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JEE Main 20206 SeptemberEvening Shift
Exam
JEE Main
Year
2020
Date
6 September
Shift
Evening Shift
Subject
Mathematics
Chapter
Application of Derivatives

JEE Main 2020 · 6 September · Evening Shift · Mathematics · Application of Derivatives

The set of all real values λ for which the function f(x) = (1 - ^ 2...

The set of all real values λ\lambda for which the function f(x)=(1−cos⁡2x).(λ+sin⁡x),xε(−π2,π2)f\left(x\right) = \left(1 - \cos^{2}x\right) . \left(\lambda + \sin x\right), xε\left(- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right), has exactly one maxima and exactly one minima, is :

Options

  1. A(−12,12)−{0}\left(- \dfrac{1}{2}, \dfrac{1}{2}\right) - \left\{0\right\}
  2. B(−32,32)\left(- \dfrac{3}{2}, \dfrac{3}{2}\right)
  3. C(−12,12)\left(- \dfrac{1}{2}, \dfrac{1}{2}\right)
  4. D(−32,32)−{0}\left(- \dfrac{3}{2}, \dfrac{3}{2}\right) - \left\{0\right\}

Correct answer

D. (−32,32)−{0}\left(- \dfrac{3}{2}, \dfrac{3}{2}\right) - \left\{0\right\}

Step-by-step solution

f(x)=sin⁡2x(λ+sin⁡x)f\left(x\right) = \sin^{2}x\left(\lambda + \sin x\right)

f′(x)=sin⁡xcos⁡x(2λ+3sin⁡x)f ' \left(x\right) = \sin x \cos x\left(2\lambda + 3\sin x\right)

sin⁡x=0\sin x = 0

(one point)

sin⁡x=−2λ3∈(−1,1)−{0}\sin x = - \dfrac{2\lambda}{3} \in \left(- 1, 1\right) - \left\{0\right\} .....(1)

λ∈(−32,32)−{0}\lambda \in \left(- \dfrac{3}{2}, \dfrac{3}{2}\right) - \left\{0\right\}

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