Question

Let \(f(x)\) be a function defined by \(f(x)=\begin{cases} \int _{0}^{x}(|t-2|+3)dt, & x>4 \\ 2x+8, & x\leq 4 \end{cases}\) then \(f(x)\) is ...

• A. continuous and differentiable everywhere
• B. neither continuous nor differentiable at \(x=4\)
• C. continuous everywhere, but not differentiable at \(x=4\)
• D. Differentiable at \(x=4\), but not continuous.

Hint:

Evaluate the integral at x = 4 to test continuity, then compare the left and right derivatives.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
At \(x=4\), continuity needs the left value, the right limit and \(f(4)\) to match. Differentiability needs equal left and right derivatives.

Step 2: Check continuity
\(f(4)=2(4)+8=16\).
Right limit: \(\int _{0}^{4}(|t-2|+3)dt=\int _{0}^{2}(5-t)dt+\int _{2}^{4}(t+1)dt=8+8=16\).
So \(f\) is continuous at \(x=4\).

Step 3: Check the derivatives
For \(x\leq 4\), \(f'(x)=2\).
For \(x>4\), \(f'(x)=|x-2|+3\), which tends to \(5\) as \(x\rightarrow 4^{+}\).

Step 4: Compare the options
The derivatives \(2\) and \(5\) differ, so \(f\) is continuous but not differentiable at \(x=4\).

Final Answer:
The function is continuous everywhere but not differentiable at \(x=4\). \[ \boxed{\text{(C) }\text{continuous everywhere, but not differentiable at }x=4} \]