Verified Solution • 2025–26 Session moderate

QJEEM20Ma7db1: JEE Main 2020 Mathematics - Class 12 JEEM20Ma7db1 Solution: Let . Then f(3) – f(1) is eqaul to :

Let f(x)=∫x(1+x)2dx(x≥0)f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} . Then f(3) – f(1) is eqaul to :

Multiple Choice Options

  • A −π12+12+34- {\pi \over {12}} + {1 \over 2} + {{\sqrt 3 } \over 4}
  • B π12+12−34{\pi \over {12}} + {1 \over 2} - {{\sqrt 3 } \over 4}
  • C −π6+12+34- {\pi \over 6} + {1 \over 2} + {{\sqrt 3 } \over 4}
  • D π6+12−34{\pi \over 6} + {1 \over 2} - {{\sqrt 3 } \over 4}

Step-by-Step Verified Solution

\int {{{\sqrt x } \over {{{(1 + x)}^2}}}} dx(x > 0)

Put x = tan2 θ\theta ⇒\Rightarrow 2xdx = 2tan θ\theta sec2 θ\theta d θ\theta

I=∫2tan⁡2θ.sec⁡2θ2dθ=∫2sin⁡2θdθ=∫(1−cos⁡2θ)dθI = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\theta )d\theta } }

=θ−sin⁡2θ2+c= \theta - {{\sin 2\theta } \over 2} + c

⇒f(x)=θ−12×2tan⁡θ1+tan⁡2θ+c\Rightarrow f(x) = \theta - {1 \over 2} \times {{2\tan \theta } \over {1 + {{\tan }^2}\theta }} + c f(x)=θ−tan⁡θ1+tan⁡2θ+c=tan⁡−1x−x1+x+cf(x) = \theta - {{\tan \theta } \over {1 + {{\tan }^2}\theta }} + c = {\tan ^{ - 1}}\sqrt x - {{\sqrt x } \over {1 + x}} + c

Now f(3)−f(1)=tan⁡−1(3)−31+3−tan⁡−1(1)+12f(3) - f(1) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {{\sqrt 3 } \over {1 + 3}} - {\tan ^{ - 1}}(1) + {1 \over 2}

=π3−34−π4+12= {\pi \over 3} - {{\sqrt 3 } \over 4} - {\pi \over 4} + {1 \over 2}

=π12+12−34= {\pi \over 12} + {1 \over 2} - {{\sqrt 3 } \over 4}

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