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\[g(1)=-2\]You need to login to perform this action.
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