\(g\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {k\sqrt {x + 1} ,}&{0 \le x \le 3}\\ {m\;x + 2,}&{3 < x \le 5} \end{array}} \right.\)

If the above function is differentiable, then the value of k + m is:

Concept:

A function g(x) is said to be differentiable at a point 'a', if:

\( \mathop {\lim }\limits_{x \to {a^ - }} g'\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} g'\left( x \right)\)

Properties:

• If a function is differentiable at any point, then it is necessarily continuous at the point.
• But the converse of this statement is not true i.e. continuity is a necessary but sufficient condition for the Existence of a finite derivative.
• Differentiability implies Continuity.
• Continuity does not necessarily imply differentiability.

Any function f is said to be continuous at a point 'a', if and only if:

\( \mathop {\lim }\limits_{x \to {a^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {a^ + }} g\left( x \right)=f(a)\)

Analysis:

\(g\left( x \right) = \left\{ {\begin{array}{*{20}{c}} {k\sqrt {x + 1} ,}&{0 \le x \le 3}\\ {m\;x + 2,}&{3 < x \le 5} \end{array}} \right.\)

For 3^-, the function is:

\(g(x)=k\sqrt {x+1}\)

For 3+, the function is:

g(x) = mx + 2

Since g(x) is differentiable, it will be continuous at x = 3, i.e.

\(\therefore \mathop {\lim }\limits_{x \to {3^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g\left( x \right)\)

2 k = 3 m + 2      ---(1)

Also, g(x) is differentiable at x = 0, i.e.

\(\mathop {\lim }\limits_{x \to {3^ - }} g'\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g'\left( x \right)\)

Equating the derivatives of both the functions for x = 3^- and x = 3⁺, we get:

\(\frac{k}{{2\sqrt {x + 1} }} = m\)

Putting x = 3:

\(\frac{k}{{2\sqrt {3 + 1} }} = m\)

K = 4 m    ---(2)

Solving (1) and (2), we get:

\(m = \frac{2}{5},\;k = \frac{8}{5}\)