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TG EAMCET 2 May 2025 Shift 2 Paper

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Question : 63 of 160

Marks: +1, -0

If the function

g(x)=\begin{cases} K \sqrt{x+1}, & 0 \leq x \leq 3 \\ m x+2, & 3 < x \leq 5 \end{cases}

is differentiable, then

K+m=

4
2
6
0

Solution:

Given,

g(x)=\begin{cases} K \sqrt{x+1}, & 0 \leq x \leq 3 \\ m x+2, & 3 < x \leq 5 \end{cases}

is differentiable

Since,

g(x)

is differentiable therefore it is continuous.

at

x=3

\;\lim\limits_{x \to 3^{-}} f(x)=\lim\limits_{x \to 3^{+}} f(x)

\Rightarrow \lim\limits_{h \to 0} f(3-h)=\lim\limits_{h \to 0} f(3+h)

\Rightarrow \lim\limits_{h \to 0} (K \sqrt{(3-h+1)})=\lim\limits_{h \to 0} (m(3+h)+2)

\Rightarrow K \sqrt{4}=m(3)+2

\Rightarrow 2 K=3 m+2

for

0 \leq x \leq 3

g'(x)=K \;\frac{1}{2 \sqrt{x+1}}

And for 3

g'(x)=m

for differentiability at

x=3

\;L g'(3)=R g'(3)

\;\Rightarrow \;\frac{K}{2 \sqrt{4}}=m

\Rightarrow K=4 m

On Solving Eqs. (i) and (ii), we get

\;2(4 m)\;=3 m+2

\Rightarrow\;8 m\;=3 m+2

\Rightarrow\;5 m\;=2 \Rightarrow m=\;\frac{2}{5}

And

\;\;K=\;\frac{8}{5}

\therefore\;\;K+m=\;\frac{8}{5}+\;\frac{2}{5}=\;\frac{10}{5}=2

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