Question:
If the function
\( g(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx + 2, & 3 \lt x \le 5 \end{cases} \)
is differentiable, then the value of \(k + m\) is :
If the function
\( g(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx + 2, & 3 \lt x \le 5 \end{cases} \)
is differentiable, then the value of \(k + m\) is :
\( g(x) = \begin{cases} k\sqrt{x+1}, & 0 \le x \le 3 \\ mx + 2, & 3 \lt x \le 5 \end{cases} \)
is differentiable, then the value of \(k + m\) is :
Show Hint
Differentiability at \(x=3\) requires both continuity and equality of the left and right derivatives there.
Updated On: Nov 11, 2026
- \(\frac{10}{3}\)
- 4
- 2
- \(\frac{16}{5}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Differentiable at \(x=3\) means continuous and equal derivatives.
Continuity: \(k\sqrt{3+1} = 3m+2 \Rightarrow 2k = 3m+2\).
Derivatives: for \(x\le 3\), \(g'(x) = \frac{k}{2\sqrt{x+1}}\), so \(g'(3^-) = \frac{k}{4}\); for \(x\gt 3\), \(g'(x)=m\).
Hence \(m = \frac{k}{4}\), i.e. \(k = 4m\).
Substituting: \(8m = 3m+2 \Rightarrow m = \frac{2}{5}\), \(k = \frac{8}{5}\).
\(k+m = \frac{8}{5}+\frac{2}{5} = 2\).
Answer: (3) 2
Continuity: \(k\sqrt{3+1} = 3m+2 \Rightarrow 2k = 3m+2\).
Derivatives: for \(x\le 3\), \(g'(x) = \frac{k}{2\sqrt{x+1}}\), so \(g'(3^-) = \frac{k}{4}\); for \(x\gt 3\), \(g'(x)=m\).
Hence \(m = \frac{k}{4}\), i.e. \(k = 4m\).
Substituting: \(8m = 3m+2 \Rightarrow m = \frac{2}{5}\), \(k = \frac{8}{5}\).
\(k+m = \frac{8}{5}+\frac{2}{5} = 2\).
Answer: (3) 2
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