Verified Solution • 2025–26 Session moderate

QJEEM20Ma08e3: JEE Main 2020 Mathematics - Class 12 JEEM20Ma08e3 Solution: The set of all real values of for which the function has…

The set of all real values of λ\lambda for which the function

f(x)=(1−cos⁡2x)(λ+sin⁡x),x∈(−π2,π2)f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right),x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

has exactly one maxima and exactly one minima, is :

Multiple Choice Options

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

Step-by-Step Verified Solution

f(x)=(1−cos⁡2x)(λ+sin⁡x)f(x) = \left( {1 - {{\cos }^2}x} \right)\left( {\lambda + \sin x} \right)

⇒\Rightarrow f(x) = sin2 x( λ\lambda + sinx) ....(1)

∴\therefore f'(x) = 2sinx cosx ( λ\lambda +sinx) + sin2x (cosx)

⇒\Rightarrow f'(x) = sin2x( 2λ+3sin⁡x2{{2\lambda + 3\sin x} \over 2} )

For maixma and minima, f'(x) = 0

∴\therefore sin2x = 0 or 2 λ\lambda + 3sinx = 0

when sin2x = 0 ⇒\Rightarrow x = 0

or when 2 λ\lambda + 3sinx = 0

⇒\Rightarrow sin x = −2λ3- {{2\lambda } \over 3}

As x∈(−π2,π2)x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

∴\therefore -1 < sinx < 1

⇒\Rightarrow -1 < −2λ3- {{2\lambda } \over 3} < 1

⇒\Rightarrow −32- {3 \over 2} < λ\lambda < 32{3 \over 2}

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

Note : If λ\lambda = 0 ⇒\Rightarrow f(x) = sin3x [from (1)]

Which is monotonic. so no maxima/minima.

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring