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.

Note that Questions are based on the NERDC curriculum (UBE compliant)

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).

See the Answer

The correct answer is B.

Solution:
x2 – 4x – 21;
x2 – 7x + 3x – 21;
x(x – 7) + 3(x – 7); (x – 7)(x + 3).

2. Solve the quadratic equation 2×2 – 3x – 5 = 0.
A. 1, .
B. -1, .
C. -1, .
D. 1, .

See the Answer

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

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.

See the Answer

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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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 + …….

See the Answer

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…

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.

See the Answer

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.

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.

See the Answer

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

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.

See the Answer

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.

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.

See the Answer

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

Further Mathematics Exam Questions and Answers for SS1

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Further Mathematics Exam Questions and Answers for SS1

Sample Further Mathematics Exam Questions and Answers

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