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Limits Continuity and Differentiability question

2023 · 25 Jan · Shift 2 · Q26
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  5. /2023 · 25 Jan · Shift 2 · Q26

Limits Continuity and Differentiability question

2023 · 25 Jan · Shift 2 · Q26

JEE MainMathematicsLimits Continuity and DifferentiabilityMCQ+4 / −1
If the function f(x)={(1+∣cos⁡x∣)λ∣cos⁡x∣,0<x<π2μ,x=π2ecot⁡6xcot⁡4x,π2<x<πf(x) = \left\{ \begin{matrix} (1 + |\cos x|)^{\frac{\lambda}{|\cos x|}} & , & 0 \lt x \lt \frac{\pi}{2} \\ \mu & , & x = \frac{\pi}{2} \\ e^{\frac{\cot 6x}{\cot 4x}} & , & \frac{\pi}{2} \lt x \lt \pi \end{matrix} \right.f(x)=⎩⎨⎧​(1+∣cosx∣)∣cosx∣λ​μecot4xcot6x​​,,,​0<x<2π​x=2π​2π​<x<π​ is continuous at x=π2x = \frac{\pi}{2}x=2π​, then 9λ+6log⁡eμ+μ6−e6λ9\lambda + 6\log_e \mu + \mu^6 - e^{6\lambda}9λ+6loge​μ+μ6−e6λ is equal to
  1. A
    11
  2. B
    10
  3. C
    8
  4. D
    2e 4^44 + 8
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