Question

Let A = $\begin{bmatrix} x^{2} & 6 & 0 \\ 1 & –5 & 1 \\ 2 & 0 & x \end{bmatrix}$ and B = $\begin{bmatrix} 4 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 8 \end{bmatrix}$. If a function is defined as f(x) = tr (AB), then $\int_{}^{}\frac{3dx}{f(x)}$ is equal to Q

• A. $\frac{1}{4}\mathcal{l}n\left| \frac{2x - 1}{2x + 5} \right| + C$
• B. $\frac{1}{4}\mathcal{l}n\left| \frac{2x + 5}{2x - 1} \right| + C$
• C. $\frac{1}{3}\mathcal{l}n\left| \frac{1 - 2x}{2x + 5} \right| + C$
• D. $\frac{1}{3}\mathcal{l}n\left| \frac{1 - 2x}{2x + 3} \right| + C$

Answer 1

Answer: (A)

$\frac{1}{4}\mathcal{l}n\left| \frac{2x - 1}{2x + 5} \right| + C$

Answer 2

AB =$\begin{bmatrix} 4x^{2} & 6 & 0 \\ 4 & –5 & 8 \\ 8 & 0 & 8x \end{bmatrix}$

̃ f(x) = tr(AB) = 4x² + 8x – 5

\ $\int_{}^{}\frac{3dx}{4x^{2} + 8x - 5}$ = $\int_{}^{}\frac{3dx}{(2x + 5)(2x - 1)}$

= $\frac{1}{2}\int_{}^{}\left( \frac{1}{(2x - 1)} - \frac{1}{(2x + 5)} \right)dx$ = $\frac{1}{4}\mathcal{l}n\left| \frac{2x - 1}{2x + 5} \right| + C$