NCERT Exemplar - MCQs

Chapter 6 Class 12 Application of Derivatives
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## (C) e (1/e)Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β  Β (D) 1/e (1/e)

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Question 30 The maximum value of (1/π₯)^π₯ is: (A) e (B) ee (C) π^(1/π) (D) γ1/πγ^(1/π) Let f (π₯) = (1/π₯)^π₯ To find maximum value, we need to differentiate f(x) For differentiating f (π₯), we use logarithmic differentiation f (π₯) = (1/π₯)^π₯ log (f(x)) = π log (π/π) Differentiating wrt π₯ π/(π(π)) fβ(x) = 1βlog (π/π) + π Γ (π/(π/π)) Γ ((βπ)/π^π ) 1/(π(π₯)) fβ(x) = log (1/π₯) + π₯ Γ (π₯) Γ ((β1)/π₯^2 ) 1/(π(π₯)) fβ(x) = log (1/π₯) + π₯^2 Γ ((β1)/π₯^2 ) 1/(π(π₯)) fβ(x) = log (1/π₯) β 1 fβ(x) = f(x) [logβ‘γ(1/π₯)β1γ ] Putting f (π₯) =(1/π₯)^π₯ fβ(x) = (π/π)^π (π₯π¨π β‘γ(π/π)βπγ ) Putting fβ(x) = 0 (1/π₯)^π₯ (logβ‘γ(1/π₯)β1γ ) = 0 Since, there is only one critical point, so it will be point of maxima Either (π/π)^π = 0 Since, it is an exponential function It can never be zero. Or (πππβ‘γ(π/π)βπγ ) = 0 log 1/π₯ = 1 Taking exponential on both sides e^logβ‘γ1/xγ = π^1 1/π₯ = e π = π^(βπ) Putting π₯ = 1/π in f (x) f (π/π) = (1/(1/π))^(1/π) f (π/π) = π^(π/π) So, the correct answer is (C)