A) \[\frac{10}{3}\]
B) 4
C) 2
D) \[\frac{16}{5}\]
Correct Answer: C
Solution :
\[\because \]g(x) is differentiable so g(x) must be continuos as well. L.H.L of g(x) at x = 3 is 2K R.H.L. of g(x) at x = 3 is 3m + 2 \[\because \] g(x) is continuous at x = 3 \[3m+2=2k\] \[3m-2k=-2\] ...(1) \[g'(x)=\left\{ \begin{matrix} \frac{k}{2\sqrt{x+1}} & 0\le x\le 3 \\ m & 3<x\le 5 \\ \end{matrix} \right.\] L.H.D at \[x=3=\frac{k}{4}\] R.H.D at \[\text{x}=\text{3}=\text{m}\] \[\because \] L.H.D = R. H.D \[\frac{k}{4}=m\] \[\] ?(2) By solving both the equations\[m=\frac{2}{5}k=\frac{8}{5}\] Hence \[k+m=\frac{2}{5}+\frac{8}{5}=\frac{10}{5}=2\]You need to login to perform this action.
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