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**Question 1**
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Integrate 1+xx3dx

**Answer Details**

∫1+xx3dx $\int \frac{1+x}{{x}^{3}}\mathrm{d}x$

= ∫(1x3+xx3)dx $\int (\frac{1}{{x}^{3}}+\frac{x}{{x}^{3}})\mathrm{d}x$

= ∫(x−3+x−2)dx $\int ({x}^{-3}+{x}^{-2})\mathrm{d}x$

= −12x2−1x+k

**Question 2**
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In how many ways can 3 seats be occupied if 5 people are willing to sit?

**Answer Details**

5 people can take 3 places in;

5P3 ways, = 5!(5−3)!
$\frac{5!}{(5-3)!}$ = 5!2!
$\frac{5!}{2!}$

= 5×4×3×2!2!
$\frac{5\times 4\times 3\times 2!}{2!}$

= 5 x 4 x 3

= 60 ways

**Question 3**
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If S = √t2−4t+4 $\sqrt{{t}^{2}-4t+4}$, find t in terms of S

**Answer Details**

S = √t2−4t+4
$\sqrt{{t}^{2}-4t+4}$

S2 = t2 - 4t + 4

t2 - 4t + 4 - S2 = 0

Using t=−b±√b2−4ac2a
$t=\frac{-b\pm \sqrt{{b}^{2}-4ac}}{2a}$

Substituting, we have;

Using t=−(−4)±√(−4)2−4(1)(4−S2)2(1)
$t=\frac{-(-4)\pm \sqrt{(-4{)}^{2}-4(1)(4-{S}^{2})}}{2(1)}$

t=4±√16−4(4−S2)2
$t=\frac{4\pm \sqrt{16-4(4-{S}^{2})}}{2}$

t=4±√16−16+4S22
$t=\frac{4\pm \sqrt{16-16+4{S}^{2}}}{2}$

t=4±√4S22
$t=\frac{4\pm \sqrt{4{S}^{2}}}{2}$

t=2(2±S)2
$t=\frac{2(2\pm S)}{2}$

Hence t = 2 + S or t = 2 - S

**Question 4**
**Report**

If a binary operation * is defined by x * y = x + 2y, find 2 * (3 * 4)

**Answer Details**

To find 2 * (3 * 4), we need to first evaluate the expression inside the parentheses, which is 3 * 4. According to the definition of the binary operation * that was given, x * y = x + 2y, we can substitute x = 3 and y = 4 to get 3 * 4 = 3 + 2 * 4 = 3 + 8 = 11. Next, we can substitute x = 2 and y = 11 into the expression x * y = x + 2y to find 2 * 11, which is 2 + 2 * 11 = 2 + 22 = 24. So, 2 * (3 * 4) = 24.

**Question 5**
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If tanθ=34 $\mathrm{tan}\theta =\frac{3}{4}$, find the value of sinθ+cosθ $\mathrm{sin}\theta +\mathrm{cos}\theta $.

**Answer Details**

tanθ=oppadj=34 $\mathrm{tan}\theta =\frac{opp}{adj}=\frac{3}{4}$

hyp2=opp2+adj2 $hy{p}^{2}=op{p}^{2}+ad{j}^{2}$

hyp=32+42−−−−−−√ $hyp=\sqrt{{3}^{2}+{4}^{2}}$

= 5

sinθ=35;cosθ=45 $\mathrm{sin}\theta =\frac{3}{5};\mathrm{cos}\theta =\frac{4}{5}$

sinθ+cosθ=35+45 $\mathrm{sin}\theta +\mathrm{cos}\theta =\frac{3}{5}+\frac{4}{5}$

= 75=125

**Question 6**
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P varies jointly as m and u, and varies inversely as q. Given that p = 4, m = 3 and u = 2 and q = 1, find the value of p when m = 6, u = 4 and q =85

**Answer Details**

P ∝
$\propto $ mu, p ∝1q
$\propto \frac{1}{q}$

p = muk ................ (1)

p = 1qk
$\frac{1}{q}k$.... (2)

Combining (1) and (2), we get

P = muqk
$\frac{mu}{q}k$

4 = m×u1k
$\frac{m\times u}{1}k$

giving k = 46=23
$\frac{4}{6}=\frac{2}{3}$

So, P = muq×23=2mu3q
$\frac{mu}{q}\times \frac{2}{3}=\frac{2mu}{3q}$

Hence, P = 2×6×43×85
$\frac{2\times 6\times 4}{3\times \frac{8}{5}}$

P = 2×6×4×53×8
$\frac{2\times 6\times 4\times 5}{3\times 8}$

p = 10

**Question 7**
**Report**

If the variance of 3+x, 6, 4, x and 7-x is 4 and the mean is 5, find the standard deviation

