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## Sample of MOUAU Post UTME Past Questions and Answers

- 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.
- Noting that, sin2θ + cos2θ = 1, simplify 1−𝑐𝑜𝑠𝜃𝑠𝑖𝑛2𝜃 (a) 11+𝑐𝑜𝑠𝜃 (b) 11−𝑐𝑜𝑠𝜃 (c) 11+𝑠𝑖𝑛𝜃 (d) 11−𝑠𝑖𝑛𝜃
- A circle has an eccentricity (a) < 1 (b) 1 (c) > 1 (d) 0.
- 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).
- Simplify 3𝑛+3− 3𝑛+23𝑛+1 −3𝑛 (a) -9 (b) 9 (c) 10 (d) -10.
- Noting that cos∝ =(90−∝),find y in terms of x in the equation cos1+12𝑥 = sin 32𝑦 (a) y = 178+𝑥3 (b) y = 𝑥−1783 (c) 178−𝑥3 (d) −(178+𝑥)3.
- 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.
- In how many ways can the letters of the word NWAFOR be permuted? (a) 7200 (b) 72 (c) 720 (d) 72000.
- If ∝,𝛽 are the roots of equation 18 + 15x – 3×2 = 0, find ∝𝛽 – ∝− 𝛽 (a) 11 (b) -11 (c) 10 (d) -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)
- 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
- What is the value of x for which x2 – 5x + 6 is minimum? (a) 52(b) −52 (c) 3 (d) -3.
- 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.
- 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}.
- Find the angle in the line 1√3𝑦−𝑥=0 makes with positive y-axis (a) 300 (b) 600 (c) 00 (d) 450.
- Find the value of p which satisfies the equation √𝑃− 6𝑝 = 1 (a) 4 (b) -4 (c) 9 (d) -9.
- 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.
- 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.
- 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
- 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.
- Find the value of k in the equation 55√2−√8=𝑘√2 (a) 4/3 (b) ¾ (c) -3/4 (d) -4/3.
- Evaluate ∫3𝑥𝑙𝑜𝑔3𝑑𝑥10 (a) 3 (b) 4 (c) 1 (d) 2.

## ANSWERS TO THESE POST-UTME SCREENING EXERCISE QUESTIONS

- C
- 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 - D
- A
- 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 - cos1+12𝑥 = sin 32𝑦

But cos = sin (90 – )

= cos1+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 - 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

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

- 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 - 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 - 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 - 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 - 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 - ∫(5𝑥4 +𝑒−𝑥)𝑑𝑥

= 5𝑥4+14+1+(−𝑒−𝑥)+𝑐

5𝑥55−𝑒−𝑥+𝑐

= 𝑥5−𝑒𝑥+𝑐

∫(5𝑥4+𝑒𝑥+𝑥5+𝑐𝐀𝐧𝐬: 𝑨 - 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 - 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 - √𝑃− 6√𝑝=1

Multiply through by √𝑃

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

- 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 - 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 - 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 - 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 - 55√2−√8=𝑘√2

= 55√2−2√2=𝑘√2

Multiply through by 2√2

5 – 2√22√2=2√2(𝑘√2)

5 – 4 (2) = 2k(2)

5 – 8 = 4k

-3 = 4k

k = -3/4Ans: C - ∫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

### Recommended:

- MOUAU Admission List 2022/2023 for UTME and Direct Entry (DE)
- MOUAU Post UTME Form 2022/2023 – www.mouau.edu.ng
- How to Succeed in the University or Polytechnic Environment
- How to Pass Post UTME | What to Read for Post UTME 2022/2023

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