We have \(\begin{bmatrix} x &-5&-1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 0 & 2\\ 0 & 2 & 1 \\2&1&3 \end{bmatrix}\)\(\begin{bmatrix} x \\ 4\\1 \end{bmatrix}=O\)
⇒ \(\begin{bmatrix} x+0-2 &0-10+-0 &2x-5-3\end{bmatrix}\) \(\begin{bmatrix} x \\ 4\\1 \end{bmatrix}=O\)
⇒ \(\begin{bmatrix} x-2 &-10&2x-8 \end{bmatrix}\) \(\begin{bmatrix} x \\ 4\\1 \end{bmatrix}=O\)
⇒ \(\begin{bmatrix} x(x-2) -40+2x-8 \end{bmatrix}=0\)
⇒ \(\begin{bmatrix} x^2 -2x-40+2x-8 \end{bmatrix}=[0]\)
⇒ \(\begin{bmatrix} x^2 -48 \end{bmatrix}=[0]\)
⇒\(\therefore x^2=48\)
⇒ \(x= 4 \sqrt3\)
Let \[ f(x)=\int \frac{7x^{10}+9x^8}{(1+x^2+2x^9)^2}\,dx \] and $f(1)=\frac14$. Given that 

A ladder of fixed length \( h \) is to be placed along the wall such that it is free to move along the height of the wall.
Based upon the above information, answer the following questions:
(iii) (b) If the foot of the ladder, whose length is 5 m, is being pulled towards the wall such that the rate of decrease of distance \( y \) is \( 2 \, \text{m/s} \), then at what rate is the height on the wall \( x \) increasing when the foot of the ladder is 3 m away from the wall?