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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, $g' (x)$ exists for all x, and $g (x) < 0$ for $x > 0$ . If $h' (1) = e$ and $g' (1) = 1$ then the possible value which g(1) can take.

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-2

-4

answer is C.

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Detailed Solution

$f (x) = \int_{- 1}^{x} e^{t^{2}} d t, h (x) = f (1 + g (x)),$

$g (x) < 0 \forall x > 0$

$h (x) = \int_{- 1}^{1 + g (x)} e^{t^{2}} d t$

Differentiating, $h' (x) = e^{{(1 + g (x))}^{2}} g' (x)$

$h' (1) = e^{{(1 + g (1))}^{2}} g' (1)$ ( By leibnetz rule)

$e^{1} = e^{{(1 + g (1))}^{2}} . (1)$

$1 + g (1) = \pm 1, g (1) = 0$ not possible $g (1) = - 2$

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