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.
The Questions are based on the current NERDC curriculum (UBE compliant)
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
See the Answer
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.
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
See the Answer
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
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
See the Answer
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
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
See the Answer
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
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)
See the Answer
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)
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
See the Answer
The correct answer is B
Explanation:
= ∫exdx + ∫secxtanxdx
= ex – secx + C
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
See the Answer
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
3. Evaluate ∫Sin1/2x dx.
A. -2cos1/2x + c
B. 2cos1/3x + c
C. cosx + c
D. -2cos3/2x – c
See the Answer
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
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
See the Answer
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
5. Evaluate ∫(sinx + 1/x) dx.
A. -cosx + Inx + c
B. -sinx + In2x + c
C. cotx + Inx + c
D. cosx – Inx2 + c
See the Answer
The correct answer is A
Explanation:
= ∫sinxdx + ∫1/xdx
= -cosx + Inx + C
