Question

If the function \(g(x) = \begin{cases} K\sqrt{x+1} & , 0 \le x \le 3 \\ mx + 2 & , 3<x \le 5 \end{cases}\) is differentiable, then \(K + m =\)

• A. 4
• B. 2
• C. 6
• D. 0

Hint:

Differentiability implies continuity. Always start by writing the continuity equation first, as it provides an essential relationship between the variables.

Answer 1

The Correct Option is B

Step 1: Check Continuity at \(x = 3\):
For \(g(x)\) to be differentiable, it must be continuous at \(x=3\). \[ \lim_{x \to 3^-} g(x) = \lim_{x \to 3^+} g(x) \] \[ K\sqrt{3+1} = m(3) + 2 \] \[ 2K = 3m + 2 \quad \dots(1) \]
Step 2: Check Differentiability at \(x = 3\):
LHD at \(x=3\): \(\frac{d}{dx}(K\sqrt{x+1}) = \frac{K}{2\sqrt{x+1}}\). At \(x=3\), LHD \(= \frac{K}{2\sqrt{4}} = \frac{K}{4}\). RHD at \(x=3\): \(\frac{d}{dx}(mx+2) = m\). Equating LHD and RHD: \[ \frac{K}{4} = m \implies K = 4m \quad \dots(2) \]
Step 3: Solve the System:
Substitute (2) into (1): \[ 2(4m) = 3m + 2 \] \[ 8m - 3m = 2 \] \[ 5m = 2 \implies m = \frac{2}{5} \] Then \(K = 4(\frac{2}{5}) = \frac{8}{5}\).
Step 4: Find \(K + m\):
\[ K + m = \frac{8}{5} + \frac{2}{5} = \frac{10}{5} = 2 \]