Verified Solution • 2025–26 Session moderate

QJEEM20Mac54c: JEE Main 2020 Mathematics - Class 12 JEEM20Mac54c Solution: If where C is a constant of integration, then the ordered pair…

If ∫dθcos⁡2θ(tan⁡2θ+sec⁡2θ)=λtan⁡θ+2log⁡e∣f(θ)∣+C\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C

where C is a constant of integration, then the ordered pair ( λ\lambda , ƒ( θ\theta )) is equal to :

Multiple Choice Options

  • A (–1, 1 – tan θ\theta )
  • B (1, 1 + tan θ\theta )
  • C (–1, 1 + tan θ\theta )
  • D (1, 1 – tan θ\theta )

Step-by-Step Verified Solution

I = ∫dθcos⁡2θ(tan⁡2θ+sec⁡2θ)\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}}

= ∫sec⁡2θdθ2tan⁡θ1−tan⁡2θ+1+tan⁡2θ1−tan⁡2θ\int {{{{{\sec }^2}\theta d\theta } \over {{{2\tan \theta } \over {1 - {{\tan }^2}\theta }} + {{1 + {{\tan }^2}\theta } \over {1 - {{\tan }^2}\theta }}}}}

= ∫(1−tan⁡2θ)sec⁡2θdθ(1+tan⁡θ)2\int {{{\left( {1 - {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \over {{{\left( {1 + {{\tan }}\theta } \right)}^2}}}}

tan θ\theta = t ⇒\Rightarrow sec2 θ\theta d θ\theta = dt

= ∫(1−t2)dt(1+t)2\int {{{\left( {1 - {t^2}} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}}

= ∫(1−t)(1+t)dt(1+t)2\int {{{\left( {1 - t} \right)\left( {1 + t} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}}

= ∫(1−t)(1+t)dt\int {{{\left( {1 - t} \right)} \over {\left( {1 + t} \right)}}} dt

= ∫(1(1+t)−t(1+t))dt\int {\left( {{1 \over {\left( {1 + t} \right)}} - {t \over {\left( {1 + t} \right)}}} \right)} dt

= log⁡e∣1+t∣{\log _e}\left| {1 + t} \right| - ∫(1+t(1+t)−1(1+t))dt\int {\left( {{{1 + t} \over {\left( {1 + t} \right)}} - {1 \over {\left( {1 + t} \right)}}} \right)} dt

= log⁡e∣1+t∣{\log _e}\left| {1 + t} \right| - t + log⁡e∣1+t∣{\log _e}\left| {1 + t} \right| + C

= 2 log⁡e∣1+t∣{\log _e}\left| {1 + t} \right| - t + C

= 2 log⁡e∣1+tan⁡θ∣{\log _e}\left| {1 + \tan \theta } \right| - tan θ\theta + C

∴\therefore λ\lambda = -1 and ƒ( θ\theta ) = 1 + tan θ\theta

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