Search any question & find its solution
Question:
Answered & Verified by Expert
If $f(x)=|x-5|+|x+5|+|x-4|+|x+4|$, then $\frac{f^{\prime}(1)-f^{\prime}(-6)}{f^{\prime}(-1)+f^{\prime}(6)}=$
Options:
Solution:
2183 Upvotes
Verified Answer
The correct answer is:
$1$
Given the function
$\begin{aligned}
& f(x)=|x-5|+|x+5|+|x-4|+|x+4| \\
& =\left\{\begin{array}{cc}
-4 x & x \leq-5 \\
-2 x+10 & -5 < x \leq-4 \\
18 & -4 < x \leq 4 \\
2 x+10 & 4 < x \leq 5 \\
4 x & 5 < x
\end{array}\right.
\end{aligned}$
Now $f^{\prime}(x)=\left\{\begin{array}{cc}-4 & x \leq-5 \\ -2 & -5 < x \leq-4 \\ 0 & -4 < x \leq 4 \\ 2 & 4 < x \leq 5 \\ 4 & 5 < x\end{array}\right.$
So, $\frac{f^{\prime}(1)-f^{\prime}(-6)}{f^{\prime}(-1)+f^{\prime}(6)}=\frac{0-(-4)}{0+4}=\frac{4}{4}=1$
$\begin{aligned}
& f(x)=|x-5|+|x+5|+|x-4|+|x+4| \\
& =\left\{\begin{array}{cc}
-4 x & x \leq-5 \\
-2 x+10 & -5 < x \leq-4 \\
18 & -4 < x \leq 4 \\
2 x+10 & 4 < x \leq 5 \\
4 x & 5 < x
\end{array}\right.
\end{aligned}$
Now $f^{\prime}(x)=\left\{\begin{array}{cc}-4 & x \leq-5 \\ -2 & -5 < x \leq-4 \\ 0 & -4 < x \leq 4 \\ 2 & 4 < x \leq 5 \\ 4 & 5 < x\end{array}\right.$
So, $\frac{f^{\prime}(1)-f^{\prime}(-6)}{f^{\prime}(-1)+f^{\prime}(6)}=\frac{0-(-4)}{0+4}=\frac{4}{4}=1$
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.