Further Mathematics Past Questions | JAMB, WAEC, NECO and Post UTME Past Questions

\(\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)

= \(\frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)

= \(\frac{3 + \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3} - \sqrt{3} - 1}\)

= \(\frac{6}{2} = 3\)

The domain of a function refers to the regions where the function is defined or has a value on a particular region.

\(\frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\) has a domain defined on all set of real numbers because the function is defined on the set of real numbers when the denominator \(\sqrt{9x^{2} + 1} \geq 0\).

\(\sqrt{9x^{2} + 1} \geq 0 \implies 9x^{2} + 1 \geq 0\) which because of the square sign has a value for all values of x, be it negative or positive.

\(f(x) = 3x^{2} - 12x + 12\) and \(f(x) = 3\)

\(\therefore f(x) = 3 = 3x^{2} - 12x + 12 \implies 3x^{2} - 12x + 12 - 3 = 0\)

\(3x^{2} - 12x + 9 = 0; 3x^{2} - 9x - 3x + 9 = 0\)

\(3x(x - 3) - 3(x - 3) = (3x - 3)(x - 3) = 0\)

3x - 3 = 0 or x - 3 = 0

\(x = 1 or 3\)

\((\sqrt{x} + 1) * (\sqrt{x} - 1) = 4  \implies  \frac{\sqrt{x} + 1}{\sqrt{x} - 1} + \frac{\sqrt{x} - 1}{\sqrt{x} + 1} = 4\)

\(\frac{(\sqrt{x} + 1)(\sqrt{x} + 1) + (\sqrt{x} - 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)}\)

= \(\frac{x + 2\sqrt{x} + 1 + x - 2\sqrt{x} + 1}{x - 1} \implies \frac{2x + 2}{x - 1} = 4\)

\(2x + 2 = 4x - 4  \therefore 4x - 2x = 2x = 2 + 4= 6\)

\(x = 3\)

\(4x^{2} + 5kx + 10 = (2x + \sqrt{10})^{2}\)

Expanding the right hand side equation, we have

\(4x^{2} + 4x\sqrt{10} + 10\)

Comparing with the left hand side, we have

\(5k = 4\sqrt{10}  \implies k = \frac{4}{5}\sqrt{10}\)

\(f(x) = 3x^{3} - 2x^{2} + 7x + 5\). 

\(x - 1 = 0, x = 1\)

\(f(1) = 3(1)^{3} - 2(1)^{2} + 7(1) + 5 = 13\)

All the statements are false except option C.

\(R \cap P = {3, 5, 7} and R \cap U = {2, 3, 5, 7, 11}\) 

\(\therefore (R \cap P) \subset (R \cap U)\)

\(\log_{3}a - 2 = 3\log_{3}b\)

Using the laws of logarithm, we know that \( 2 = 2\log_{3}3 = \log_{3}3^{2}\)

\(\therefore \log_{3}a - \log_{3}3^{2} = \log_{3}b^{3}\)

= \(\log_{3}(\frac{a}{3^{2}}) = \log_{3}b^{3}   \implies  \frac{a}{9} = b^{3}\)

\(\implies a = 9b^{3}\)

Note: Given the sum of the roots and its product, we can get the equation using the formula:

\(x^{2} - (\alpha + \beta)x + (\alpha\beta) = 0\). This will be used later on in the course of our solution.

Given equation: \(2x^{2} - 5x + 6 = 0; a = 2, b = -5, c = 6\).

\(\alpha + \beta = \frac{-b}{a} = \frac{-(-5)}{2} = \frac{5}{2}\)

\(\alpha\beta = \frac{c}{a} = \frac{6}{2} = 3\)

Given the roots of the new equation as \((\alpha + 1)\) and \((\beta + 1)\), their sum and product will be

\((\alpha + 1) + (\beta + 1) = \alpha + \beta + 2 = \frac{5}{2} + 2 = \frac{9}{2} = \frac{-b}{a}\)

\((\alpha + 1)(\beta + 1) = \alpha\beta + \alpha + \beta + 1 = 3 + \frac{5}{2} + 1 = \frac{13}{2} = \frac{c}{a}\)

The new equation is given by: \(x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0\)

= \(x^{2} - (\frac{9}{2})x + \frac{13}{2} = 2x^{2} - 9x + 13 = 0\)

\(\frac{3x - 1}{(x - 2)^{2}} = \frac{A}{(x - 2)} + \frac{Bx}{(x - 2)^{2}}\)

\(\frac{3x - 1}{(x - 2)^{2}} = \frac{A(x - 2) + Bx}{(x - 2)^{2}}\)

Comparing, we have

\(3x - 1 = Ax - 2A + Bx  \implies -2A = -1;  A + B = 3\)

\(\therefore A = \frac{1}{2}; B = \frac{5}{2}\)

= \(\frac{1}{2(x - 2)} + \frac{5x}{2(x - 2)^{2}}\)