Further Mathematics Questions and Answers
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
Test your knowledge of advanced level mathematics with this aptitude test. This test comprises Further Maths questions and answers from past JAMB and WAEC examinations.
For what values of x is \(\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}\) undefined?
6 or 3
-18 or -9
-7 or 5
-5 or 7
Correct answer is C
An fraction is undefined when the denominator has value = 0.
\(\frac{x^{2} - 9x + 18}{x^{2} + 2x - 35}\) is undefined when \(x^{2} + 2x - 35 = 0\)
\(x^2 + 7x - 5x - 35 = 0 \implies (x + 7)(x - 5) = 0\)
\(x = \text{-7 or 5}\)
If \(\sin x = -\sin 70°, 0° < x < 360°\), determine the two possible values of x.
110°, 250°
110°, 290°
200°, 250°
250°, 290°
Correct answer is D
The value of the sine of an angle is negative in the third and fourth quadrant. Hence options A and B are not the options.
\(\sin 250 = -\sin (250 - 180) = - \sin 70\)
\(\sin 290 = - \sin (360 - 290) = - \sin 70\)
If (x - 3) is a factor of \(2x^{2} - 2x + p\), find the value of constant p.
-12
-6
3
6
Correct answer is A
Using remainder theorem, since x - 3 is a factor, then
given \(2x^{2} - 2x + p\), f(3) = 0
\(2(3^{2}) - 2(3) + p = 0 \implies 18 - 6 = -p\)
\(p = -12\)
The roots of a quadratic equation are \((3 - \sqrt{3})\) and \((3 + \sqrt{3})\). Find its equation.
\(x^{2} - 6x - 9 = 0\)
\(x^{2} - 6x + 6 = 0\)
\(x^{2} + 6x - 9 = 0\)
\(x^{2} + 6x + 6 = 0\)
Correct answer is B
\((x - \alpha)(x - \beta) = 0\)
\((x - (3 - \sqrt{3}))(x - (3 + \sqrt{3})) = 0\)
\((x^{2} - (3 - \sqrt{3})x - (3 + \sqrt{3})x + (9 + 3\sqrt{3} - 3\sqrt{3} - 3) = 0\)
\(x^{2} - 3x - x\sqrt{3} - 3x + x\sqrt{3} + 6 = 0\)
\(x^{2} - 6x + 6 = 0\)
\(\frac{560}{243}\)
\(\frac{841}{243}\)
\(\frac{1120}{243}\)
\(\frac{4481}{243}\)
Correct answer is C
\(^{10}C_{7 - 1} (2^{10 - 6}) (\frac{-1}{3})^{6}\)
\(\frac{10!}{(10 - 6)! 6!} \times 16 \times \frac{1}{243} \)
= \(210 \times 16 \times \frac{1}{729} \)
= \(\frac{1120}{243}\)