JAMB UTME - Mathematics - 2015

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$\int \left[\right(x^2+4x^2+1)÷x^2]dx=\int \left(x^3/x^2\right)dx+\int \left(1/x^2\right)dx$
⇒$\int xdx+\int 4^{-1}or+\int \left(1/x^2\right)dx$
= $x^2/2-4x-1/x^2+C$

Akọwa Nkọwa

dy/dx $(4x^3+3x^2+2x+1)$
= $12x^2+6x+2$
⇒ $12x^2+6+2$
Divide both sides by 2
$\frac{12x^2}{2}-\frac{6x}{2}-\frac{2}{2}$
= $6x^2+3x+1$
dy/dx
= $6x^2+3x+1$

Akọwa Nkọwa

$[1÷(x^2+3x+2\left)\right]+[1÷(x^2+5x+6\left)\right]$
= $1÷(x^2+3x+2)+[1÷(x^2+5x+6\left)\right]$
= $[1÷((x^2+x)+(2x+2)\left)\right]+[1÷((x^2+3x)+(2x+6)\left)\right]$
= [1 ÷ (x(x + 2) + 2(x +1))] + [1 ÷ (x(x + 3) +2(x + 3) )]
= [1 ÷ (x + 1)(x + 2)] + [1 ÷ ((x + 3) + (x + 2))]
=((x + 3) + (x + 1)) ÷ (x + 1)(x + 2)(x + 3)
Using the L.C.M
=((x + x + 3 + 1)) ÷ (x + 1)(x + 2)(x + 3)
=(2x+4)/(x+1)(x+2)(x+3) =2(x+2)/(x+1)(x+2)(x+3)
= $\frac{2}{(x+1)(x+3)}$

Akọwa Nkọwa

The power set of a set is the set of all possible subsets of that set, including the empty set and the set itself. The formula for finding the power set of a set is 2^n, where n is the number of elements in the original set. In this case, Z has four elements, so the number of subsets is 2^4 = 16. Therefore, the power set of Z has 16 elements, which is the answer.

Akọwa Nkọwa

[1 ÷ (x+1)] + [1 ÷ (x − 1)]
= ((x − 1) + [(x + 1)) ÷ (x+1)(x − 1)]
Using the L.C.M.
= (x − 1 + x + 1) ÷ (x + 1)(x − 1)
= (x + 2 − 1 + 1) ÷ (x + 1)(x − 1)
= 2x ÷ (x + 1)(x − 1) =2x ÷ (x + 1)(x − 1)

Akọwa Nkọwa

$[1÷{64}^{(x+2)}]=[4^{(x-3)}÷{16}^x]$
${64}^{-(x+2)}=\left[4^{(x-3)}\right]÷\left[{16}^x\right]$
Break down 4, 16 and 64 into smaller index numbers:
$2^{-6(x+2)}=2^{2(x-3)}÷2^{4\left(x\right)}$
$2^{-6x-12}=2^{2x-4x-6}$
$2^{-6x-12}=2^{-2x-6}$
− 6x − 12 = − 2x − 6
Collect the like terms:
−6x + 2x = −6 + 12
−4x =6
x = $\frac{6}{4}$
x = $\frac{-3}{2}$

Akọwa Nkọwa

1st G.P. = a =2
2nd G.P. = ar − 1 = 54
2(r) = 54
r = 54/2 = 27
r = 27
3rd term = ar2 = (2) (27)2
2 × 27 × 27
= 1458

Akọwa Nkọwa

6½ − 2¾ + 1½
73/12 − 11/4 + 3/2
[(73 − 33 + 18) ÷ (12)]
58/12
29/6
$4\frac{5}{6}$

Akọwa Nkọwa

(159.75)10 = (y)6
$\begin{array}{cc}6& 159\\ 6& 26\text{R 3}\\ 6& 4\text{R 2}\\ & 0\text{R 4}\end{array}\uparrow$
15910 = 4236
Then Convert 0.75 to base 6
0.75 x 6 =4.5 i.e 4.5 - 3.00 = 1.5
0.5 x 6 = 3.00 423 + 1.5 =424.5
159.75 = (424.5)6
(159)10 = (424.5)6

