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## Saample of MOUAU Post UTME Past Questions and Answers

1. The probability of an event A given by P(A) is a number between (a) -1 and 1 (b) 0 and ½ (c) 0 and 1 (d) -1 and 0.
2. Noting that, sin2θ + cos2θ = 1, simplify 1−𝑐𝑜𝑠𝜃𝑠𝑖𝑛2𝜃 (a) 11+𝑐𝑜𝑠𝜃 (b) 11−𝑐𝑜𝑠𝜃 (c) 11+𝑠𝑖𝑛𝜃 (d) 11−𝑠𝑖𝑛𝜃
3. A circle has an eccentricity (a) < 1 (b) 1 (c) > 1 (d) 0.
4. It two elements A and B are independent then P(A and B) is (a) P(A ∩ B) (b) (A B) (c) P(A) (d) P(B).
5. Simplify 3𝑛+3− 3𝑛+23𝑛+1 −3𝑛 (a) -9 (b) 9 (c) 10 (d) -10.
6. Noting that cos∝ =(90−∝),find y in terms of x in the equation cos􁉀1+12𝑥􁉁 = sin 􁉀32𝑦􁉁 (a) y = 178+𝑥3 (b) y = 𝑥−1783 (c) 178−𝑥3 (d) −(178+𝑥)3.
7. For what values of x is x–1< – 1? (a) 0 < x < 1 (b) x < -1, x > 0 (c) x > 1, x < 0 (d) -1 < x < 0.
8. In how many ways can the letters of the word NWAFOR be permuted? (a) 7200 (b) 72 (c) 720 (d) 72000.
9. If ∝,𝛽 are the roots of equation 18 + 15x – 3×2 = 0, find ∝𝛽 – ∝− 𝛽 (a) 11 (b) -11 (c) 10 (d) -10.
10. Resolve 1(1−𝑥2) into partial fractions (a) 12(1+𝑥)− 12(1−𝑥) (b) 12(1+𝑥)+ 12(𝑥−1) (c) 12(𝑥+1)+ 12(1−𝑥) (d) 12(1−𝑥2)
11. Given that the sum of infinity 𝑆∞ = a + ar + ar2 + ….. = 𝑎1−𝑟 , to what sum does the infinite series 1 – 23+ 49− 827+⋯ coverage (a) – 35 (b) 53 (c) − 53 (d) 35
12. What is the value of x for which x2 – 5x + 6 is minimum? (a) 52(b) −52 (c) 3 (d) -3.
13. Integrate 5×4 + e-x with respect to x (a) −𝑒−𝑥+5𝑥+𝑘 (b) 𝑒−𝑥+𝑥5+𝐾 (c) −𝑒−𝑥−𝑥−5+𝐾 (d) −𝑒−𝑥+𝑥4+𝐾.
14. If X = {2, 3, 6, 7, 8} and Y = {6, 7, 10, 3, 17}, find Y – {X Y). (a) { } (b) {10, 17} (c) {2, 3, 6, 7, 8, 10, 17} (d) {3, 6, 7}.
15. Find the angle in the line 1√3𝑦−𝑥=0 makes with positive y-axis (a) 300 (b) 600 (c) 00 (d) 450.
16. Find the value of p which satisfies the equation √𝑃− 6𝑝 = 1 (a) 4 (b) -4 (c) 9 (d) -9.
17. Find the area of circle 4×2 + 4y2 – 400 = 0 (a) 10𝜋 𝑠𝑞.𝑢𝑛𝑖𝑡𝑠 (b) 40𝜋 𝑠𝑞.𝑢𝑛𝑖𝑡𝑠 (c) 400𝜋 𝑠𝑞.𝑢𝑛𝑖𝑡𝑠 (d) 100𝜋 𝑠𝑞.𝑢𝑛𝑖𝑡𝑠.
18. Let the mean of x, y-1, z5 be 6 find the mean of 10, y-1, 12, x z5. (a) 7 (b) 8 (c) 9 (d) 10.
19. What is the addition of y and x- intercepts of the line 23+ 32𝑦+9=0? (a) -19.5 (b) 19.5 (c) 20.5 (d) -20.5
20. Given that h(x) = 3 + 2x and f(x) = 1 – x, find h(– f (x)). (a) 1 – 2x (b) 1 + 2x (c) 2x – 1 (d) -1 – 2x.
21. Find the value of k in the equation 55√2−√8=𝑘√2 (a) 4/3 (b) ¾ (c) -3/4 (d) -4/3.
22. Evaluate ∫3𝑥𝑙𝑜𝑔3𝑑𝑥10 (a) 3 (b) 4 (c) 1 (d) 2.

