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JEE Main & Advanced Sample Paper JEE Main Sample Paper-9

  • question_answer
    Let \[f(x)=\int\limits_{-1}^{x}{{{e}^{{{t}^{2}}}}}dt\]and \[h(x)=f(1+g(x)),\] where g(x) is defined for ail x, g'(x) exist for all x, and \[g(x)\le 0\] for x > 0. If h'(1) = 1, then the possible value which g(l) can take is

    A)  0                            

    B)  -1     

    C)  2                            

    D)  -4

    Correct Answer: A

    Solution :

    Given \[f(x)=\int\limits_{-1}^{x}{{{e}^{{{t}^{2}}}}}dt\]and \[h(x)=f(1+g(x));\] \[g(x)\le 0\]for \[x>0.\] Now,\[h(x)=\int\limits_{-1}^{x}{{{e}^{{{t}^{2}}}}}dt\] On differentiating, \[h'(x)={{e}^{{{(1+g(x))}^{2}}}}.g'(x)\]\[h'(1)=e\](given) \[{{e}^{{{(1+g(1))}^{2}}}}.g'(1)=e\] \[\therefore \]\[\left( 1+g{{(1)}^{2}} \right)=1\] \[1+g(1)=\pm 1\]\[\Rightarrow g(1)=0\]or\[g(1)=-2\]


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