Search any question & find its solution
Question:
Answered & Verified by Expert
Let $\mathrm{f}(\mathrm{x})=[|\mathrm{x}|-\mid \mathrm{x}-1]]^{2}$
Which of the following equations is/are correct?
$1 \quad \mathrm{f}(-2)=\mathrm{f}(5)$
$2 \quad f^{\prime \prime}(-2)+f^{\prime \prime}(0.5)+f^{\prime \prime}(3)=4$
Select the correct answer using the code given below:
Options:
Which of the following equations is/are correct?
$1 \quad \mathrm{f}(-2)=\mathrm{f}(5)$
$2 \quad f^{\prime \prime}(-2)+f^{\prime \prime}(0.5)+f^{\prime \prime}(3)=4$
Select the correct answer using the code given below:
Solution:
2087 Upvotes
Verified Answer
The correct answer is:
1 only
Given $f(x)=(|x|-|x+1|)^{2}$
$f(x)=\left\{\begin{array}{cc}1 & x \leq 0 \\ (2 x-1)^{2} & 0 < x < 1 \\ 1 & x \geq 1\end{array}\right.$
For $x=-2$
$f(x)=1$ so $f(-2)=1$
For $x=5$ $f(x)=1 \Rightarrow f(5)=1$
Hence $f(-2)=f(5)$
Now, for $x=-1$
$f^{\prime \prime}(x)=0$
$f^{\prime \prime}(-2)=0$
For $x=0.5$
$f^{\prime \prime}(x)=8 \Rightarrow f^{\prime \prime}(0.5)=8$
For $x=3$
$f^{\prime \prime}(x)=0 \Rightarrow f^{\prime \prime}(3)=0$
$\Rightarrow f^{\prime \prime}(-2)+f^{\prime \prime}(0.5)+f^{\prime \prime}(3)=8 \neq 4$
Only statement 1 is correct.
$f(x)=\left\{\begin{array}{cc}1 & x \leq 0 \\ (2 x-1)^{2} & 0 < x < 1 \\ 1 & x \geq 1\end{array}\right.$
For $x=-2$
$f(x)=1$ so $f(-2)=1$
For $x=5$ $f(x)=1 \Rightarrow f(5)=1$
Hence $f(-2)=f(5)$
Now, for $x=-1$
$f^{\prime \prime}(x)=0$
$f^{\prime \prime}(-2)=0$
For $x=0.5$
$f^{\prime \prime}(x)=8 \Rightarrow f^{\prime \prime}(0.5)=8$
For $x=3$
$f^{\prime \prime}(x)=0 \Rightarrow f^{\prime \prime}(3)=0$
$\Rightarrow f^{\prime \prime}(-2)+f^{\prime \prime}(0.5)+f^{\prime \prime}(3)=8 \neq 4$
Only statement 1 is correct.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.