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Question: Let f(x) = \(\int_{–1}^{x}e^{t^{2}}\)dt and h(x) = f(1 + g(x)), where g(x) is defined for all x, g'(...

Let f(x) = ∫–1xet2\int_{–1}^{x}e^{t^{2}}dt 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, g'(1) = 1 then the possible values which g(1) can take –

A

0

B

– 1

C

2

D

– 4

Answer

0

Explanation

Solution

Given f(x) = ∫–1xet2\int_{–1}^{x}e^{t^{2}}dt; h(x) = f(1 + g(x));

g (x) ≤ 0 for x > 0 Now, h(x) = ∫–11+g(x)et2\int_{–1}^{1 + g(x)}e^{t^{2}} dt

On differentiating wrt x, we get

h'(x) = e(1+g(x))2e^{(1 + g(x))^{2}} . g'(x)

h'(1) = e (given)

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

∴ (1 + g(1))2 = 1

1 + g(1) = ± 1 ⇒ g(1) = 0 or g(1) = – 2.