Definite IntegrationmediumFree

Quadratic in x with definite-integral coefficients | JEE

JEE Maths question with a full step-by-step solution.

Question

x

satisfies

\left(\displaystyle\int_{0}^{1}\dfrac{dt}{t^{2}+2t\cos\alpha+1}\right)x^{2}-\left(\displaystyle\int_{-3}^{3}\dfrac{t^{2}\sin 2t}{t^{2}+1}\,dt\right)x-2=0

for

0<\alpha<\pi

, then the value of

x

\pm\sqrt{\dfrac{\alpha}{2\sin\alpha}}

\pm\sqrt{\dfrac{2\sin\alpha}{\alpha}}

\pm\sqrt{\dfrac{\alpha}{\sin\alpha}}

\pm 2\sqrt{\dfrac{\sin\alpha}{\alpha}}

correct

Solution

Step 1: The middle coefficient vanishes. The integrand

\dfrac{t^{2}\sin 2t}{t^{2}+1}

is odd in

t

, so over the symmetric interval

[-3,3]

\int_{-3}^{3}\dfrac{t^{2}\sin 2t}{t^{2}+1}\,dt=0.

Step 2: Evaluate the leading coefficient. Complete the square:

t^{2}+2t\cos\alpha+1=(t+\cos\alpha)^{2}+\sin^{2}\alpha

, hence

\int_{0}^{1}\dfrac{dt}{(t+\cos\alpha)^{2}+\sin^{2}\alpha}=\dfrac{1}{\sin\alpha}\Big[\tan^{-1}\dfrac{t+\cos\alpha}{\sin\alpha}\Big]_{0}^{1}.

Step 3: Simplify the bracket. Using

\dfrac{1+\cos\alpha}{\sin\alpha}=\cot\dfrac{\alpha}{2}

and

\dfrac{\cos\alpha}{\sin\alpha}=\cot\alpha

\tan^{-1}\!\big(\cot\tfrac{\alpha}{2}\big)-\tan^{-1}\!\big(\cot\alpha\big)=\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right)-\left(\dfrac{\pi}{2}-\alpha\right)=\dfrac{\alpha}{2}.

Therefore the leading coefficient is

\dfrac{1}{\sin\alpha}\cdot\dfrac{\alpha}{2}=\dfrac{\alpha}{2\sin\alpha}.

Step 4: The equation reduces to

\dfrac{\alpha}{2\sin\alpha}\,x^{2}-2=0

, so

x^{2}=\dfrac{4\sin\alpha}{\alpha}\ \Rightarrow\ x=\pm 2\sqrt{\dfrac{\sin\alpha}{\alpha}}.

Correct answer: (4)

Still stuck on this question?Ask your doubt on WhatsApp

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.