Let $f(x)=\int_{-1}^x e^{t^2} d t$ and $h(x)=f(1+g(x))$ where $g(x)$ is defined for all $x, g^{\prime}(x)$ exists for all $x$, and $g(x) 0$. If $h^{\prime}(1)=e$ and $g^{\prime}(1)=1$, then the possible values which $g(1)$ can take

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

Let

f(x)=\int_{-1}^x e^{t^2} d t

and

h(x)=f(1+g(x))

where

g(x)

is defined for all

x, g^{\prime}(x)

exists for all

x

, and

g(x)<0

for

x>0

. If

h^{\prime}(1)=e

and

g^{\prime}(1)=1

, then the possible values which

g(1)

can take

View answer and explanation

Correct option: C

-2

Option C is correct because by differentiating h(x) and using the given conditions, we can solve for g(1), which is -2.

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