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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2023 · Paper 1 · Mathematics · Application of Derivatives

The interval in which the function f(x)=(4 x^2+1)/(x) is decreasing...

The interval in which the function f(x)=4x2+1xf(x)=\frac{4 x^2+1}{x} is decreasing is :

Options

  1. A(−12,12)\left(-\frac{1}{2}, \frac{1}{2}\right)
  2. B[−12,12]\left[-\frac{1}{2}, \frac{1}{2}\right]
  3. C(−1,1)(-1,1)
  4. D[−1,1][-1,1]

Correct answer

A. (−12,12)\left(-\frac{1}{2}, \frac{1}{2}\right)

Step-by-step solution

Given f(x)=4x2+1xf(x)=\frac{4 x^2+1}{x} Thus f′(x)=4−1x2f^{\prime}(x)=4-\frac{1}{x^2}
f(x)\mathrm{f}(\mathrm{x}) will be decreasing if \mathrmf′(x)<0\mathrmf^{\prime}(\mathrm{x}) \lt 0
Thus 4−1x2<0⇒1x2>4⇒−12<x<124-\frac{1}{x^2} \lt 0 \Rightarrow \frac{1}{x^2} \gt 4 \Rightarrow \frac{-1}{2} \lt x \lt \frac{1}{2}
Thus interval in which f(x)f(x) is decreasing, is \left(-\frac{1}{2}, \frac{1}{2}\right)$.
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