Question ID: #879
Let $f(x)=\int\frac{7x^{10}+9x^{8}}{(1+x^{2}+2x^{9})^{2}}dx$, $x>0$, $\lim_{x\rightarrow0}f(x)=0$ and $f(1)=\frac{1}{4}$. If $A=\begin{bmatrix}0&0&1\\ \frac{1}{4}&f'(1)&1\\ \alpha^{2}&4&1\end{bmatrix}$ and $B=adj(adj~A)$ be such that $|B|=81$, then $\alpha^{2}$ is equal to
- (1) 2
- (2) 3
- (3) 1
- (4) 4
Solution:
$$f(x) = \int \frac{7x^{10} + 9x^8}{(1+x^2+2x^9)^2} dx$$
Divide numerator and denominator by $x^{18}$:
$$f(x) = \int \frac{\frac{7}{x^8} + \frac{9}{x^{10}}}{(\frac{1}{x^9} + \frac{1}{x^7} + 2)^2} dx$$
Let $t = \frac{1}{x^9} + \frac{1}{x^7} + 2 \Rightarrow dt = (-\frac{9}{x^{10}} – \frac{7}{x^8})dx \Rightarrow -dt = (\frac{9}{x^{10}} + \frac{7}{x^8})dx$
$$f(x) = \int \frac{-dt}{t^2} = \frac{1}{t} + C = \frac{1}{\frac{1}{x^9} + \frac{1}{x^7} + 2} + C = \frac{x^9}{1+x^2+2x^9} + C$$
Given $\lim_{x\to0} f(x) = 0 \Rightarrow \frac{0}{1} + C = 0 \Rightarrow C=0$
$$f(x) = \frac{x^9}{1+x^2+2x^9}$$
Now differentiate with respect to $x$ at $x=1$:
$$f'(x) = \frac{(1+x^2+2x^9)(9x^8) – x^9(2x+18x^8)}{(1+x^2+2x^9)^2}$$
$$f'(1) = \frac{(1+1+2)(9) – 1(2+18)}{(1+1+2)^2} = \frac{36 – 20}{16} = \frac{16}{16} = 1$$
$$A = \begin{bmatrix} 0 & 0 & 1 \\ \frac{1}{4} & 1 & 1 \\ \alpha^2 & 4 & 1 \end{bmatrix}$$
$$|A| = 1\left(\frac{1}{4}(4) – 1(\alpha^2)\right) = 1 – \alpha^2$$
$$|B| = |adj(adj~A)| = |A|^{(3-1)^2} = |A|^4$$
$$|A|^4 = 81 \Rightarrow (1-\alpha^2)^4 = 3^4 \Rightarrow |1-\alpha^2| = 3$$
$$1 – \alpha^2 = -3 \Rightarrow \alpha^2 = 4$$
Ans. (4)
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