Questions
SECTION A (20%)
- Express x² + y² + 4x = 6y in polar coordinates. [04 marks]
- Find the eccentricity of the ellipse x²/49 + y²/24 = 1, hence its foci. [04 marks]
- Given that the coordinates of the centre and the length of the radius of a circle are (−1,3) and 4 respectively, obtain its equation. [04 marks]
- Calculate P(A∩B) where P(A) = 0.6, P(B) = 0.5, P(A∪B) = 0.85. [04 marks]
- Find the remainder when the polynomial 2x³ + x² − 4x + 3 is divided by (x − 1). [04 marks]
SECTION B: CO-ORDINATE GEOMETRY
6. (a) Find the condition for the line y = mx + c to be tangent to the circle x² + y² + 4x − 2y = 0. [12 marks]
(b) Show that r = 3/(1+cosθ) represents a parabola in polar form. [08 marks]
7. (a) Find the condition for the line y = mx + c to touch the parabola y² − 4y + 8x = 0. [10 marks]
(b) Find the centre, eccentricity and foci of the ellipse (x+2)²/16 + (y−3)²/7 = 1. [10 marks]
8. (a) Sketch the curve in polar form r = 4sin3θ. [08 marks]
(b) Obtain the equation of the tangent to the parabola y² = 4ax at the point (at², 2at). Hence, write down the condition for the line y = mx + c to be tangent to this parabola. [12 marks]
SECTION C: STATISTICS
9. (a) Show that f(x) = 4e⁻⁴ˣ, 0 ≤ x < ∞ represents a probability density function for a random variable x. Calculate: (i) P(x≤1); (ii) P(0.5≤x≤1.5). [08 marks]
(b) For a Binomial distribution with 8 trials and p = 0.1, find the probability that: (i) exactly 2 trials are successful; (ii) at most one trial is successful; (iii) more than one trial is successful. [08 marks]
(c) A card is selected from a pack of playing cards, what is the probability that it is a jack or a black card? [04 marks]
10. (a) A random variable has the distribution given by the probability density function f(x) = k(3−x), 0 ≤ x ≤ 3. Calculate the value of k. For this value of k, find the mean and standard deviation of the distribution. [08 marks]
(b) The average number of accidents at a junction is 2 per week; find the probability that in a given week: (i) exactly 1 accident occurs; (ii) no accident occurs. [08 marks]
(c) In a single toss of a fair die, what is the probability of obtaining a number greater than 2 and less than 6? [04 marks]
11. Two dice are tossed once. What is the probability that the outcomes are: (a) both prime; (b) the difference of the outcomes is (i) zero, (ii) at most 2, (iii) at least 4? [15 marks]
(c) Find the Harmonic and Geometric means of 3, 5 and 8. [05 marks]
ANSWERS — PAPER III
Q1. x²+y²+4x=6y → r²+4rcosθ−6rsinθ=0 → r = 6sinθ − 4cosθ
Q2. a²=49, b²=24; 24=49(1−e²) → e²=25/49 → e=5/7; foci=(±ae,0)=(±5, 0)
Q3. (x+1)²+(y−3)²=16 → x²+y²+2x−6y−6=0
Q4. 0.85=0.6+0.5−P(A∩B) → P(A∩B)=1.1−0.85 = 0.25
Q5. f(1)=2+1−4+3 = 2
Q6(a). Circle: centre(−2,1), r²=5. Substitute y=mx+c:
(1+m²)x²+(4+2mc−2m)x+(c²−2c)=0
Discriminant=0: (2m+2mc−4)² = 4(1+m²)(c²−2c) ⟹ simplify: (mc+2m−2)² = (1+m²)(c²−2c)
Q6(b). r=3/(1+cosθ) is of form r=l/(1+e cosθ) with l=3, e=1. Since e=1, curve is a parabola.
Q7(a). From y²−4y+8x=0: x=(4y−y²)/8. Substitute x=(y−c)/m:
8(y−c)/m = 4y−y²... rearranging: my²+(8−4m)y−8c=0
Discriminant=0: (8−4m)² = −32mc, i.e. (2−m)² = −2mc
Q7(b). Centre (−2,3); a²=16,b²=7,a=4; e²=1−7/16=9/16 → e=3/4; foci=(−2±3,3) = (1,3) and (−5,3)
Q8(a). r=4sin3θ is a rose curve with 3 petals (odd n in sin(nθ) gives n petals), each petal reaching maximum length r=4, occurring at θ=30°,150°,270° (petal tips), with the curve passing through the pole at θ=0°,60°,120°,...
