\le t\le 1.λ=21​t2−23​,−1≤t≤1.
  1. Find the range

Since t∈[−1,1]t\in[-1,1]t∈[−1,1], we have

0≤t2≤1.0\le t^2\le 1.0≤t2≤1.

Therefore,

λ=12t2−32\lambda=\frac12 t^2-\frac32λ=21​t2−23​

will vary from

12(0)−32=−32\frac12(0)-\frac32=-\frac3221​(0)−23​=−23​

to

12(1)−32=−1.\frac12(1)-\frac32=-1.21​(1)−23​=−1.

So the range is

[−32,−1].\boxed{\left[-\frac32,-1\right]}.[−23​,−1]​.
  1. Check options
  • A: [−2,−1][-2,-1][−2,−1] ❌
  • B: [−32,−1]\left[-\frac32,-1\right][−23​,−1] ✅
  • C: [−2,−32]\left[-2,-\frac32\right][−2,−23​] ❌
  • D: [−1,−12]\left[-1,-\frac12\right][−1,−21​] ❌

So the correct option is

B.\boxed{\text{B}}.B​.
  1. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.

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