Akọwa Nkọwa

P(A outcome) = $\frac{1}{6}$, P(B outcome) = ¼
P(A ∪ B) = 1/12, P(A ∩ B)
P(A ∩ B) = P(A) + P(B) − P(A ∪ B)
1/6 + 1/4 − 1/12
= (2 + 3 − 1) ÷ 12
= 4/12
= 1/3
P(A ∩ B) = 1/3

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Akọwa Nkọwa

(0.09)2 = 0.09 × 0.09
= 0.0081
= 0.008100 to 4 significant figures
Please, start counting first from the non-zero digits i.e. 8

Akọwa Nkọwa

The loss percentage can be calculated as follows: Loss percentage = (Loss / Cost price) x 100 Here, the cost price of the car is ₦800, and the selling price is ₦520. Since the selling price is less than the cost price, the man has incurred a loss. Loss = Cost price - Selling price Loss = ₦800 - ₦520 Loss = ₦280 Substituting the values in the formula: Loss percentage = (280 / 800) x 100 Loss percentage = 0.35 x 100 Loss percentage = 35% Therefore, the man incurred a loss of 35%.

Akọwa Nkọwa

To find A∩B (the intersection of A and B), we need to determine the elements that are present in both A and B. Looking at the sets A and B, we can see that 3 and 4 are present in both sets. Additionally, ⅓ is also present in both sets. Therefore, the intersection of A and B is {3, 4, ⅓}. Hence, option C: {3, 4, ⅓} is the correct answer.

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Akọwa Nkọwa

Let u= x2
du = 2xdx
Let v = sinx
∫dv = ∫ sin x i.e. v = cos x
∫ udv = uv − ∫ vdu
∫ x2 Sin xdx = − x2 Cos x − ∫ (− Cos x)2 × dx
x2 Cos x + 2 ∫ x cos x1 dx
= − x2 Cos x + 2 ∫ x Cos x2 dx
2 ∫ u Cos xdx
Let u = x
du = dx
dv = Cos x
∫dv = ∫ Cos x
∫ x2 Sin x1 dx = − x2 Cos x + 2Sin x − 2Cos x + C
∫ x2 Sin xdx
= − x2 Cos x + 2xSin + 2Cos x + C

Akọwa Nkọwa

A regular polygon is a polygon with equal sides and equal angles. The formula to find the interior angles of a regular polygon is: Interior Angle = (n - 2) x 180 / n where n is the number of sides of the polygon. We are given that the interior angles of the regular polygon are 120°. Substituting this value into the formula, we get: 120 = (n - 2) x 180 / n Simplifying this equation, we get: 120n = 180n - 360 360 = 60n n = 6 Therefore, the regular polygon has 6 sides.

Akọwa Nkọwa

Let assume the cost price is 100%
Marked up price + cost price = 20 + 100 = 120%
Discount at 20% = 20/100 × 120 % of cost price
Selling price = cost price − gain
= (120 − 24)% of cost price
= 96% of cost price
Loss = (100− 96)% of cost price
= 4% of cost price
∴ He will make a 4% loss.

Akọwa Nkọwa

The formula for the volume of a cylinder is V = πr^2h, where "r" is the radius of the base, "h" is the height, and π (pi) is a mathematical constant approximately equal to 3.14. In this case, the height of the cylinder is 4cm, and the radius is 5cm. So we can substitute these values into the formula to find the volume: V = πr^2h V = π(5cm)^2(4cm) V = π(25cm^2)(4cm) V = π(100cm^3) V = 100π cm^3 Therefore, the volume of the cylinder is 100π cubic centimeters. We can leave the answer in terms of π or use the approximate value of 3.14 to get an estimate: V ≈ 100(3.14) cm^3 V ≈ 314 cm^3 So the answer is option C: 314cm^2.