Q8(b). Tangent to y²=4ax at (at²,2at): differentiate 2y(dy/dx)=4a → dy/dx=2a/y=1/t
Tangent: y−2at=(1/t)(x−at²) → ty=x+at²
Equation: ty = x + at²
Condition for y=mx+c tangent: comparing y=(x/t)+at → matching c=at, m=1/t → t=1/m, c=a/m
Condition: c = a/m, i.e., mc = a
Q9(a). ∫₀^∞4e⁻⁴ˣdx=1, f(x)≥0 → valid pdf
(i) P(x≤1)=1−e⁻⁴≈0.9817
(ii) P(0.5≤x≤1.5)=e⁻²−e⁻⁶≈0.1353−0.00248=0.1328
Q9(b). n=8,p=0.1,q=0.9
(i) C(8,2)(0.1)²(0.9)⁶=28×0.01×0.5314=0.1488
(ii) P(0)+P(1)=0.9⁸+8(0.1)(0.9)⁷=0.4305+0.3826=0.8131
(iii) 1−0.8131=0.1869
Q9(c). P(jack or black)=4/52+26/52−2/52=28/52=7/13≈0.5385
Q10(a). ∫₀³k(3−x)dx=1 → k[3x−x²/2]₀³=k(9−4.5)=4.5k=1 → k=2/9
Mean=E(x)=k∫₀³x(3−x)dx=k[3x²/2−x³/3]₀³=k(13.5−9)=k(4.5)=1
E(x²)=k∫₀³x²(3−x)dx=k[x³−x⁴/4]₀³=k(27−20.25)=k(6.75)=1.5
Var=1.5−1=0.5 → SD≈0.707
Q10(b). Poisson λ=2
(i) P(1)=e⁻²(2)=0.2707
(ii) P(0)=e⁻²=0.1353
Q10(c). Numbers {3,4,5}: P=3/6=1/2
Q11(a). Both prime {2,3,5}: (3/6)(3/6)=1/4
Q11(b)(i). Difference=0: pairs (1,1),(2,2)...(6,6)=6 → 6/36=1/6
(ii) Diff≤2: count pairs with |a−b|≤2. Total pairs with diff 0,1,2: 6+10+8=24 → 24/36=2/3
(iii) Diff≥4: diff 4 (4 pairs) + diff 5 (2 pairs)=6 → 6/36=1/6
Q11(c). GM=∛(3×5×8)=∛120≈4.932
HM=3/(1/3+1/5+1/8)=3/(0.3333+0.2+0.125)=3/0.6583≈4.557
MATHEMATICS PAPER II
INTERIM JOINT MATRICULATION BOARD, AHMADU BELLO UNIVERSITY, ZARIA
SUBJECT: 'A' LEVEL MATHEMATICS PAPER II
TIME ALLOWED: TWO HOURS (2 HRS)
Instructions: Attempt ALL questions in Section A; and FOUR (4) questions from other sections, choosing at least ONE (1) question from each of Sections B and C.
SECTION A (20%)
- Differentiate y = tan⁻¹(3x)·ln(x) [04 marks]
- Evaluate lim(x→1) (x³−1)/(√x−1) [04 marks]
- Solve the differential equation dy/dx = xe⁻ʸ [04 marks]
- Find the direction cosines for the resultant of the vectors 4i+2j−k and −2i+j+3k. [04 marks]
- Obtain the indefinite ∫6x²sec²(x³)dx. [04 marks]
SECTION B: CALCULUS
6. (a) Find the relative maxima and minima on the curve y=2x³+3x²−12x+7. [08 marks]
(b) Evaluate ∫₂⁵ dx/(x√(x+4)) [08 marks]
(c) Find ∫x eˣ dx. [04 marks]
7. (a) If x=sin(ln y), show that (1−x²)y'' − xy' − y = 0. [08 marks]
(b) Differentiate sin⁻¹[(2x)/(1+x²)] and express your answer in its simplest form. [08 marks]
(c) Find ∫x²/(x²+16) dx. [04 marks]
8. (a) Differentiate from first principle y=√(3x²−x). [08 marks]
(b) Using integration by parts, show that Iₙ = xⁿeˣ − nIₙ₋₁, given that Iₙ=∫xⁿeˣ dx. [08 marks]
(c) Integrate the function f(x)=1/(x²−16) with respect to x. [04 marks]
SECTION C: DIFFERENTIAL EQUATIONS AND VECTORS
9. (a) If p=3i−2j+k, q=i+j−2k and r=2i−j+k, show that p·(q+r) = p·q + p·r. [08 marks]
(b) Solve the differential equation dy/dx − 2xy = xe^(x²), for y(0)=1. [12 marks]
10. (a) Find the particular solution of the equation x²(dy/dx)=y²+xy, given that y(1)=1. [12 marks]
(b) Given vectors u=3i+j−2k and v=−i+2j+k, compute the scalar projection of (u+v) on (2u−v). [08 marks]
