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Trace of AB then Integrate 3/f(x) | JEE

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

Question
Let A=[x2601−5120x]A=\begin{bmatrix} x^{2}&6&0\\ 1&-5&1\\ 2&0&x\end{bmatrix} and B=[400010008]B=\begin{bmatrix} 4&0&0\\ 0&1&0\\ 0&0&8\end{bmatrix}. If f(x)=trace⁡(AB)f(x)=\operatorname{trace}(AB), then ∫3 dxf(x)\displaystyle\int \dfrac{3\,dx}{f(x)} is
A14ln⁡∣2x−12x+5∣+c\dfrac{1}{4}\ln\left|\dfrac{2x-1}{2x+5}\right|+ccorrect
B14ln⁡∣2x+52x−1∣+c\dfrac{1}{4}\ln\left|\dfrac{2x+5}{2x-1}\right|+c
C13ln⁡∣1−2x2x+5∣+c\dfrac{1}{3}\ln\left|\dfrac{1-2x}{2x+5}\right|+c
D13ln⁡∣1−2x2x+3∣+c\dfrac{1}{3}\ln\left|\dfrac{1-2x}{2x+3}\right|+c
Solution
Step 1: Compute the trace of ABAB. Since B=diag⁡(4,1,8)B=\operatorname{diag}(4,1,8) scales column jj of AA by the jjth diagonal entry, the diagonal entries of ABAB are
(AB)11=4x2,(AB)22=−5,(AB)33=8x(AB)_{11}=4x^{2},\qquad (AB)_{22}=-5,\qquad (AB)_{33}=8x
∴f(x)=trace⁡(AB)=4x2−5+8x=4x2+8x−5\therefore f(x)=\operatorname{trace}(AB)=4x^{2}-5+8x=4x^{2}+8x-5
 Step 2: Set up the integral and pull out the leading coefficient:
∫3 dx4x2+8x−5=34∫dxx2+2x−54\int\dfrac{3\,dx}{4x^{2}+8x-5}=\dfrac{3}{4}\int\dfrac{dx}{x^{2}+2x-\tfrac{5}{4}}
 Step 3: Complete the square in the denominator:
x2+2x−54=(x+1)2−1−54=(x+1)2−94=(x+1)2−(32)2x^{2}+2x-\dfrac{5}{4}=(x+1)^{2}-1-\dfrac{5}{4}=(x+1)^{2}-\dfrac{9}{4}=(x+1)^{2}-\left(\dfrac{3}{2}\right)^{2}
 Step 4: Apply ∫duu2−a2=12aln⁡∣u−au+a∣\displaystyle\int\dfrac{du}{u^{2}-a^{2}}=\dfrac{1}{2a}\ln\left|\dfrac{u-a}{u+a}\right| with u=x+1, a=32u=x+1,\ a=\dfrac{3}{2}:
=34⋅12⋅32ln⁡∣(x+1)−32(x+1)+32∣+c=34⋅13ln⁡∣x−12x+52∣+c=14ln⁡∣2x−12x+5∣+c=\dfrac{3}{4}\cdot\dfrac{1}{2\cdot\tfrac{3}{2}}\ln\left|\dfrac{(x+1)-\tfrac{3}{2}}{(x+1)+\tfrac{3}{2}}\right|+c=\dfrac{3}{4}\cdot\dfrac{1}{3}\ln\left|\dfrac{x-\tfrac{1}{2}}{x+\tfrac{5}{2}}\right|+c=\dfrac{1}{4}\ln\left|\dfrac{2x-1}{2x+5}\right|+c
 Correct answer: (1)
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