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  • <img alt="\mathrm{\text { If the function } g(x)=\left\{\begin{array}{ll} k \sqrt{x+1}, & 0 \leq x \leq 3 \\ m x+2, & 3<x \leq 5 \end{array}\right. \text { is differentiable, }}" src="ht

\mathrm{\text { If the function } g(x)=\left\{\begin{array}{ll} k \sqrt{x+1}, & 0 \leq x \leq 3 \\ m x+2, & 3<x \leq 5 \end{array}\right. \text { is differentiable, }}

 then the value of g(5)  is 

Option: 1

\frac{10}{3}


Option: 2

2


Option: 3

4


Option: 4

\frac{16}{5}


Answers (1)

best_answer

\mathrm{\text { As } g(x)=\left\{\begin{array}{ll} k \sqrt{x+1}, & 0 \leq x \leq 3 \\ m x+2, & 3<x \leq 5 \end{array}\right. \text { is differentiable at }}

x = 3, it must be continuous at x = 3.

\mathrm{\text { Hence, } \lim _{x \rightarrow 3^{+}} g(x)=g(3) \Rightarrow 3 m+2=2 k}

\mathrm{\text { Again, } g^{\prime}\left(3^{+}\right)=\lim _{h \rightarrow 0} \frac{g(3+h)-g(3)}{h}}

\mathrm{\begin{aligned} & =\lim _{h \rightarrow 0} \frac{m(3+h)+2-2 k}{h} \\ & =\lim _{h \rightarrow 0} \frac{m h}{h}=m \quad(\text { as } 3 m+2=2 k) \end{aligned}}

\mathrm{\text { Also, } g^{\prime}\left(3^{-}\right)=\lim _{h \rightarrow 0} \frac{g(3-h)-g(3)}{-h}}

\mathrm{\begin{aligned} & =\lim _{h \rightarrow 0} \frac{k \sqrt{4-h}-2 k}{-h}=\lim _{h \rightarrow 0} \frac{k(\sqrt{4-h}-2)}{-h} \\ & =\lim _{h \rightarrow 0} \frac{k((4-h)-4)}{-h} \cdot \frac{1}{\sqrt{4-h}+2}=\lim _{h \rightarrow 0} \frac{k}{\sqrt{4-h}+2}=\frac{k}{4} \end{aligned}}

g(x) is differentiable at x = 3 \ m = k/4
Now, 3m + 2 = 2k yields m = 2/5, k = 8/5
 g(5) = 5m + 2 = 4

 

 

Posted by

Suraj Bhandari

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