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Determinant with A=∫t/(1+t²)dt, B=∫1/(t(1+t²))dt | JEE

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

Question

A=\displaystyle\int_{1}^{\sin\theta}\dfrac{t\,dt}{1+t^{2}}

and

B=\displaystyle\int_{1}^{\operatorname{cosec}\theta}\dfrac{dt}{t\,(1+t^{2})}

, then

\begin{vmatrix}A & A^{2} & B\\[2pt] e^{A+B} & B^{2} & -1\\[2pt] 1 & A^{2}+B^{2} & -1\end{vmatrix}

equals

\sin\theta

\operatorname{cosec}\theta

0

correct

1

Solution

Step 1: Relate

B

A

by substitution. In

B

put

t=\dfrac{1}{u}

, so

dt=-\dfrac{du}{u^{2}}

. The limits

t:1\to\operatorname{cosec}\theta

become

u:1\to\sin\theta

, and

\dfrac{1}{t(1+t^{2})}\,dt=\dfrac{1}{\frac1u\!\left(1+\frac{1}{u^{2}}\right)}\left(-\dfrac{du}{u^{2}}\right)=-\dfrac{u}{1+u^{2}}\,du.

Hence

B=\displaystyle\int_{1}^{\sin\theta}\!\left(-\dfrac{u}{1+u^{2}}\right)du=-A.

Step 2: Substitute

A+B=0

, so

e^{A+B}=e^{0}=1

, and

B=-A

gives

B^{2}=A^{2}

and

A^{2}+B^{2}=2A^{2}

. The determinant becomes

\begin{vmatrix}A & A^{2} & -A\\[2pt] 1 & A^{2} & -1\\[2pt] 1 & 2A^{2} & -1\end{vmatrix}.

Step 3: Expand along the first row:

A\big(A^{2}(-1)-(-1)(2A^{2})\big)-A^{2}\big(1\cdot(-1)-(-1)\cdot 1\big)+(-A)\big(1\cdot 2A^{2}-A^{2}\cdot 1\big).

Step 4: Simplify each term:

A\big(-A^{2}+2A^{2}\big)-A^{2}(0)+(-A)\big(2A^{2}-A^{2}\big)=A\cdot A^{2}-0-A\cdot A^{2}=A^{3}-A^{3}=0.

Correct answer: (3)

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