Let f(x)= 7x^10+9x^8(1+x^2+2x^9)^2,dx, x0, with _x0f(x)=0 and f(1)=14.

Question

Let $f(x)=\displaystyle\int \dfrac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}\,dx,\ x>0$, with $\lim\limits_{x\to0}f(x)=0$ and $f(1)=\dfrac14$. Let
$$A=\begin{pmatrix}0&0&1\ \tfrac14 & f'(1) & 1\ \alpha^2 & 4 & 1\end{pmatrix},\qquad B=\operatorname{adj}(\operatorname{adj}A).$$
If $|B|=81$, then $\alpha^2$ is equal to

Accepted Answer

Substituting $t=\dfrac{1}{x^9}+\dfrac1{x^7}+2$, one finds $f(x)=\dfrac{x^9}{2x^9+x^2+1}+C$. The condition $\lim_{x	o0}f(x)=0$ gives $C=0$, so $f(x)=\dfrac{x^9}{2x^9+x^2+1}$.

Differentiating and evaluating at $x=1$: $f'(1)=\dfrac{9(4)-1(20)}{16}=\dfrac{16}{16}=1$.

So $A=\begin{pmatrix}0&0&1\ 	frac14&1&1\ \alpha^2&4&1\end{pmatrix}$, and expanding, $|A|=1\big(4\cdot	frac14-\alpha^2\big)=1-\alpha^2$.

For a $3	imes3$ matrix, $|\operatorname{adj}(\operatorname{adj}A)|=|A|^{(n-1)^2}=|A|^4$. Given $|B|=81=3^4$, $|A|^4=81\Rightarrow|A|=\pm3$.

So $1-\alpha^2=\pm3$. Taking $1-\alpha^2=-3$ gives $\alpha^2=4$ (the case $1-\alpha^2=3\Rightarrow\alpha^2=-2$ is rejected as $\alpha^2\ge0$).

So $\alpha^2=4$.

Solutions · 1