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JEE Main & Advanced JEE Main Solved Paper-2015

  • question_answer
    If the function. \[g\left( x \right)=\left\{ \begin{align}   & k\sqrt{x+1},\,\,\,\,\,\,0\le x\le 3 \\  & mx+2,\,\,\,\,\,\,\,\,3<x\le 5 \\ \end{align} \right.\]is differentiable, then the value of k + m is : [JEE Main Solved Paper-2015 ]

    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 \[k+m=\frac{2}{5}+\frac{8}{5}=\frac{10}{5}=2\]                


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