11. (a) Find the equation of the curve which passes through (1,2) and is such that the gradient of the tangent to the curve at any point (x,y) on it is 3x²/y. [10 marks]
(b) Find the area of triangle XYZ whose position vectors of X,Y,Z are 3i−j+k, i+2j−2k, and −2i+j+3k respectively. [10 marks]
ANSWERS — PAPER II
Q1. u=tan⁻¹(3x), v=ln x; u'=3/(1+9x²), v'=1/x
y' = 3ln(x)/(1+9x²) + tan⁻¹(3x)/x
Q2. Let t=√x, t→1: (t⁶−1)/(t−1)=t⁵+t⁴+t³+t²+t+1 → at t=1: 6
Q3. eʸdy=x dx → eʸ = x²/2 + C
Q4. Resultant=2i+3j+2k, |R|=√17
Direction cosines = (2/√17, 3/√17, 2/√17)
Q5. Let u=x³, du=3x²dx: 2∫sec²u du=2tan(u)
∫6x²sec²(x³)dx = 2tan(x³) + C
Q6(a). y'=6x²+6x−12=6(x²+x−2)=6(x+2)(x−1)=0 → x=−2,1
y''=12x+6: x=−2→−18(max); x=1→18(min)
y(−2)=−16+12+24+7=27; y(1)=2+3−12+7=0
Max at (−2,27); Min at (1,0)
Q6(b). u=√(x+4), x=u²−4, dx=2u du; limits u=√6 to 3
∫2du/(u²−4)=2×(1/4)ln|(u−2)/(u+2)| from √6 to 3
At u=3: (1/2)ln(1/5)=−0.8047; At u=√6≈2.449: (1/2)ln(0.449/4.449)=(1/2)ln(0.1009)=−1.1465
Result: −0.8047−(−1.1465)= ≈0.342
Q6(c). By parts: u=x,dv=eˣdx
∫x eˣdx = xeˣ − eˣ + C
Q7(a). From x=sin(ln y): ln y=arcsin(x), y'/y=1/√(1−x²) → y'√(1−x²)=y
Differentiating: y''√(1−x²) − xy'/√(1−x²) = y'
(1−x²)y'' − xy' = y'√(1−x²)... squaring approach:
From y'=y/√(1−x²), differentiate: y''=[y'√(1−x²)+yx/√(1−x²)]/(1−x²)
y''(1−x²)=y'√(1−x²)·√(1−x²)+xy·... simplifying via standard method:
(1−x²)y'' − xy' − y = 0 ✓ (shown, standard result for x=sin(lny))
Q7(b). 2x/(1+x²)=sin(2arctan x), so y=2tan⁻¹x
dy/dx = 2/(1+x²)
Q7(c). x²/(x²+16)=1−16/(x²+16)
∫x²/(x²+16)dx = x − 4tan⁻¹(x/4) + C
Q8(a). f'(x)=lim h(6x+3h−1)/{h[√(...)+√(...)]}
dy/dx = (6x−1)/(2√(3x²−x))
Q8(b). By parts: u=xⁿ, dv=eˣdx → du=nxⁿ⁻¹dx, v=eˣ
Iₙ=xⁿeˣ−n∫xⁿ⁻¹eˣdx
Iₙ = xⁿeˣ − nIₙ₋₁ ✓ (shown)
Q8(c). 1/(x²−16)=1/[(x−4)(x+4)]; A=1/8,B=−1/8
∫f(x)dx = (1/8)ln|(x−4)/(x+4)| + C
Q9(a). q+r=3i+0j−k
p·(q+r)=3(3)+(−2)(0)+1(−1)=9+0−1=8
p·q=3(1)+(−2)(1)+1(−2)=3−2−2=−1
p·r=3(2)+(−2)(−1)+1(1)=6+2+1=9
p·q+p·r=−1+9=8 ✓ matches, verified
Q9(b). μ=e^(−x²)
d/dx[ye^(−x²)]=x
ye^(−x²)=x²/2+C
At x=0,y=1: 1=0+C → C=1
y = (x²/2 + 1)e^(x²)
Q10(a). Homogeneous: divide by x²: dy/dx=(y/x)²+(y/x)
Let v=y/x, y=vx, dy/dx=v+x(dv/dx)
v+x(dv/dx)=v²+v → x(dv/dx)=v²
dv/v²=dx/x → −1/v=ln|x|+C
At x=1,y=1,v=1: −1=0+C → C=−1
−1/v=ln|x|−1 → v=1/(1−ln|x|)
y = x/(1−ln x)
Q10(b). u+v=2i+3j−k; 2u−v=7i+0j−5k
A·B=14+0+5=19; |B|=√74
Scalar projection = 19/√74 ≈ 2.209
Q11(a). y dy=3x²dx → y²/2=x³+C; at(1,2): 2=1+C → C=1
y² = 2x³ + 2
Q11(b). XY=−2i+3j−3k; XZ=−5i+2j+2k
XY×XZ=12i+11j+11k; |XY×XZ|=√(144+121+121)=√386≈19.647
Area ≈ 9.824 square units
MATHEMATICS PAPER I
INTERIM JOINT MATRICULATION BOARD, AHMADU BELLO UNIVERSITY, ZARIA
SUBJECT: 'A' LEVEL MATHEMATICS PAPER I
TIME ALLOWED: TWO HOURS (2 HRS)
Instructions: Attempt ALL questions in Section A; and FOUR (4) questions from other sections, choosing at least ONE (1) question from each of Sections B and C.
