JEE MainMathematicsModerateContinuity and differentiability

If $g(x) = \left\{ \begin{matrix}
k\sqrt{x + 1}, - \frac{1}{2} \leq x \leq 2 \\
mx + 2,2 < x \leq 5
\end{matrix}\ \ \right.\ $is continuous and differentiable then the
value of $\sqrt{3}k + 4m$ is.

Show answer and solution

k\sqrt{3} = 2m + 2

(for continuity at x = 2)

m = \frac{k}{2\sqrt{3}}

(for differentiability at x = 2)

k\sqrt{3} = \frac{k}{\sqrt{3}} + 2 \Rightarrow

2k = 2\sqrt{3}

m = \frac{1}{2}

Source: Grass-G / M-19 Mathematics Paper 28 Nov Fourth.docx

Need a video for this idea?

Scan any doubt from your notes and get a whiteboard explanation.