JEE MAIN - Mathematics (2021 - 17th March Morning Shift - No. 3)
Which of the following statements is correct for the function g($$\alpha$$) for $$\alpha$$ $$\in$$ R such that
$$g(\alpha ) = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sin }^\alpha }x} \over {{{\cos }^\alpha }x + {{\sin }^\alpha }x}}dx} $$
$$g(\alpha ) = \int\limits_{{\pi \over 6}}^{{\pi \over 3}} {{{{{\sin }^\alpha }x} \over {{{\cos }^\alpha }x + {{\sin }^\alpha }x}}dx} $$
$$g(\alpha )$$ is a strictly increasing function
$$g(\alpha )$$ is an even function
$$g(\alpha )$$ has an inflection point at $$\alpha$$ = $$-$$$${1 \over 2}$$
$$g(\alpha )$$ is a strictly decreasing function
Explanation
$$g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{{{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)} \over {{{\cos }^\alpha }\left( {{\pi \over 2} - x} \right)x + {{\sin }^\alpha }\left( {{\pi \over 2} - x} \right)}}dx} $$
$$ = \int\limits_{\pi /6}^{\pi /3} {{{{{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} $$
$$ \therefore $$ $$2.g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{si{n^\alpha }x + {{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} = \int\limits_{\pi /6}^{\pi /3} {dx} = {\pi \over 3} - {\pi \over 6} = {\pi \over 6}$$
$$ \Rightarrow $$ $$g(\alpha ) = {\pi \over {12}}$$ i.e. a constant function hence an even function.
$$ = \int\limits_{\pi /6}^{\pi /3} {{{{{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} $$
$$ \therefore $$ $$2.g(\alpha ) = \int\limits_{\pi /6}^{\pi /3} {{{si{n^\alpha }x + {{\cos }^\alpha }x} \over {{{\sin }^\alpha }x + {{\cos }^\alpha }x}}dx} = \int\limits_{\pi /6}^{\pi /3} {dx} = {\pi \over 3} - {\pi \over 6} = {\pi \over 6}$$
$$ \Rightarrow $$ $$g(\alpha ) = {\pi \over {12}}$$ i.e. a constant function hence an even function.
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