JEE MAIN - Mathematics (2020 - 9th January Morning Slot - No. 1)
If ƒ'(x) = tan–1(secx + tanx), $$ - {\pi \over 2} < x < {\pi \over 2}$$,
and ƒ(0) = 0, then ƒ(1) is equal to :
and ƒ(0) = 0, then ƒ(1) is equal to :
$${1 \over 4}$$
$${{\pi - 1} \over 4}$$
$${{\pi + 1} \over 4}$$
$${{\pi + 2} \over 4}$$
Explanation
ƒ'(x) = tan–1(secx + tanx)
$$ \Rightarrow $$ ƒ'(x) = $${\tan ^{ - 1}}\left( {{{1 + \sin x} \over {\cos x}}} \right)$$ = $${\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)$$
= $${\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)$$
As $$ - {\pi \over 2} < x < {\pi \over 2}$$
$$ \Rightarrow $$ $$0 < {\pi \over 4} + {x \over 2} < {\pi \over 2}$$
$$ \therefore $$ ƒ'(x) = $${\pi \over 4} + {x \over 2}$$
$$ \Rightarrow $$ f(x) = $${\pi \over 4}x + {{{x^2}} \over 4} + c$$
$$ \because $$ ƒ(0) = 0 $$ \Rightarrow $$ c = 0
$$ \therefore $$ f(x) = $${\pi \over 4}x + {{{x^2}} \over 4}$$
$$ \Rightarrow $$ f(1) = $${{{\pi + 1} \over 4}}$$
$$ \Rightarrow $$ ƒ'(x) = $${\tan ^{ - 1}}\left( {{{1 + \sin x} \over {\cos x}}} \right)$$ = $${\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)$$
= $${\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)$$
As $$ - {\pi \over 2} < x < {\pi \over 2}$$
$$ \Rightarrow $$ $$0 < {\pi \over 4} + {x \over 2} < {\pi \over 2}$$
$$ \therefore $$ ƒ'(x) = $${\pi \over 4} + {x \over 2}$$
$$ \Rightarrow $$ f(x) = $${\pi \over 4}x + {{{x^2}} \over 4} + c$$
$$ \because $$ ƒ(0) = 0 $$ \Rightarrow $$ c = 0
$$ \therefore $$ f(x) = $${\pi \over 4}x + {{{x^2}} \over 4}$$
$$ \Rightarrow $$ f(1) = $${{{\pi + 1} \over 4}}$$
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