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WAEC May/June 2023 FREE FURTHER MATHEMATICS QUESTION AND ANSWER ROOM [School Candidates].
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Further Mathematics/Mathematics (Elective) 2 (Essay) – 09:30am – 12:00pm.
Further Mathematics/Mathematics (Elective) 1 (Objective) – 3:00pm – 4:30pm.
2023 WAEC FURTHER MATHEMATICS OBJECTIVES (OBJ) ANSWERS
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2023 Further Mathematics Questions and Answers (Obj/Theory).
Past Essay (Theory) answers
X-3|2 2|+4|5 2 |+3|5 2|= -24
To resolve this equation, we need to simplify the expressions within the absolute value signs.
|2 2| equals 2 - 2, which is 0.
|5 2 | equals 5 - 2, which is 3.
|5 2| equals 5 - 2, which is 3.
Now we substitute these values back into the equation:
(X - 3)(0) + 4(3) + 3(3) = -24
0 + 12 + 9 = -24
21 = -24
Since 21 is not equal to -24, there is no solution to this equation.
Log 3x - 3logx³ + 2 = 0
To resolve this equation, let's simplify it step by step:
Log 3x - 3log3³/log3x + 2 = 0
Using the property of logarithms, we can rewrite log3³ as 3log3:
Log 3x - 3 * 3log3 / log3x + 2 = 0
Now let's simplify further:
Log3x - 9 / log3x + 2 = 0
Let's introduce a substitution to make it easier. Let p = log3x:
p - 9 / p + 2 = 0
To eliminate the fraction, let's multiply through by (p + 2):
p(p + 2) - 9 = 0
p² + 2p - 9 = 0
Now let's solve this quadratic equation:
(p + 3)(p - 3) = 0
p = -3 or p = 3
But we remember that p = log3x, so:
log3x = -3 or log3x = 3
Now let's solve for x:
When log3x = -3:
3x = 3^(-3)
3x = 1/27
x = 1/27
When log3x = 3:
3x = 3^3
3x = 27
x = 27/3
x = 9
So the solutions to the equation are x = 1/27 and x = 9.
3a) U = x - 2, therefore x = u + 2.
Now let's substitute this back into the expression:
(x^2 + 5) / (x - 2)^4
Substituting x = u + 2:
((u + 2)^2 + 5) / (u + 2 - 2)^4
((u + 2)^2 + 5) / u^4
(u^2 + 4u + 4 + 5) / u^4
(u^2 + 4u + 9) / u^4
3b) Simplifying further:
(u^2 + 4u + 9) / u^4
So, the expression simplifies to (u^2 + 4u + 9) / u^4.
S12 = 12/6[2a + 11d] = 168
To solve this equation, we use the formula for the sum of an arithmetic series:
Sn = n/2[2a + (n - 1)d]
Given that S12 = 168, we have:
12/2[2
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