Question

Let \(f(x)\) be defined by: \[ f(x)= \begin{cases} \displaystyle \int_x^6 (|t-2|+3)\,dt, & x>4 2x+8, & x\le4 \end{cases} \] Then at \(x=4\), \(f(x)\) is:

• A. Continuous but not differentiable
• B. Differentiable
• C. Discontinuous
• D. None of these

Hint:

A function cannot be differentiable unless it is first continuous at that point.

Answer 1

The Correct Option is A

Concept: To check continuity and differentiability at a point:
• Compare LHL, RHL and function value.
• Compare left derivative and right derivative.

Step 1: Find \(f(4)\). Since: \[ x\le4 \] we use: \[ f(x)=2x+8 \] Therefore: \[ f(4)=2(4)+8=16 \]

Step 2: Find right hand limit. For \(x>4\): \[ f(x)=\int_x^6(|t-2|+3)\,dt \] Since \(t>2\) in interval \([4,6]\): \[ |t-2|=t-2 \] Thus: \[ f(x)=\int_x^6(t+1)\,dt \] \[ = \left[\frac{t^2}{2}+t\right]_x^6 \] \[ = 24-\frac{x^2}{2}-x \] Now: \[ \lim_{x\to4^+}f(x) = 24-8-4 = 12 \] Since: \[ 12\ne16 \] function is discontinuous.

Step 3: Conclusion. As LHL and RHL are unequal: \[ \boxed{\text{Function is discontinuous}} \] Hence correct option is: \[ \boxed{(3)} \]