**Answer Details**

To find the standard deviation of the given set of numbers, we need to find the variance first, using the formula: Variance = (sum of squares of deviations from the mean) / (number of observations) Let's start by finding the mean of the given set of numbers: Mean = (3 + x + 6 + 4 + 7 - x) / 5 = 20 / 5 = 4 Next, we can substitute this mean into the formula for variance, which gives us: 4 = [(3 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (x - 4)^2 + (7 - x - 4)^2] / 5 Simplifying this equation, we get: 20 = (3 - 4)^2 + (6 - 4)^2 + (4 - 4)^2 + (x - 4)^2 + (7 - x - 4)^2 20 = 1 + 4 + 0 + (x - 4)^2 + (3 - x)^2 Simplifying further, we get: 15 = 2(x^2 - 8x + 13) Expanding and simplifying this equation, we get: 2x^2 - 16x + 26 = 15 2x^2 - 16x + 11 = 0 We can solve this quadratic equation using the quadratic formula: x = [16 ± sqrt(16^2 - 4(2)(11))] / (2(2)) x = [16 ± sqrt(176)] / 4 x = [16 ± 4sqrt(11)] / 4 x = 4 ± sqrt(11) Now that we have found the possible values of x, we can calculate the standard deviation using the formula: Standard deviation = sqrt(variance) We already know the variance is 4, so: Standard deviation = sqrt(4) = 2 Therefore, the answer is (B) 2.

**Question 8**
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In the diagram, find the size of the angle marked ao

**Answer Details**

2 x s = 280o(Angle at centre = 2 x < at circum)

S = 280o2 $\frac{{280}^{o}}{2}$

= 140

< O = 360 - 280 = 80o

60 + 80 + 140 + a = 360o

(< in a quad); 280 = a = 360

a = 360 - 280

a = 80o

**Question 9**
**Report**

If the midpoint of the line PQ is (2,3) and the point P is (-2, 1), find the coordinate of the point Q.

**Answer Details**

The midpoint formula states that the midpoint of a line segment with endpoints (x1,y1) and (x2,y2) is: ((x1+x2)/2, (y1+y2)/2) In this problem, we are given the midpoint of the line segment PQ, which is (2,3), and one endpoint, P, which is (-2,1). Let Q have coordinates (x,y), then we can use the midpoint formula to find the coordinates of Q: ((x + (-2))/2, (y + 1)/2) = (2,3) Simplifying this expression, we get: (x-2, y+1) = (4,6) Now we can solve for x and y by equating the x-coordinates and y-coordinates separately: x - 2 = 4 => x = 6 y + 1 = 6 => y = 5 Therefore, the coordinates of point Q are (6,5). So, the correct answer is (6,5).

**Question 10**
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The remainder when 6p3 - p2 - 47p + 30 is divided by p - 3 is

**Answer Details**

Let f(p) = 6p3 - p2 - 47p + 30

Then by the remainder theorem,

(p - 3): f(3) = remainder R,

i.e. f(3) = 6(3)3 - (3)2 - 47(3) + 30 = R

162 - 9 - 141 + 30 = R

192 - 150 = R

R = 42

**Question 11**
**Report**

The mean of seven numbers is 10. If six of the numbers are 2, 4, 8, 14, 16 and 18, find the mode.

**Answer Details**

To find the mode of the seven numbers, we need to determine the number that appears most frequently among the seven numbers. We are given that the mean (average) of the seven numbers is 10. To find the sum of the seven numbers, we can multiply the mean by 7: mean = (sum of seven numbers) / 7 Rearranging this formula, we get: (sum of seven numbers) = mean x 7 Substituting the given mean of 10, we get: (sum of seven numbers) = 10 x 7 = 70 We are also given that six of the seven numbers are 2, 4, 8, 14, 16, and 18. To find the seventh number, we can subtract the sum of these six numbers from the sum of the seven numbers: (seventh number) = (sum of seven numbers) - (sum of six numbers) Substituting the values we know, we get: (seventh number) = 70 - (2 + 4 + 8 + 14 + 16 + 18) (seventh number) = 8 So the seven numbers are: 2, 4, 8, 14, 16, 18, and 8. To find the mode, we need to determine which number appears most frequently among these seven numbers. We can see that the number 8 appears twice, while all the other numbers appear only once. Therefore, the mode of these seven numbers is 8.

**Question 12**
**Report**

If log104 = 0.6021, evaluate log1041/3

**Answer Details**

log1041/3 = 1/3 log104

= 1/3 x 0.6021

= 0.2007

**Question 13**
**Report**

In the diagram given, find the value of x.

