If the function \(g(x) = \begin{cases} k\sqrt{x + 3}, & 0 \le x \le 1\\ mx + 1, & 1 \lt x \le 3\end{cases}\) is differentiable at \(x = 1\), then the value of \(k + m\) is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle\frac{5}{7}\)
A function differentiable at x = 1 is also continuous there, and its left and right derivatives at x = 1 are equal.
Continuity at x = 1 (left value \(k\sqrt{4} = 2k\), right value m + 1):
\(2k = m + 1\)
Left derivative, from \(k\sqrt{x + 3}\):
\( \displaystyle \frac{d}{dx}\left(k\sqrt{x + 3}\right) = \frac{k}{2\sqrt{x + 3}} \)
\( \displaystyle \text{at } x = 1: \ \frac{k}{2 \times 2} = \frac{k}{4} \)
Right derivative, from mx + 1, is m. Equating:
\( \displaystyle m = \frac{k}{4} \)
Substituting in 2k = m + 1:
\( \displaystyle 2k = \frac{k}{4} + 1 \)
\( \displaystyle \frac{7k}{4} = 1 \)
\( \displaystyle k = \frac{4}{7}, \quad m = \frac{1}{7} \)
\( \displaystyle k + m = \frac{5}{7} \)