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CGPET 2015 Solved Paper

Section: Mathematics

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Question : 132 of 150

Marks: +1, -0

If

f(x)=\int\limits_{2}^{x} \; \frac{dt}{\sqrt{1+t^4}}

g

is the inverse of

f

. Then, the value of

g'(0)

$1$
$17$
$\sqrt{17}$
None of the above

Solution:

Given,

f(x)=\int\limits_{2}^{x} \; \frac{dt}{\sqrt{1+t^4}}

Using Leibnitz's rule, we have

\frac{d}{dx} f(x) = f'(x) = \frac{1}{\sqrt{1+x^4}} \frac{d}{dx}(x) - 0

⇒

f'(x) = \frac{1}{\sqrt{1+x^4}}

It is given that, g is inverse of f.

i.e.

f^{-1} = g

Then,

(gof)(x) = x \; [\because f^{-1} = g]

On differentiating, we have

g'(f(x)) \times f'(x) = 1

[by chain rule]

⇒

g'(f(x)) = \frac{1}{f'(x)}

∵

f(x) = 0, \text{ if } x = 2

∴

g'(f(2)) = \frac{1}{f'(2)}

⇒

g'(0) = \frac{1}{f'(2)} = \frac{1}{\frac{1}{\sqrt{1+2^{4}}}}

⇒

g'(0) = \sqrt{17}

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