Akọwa Nkọwa

To solve the equation x^2 - 2x - 3 = 0, we can use the quadratic formula which is: x = (-b ± sqrt(b^2 - 4ac)) / 2a where a, b, and c are the coefficients of the quadratic equation ax^2 + bx + c = 0. In this case, a = 1, b = -2, and c = -3. Substituting these values into the formula, we get: x = (-(-2) ± sqrt((-2)^2 - 4(1)(-3))) / 2(1) Simplifying this expression, we get: x = (2 ± sqrt(16)) / 2 x = (2 ± 4) / 2 So, x can be either (2 + 4)/2 = 3 or (2 - 4)/2 = -1. Therefore, the answer is: x = 3 or -1.

Akọwa Nkọwa

To simplify the expression 2.04 × 3.7, we can use the distributive property of multiplication over addition. That is: 2.04 × 3.7 = (2 + 0.04) × 3.7 We can now expand this expression as: (2 + 0.04) × 3.7 = 2 × 3.7 + 0.04 × 3.7 Simplifying further: 2 × 3.7 = 7.4 0.04 × 3.7 = 0.148 So, substituting these values back into the expression: 2.04 × 3.7 = 7.4 + 0.148 = 7.548 Rounding this answer to two decimal places, we get: 2.04 × 3.7 = 7.55 Therefore, the answer is option D - 7.55.

Akọwa Nkọwa

Area of a circle (A) = πr2
Radius = 4cm
π = 3.142
A = 3.142 × (4)2
= 3.142 × 16
= 50.272
A = 50.3cm2 (1dp)

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Akọwa Nkọwa

To solve the equation 5(x - 2) = (1 ÷ 125)(x + 3), we need to isolate the variable "x" on one side of the equation. First, let's simplify the right side of the equation: (1 ÷ 125)(x + 3) = (1 ÷ 125)x + (1 ÷ 125)3 = x/125 + 3/125 = (x + 3)/125 Now, let's simplify the left side of the equation: 5(x - 2) = 5x - 10 Now, we have: 5x - 10 = (x + 3)/125 To isolate "x" on one side, we'll multiply both sides by 125: 125 * 5x - 125 * 10 = 125 * (x + 3) / 125 Expanding the right side: 125 * 5x - 125 * 10 = x + 3 Combining like terms on the left side: 625x - 1250 = x + 3 Subtracting "x" from both sides: 624x - 1250 = 3 Adding 1250 to both sides: 624x = 1253 Dividing both sides by 624: x = 2 So the solution to the equation is x = 2, which is not equal to any of the options (-7/4, 4/7, or 7/4).

Akọwa Nkọwa

${27}^x÷{81}^{(x+2y)}=9$
$\left(27\right)x=9\times {81}^{(x+2y)}$
$(3^3{)}^x=3^2\times 3^{4(x+2y)}$
$=3^{(2+4x+8y)}$
$3^{3x}=3^{(2+4x+8y)}$
$3x=2+4x+8y$
$3x-4x-8y=2........\left(1\right)$
$x+4y=0........\left(2\right)$
$-4y=2$
$y=(-2)÷4=-\frac{1}{2}$
$y=-\frac{1}{2}$
Substitute the value of y into equation (2)
i.e $x+4y=0$
$x+4(-\frac{1}{2})=0$
$x-2=0$
$x=2$
$\therefore x=2,y=-\frac{1}{2}$
Method II
${27}^x÷{31}^{(x+2y)}=9$
$3^{3x}\times 3^{(-4x-8y)}=3^2$
$3^{(3x-8y)}=3^2$
$-x-8y=2........\left(1\right)$
$x+4y=0........\left(2\right)$
$-4y=2$
$y=-\frac{2}{4}=-\frac{1}{2}$
$y=-\frac{1}{2}$
Substitute the value of y into equation 2
$x+4y=0$
$x+4(-1)÷2)=0$
$x-2=0$
$x=2$
$x=2,y=-\frac{1}{2}$

Akọwa Nkọwa

The expression x^2 - 2x - 15 can be factored into the product of two binomials. To find these binomials, we can use the "ac method," where we try to find two numbers that multiply to -15 and add up to -2. These numbers are -5 and 3. We can then write the expression as:

(x - 5)(x + 3)

So, the expression x^2 - 2x - 15 can be factored as (x - 5)(x + 3).

