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NDA 2016
Exam
NDA
Year
2016
Subject
Mathematics
Chapter
Functions

NDA 2016 · Mathematics · Functions

Paragraph: Let f(x)=[|x|-|x-1|]^2 Question: Which of the following...

Paragraph:  Let f(x)=[∣x∣−∣x−1∣]2\text{ Let } f(x)=[|x|-|x-1|]^2Question: Which of the following equations is/are correct? 1. f(−2)=f(5)2.f′′(−2)+f′′(0.5)+f′′(3)=4\mathrm{f}(-2)=\mathrm{f}(5) 2. \mathrm{f}^{\prime \prime}(-2)+\mathrm{f}^{\prime \prime}(0.5)+\mathrm{f}^{\prime \prime}(3)=4 Select the correct answer using the code given below:

Options

  1. A1 only
  2. B2 only
  3. CBoth 1 and 2
  4. DNeither 1 nor 2

Correct answer

A. 1 only

Step-by-step solution

Given f(x)=(|x|-|x+1|)^2f(x)=\left\{1x≤0(2x−1)20<x<11x≥1\begin{array}{cc} 1 & x \leq 0 \\ (2 x-1)^2 & 0 < x < 1 \\ 1 & x \geq 1 \end{array}\right. For x=−2x=-2f(x)=1 so f(−2)=1 For x=5f(x)=1⇒f(5)=1\begin{aligned} & f(x)=1 \text{ so } f(-2)=1 \\ & \text{ For } x=5 \\ & f(x)=1 \Rightarrow f(5)=1 \end{aligned}Hence $f(-2)=f(5)$ Now, for $x=-1\begin{aligned} & f^{\prime \prime}(x)=0 \\ & f^{\prime \prime}(-2)=0 \end{aligned}$ For $x=0.5f^{\prime \prime}(x)=8 \Rightarrow f^{\prime \prime}(0.5)=8$ For $x=3\begin{aligned} & f^{\prime \prime}(x)=0 \Rightarrow f^{\prime \prime}(3)=0 \\ & \Rightarrow f^{\prime \prime}(-2)+f^{\prime \prime}(0.5)+f^{\prime \prime}(3)=8 \neq 4 \end{aligned}$$$ Only statement 1 is correct.
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