These are the solutions to the 2024 WASSCE Elective Mathematics Paper 1. The questions without the solutions may be found at WASSCE 2024 Elective Maths Paper 1.
1. If \( \displaystyle \frac{5}{\sqrt 2} – \frac{\sqrt 8}{8} = m\sqrt 2 \), find the value of \( m \).
A. \( \displaystyle \frac{12}{5} \)
B. \( \displaystyle \frac{9}{4} \)
C. \( \displaystyle \frac{7}{3} \)
D. \( \displaystyle \frac{5}{2} \)
Answer: B
Solution
Writing the expression over a common denominator gives
$$ \frac{5}{\sqrt 2} – \frac{\sqrt 8}{8} = \frac{5(8) – \sqrt 8 \sqrt 2}{8 \sqrt 2} = \frac{40 – \sqrt{16}}{8 \sqrt 2} = \frac{40 – 4}{8 \sqrt 2} = \frac{36}{8 \sqrt 2} $$
Rationalizing the denominator in the simplified expression gives
$$ \frac{36}{8 \sqrt 2} = \frac{36}{8 \sqrt 2} \cdot \frac{\sqrt 2}{\sqrt 2} = \frac{36 \sqrt 2}{16} $$
Writing the fraction in its lowest terms gives \( \displaystyle \frac{9\sqrt2}{4} \) or \( \displaystyle \frac{9}{4} \sqrt 2 \).
Thus, m is 9/4.
2. Given that \( f:x \to \sqrt x \) and \( g:x \to 25 – x^2 \), find the value of \( f \circ g(3) \).
A. 4
B. 3
C. 2
D. 1
Answer: A
Solution
To evaluate \( f \circ g(3) \), we first evaluate \( g(3) \) and then put the result in \( f \).
$$ g(3) = 25 – 3^2 = 25 – 9 = 16 $$
Hence, \( f \circ g(3) = f(g(3)) = \sqrt{16} = 4 \).
3. If \( \displaystyle \sqrt x – \frac{6}{\sqrt x} = 1 \), find the value of \( x \).
A. 9
B. 16
C. 25
D. 36
Answer: A
Solution
Let \( k = \sqrt x \).
Then the expression may be written as \( \displaystyle k – \frac{6}{k} = 1 \).
Multiplying both sides by \( k \) and solving gives
$$ \begin{align} k^2 – 6 &= k \\ k^2 – k – 6 &= 0 \\ (k + 2)(k – 3) &= 0 \end{align} $$
The solutions are k = -2 and k = 3.
But since \( k = \sqrt x \) and the function \( \sqrt x \) only gives positive numbers as outputs, we rule out k = -2, giving the answer as k = 3, so that \( \sqrt x = 3 \) and x = 9.
4. If \( 5x + 7 \equiv P(x + 3) + Q(x – 1) \), find the value of \( P \).
A. 3
B. 2
C. 1
D. 0
Answer: A
Solution
We take advantage of the fact that the expression is an identity, that is, it is true for all values of x. Setting x = 1 will conveniently eliminate Q from the expression since it will then be multiplying 0.
Setting x = 1 gives
$$ \begin{align} 5(1) + 7 &= P(1 + 3) + Q(1 – 1) \\ 5 + 7 &= P(4) + Q(0) \\ 12 &= 4P \end{align} $$
which simplifies to P = 3.
5. A binary operation \( * \) is defined on the set of real numbers, R by \( \displaystyle x * y = \frac{y^2 – x^2}{2xy}, x, y \neq 0 \), where \( x \) and \( y \) are real numbers. Evaluate \( -3 * 2 \).
A. 13/12
B. 5/12
C. -5/12
D. -13/12
Answer: B
Solution
We proceed by substituting x = -3 and y = 2 into the expression for the binary operation and simplifying:
$$ \begin{align} -3 * 2 &= \frac{2^2-(-3)^2}{2(-3)(2)} \\ &= \frac{4-9}{-12} \\ &= \frac{-5}{-12} = \frac{5}{12} \end{align} $$
Leave a Reply