## ANSWERS TO THESE POST-UTME SCREENING EXERCISE

1. C
2. 1−cos𝜃𝑠𝑖𝑛2𝜃
Recall that: sin2θ + cos2 θ = 1
sin2θ = 1 – cos2 θ
1−cos𝜃1− 𝑐𝑜𝑠2𝜃
But 1 – cos2 θ = (1 – cos θ)(1 + cos θ) 1 1−cos𝜃1− 𝑐𝑜𝑠2𝜃
= 1−cos𝜃(1−cos𝜃)(1+𝑐𝑜𝑠 𝜃)
= 11+𝑐𝑜𝑠 𝜃Ans:A
3. D
4. A
5. 3𝑛+3− 3𝑛+23𝑛+1 −3𝑛 = 3𝑛 𝑋 33−3𝑛𝑋 323𝑛 𝑋 31− 3𝑛 = 3𝑛+3− 3𝑛+23𝑛 (3−1)
= 3𝑛𝑋 32𝑋 23𝑛 𝑋 2 = 32 = 9 Ans:B
6. cos􁉀1+12𝑥􁉁 = sin 􁉀32𝑦􁉁
But cos = sin (90 – )
= cos􁉀1+12𝑥􁉁 = sin [90−􁉀1+ 12𝑥􁉁]
Cos 􁉀1+12𝑥􁉁 = sin 􁉀32𝑦􁉁
= sin 􁉂90−􁉀1+ 12𝑥􁉁􁉃=sin(32𝑦)
= 90−􁉀1+ 12𝑥􁉁=32𝑦
= 90−􁉀2+𝑥2𝑥􁉁=32𝑦
Multiply through by 2
180 – (2 + x) = 3y
180 – 2 – x = 3y
178 – x = 3y
y = 1/3 (178 – x)
y = 178−𝑥3 Ans: C
7. 𝑥−1=<−1
1𝑥<−1
Multiply through by 𝑥2
1𝑥𝑋 𝑧𝑥2<−1 𝑋 𝑥2
𝑥<−𝑥2
𝑥+𝑥2<0
𝑥 (1+𝑥)<0
𝑥=0 𝑜𝑟 1+𝑥= 0 then 𝑥=0 𝑜𝑟−1
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Factor
x < – 1 – 1 < x < 0 x > 0
x
1 + x
-ve
-ve
-ve
+ve
+ve
+ve
x(1 + x)
+ve
-ve
+ve
Since 𝑥(1 + 𝑥) < 0 (𝑖.𝑒.𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒) 𝑡ℎ𝑒 𝑠𝑜𝑙𝑢𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑒 𝑖𝑛𝑒𝑞𝑢𝑎𝑙𝑖𝑡𝑦 𝑖𝑠 – 1 < 𝑥 < 0 Ans: D

1. The word NWAFOR has six (6) distinct letters. n = 6
The number of ways of arranging n distinct object is n!
No of ways = n! = 6! = 720 Ans: C
2. 18 + 15𝑥 – 3𝑥2 = 0
a = – 3, b = 15, c = 18
𝛼+ 𝛽= −𝑏𝑎= −15−3=5
𝛼𝛽= 𝑐𝑎= 18−3= −6
𝛼𝛽− 𝛼−𝛽=𝛼𝛽−(𝛼+𝛽)
= –6 – (5) = – 11 Ans:B
3. 11−𝑥2
𝐵𝑢𝑡 1 – 𝑥2 = (1 – 𝑥)(1 + 𝑥)
11−𝑥2= 1(1−𝑥)(1+𝑥) .
A linear factor of the form ax + b always gives a partial fraction of 𝐴𝑎𝑥_𝑏
1(1−𝑥)(1+𝑥)= 𝐴1−𝑥+ 𝐵1+𝑥
1 (1−𝑥)(1+𝑥)
= 𝐴(1+𝑥)+𝐵(1−𝑥)(1−𝑥)(1+𝑥)
1 = A(1 + x) + B(1 – x)
Let x = 1
1 = A(1 + 1) + B(1 – 1)
1 = 2A + B(0)
1 = 2A
A = ½
𝐿𝑒𝑡 𝑥 = –1
1 = A(-1 + 1) + B[1 – (-1)]
1 = A(0) + B(1 + 1)
B = ½
1(1−𝑥)(1+𝑥)= 𝐴1−𝑥+ 𝐵1+𝑥
= 121−𝑥+ 121+𝑥
12(1−𝑥)+12(1+𝑥)
= 1(1−𝑥)(1+𝑥)= 11−𝑥2= 12(1−𝑥)+12(1+𝑥) Ans: C
4. 1−23+49−827+⋯
T1 = 1, T2 = −23,𝑇3= 49
For a given series to be an A.P
𝑇2−𝑇1=𝑇3−𝑇2
For a given series to be a G.P
𝑇2𝑇1=𝑇32
The series is a G.P
r = 𝑇2𝑇1= −231
r = −23
𝑆∞=𝑎1−𝑟
a = T1 = 1
𝑆∞=11−􀵫−23􀵯=111+23
= 13+23=153= 35 Ans: D
5. 𝐿𝑒𝑡 𝑦 = 𝑥2 – 5𝑥 + 6
Minimum and maximum are turning point. At turning 𝑑𝑦𝑑𝑥=0
𝑑𝑦𝑑𝑥 = 2𝑥−5=0
2𝑥 – 5 = 0
x = 52 Ans: A
6. ∫(5𝑥4 +𝑒−𝑥)𝑑𝑥
= 5𝑥4+14+1+(−𝑒−𝑥)+𝑐
5𝑥55−𝑒−𝑥+𝑐
= 𝑥5−𝑒𝑥+𝑐
∫(5𝑥4+𝑒𝑥+𝑥5+𝑐𝐀𝐧𝐬: 𝑨
7. X = {2, 3, 6, 7, 8} Y = {6, 7, 10, 3, 17}
The intersect of two sets X and Y is a set that contain elements that are common to both sets. 𝑋∩𝑌={3,6,7}
The difference of two sets A and B (i.e. A – B) is a set, which contain only elements that are formed in set A but not in set B.
Y – (𝑋∩𝑌) = {6, 7, 10, 3, 17} – {3, 6, 7} = {10, 17}
Y – (𝑋∩𝑌) = {10, 17} Ans: B
8. 1√3𝑦−𝑥=0
𝑦√3−𝑥=0
Multiply through with √3
y – x√3=0
y = x√3
Divide through by 𝑥
𝑦𝑥= √31
but tan θ = 𝑦𝑥= √31
θ = tan-1 (√3) = 600
The angle 600 is the angle the line makes with the positive x-axis
0yxB
Θ + 𝛽 = 90
60 + 𝛽=90
𝛽=90−60
𝛽 = 300
Note that the angle the line 1√3𝑦−𝑥=0 makes with the positive y-axis is given by tan 𝛽=𝑥𝑦Ans: A
9. √𝑃− 6√𝑝=1
Multiply through by √𝑃
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P – 6 = √𝑃
Square both side (P – 6)2 = (√𝑃)2
P2 – 12P + 36 = P
P2 – 12P – P + 36 = 0
P = 9 or 4
Check to see if 9 or 4 satisfied the equation
√𝑃−6√𝑃=1
When P = 9
√9−6√𝑃=1
3−63=1
3 – 2 = 1
1 = 1
Hence the value p = 9 satisfied the equation when p = 4
√4−6√4=1
2−62=1
2 – 3 = 1
-1 1
Hence the value p = 4 does not satisfy the equation ∴𝑝=9Ans: C

