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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Application of Derivatives

NDA 2024 · Mathematics · Application of Derivatives

Let f(x)=(x)/( x) ;(x > 1) Consider the following statements: 1. f^...

Let f(x)=xln⁡x;(x>1)f(x)=\frac{x}{\ln x} ;(x > 1) Consider the following statements: 1. f′′(e)=1e2.f(x)\quad f^{\prime \prime}(e)=\frac{1}{e} 2. \mathrm{f}(\mathrm{x}) attains local minimum value at x=e3.Alocalminimumvalueofx=e 3. A local minimum value off(x)isiseWhichofthestatementsgivenabovearecorrect?Which of the statements given above are correct?

Options

  1. A1 and 2 only
  2. B2 and 3 only
  3. C1 and 3 only
  4. D1,2 and 3

Correct answer

D. 1,2 and 3

Step-by-step solution

f′(x)=(ln⁡x)2×1x−(ln⁡x−1)×2ln⁡xx(ln⁡x)4=(ln⁡x)2−2(ln⁡x)2×2ln⁡xx(ln⁡x)4\begin{aligned} f^{\prime}(x) & =\frac{(\ln x)^2 \times \frac{1}{x}-(\ln x-1) \times 2 \frac{\ln x}{x}}{(\ln x)^4} \\ & =\frac{(\ln x)^2-2(\ln x)^2 \times 2 \ln x}{x(\ln x)^4}\end{aligned}\Rightarrow \quad f^{\prime \prime}(e)=\frac{1-2+2}{e \times 1}=\frac{1}{e}( > 0)\RightarrowStatement(1)iscorrect.SinceStatement (1) is correct. Sincef^{\prime \prime}(e) > 0(true)(true)f(x)attainslocalminimaatattains local minima atx=e\RightarrowStatement(2)iscorrect.AlocalminimumvalueoccursatStatement (2) is correct. A local minimum value occurs atx=ef(x)atatx=e,is, isf(e)=\frac{e}{\ln e}=e\Rightarrow$ Statement (3) is correct. Hence, Statements (1), (2) and (3) are correct.
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