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BITSAT 2015
Exam
BITSAT
Year
2015
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2015 · Mathematics · Application of Derivatives

The function f(x)= x 2 + 2 x has a local minimum at

The function f(x)=x2+2x\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}+\frac{2}{\mathrm{x}} has a local minimum at

Options

  1. Ax=2x=2
  2. Bx=−2x=-2
  3. Cx=0x=0
  4. Dx=1x=1

Correct answer

D. x=1x=1

Step-by-step solution

x2+2x\frac{x}{2}+\frac{2}{x} is of the form x+1x≥2x+\frac{1}{x} \geq 2 and equality holds for x=1x=1
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