1. 4×2 + 4y2 – 400 = 0
Divide through by 4
x2 + y2 – 100 = 0
x2 + y2 = 100
x2 + y2 + 102……………(i)
The general equation of a circle is given by x2 + y2 = r2 ………………………….(ii)
From equation i and ii
𝑟2= 102
r = 10
Area of a circle (A) = 𝜋r2
A = 𝜋(10)2
A = 100𝜋 Ans:D
2. 𝑥=Σ𝑥𝑛
For the numbers: x, y-1& z5
𝑥= 𝑥+𝑦−1+𝑧53=6
x + y-1 + z5 = 3 x 6 = 18
x + y-1 + z5 = 18……………………(i)
𝑥=Σ𝑥𝑛
For the numbers: 10, y-1, 12, x, z5
𝑥=10+𝑦−1+12+𝑥+𝑦55
𝑥=10+12+𝑥+𝑦−1+𝑦55
But x + y-1 + z5 = 18
𝑥=10+12+185
𝑥=405=8 Ans: B
3. 23+ 32𝑦+9=0?
2𝑥3+ 3𝑦2=−9
4𝑥+9𝑦6=−9
4x + 9x = -9 x 6
Divide through by 36
4𝑥36+ 9𝑦36=−9𝑋636
𝑥9+𝑦4=−32
Multiply through by 23
23𝑥𝑥9+23𝑥𝑦4=−32𝑥23
2𝑥27+𝑦6=−1
Multiply through by -1
−2𝑥27+𝑦6=−1
The above equation can be written as shown below
−𝑥276−𝑦6=1………….(𝑖)
The double intercept form of the equation of a straight line is
𝑥𝑎+𝑦𝑏=1……………(𝑖𝑖)
a = −272,𝑏=−6
a + b = −272−61
= −27−122
= −392 = -19.5 Ans: A
4. h(x) = 3 + 2x
f(x) = 1 – x = – (x – 1)
-f(x) = -[(x – 1)]
= x – 1
h[- f(x)] = h(x – 1)
= 3 + 2(x – 1)
= 3 + 2x – 2
h[-f(x)] = 2x + 1 Ans: B
5. 55√2−√8=𝑘√2
= 55√2−2√2=𝑘√2
Multiply through by 2√2
5 – 2√2􀵫2√2􀵯=2√2(𝑘√2)
5 – 4 (2) = 2k(2)
5 – 8 = 4k
-3 = 4k
k = -3/4Ans: C
6. ∫3𝑥𝑙𝑜𝑔3𝑑𝑥10
∫3𝑥𝑙𝑜𝑔3𝑑𝑥10
∫3𝑥𝑙𝑜𝑔 3𝑒 𝑑𝑥10
= ∫3𝑥 𝐼𝑛 3 𝑑𝑥10
= ∫𝑎𝑥𝐼𝑛 𝑎 𝑑𝑥=0𝑎𝑥
∫3𝑥𝐼𝑛 3 𝑑𝑥=[3𝑥]0110
= 31−30
= 3 – 1 = 2 Ans: D

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