Akọwa Nkọwa

√30 × √40
= √(30 × 40)
= √(3 × 10 × 4 × 10)
= √(400 × 3)
= √400 × √3
= 20 × √3
= 20√3

Akọwa Nkọwa

The correct answer is ₦20. To find the duty to be added to a bill of ₦80, we need to calculate 25% of ₦80. 25% means 25 out of 100, so we can calculate 25% of ₦80 by multiplying ₦80 by 25/100. ₦80 * 25/100 = ₦20 So the duty to be added to a bill of ₦80 is ₦20.

Akọwa Nkọwa

The formula for the volume of a cone is given by: V = (1/3)πr²h where V is the volume, r is the radius of the base of the cone, h is the height of the cone, and π is a constant value of approximately 3.14. Substituting the given values into the formula, we have: V = (1/3)π(5cm)²(6cm) Simplifying this expression, we get: V = (1/3)π(25cm²)(6cm) V = (1/3)π(150cm³) V = 50π cm³ Using a calculator to approximate π to two decimal places (π ≈ 3.14), we have: V ≈ 157 cm³ Therefore, the correct option is: 157cm3

Akọwa Nkọwa

To rationalize the expression 5 ÷ (2 − √3), we need to eliminate the square root in the denominator. We can do this by multiplying both the numerator and denominator by the conjugate of the denominator, which is 2 + √3. This is because when we multiply the conjugate of the denominator by the original denominator, we get a difference of squares that simplifies the expression: 5 ÷ (2 − √3) * (2 + √3) / (2 + √3) Simplifying the numerator by multiplying the terms inside the brackets, we get: 5(2 + √3) / (2^2 - (√3)^2) Simplifying further, we get: 5(2 + √3) / (4 - 3) 5(2 + √3) / 1 Therefore, the rationalized form of the expression is 5(2 + √3), which is equivalent to option C.

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S = {0, 1, 2, 3, 4, 5}
T = {− 1, 0, 1, 2}
S $\cup$ T = {− 1, 0, 1, 2, 3, 4, 5 }
⇒ − 2 < x ≤ 5

Akọwa Nkọwa

Let x represent the first number;
Then, the other is (5 − x) , since their sum is 5 and their product is 14
x(5 − x) = 14
5x − x2 = 14
x2 − 5x − 14 = 0
(x2 − 7x) + (2x − 14) = 0
x(x − 7) + 2(x − 7) = 0
(x − 7)(x + 2) = 0
Either (x − 7) = 0 or (x + 2) = 0
x = 7 or x = − 2
The two numbers are 7 and − 2

Akọwa Nkọwa

The trinomial = $a^2-4a+4$
$a^2-b^2-4a+4=(a^2-4a+4)-b^2$
$(a^2-2a-2a+4)-b^2$
$a(a-2)-2(a-2)-b^2$
$(a-2{)}^2-b^2$
$(a-2+b)(a-2-b)$

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Since the extension of a stretched string is directly proportional to its tension, we can use the formula:

extension = k * tension

where k is a constant of proportionality. We can find k by using the known values of tension and extension:

k = extension / tension
k = 2 cm / 8 N = 0.25 cm/N

Now that we have found k, we can use it to find the extension produced by a tension of 12 Newtons:

extension = k * tension
extension = 0.25 cm/N * 12 N = 3 cm

So, the extension produced by a tension of 12 Newtons is 3 cm.

