If f(x)={ax+5,x≤22x+k,x>2 is differentiable at x=2 then k=

Solution:

$f (x)$ is differentiable at $x = 2 \Rightarrow f (x)$ is continuous at $x = 2$

$\begin{array}{l} \underset{x \to 2^{-}}{L i m} a x + 5 = \underset{x \to 2^{+}}{L i m} 2 x + k \\ \Rightarrow 2 a + 5 = 4 + k \to (1) \\ f (x) is differentiable at x = 2 \\ \underset{x \to 2^{-}}{L t} a (1) = a \\ \underset{x \to 2^{+}}{L t} 2 (1) = 2 \\ ∴ a = 2 \\ F r o m (1) \Rightarrow 2 (2) + 5 = 4 + k \\ \Rightarrow k = 5 \end{array}$

If $f (x) = {\begin{cases} a x + 5, x \leq 2 \\ 2 x + k, x > 2 \end{cases}$ is differentiable at $x = 2$ then $k =$

Solution:

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