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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

BITSAT 2023 · Paper 2 · Mathematics · Application of Derivatives

The function f(x)= (1+x)-(2 x)/(2+x) is increasing on

The function f(x)=log⁡(1+x)−2x2+xf(x)=\log (1+x)-\frac{2 x}{2+x} is increasing on

Options

  1. A(0,∞)(0, \infty)
  2. B(−∞,0)(-\infty, 0)
  3. C(−∞,∞)(-\infty, \infty)
  4. D
    None of these

Correct answer

A. (0,∞)(0, \infty)

Step-by-step solution

Given f(x)=log⁡(1+x)−2x2+xf(x)=\log (1+x)-\frac{2 x}{2+x}f′(x)=11+x−(2+x)(2)−2x(2+x)2=11+x−4(2+x)2=(2+x)2−4−4x(1+x)(2+x)2=x2(1+x)(2+x)2>0 for all x∈(0,∞)\begin{aligned} \\ & f^{\prime}(x)=\frac{1}{1+x}-\frac{(2+x)(2)-2 x}{(2+x)^2} \\ \\ & =\frac{1}{1+x}-\frac{4}{(2+x)^2} \\ \\ & =\frac{(2+x)^2-4-4 x}{(1+x)(2+x)^2}=\frac{x^2}{(1+x)(2+x)^2} > 0 \text{ for all } \\ \\ & x \in(0, \infty) \\ \end{aligned}$Thus,givenfunctionThus, given function\mathrm{f}(\mathrm{x})isincreasingonis increasing on(0,,\infty)$.
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