Verified Solution • 2025–26 Session moderate

QJEEM23Ma6494: JEE Main 2023 Mathematics - Class 12 JEEM23Ma6494 Solution: If the local maximum value of the function , is , then is equal …

If the local maximum value of the function f(x)=(3e2sin⁡x)sin⁡2x,x∈(0,π2)f(x)=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, x \in\left(0, \frac{\pi}{2}\right) , is ke\frac{k}{e} , then (ke)8+k8e5+k8\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8} is equal to

Multiple Choice Options

  • A e3+e6+e10e^{3}+e^{6}+e^{10}
  • B e3+e5+e11e^{3}+e^{5}+e^{11}
  • C e3+e6+e11e^{3}+e^{6}+e^{11}
  • D e5+e6+e11e^{5}+e^{6}+e^{11}

Step-by-Step Verified Solution

 Let y=(3e2sin⁡x)sin⁡2x\text { Let } y=\left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^2 x}

ln⁡y=sin⁡2x⋅ln⁡(3e2sin⁡x)\ln \mathrm{y}=\sin ^2 \mathrm{x} \cdot \ln \left(\frac{\sqrt{3 \mathrm{e}}}{2 \sin \mathrm{x}}\right)

1yy′=ln⁡(3e2sin⁡x)2sin⁡xcos⁡x+sin⁡2x2sin⁡x3e3e2(−cosec⁡xcot⁡x)\frac{1}{y} y^{\prime}=\ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x+\sin ^2 x \frac{2 \sin x}{\sqrt{3 e}} \frac{\sqrt{3 e}}{2}(-\operatorname{cosec} x \cot x)

For maxima or minima, dydx=0\frac{d y}{d x}=0

⇒ln⁡(3e2sin⁡x)2sin⁡xcos⁡x−sin⁡xcos⁡x=0\Rightarrow \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right) 2 \sin x \cos x-\sin x \cos x=0

⇒sin⁡xcos⁡x[2ln⁡(3e2sin⁡x)−1]=0\Rightarrow \sin x \cos x\left[2 \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)-1\right]=0

⇒ln⁡(3e4sin⁡2x)=1\Rightarrow \ln \left(\frac{3 \mathrm{e}}{4 \sin ^2 \mathrm{x}}\right)=1

⇒3e4sin⁡2x=e\Rightarrow \frac{3 e}{4 \sin ^2 x}=e

⇒sin⁡2x=34\Rightarrow \sin ^2 x=\frac{3}{4}

⇒sin⁡x=32( as x∈(0,π2))\Rightarrow \sin \mathrm{x}=\frac{\sqrt{3}}{2} \quad\left(\text { as } \mathrm{x} \in\left(0, \frac{\pi}{2}\right)\right)

⇒ Local max value =(3e3)3/4=e3/8=ke\Rightarrow \text { Local max value }=\left(\frac{\sqrt{3 \mathrm{e}}}{\sqrt{3}}\right)^{3 / 4}=\mathrm{e}^{3 / 8}=\frac{\mathrm{k}}{\mathrm{e}}

⇒k8=e11\Rightarrow \mathrm{k}^8=\mathrm{e}^{11}

∴\therefore (ke)8+k8e5+k8=e3+e6+e11\left(\frac{k}{e}\right)^8+\frac{k^8}{e^5}+k^8=e^3+e^6+e^{11}

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