**Answer Details**

In the diagram above, < CDE = < CED = 15o

(base < s of isos. △)

< ECD = 180o - (15 + 15)o

= 180o - 30o = 150o

But x + 110o = 150o

(Sum of opp. interior < s of a△ = opp. exterior < )

x = 150o - 110o = 40o

**Question 14**
**Report**

If P = ∣∣∣5321∣∣∣
$\left|\begin{array}{cc}5& 3\\ 2& 1\end{array}\right|$ and Q = ∣∣∣4235∣∣∣
$\left|\begin{array}{cc}4& 2\\ 3& 5\end{array}\right|$, find 2P + Q

**Answer Details**

Given that P = |5321| and Q = |4235|, we need to find 2P + Q. First, we need to evaluate P and Q. The vertical bars indicate the absolute value of the number, which means the distance of the number from zero on the number line. Therefore, we have: P = |5321| = 5321 Q = |4235| = 4235 Now, we can substitute these values in 2P + Q and simplify: 2P + Q = 2(5321) + 4235 = 10642 + 4235 = 14877 Therefore, the value of 2P + Q is 14877. Hence, the answer is option B, i.e., |14877|.

**Question 15**
**Report**

Find the inverse ∣∣∣5364∣∣∣

**Answer Details**

Let A = (5364)
$\left(\begin{array}{cc}5& 3\\ 6& 4\end{array}\right)$

Then |A| = (5364)
$\left(\begin{array}{cc}5& 3\\ 6& 4\end{array}\right)$ = 20 - 18 = 2

Hence A-1 = 1|A|(4−3−65)
$\frac{1}{|A|}\left(\begin{array}{cc}4& -3\\ -6& 5\end{array}\right)$

= 12(4−3−65)
$\frac{1}{2}\left(\begin{array}{cc}4& -3\\ -6& 5\end{array}\right)$

= (4×1/2−3×1/2−6×1/25×1/2)
$\left(\begin{array}{cc}4\times 1/2& -3\times 1/2\\ -6\times 1/2& 5\times 1/2\end{array}\right)$

= ∣∣ ∣∣2−32−352∣∣ ∣∣

**Question 16**
**Report**

Calculate the time taken for ₦3000 to earn ₦600 if invested at 8% simple interest

**Answer Details**

To calculate the time taken for ₦3000 to earn ₦600 if invested at 8% simple interest, we can use the formula: time = (interest / principal) / rate where: interest = ₦600 principal = ₦3000 rate = 8% So, substituting the values in the formula, we get: time = (600 / 3000) / 0.08 time = (0.2) / 0.08 time = 2.5 years So, it takes approximately 2.5 years for ₦3000 to earn ₦600 if invested at 8% simple interest.

**Question 17**
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If y = (2x + 2)3, find δxδy

**Answer Details**

y = (2x +2)3

Then δyδx
$\frac{\delta y}{\delta x}$ = 3(2x +2)22

=6(2x +2)2

**Question 18**
**Report**

Find the equation of the perpendicular bisector of the line joining P(2, -3) to Q(-5, 1)

**Answer Details**

Given P(2, -3) and Q(-5, 1)

Midpoint = (2+(−5)2,−3+12) $(\frac{2+(-5)}{2},\frac{-3+1}{2})$

= (−32,−1) $(\frac{-3}{2},-1)$

Slope of the line PQ = 1−(−3)−5−2 $\frac{1-(-3)}{-5-2}$

= −47 $-\frac{4}{7}$

The slope of the perpendicular line to PQ = −1−47 $\frac{-1}{-\frac{4}{7}}$

= 74 $\frac{7}{4}$

The equation of the perpendicular line: y=74x+b $y=\frac{7}{4}x+b$

Using a point on the line (in this case, the midpoint) to find the value of b (the intercept).

−1=(74)(−32)+b $-1=(\frac{7}{4})(\frac{-3}{2})+b$

−1+218=138=b $-1+\frac{21}{8}=\frac{13}{8}=b$

∴ $\therefore $ The equation of the perpendicular bisector of the line PQ is y=74x+138 $y=\frac{7}{4}x+\frac{13}{8}$

≡8y=14x+13⟹8y−14x−13=0

**Question 19**
**Report**

Find the length of a chord which subtends an angle of 90° at the centre of a circle whose radius is 8 cm.

**Answer Details**

Length of chord = 2rsin(θ2) $2r\mathrm{sin}(\frac{\theta}{2})$

= 2×8×sin(902) $2\times 8\times \mathrm{sin}(\frac{90}{2})$

= 16×2√2 $16\times \frac{\sqrt{2}}{2}$

= 82–√cm

**Question 20**
**Report**

Simplify 3−5n91−n×27n+1

**Answer Details**

3−5n91−n×27n+1
$\frac{{3}^{-5n}}{{9}^{1-n}}\times {27}^{n+1}$

3−5n32(1−n)×33(n+1)
$\frac{{3}^{-5n}}{{3}^{2(1-n)}}\times {3}^{3(n+1)}$

3−5n÷32(1−n)×33(n+1)
${3}^{-}$