Let $f(x)=\int \frac{7 x^{10}+9 x^8}{\left(1+x^2+2 x^9\right)^2} d x, x>0, \lim\limits_{x \rightarrow 0} f(x)=0$ and $f(1)=\frac{1}{4}$.

If $\mathrm{A}=\left[\begin{array}{ccc}0 & 0 & 1 \\ \frac{1}{4} & f^{\prime}(1) & 1 \\ \alpha^2 & 4 & 1\end{array}\right]$ and $\mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})$ be such that $|\mathrm{B}|=81$, then $\alpha^2$ is equal to

Let the length of the latus rectum of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,(a>b)$, be 30 . If its eccentricity is the maximum value of the function $f(t)=-\frac{3}{4}+2 t-t^2$, then $\left(a^2+b^2\right)$ is equal to

Let $f$ be a function such that $3 f(x)+2 f\left(\frac{m}{19 x}\right)=5 x, x \neq 0$, where $m=\sum\limits_{i=1}^9(i)^2$. Then $f(5)-f(2)$ is equal to

Let $\vec{a}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to