Further Mathematics Exam Questions and Answers for SS1
These Further Mathematics questions and answers were pulled from our book (Further Mathematics questions for SS 1); Compiled to serve as a reference material to help teachers draw up test and exam questions faster. It could also help students assess their level of exam preparation. Each sample question includes correct answers.
[su_note note_color=”#fcf2e0″]Note that Questions are based on the NERDC curriculum (UBE compliant)[/su_note]
Sample Further Mathematics Exam Questions and Answers
TOPIC: ALGEBRAIC EQUATIONS
DIRECTION: Choose the correct answer from the lettered options.
1. Factorise x2 – 4x – 21.
A. (x – 3)(x + 7).
B. (x – 7)(x + 3).
C. (x + 7)(x + 3).
D. (x – 7)(x – 3).
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is B.
Solution:
x2 – 4x – 21;
x2 – 7x + 3x – 21;
x(x – 7) + 3(x – 7); (x – 7)(x + 3).
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2. Solve the quadratic equation 2×2 – 3x – 5 = 0.
A. 1, .
B. -1, .
C. -1, .
D. 1, .
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is C.
Solution:
2×2 – 3x – 5 = 0; 2×2 + 2x – 5x – 5 = 0; 2x(x + 1) – 5(x + 1) = 0; (x + 1)(2x – 5) = 0; x = -1
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3. If a, b are the roots of 2×2 + x – 3 = 0, constant equation whose root is 5a, 5b.
A. 4×2 + 160x + 375.
B. 4×2 – 160x + 375.
C. 4×2 – 160x – 375.
D. 4×2 + 160x – 375.
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A.
Solution:
The sum of roots = 5 + 5;
The product of root = 5.5;
Note: + = -b/a = -1/2, . = c/a = -3/2;
5[ + ] = 5-1/2; 25. = -75/2;
[x – (-5/2)][x – (-75/2)]; (2x + 5)(2x + 75);
4×2 + 160x + 375.
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Click here to get the complete Further Mathematics questions for SS 1
TOPIC: BINOMIAL THEOREM
DIRECTION: Choose the correct answer from the lettered options.
1. Use the substitution y = x – x2 to deduce the expansion of
(1 + x – x2)8, in ascending powers of x as far as (1 + y)8 the term in x4.
A. 1 + 8x + 20×2 – 196×4 + …..
B. 1 + 8x + 20×2 + 112×3 – 126×4 + …….
C. 1 + 8x + 20×2 – 126×4 + …..
D. 1 + 8x + 20×2 – 196×4 + …….
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is C.
Solution:
(1 + y)8; 1 + 8C1y + 8C2y2 + 8C3y3 + 8C4y4 + 8C5y5 + 8C6y6 + 8C7y7 + y8; 1 + 8y + 28y2 + 56y3 + 70y4 + 56y5 + 28y6 + 8y7 + y6; 1 + 8(x – x2) + 28(x – x2)2 + 56(x – x2)3 + 70(x – x2)4 + …… = 1 + 8x + 20×2 – 126×4…
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2. Find the exact value of (1.02)5, using Pascal’s triangle (Leave your answer in 3 sig. fig.).
A. 1.10.
B. 1.10141.
C. 1.104.
D. 1.10408.
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A.
Solution:
(1 + 0.02)5; 1 + 5(0.02) + 10(0.02)2 + 10(0.02)3 + 5(0.02)4 + 0.025 = 1 + 0.10 + 0.004 + 0.00008 + 0.0000008 + 0.0000000032 = 1.1040808032 » 1.10.
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3. Calculate the value of y if the coefficient of x4 in (2x + 3y)5 is 960.
A. 4.
B. 20.
C. 160.
D. 40.
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A.
Solution:
Using Pascal’s triangle; (2x)5 + 5(2x)4(3y) + 10(2x)3(3y)2 + 10(2x)2(3y)3 + 5(2x)(3y)4 (3y)5; 32×5 + 240x4y + 720x3y2 + 1080x2y3 + 810xy4 + 243y5; coefficient of x4 is 240; 240y = 960; y = 960/240 = 4
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4. If the first three terms of the expansion of (1 + bx)n are
1 + 8x + 16×2, find the values of n and b.
A. 2 and 4.
B. 4 and 2.
C. 2 and 2.
D. and 32.
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A.
Solution:
1 + nbx + n(n – 1)b2x2/2 = 1 + 8x + 16×2; nb = 8—(i);n(n – 1)b2/2 = 16—(ii); solve for b in equation (i) and substitute in equation (ii); b = 8/n–(iii);n(n -1)64/2n = 16; 32n – 32 = 16n; n = 2 substitute the value in equation (iii); b = 8/2 = 4; Therefore the answer is 2 and 4.
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5. Write down and simplify all the terms of the binomial expansion of (1 – x)6.
A. 1 + 6x + 15×2 + 20×3 + 15×4 + 6×5 + x6.
B. 1 – 6x + 15×2 – 20×3 + 15×4 – 6×5 + x6.
C. 1 + 6x + 15×2 – 20×3 + 15×4 – 6×5 + x6.
D. 1 – 6x + 15×2 – 20×3 + 15×4 + 6×5 + x6.
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is B.
Solution:
1 + 6C1(-x) + 6C2(-x)2 + 6C3(-x)3 + 6C4(-x)4 +
6C5(-x)5 + 6C6(-x)6; 1 – 6x + 15×2 – 20×3 + 15×4 – 6×5 + x6
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[su_column size=”4/5″ center=”no” class=””]Want more questions like this? Get the Complete Further Mathematics Exam Questions and Answers for SS 1, with even more questions and answers
