MHT CET · Maths · Differentiation
If \(f(x)=|x-3|\), then \(f^{\prime}(3)\) is
- A \(-1\)
- B 1
- C 0
- D does not exist
Answer & Solution
Correct Answer
(D) does not exist
Step-by-step Solution
Detailed explanation
Given, \(f(x)=|x-3|\) Redefine this function:
\(
f(x)=\left\{\begin{array}{cc}
3-x, & x < 3 \\
0, & x=3 \\
x-3, & x>3
\end{array}\right.
\)
\(f^{\prime}(x)=\left\{\begin{aligned}-1, & x < 3 \\ 0, & x=3 \\ 1, & x>3 \end{aligned}\right.\)
It is clear that, \(L f^{\prime}(3) \neq R f^{\prime}(3)\). \(\therefore f^{\prime}(3)\) does not exist.
\(
f(x)=\left\{\begin{array}{cc}
3-x, & x < 3 \\
0, & x=3 \\
x-3, & x>3
\end{array}\right.
\)
\(f^{\prime}(x)=\left\{\begin{aligned}-1, & x < 3 \\ 0, & x=3 \\ 1, & x>3 \end{aligned}\right.\)
It is clear that, \(L f^{\prime}(3) \neq R f^{\prime}(3)\). \(\therefore f^{\prime}(3)\) does not exist.
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