Verified Solution • 2025–26 Session moderate

QJEEM20Madfe9: JEE Main 2020 Mathematics - Class 12 JEEM20Madfe9 Solution: If ƒ'(x) = tan –1 (secx + tanx), , and ƒ(0) = 0, then ƒ(1) is…

If ƒ'(x) = tan–1(secx + tanx), - {\pi \over 2} < x < {\pi \over 2} ,
and ƒ(0) = 0, then ƒ(1) is equal to :

Multiple Choice Options

  • A 14{1 \over 4}
  • B π−14{{\pi - 1} \over 4}
  • C π+14{{\pi + 1} \over 4}
  • D π+24{{\pi + 2} \over 4}

Step-by-Step Verified Solution

ƒ'(x) = tan–1(secx + tanx)

⇒\Rightarrow ƒ'(x) = tan⁡−1(1+sin⁡xcos⁡x){\tan ^{ - 1}}\left( {{{1 + \sin x} \over {\cos x}}} \right) = tan⁡−1(1+tan⁡x21−tan⁡x2){\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)

= tan⁡−1(tan⁡(π4+x2)){\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) = π4+x2{\pi \over 4} + {x \over 2}

⇒\Rightarrow f(x) = π4x+x24+c{\pi \over 4}x + {{{x^2}} \over 4} + c

∵\because ƒ(0) = 0 ⇒\Rightarrow c = 0

∴\therefore f(x) = π4x+x24{\pi \over 4}x + {{{x^2}} \over 4}

⇒\Rightarrow f(1) = π+14{{{\pi + 1} \over 4}}

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