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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Differentiation

NDA 2023 · Mathematics · Differentiation

Let a function f be defined on ℝ-\ 0\ and 2 f(x)+f ((1)/(x))=x+3 If...

Let a function f be defined on R−{0}and\mathbb{R}-\{0\} and2 f(x)+f(1x)=x+3\left(\frac{1}{x}\right)=x+3 If f is differentiable, then what is (0.5)\quad(0.5) equal to?

Options

  1. A14\frac{1}{4}
  2. B23\frac{2}{3}
  3. C2
  4. D4

Correct answer

C. 2

Step-by-step solution

 Since, 2f(x)+f(1x)=x+3⇒2f′(x)+f′(1x)(−1x2)=1 Put x=2,12⇒2f′(2)−14f′(12)=1...(i) and, 2f′(12)−4f′(2)=1...(ii)\begin{aligned} & \text{ Since, } 2 \mathrm{f}(\mathrm{x})+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\mathrm{x}+3 \\ & \Rightarrow 2 \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime}\left(\frac{1}{\mathrm{x}}\right)\left(-\frac{1}{\mathrm{x}^2}\right)=1 \\ & \text{ Put } \mathrm{x}=2, \frac{1}{2} \\ & \Rightarrow 2 \mathrm{f}^{\prime}(2)-\frac{1}{4} \mathrm{f}^{\prime}\left(\frac{1}{2}\right)=1...(i) \\ & \text{ and, } 2 \mathrm{f}^{\prime}\left(\frac{1}{2}\right)-4 \mathrm{f}^{\prime}(2)=1...(ii) \end{aligned} after solving (i) \& (ii), we get f′(12)=2,f′(2)=34\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=2, \mathrm{f}^{\prime}(2)=\frac{3}{4} So, f′(0.5)=2\mathrm{f}^{\prime}(0.5)=2
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