JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer Let\[f(x)={{e}^{x}}-x\] and \[g(x){{x}^{2}}-x,\forall x\in R.\]
    Then the set of all \[x\in R,\]where the function \[h(x)=(fog)(x)\]is increasing is :
    [JEE Main 10-4-2019 Morning]

    A) \[\left[ -1,\frac{-1}{2} \right]\cup \left[ \frac{1}{2},\infty  \right)\]      

    B) \[\left[ 0,\frac{1}{2} \right]\cup \left[ 1,\infty  \right)\]

    C) \[\left[ \frac{-1}{2},0 \right]\cup \left[ 1,\infty  \right)\]

    D)   \[\left[ 0,\infty  \right)\]

    Correct Answer: B

    Solution :

    \[h\left( x \right)=f\left( g\left( x \right) \right)\] \[\Rightarrow h'(x)=f'(g(x)).g'(x)\]and\[f'(x)={{e}^{x}}-1\] \[\Rightarrow h'(x)=({{e}^{g(x)}}-1)g'(x)\] \[\Rightarrow h'(x)=\left( {{e}^{{{x}^{2}}-x}}-1 \right)(2x-1)\ge 0\] Case-I \[{{e}^{{{x}^{2}}-x}}\ge 1\]and\[2x-1\ge 0\] \[\Rightarrow x\in [1,\infty )\]                                    ......(1) Case-II \[{{e}^{{{x}^{2}}-x}}\le 1\]and \[2x-1\le 0\] \[\Rightarrow \]\[x\in \left[ 0,\frac{1}{2} \right]\]                                            .....(2) Hence,\[x\in \left[ 0,\frac{1}{2} \right]\cup [1,\infty )\]

You need to login to perform this action.
You will be redirected in 3 sec spinner