Waec Past Questions And Answers On Further Mathematics Pdf

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

Further Mathematics an advanced secondary school mathematics subject. The subject usually picks up where the regular maths class leaves off and covers topics like mathematical induction, complex numbers, polar curves, and conic sections, differential equations, matrices, and statistical inference.

Question 1

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

Options

The correct answer is B.

Explanation:

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

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

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

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

Comments (1)

Question 2

Find the domain of \(g(x) = \frac{4x^{2} - 1}{\sqrt{9x^{2} + 1}}\)

Options

\({x : x \in R, x = \frac{1}{2}}\)

\(x: x \in R, x\neq \frac{1}{3}\)

\(x : x \in R, x = \frac{1}{3}\)

\(x: x \in R\)

The correct answer is D.

Explanation:

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.

Comments (1)

Question 3

Given that \(f(x) = 3x^{2} -Â 12x + 12\) and \(f(x) = 3\), find the values of x.

Options

1, 3

The correct answer is A.

Explanation:

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

\(x = 1 or 3\)

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Question 4

A binary operation * is defined on the set of real numbers, by \(a * b = \frac{a}{b} + \frac{b}{a}\). If \((\sqrt{x} + 1) * (\sqrt{x} - 1) = 4\), find the value of x.

Options

The correct answer is D.

Explanation:

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

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

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

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

\(x = 3\)

Comments (1)

Question 5

If \(4x^{2} + 5kx + 10\) is a perfect square, find the value of k.

Options

\(\frac{5\sqrt{10}}{4}\)

\(\frac{4\sqrt{10}}{5}\)

The correct answer is D.

Explanation:

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

Expanding the right hand side equation, we have

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

Comparing with the left hand side, we have

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

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Question 6

If the polynomial \(f(x) = 3x^{3} - 2x^{2} + 7x + 5\) is divided by (x - 1), find the remainder.

Options

The correct answer is D.

Explanation:

\(f(x) = 3x^{3} - 2x^{2} + 7x + 5\).

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

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

Comments (2)

Question 7

\(P = {1, 3, 5, 7, 9}, Q = {2, 4, 6, 8, 10, 12}, R = {2, 3, 5, 7, 11}\) are subsets of \(U = {1, 2, 3, ... , 12}\). Which of the following statements is true?

Options

\(Q \cap R = \varnothing\)

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

The correct answer is C.

Explanation:

All the statements are false except option C.

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

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

Comments (1)

Question 8

If \(\log_{3}a - 2 = 3\log_{3}b\), express a in terms of b.

Options

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

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

The correct answer is C.

Explanation:

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

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

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

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

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

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Question 9

If \(\alpha\) and \(\beta\) are the roots of \(2x^{2} - 5x + 6 = 0\), find the equation whose roots are \((\alpha + 1)\) and \((\beta + 1)\).

Options

\(2x^{2} - 9x + 15 = 0\)

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

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

\(2x^{2} - 9x - 15 = 0\)

The correct answer is B.

Explanation:

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: \(x^{2} - (\frac{-b}{a})x + (\frac{c}{a}) = 0\)

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

Comments (1)

Question 10

Resolve \(\frac{3x - 1}{(x - 2)^{2}}, x \neq 2\) into partial fractions.

Options

\(\frac{x}{2(x - 2)} - \frac{5}{(x - 2)^{2}}\)

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

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

\(\frac{-1}{2(x - 2)} + \frac{8x}{2(x - 2)^{2}}\)

The correct answer is C.

Explanation:

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

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

Comparing, we have

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

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

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

Comments (1)

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