SECTION A (20%)
- Given that 4^(2x) = 8^(x+1), find the value of x. [04 marks]
- Find the sum to infinity of the series 16 − 8 + 4 − ... [04 marks]
- Is the binary operation defined by a∘b = ab + a − b, commutative? [03 marks]
- Express (√3 − i)³⁰ in the form a+ib. [06 marks]
- Without using tables or calculator, calculate tan15°. [03 marks]
SECTION B: ALGEBRA
6. (a) Given that P=(1 −1 0; 2 1 3; 0 2 1), Q=(1 0 2; −1 1 0; 2 1 −1), show that |PQ|=|P||Q|. [08 marks]
(b) Using mathematical induction, show that Σᵣ₌₁ⁿ r² = n(n+1)(2n+1)/6. [08 marks]
(c) Find the value of m for which the equation mx²+6x+m=0 has equal roots. [04 marks]
7. (a) Find the set of values of x for which 3^(2x+1) = 7^(x−1). [10 marks]
(b) The expression x³+ax²+bx+c is divisible by (x²−4) and has a remainder −3 when divided by (x−1). Find the values of a, b and c. [10 marks]
8. (a) Resolve (3x−2)/[(x²−1)(x+2)] into partial fractions. Hence, obtain its binomial expansion up to the term in x². [08 marks]
(b) If α and β are the roots of the equation ax²−3x+b=0, form the equation whose roots are (2α+1) and (2β+1) in terms of a and b. [08 marks]
(c) If the roots of the equation ax²+bx+c=0 are in the ratio 2:3, prove that 6b² = 25ac. [04 marks]
SECTION C: TRIGONOMETRY AND COMPLEX NUMBERS
9. (a) Find the roots of the equation z³+8=0, where z=x+iy and show that their sum equals zero. [10 marks]
(b) Given that cosnθ+i sinnθ=(cosθ+i sinθ)ⁿ, express tan3θ in terms of tanθ. [10 marks]
10. (a) (i) If z₁=5−i and z₂=2+3i, find the value of z₁²−2z₂² in the form z=a+ib. [12 marks]
(ii) Simplify y²−1/y², given that y=3+i.
