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JEE Main 20202 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
2 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Trigonometric Ratios & Identities

JEE Main 2020 · 2 September · Evening Shift · Mathematics · Trigonometric Ratios & Identities

If the equation ^ 4 θ + ^ 4 θ + λ = 0 has real solutions for θ then...

If the equation cos⁡4θ+sin⁡4θ+λ=0\cos^{4}\theta + \sin^{4}\theta + \lambda = 0 has real solutions for θ\theta then λ\lambda lies in interval

Options

  1. A(−54,−1)\left(- \dfrac{5}{4}, - 1\right)
  2. B[−1,−12]\left[- 1, - \dfrac{1}{2}\right]
  3. C(−12,−14]\left(- \dfrac{1}{2}, - \dfrac{1}{4}\right]
  4. D[−32,−54]\left[- \dfrac{3}{2}, - \dfrac{5}{4}\right]

Correct answer

B. [−1,−12]\left[- 1, - \dfrac{1}{2}\right]

Step-by-step solution

−λ=sin⁡4θ+cos⁡4θ- \lambda = \sin^{4}\theta + \cos^{4}\theta

=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ= {\left(\sin^{2}\theta + \cos^{2}\theta\right)}^{2} - 2\sin^{2}\theta\cos^{2}\theta

=1−4sin⁡2θcos⁡2θ2= 1 - \dfrac{4\sin^{2}\theta\cos^{2}\theta}{2}

=1−sin⁡22θ2= 1 - \dfrac{\sin^{2}2\theta}{2}

⇒λ=sin⁡22θ2−1⇒ \lambda = \dfrac{\sin^{2}2\theta}{2} - 1

λ∈[−1,−12]\lambda \in \left[- 1, - \dfrac{1}{2}\right]

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