If g(x) = \(\int^x_0\sqrt{1-t^2}\ dt\), then the domain of g'(x) is:

Question

Question

This question was previously asked in

AAI ATC Junior Executive 27 July 2022 Shift 3 Official Paper

Answer 1

Concept:

Leibnitz integral rule,

\({d\over dx} [\int_{a(x)}^{b(x)} f(t,x)dt] = \int_a^b {\partial f \over \partial x} dt + b'f(b(x),x) - a'f(a(x),x)\)

Calculation:

Given, g(x) = \(\int^x_0\sqrt{1-t^2}\ dt\),

By Leibnitz rule,

g'(x) = \(\int^x_0 0\ dt + (x)'\sqrt{1-x^2} - (0)'\sqrt{1-0^2} \)

⇒ g'(x) = \(0 + 1.\sqrt{1-x^2} + 0\)

⇒ g'(x) = \(\sqrt{1-x^2}\)

If g'(x) is defined, then 1 - x² ≥ 0

⇒ x ∈ [-1, 1]

∴ The correct option is (3).

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