(a) Given that \(\begin{vmatrix} 5 & 2 & -3 \\ -1... | WAEC SSCE - Further Mathematics - 2007

Question 1 Report

(a) Given that \(\begin{vmatrix} 5 & 2 & -3 \\ -1 & k & 6 \\ 3 & 9 & (k + 2) \end{vmatrix} = -207\), find the values of the constant k.

(b) The equation of a curve is \(x(y^{2} + 1) - y(x^{2} + 1) + 4 = 0\). Find the:

(i) gradient of the curve at any point (x, y).

(ii) equation of the tangent to the curve at the point (-1, -3).

Answer

(a) Find k given the determinant equals \(-207\).

Expand along the first row:

\[5\big(k(k+2)-54\big)-2\big(-(k+2)-18\big)+(-3)\big(-9-3k\big)\]\[=5(k^{2}+2k-54)-2(-k-20)-3(-9-3k)\]\[=5k^{2}+10k-270+2k+40+27+9k\]\[=5k^{2}+21k-203\]

Set equal to \(-207\):

\[5k^{2}+21k-203=-207\;\Rightarrow\;5k^{2}+21k+4=0\]\[k=\frac{-21\pm\sqrt{441-80}}{10}=\frac{-21\pm 19}{10}\]

So \(k=-\dfrac{1}{5}\) or \(k=-4\).

(b) Curve \(x(y^{2}+1)-y(x^{2}+1)+4=0\).

Expand: \(xy^{2}+x-x^{2}y-y+4=0\). Differentiate implicitly:

\[y^{2}+2xy\,y'+1-2xy-x^{2}y'-y'=0\]\[y'(2xy-x^{2}-1)=-(y^{2}-2xy+1)\]

(i) Gradient at any point:

\[\frac{dy}{dx}=\frac{y^{2}-2xy+1}{x^{2}-2xy+1}\]

(ii) At \((-1,-3)\):

\[\frac{dy}{dx}=\frac{9-2(-1)(-3)+1}{1-2(-1)(-3)+1}=\frac{9-6+1}{1-6+1}=\frac{4}{-4}=-1\]

Tangent through \((-1,-3)\):

\[y+3=-1(x+1)\;\Rightarrow\;y=-x-4\quad\text{i.e. }x+y+4=0\]

Read lesson note on Matrices And Linear Transformation (WAEC)

Read lesson note on Surds (WAEC)

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