Verified Solution • 2025–26 Session moderate

QJEEM21Ma8ba3: JEE Main 2021 Mathematics - Class 12 JEEM21Ma8ba3 Solution: Let a curve y = y(x) be given by the solution of the…

Let a curve y = y(x) be given by the solution of the differential equation cos⁡(12cos⁡−1(e−x))dx=e2x−1dy\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy . If it intersects y-axis at y = −- 1, and the intersection point of the curve with x-axis is ( α\alpha , 0), then e α\alpha is equal to __________________.

Step-by-Step Verified Solution

cos⁡(12cos⁡−1(e−x))dx=e2x−1dy\cos \left( {{1 \over 2}{{\cos }^{ - 1}}({e^{ - x}})} \right)dx = \sqrt {{e^{2x}} - 1} dy

Put cos −- 1(e −- x) θ\theta , θ\theta ∈\in [0, π\pi ]

cos⁡θ=e−x⇒2cos⁡2θ2−1=e−x\cos \theta = {e^{ - x}} \Rightarrow 2{\cos ^2}{\theta \over 2} - 1 = {e^{ - x}}

cos⁡θ2=e−x+12=ex+12cx\cos {\theta \over 2} = \sqrt {{{{e^{ - x}} + 1} \over 2}} = \sqrt {{{{e^x} + 1} \over {2{c^x}}}}

ex+12cxdx=e2x−1dy\sqrt {{{{e^x} + 1} \over {2{c^x}}}} dx = \sqrt {{e^{2x}} - 1} dy

12∫dxexex−1=∫dy{1 \over {\sqrt 2 }}\int {{{dx} \over {\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} }

Put ex=t,dtdx=ex{e^x} = t,{{dt} \over {dx}} = {e^x}

12∫dxexexex−1=∫dy{1 \over {\sqrt 2 }}\int {{{dx} \over {{e^x}\sqrt {{e^x}} \sqrt {{e^x} - 1} }} = \int {dy} }

∫dttt2−t=2y\int {{{dt} \over {t\sqrt {{t^2} - t} }} = \sqrt 2 y}

Put t=1z,dtdz=−1z2t = {1 \over z},{{dt} \over {dz}} = - {1 \over {{z^2}}}

∫−dzz21z1z2−1z=2y\int {{{ - {{dz} \over {{z^2}}}} \over {{1 \over z}\sqrt {{1 \over {{z^2}}} - {1 \over z}} }} = \sqrt {2y} }

−∫dz1−z=2y- \int {{{dz} \over {\sqrt {1 - z} }} = \sqrt 2 y}

−2(1−z)1/2−1=2y+c{{ - 2{{(1 - z)}^{1/2}}} \over { - 1}} = \sqrt 2 y + c

2(1−1t)1/2=2y+c2{\left( {1 - {1 \over t}} \right)^{1/2}} = \sqrt 2 y + c

2(1−e−x)1/2=2y+c⟶(0,−1)⇒c=2

2(1−e−x)1/2=2(y+1)2{(1 - {e^{ - x}})^{1/2}} = \sqrt 2 (y + 1) , passes through ( α\alpha , 0)

2(1−e−α)1/2=22{(1 - {e^{ - \alpha }})^{1/2}} = \sqrt 2

1−e−α=12⇒1−e−α=12\sqrt {1 - {e^{ - \alpha }}} = {1 \over {\sqrt 2 }} \Rightarrow 1 - {e^{ - \alpha }} = {1 \over 2}

e−α=12⇒eα=2{e^{ - \alpha }} = {1 \over 2} \Rightarrow {e^\alpha } = 2

Related Learning Resources & Exam Preparation

JEE Main PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring