Integral Calculus
Integral Calculus
star_batch_jee_advanced_2025
Grade 12
Question:
If 'A' is a square matrix and $e^A$ is defined as $e^A = I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + ...... = I\begin{bmatrix} f(x) & g(x) \\ g(x) & f(x) \end{bmatrix}$ where $A = \begin{bmatrix} x & x \\ x & x \end{bmatrix}$ and $0 < x < 1$, I is an identity matrix, then:
$\int \frac{g(x)}{f(x)}dx = \ln(e^x + e^{-x}) + c$
$f(x) \in \mathbb{R}^* \text{ & } g(x) \in \mathbb{R}^*$
$f(x) \in \mathbb{R}^* \text{ & } g(x) \in \mathbb{R}$
$\int(g(x)+1)\sin x\,dx = \frac{e^{2x}}{5}(2\sin x - \cos x) + c$
Step-by-Step Solution
Key Concept: Matrix exponential expansion reveals that the series converges to $e^A$, where individual entries follow standard exponential and hyperbolic function patterns.
The solution uses matrix exponential expansion. Starting with $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ and $A = \begin{bmatrix} x & x \\ x & x \end{bmatrix}$, compute $A^2 = \begin{bmatrix} 2x^2 & 2x^2 \\ 2x^2 & 2x^2 \end{bmatrix}$ and $A^3 = \begin{bmatrix} 4x^3 & 4x^3 \\ 4x^3 & 4x^3 \end{bmatrix}$. The sum $I + A + \frac{A^2}{2!} + \frac{A^3}{3!} + \ldots$ produces a matrix whose entries give $f(x) = 1 + e^{2x}$ and $g(x) = e^{2x} - 1$. Then $\int \frac{g(x)}{f(x)} dx = \ln(e^x + e^{-x}) + c$ and $\int (g(x) + 1)\sin x\, dx = \frac{e^{2x}}{5}(2\sin x - \cos x) + c$.
Correct Answer: 1,3,4