These Further Mathematics questions and answers were pulled from our book (Further Mathematics questions for SS3); 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″]The Questions are based on the current NERDC curriculum (UBE compliant)[/su_note]
Sample Further Mathematics Exam Questions and Answers
CONIC SECTIONS
1. Find the focus and -directix of the parabola y2 = 16x.
A. focus = (4, 0); directix, x = -4
B. focus = (3, 0); directix, x = -2
C. focus = (4, 1); directix, x = 4
D. focus = (-1, 0); directix, x = 3
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A
Explanation:
By comparing y2 = 16x with y2 = 4ax
Note: If the vertex of a parabola y2 = 4ax is translated to the point (x1,
y), the equation of the corresponding parabola
(y – y1)2 = 4a(x – x1)
=> 4a = 16; a = 4
Hence focus = (4, 0) while directrix x = -4.
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2. Find the equation of the tangent to the parabola y2 = 12x at the point (3, 6).
A. y + x – 2 = 0
B. 2y + x – 3 = 0
C. y – x – 3 = 0
D. y – 3x – 5 = 0
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is C
Explanation:
By comparing y2 = 12x with y2 = 4a x a = 3
The equation of the tangent at (x1, y1) is
y,y1 = 2a(x + x1)
Equation of the tangent is
6y = 2×3(x + 3)
6y = 6(x + 3)
Thus, 6y – 6x – 18 = 0
Or y – x – 3 = 0
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3. Find the equations of the tangent and normal to the ellipse 4×2 + 25y2 = 100 at the point (-3, 8/5).
A. 50x + 15y + 126 = 0
B. 25x – 15y + 138 = 0
C. 30x + 12y + 126 = 0
D. 50x + 5y – 114 = 0
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A
Explanation:
Writing 4×2 + 25y2 = 100, we have
x2/25 + y2/4 = 1; a = 5 and b = 2
Equation of tangent @ (x1, y1) is
xx1/a2 + yy1/b2 = 1
Hence, the equation of the tangent @ (-3, 8/5) is
x(-3)/52 + y(8/5)/22 = 1
-3x/25 + 8y/20 = 1
-12x + 40y = 100
-3x + 10y = 25
10y – 3x – 25 = 0
Equation of the normal @ (x1, y1) is
a2xy1 – b2x1y = (a2 – b2)x1y1
Hence, the equation of the normal at (-3, 8/5) i
25x(8)/5 – 4(-3)y = (25 – 4)(-3)(8/5)
40x + 12y = 21(-24/5)
200x + 60y = -504
200x + 60y + 504 =
50x + 15y + 126 = 0
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4. Find the centre and the radius of the circle 36×2 + 36y2
– 24x – 36y – 23 = 0
A. Centre of circle = (1/3, 1/2); radius = 1 unit
B. Centre of circle = (1/3, 1/5); radius = 2 units
C. Centre of circle = (1/2, 1/5); radius = 2 units
D. Centre of circle = (1/3, 1/4); radius = 5 units
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A
Explanation:
Divide through by 36
x2 + y2 – 2/3x – y –
23/36 = 0
x2 – 2/3x + y2 – y =
23/36
x2 – 2/3x + 1/9 + y2 –
y + 1/4 =
23/36 + 1/9 + 1/4
(x – 1/3)2 + (y – 1/2)2 = 1
Hence, the center of the circle is (1/3, 1/2)
The radius of the circle is 1 unit
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5. Find four vertices and foci of the ellipse,
x2/9 + y2/25 = 1
A. F1(0, 2) and F2(0, -2)
B. F1(0, -2) and F2(0, 4)
C. F1(0, 4) and F2(0, -4)
D. F1(0, 1) and F2(0, 3)
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is C
Explanation:
By comparing x2/9 + y2/25 = 1 with x2/b2 + y2/a2 =
We have b2 = 9 ; b ± 3
a2 = 25 ; a ± 5
Hence the four vertices are
V1(0, 5), V2(0, -5), V3(3, 0) & V4(-3, 0)
since c2 = a2 – b2 ; 25 – 9 = 16
c = ± 4
Hence, the foci are F1(0, 4) and F2(0, -4)
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Click here to get the complete Further Mathematics questions for SS3
INTEGRATION
1. Evaluate ∫ (ex – secx tanx) dx.
A. e2x – tanx + c
B. ex – secx + c
C. 2ex – cosecx + c
D. ex – tanx + c
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is B
Explanation:
= ∫exdx + ∫secxtanxdx
= ex – secx + C
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2. Evaluate ∫x√(1 + x2) dx.
A. 1/2(1 + 2×2)3/2 + c
B. 3/2(1 + 5×2)1/2 + c
C. (1 + 3×2)1/4 + c
D. 1/3(1 + x2)3/2 + c
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is D
Explanation:
Let U = 1 + x2
dU/dx = 2x
dx/dU = 1/2x
Therefore ∫x√1 +
x2dx = ∫x√U.dU/2x
= 1/2∫U1/2dU
= 1/2(U3/2/(3/2)) + C
= 1/3U3/2 + C
= 1/3(1 + x2)3/2 + C
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3. Evaluate ∫Sin1/2x dx.
A. -2cos1/2x + c
B. 2cos1/3x + c
C. cosx + c
D. -2cos3/2x – c
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A
Explanation:
Let U = 1/2x
dU/dx = 1/2; dx/dU = 2
Therefore ∫Sin1/2xdx = ∫SinU.dx/dU.dU
= 2∫SinUdU
= -2CosU + C
= -2CosU1/2x + C
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4. Evaluate ∫3/(4x – 1)3 dx.
A. 1/5(x – 1)2 + c
B. -3/8(4x – 1)2 + c
C. 2/7(3x – 1)+ c
D. (2x – 1)2 + c
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is B
Explanation:
Let U = 4x – 1
dU/dx = 4; dx/dU = 1/4
Therefore ∫3dx/(4x – 1)3 = ∫3/U3.dx/dU.dU
= 3/4∫U-3dU
= 3/4.U-2/-2 + C
= -3/8.U-2 + C
= -3/8.1/(4x – 1)2 + C
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5. Evaluate ∫(sinx + 1/x) dx.
A. -cosx + Inx + c
B. -sinx + In2x + c
C. cotx + Inx + c
D. cosx – Inx2 + c
[su_accordion][su_spoiler title=”See the Answer” open=”no” style=”default” icon=”plus” anchor=”” class=””]The correct answer is A
Explanation:
= ∫sinxdx + ∫1/xdx
= -cosx + Inx + C
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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 SS3, with even more questions and answers
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