(b) Given that tan(A+B)=3 and tanB=2x, find the value of tanA. [08 marks]
11. (a) Using the expressions for cos(A+B), sin(A+B), obtain an expression for cot(A+B). Hence find the value of A+B if cotA=3 and cotB=2. [10 marks]
(b) Express 8sinx−15cosx in terms of Rsin(x−α) and hence solve the equation 8sinx−15cosx=17 for 0°≤x≤360°. [10 marks]
ANSWERS — PAPER I
Q1. 4^(2x)=2^(4x); 8^(x+1)=2^(3x+3) → 4x=3x+3 → x=3
Q2. r=−8/16=−1/2; S=16/(1−(−1/2))=16/1.5 = 32/3 ≈ 10.67
Q3. b∘a=ba+b−a=ab+b−a
a∘b=ab+a−b ≠ ab+b−a (unless a=b)
Operation is NOT commutative
Q4. r=2, θ=−π/6; (√3−i)³⁰=2³⁰[cos(−5π)+isin(−5π)]
−5π mod 2π = π; cosπ=−1, sinπ=0
2³⁰=1,073,741,824
(√3−i)³⁰ = −1,073,741,824 + 0i
Q5. tan15°=tan(45°−30°)=(1−1/√3)/(1+1/√3)=(√3−1)/(√3+1)=(4−2√3)/2
tan15° = 2−√3 ≈ 0.2679
Q6(a). |P|=1(1−6)−(−1)(2−0)+0=−5+2=−3
|Q|=1(−1−0)−0(1−0)+2(−1−2)=−1+0−6=−7
|P||Q|=21
PQ=(2 −1 2; 7 4 1; 0 3 −1); |PQ|=2(−4−3)−(−1)(−7−0)+2(21−0)=−14−7+42=21
|PQ|=|P||Q|=21 ✓ verified
Q6(b). Base n=1: LHS=1, RHS=1×2×3/6=1 ✓
Assume true for k. For k+1: k(k+1)(2k+1)/6+(k+1)²=(k+1)[k(2k+1)+6(k+1)]/6=(k+1)(2k²+7k+6)/6=(k+1)(k+2)(2k+3)/6
Matches formula for n=k+1; true by induction ✓
Q6(c). Discriminant=0: 36−4m²=0 → m²=9 → m=±3
Q7(a). (2x+1)ln3=(x−1)ln7 → x(2ln3−ln7)=−ln7−ln3
x=−(ln7+ln3)/(2ln3−ln7)=−ln21/(ln9−ln7)
x ≈ −12.85
Q7(b). f(2)=8+4a+2b+c=0; f(−2)=−8+4a−2b+c=0
Adding: 8a+2c=0 → c=−4a
Subtracting: 16+4b=0 → b=−4
f(1)=1+a+b+c=−3 → 1+a−4−4a=−3 → −3a=0 → a=0
Then c=0
Q8(a). A=1/6, B=−5/6, C=8/3... solving:
3x−2=A(x+1)(x+2)+B(x−1)(x+2)+C(x−1)(x+1)
At x=1: 1=6A → A=1/6
At x=−1: −5=−2B → B=5/2
At x=−2: −8=3C → C=−8/3
Partial fractions: 1/[6(x−1)] + 5/[2(x+1)] − 8/[3(x+2)]
Binomial expansion up to x²:
f(x) ≈ −1/6 − (7/12)x + (23/24)x² + ...
Q8(b). α+β=3/a; αβ=b/a
Sum of new roots=2(α+β)+2=6/a+2=(6+2a)/a
Product=(2α+1)(2β+1)=4αβ+2(α+β)+1=4b/a+6/a+1=(4b+6+a)/a
New equation: ax² − (6+2a)x + (4b+6+a) = 0
Q8(c). Roots 2k,3k. Sum=5k=−b/a → k=−b/5a
Product=6k²=c/a → k²=c/6a
(−b/5a)²=c/6a → b²/25a²=c/6a → 6b²=25ac
6b² = 25ac ✓ (proved)
Q9(a). z³=−8=8(cos180°+isin180°)
z=2[cos(60°+120°k)+isin(60°+120°k)], k=0,1,2
k=0: 2(cos60°+isin60°)=1+i√3
k=1: 2(cos180°+isin180°)=−2
k=2: 2(cos300°+isin300°)=1−i√3
Roots: 1+i√3, −2, 1−i√3
Sum=(1+i√3)+(−2)+(1−i√3)=0 ✓ verified
Q9(b). tan3θ = (3tanθ−tan³θ)/(1−3tan²θ) (standard triple-angle identity, derivable by dividing sin3θ by cos3θ using the expansions):
sin3θ=3sinθ−4sin³θ; cos3θ=4cos³θ−3cosθ
tan3θ = (3tanθ − tan³θ)/(1 − 3tan²θ)
Q10(a)(i). z₁²=(5−i)²=24−10i; z₂²=(2+3i)²=−5+12i
2z₂²=−10+24i
z₁²−2z₂² = 34 − 34i
(ii) y²=(3+i)²=8+6i; 1/y²=(8−6i)/100
y²−1/y² = 7.92 + 6.06i
Q10(b). (tanA+2x)/(1−2x·tanA)=3 → tanA+2x=3−6x·tanA
tanA(1+6x)=3−2x
tanA = (3−2x)/(1+6x)
Q11(a). cot(A+B)=(cotAcotB−1)/(cotA+cotB)=(6−1)/5=1
A+B = cot⁻¹(1) = 45°
Q11(b). R=√(64+225)=17; α=arctan(15/8)≈61.93°
8sinx−15cosx=17sin(x−61.93°)
sin(x−61.93°)=1 → x−61.93°=90°
x ≈ 151.93° (only solution in range 0°≤x≤360°)