Akọwa Nkọwa

One bag with 3 blue and 5 red
Pr (H) = 3/8Pr (r) 5/8
Another bag with 2 blue and 4 red (note one ball is drawn from each bag)
i.e. prob.=8 − 1 = 7
prob. (2 ball both are blue=pr (1st blue and 2nd blue)
prob. 2 balls both blue = 3/8 × 2/7
= $\frac{3}{28}$

Akọwa Nkọwa

132x = 708
$1\times x^2+3\times x^1+2\times x^0$
$7\times 8^1+0\times 8^0$
$x^2+3x+2-56=0$
$x^2+3x-54=0$
x(x + 9)−6(x + 9) = 0
(x + 9)(x − 6) = 0
Either (x + 9) = 0 or (x − 6) = 0
x = − 9 or +6
The positive values for x = 6
The base number for 132x = 1326

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To evaluate log base 7 of 17, we need to find the exponent that 7 must be raised to in order to obtain 17. In other words, we need to solve the equation 7^x = 17. This cannot be done algebraically, so we must use numerical methods or a calculator to approximate the solution. Using a calculator, we find that 7^1.455 ≈ 16.999, and 7^1.456 ≈ 17.035. Therefore, log base 7 of 17 is approximately 1.455. So, the answer is: 1.455.

Akọwa Nkọwa

The probability of either event A or B or both can be calculated using the formula for the union of two events: P(A or B) = P(A) + P(B) - P(A and B) where P(A) is the probability of event A, P(B) is the probability of event B, and P(A and B) is the probability of both events happening at the same time. Plugging in the given probabilities, we get: P(A or B) = 1/5 + 1/3 - 1/15 = 7/15 So, the probability of either event A or B or both is 7/15.

Akọwa Nkọwa

If the man sold the car for ₦420,000 at a gain of 15%, then we can find the cost price of the car using the following formula: Selling price = Cost price + Profit We know the selling price is ₦420,000 and the profit is 15% of the cost price. Let's call the cost price "x." Profit = 15% of x = 0.15x Substituting the values in the formula: ₦420,000 = x + 0.15x ₦420,000 = 1.15x To solve for x, we need to divide both sides by 1.15: x = ₦420,000 / 1.15 x = ₦365,217.39 Therefore, the cost price of the car was ₦365,217.39. So, option B - ₦365,217 is the correct answer.

Akọwa Nkọwa

To solve for x in the equation √(2x + 2) − √x = 1, we can use algebraic manipulation to isolate x on one side of the equation. Here's how: 1. Start by adding √x to both sides of the equation: √(2x + 2) = √x + 1 2. Square both sides of the equation to eliminate the square root: (√(2x + 2))^2 = (√x + 1)^2 Simplifying the left side: 2x + 2 = x + 1 + 2√x + 1 3. Rearrange terms: x - 2√x + 1 = 0 4. Factor the left side of the equation: (√x - 1)^2 = 0 5. Solve for x by taking the square root of both sides: √x - 1 = 0 √x = 1 x = 1^2 = 1 Therefore, the solution to the equation √(2x + 2) − √x = 1 is x = 1.

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11012 + 101112 + 1112 = 101011

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To find the curved surface area of a cone, we need to use the formula: `Curved Surface Area = πrs` where `r` is the radius of the base of the cone and `s` is the slant height of the cone. To find the slant height of the cone, we can use the Pythagorean theorem, which tells us that: `h^2 + r^2 = s^2` where `h` is the height of the cone. Substituting `r = 3cm` and `h = 4cm`, we get: `4^2 + 3^2 = s^2` `16 + 9 = s^2` `25 = s^2` `sqrt(25) = s` `s = 5cm` Now, substituting `r = 3cm` and `s = 5cm` into the formula for curved surface area, we get: `Curved Surface Area = πrs` `= 3.14 × 3 × 5` `= 47.1cm^2` Therefore, the curved surface area of the cone is `47.1cm^2`. Hence, option C, "47.1cm^2" is the correct answer.

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Akọwa Nkọwa

The area of an ellipse (e) = $\frac{\pi ab}{4}$
Let a rep. the length of its major axis = 14cm
Let b rep. the length of its minor axis = ?
π = 3.142
Area of an ellipse = 132cm2
132 = (π14 × b) ÷ 4
132 = (3.142 × 14b) ÷ 4
3.142 × 14b =132 × 4
b = (1324 × 4) ÷ (3.142 × 14)
= 528/43.988
= 12cm
The length of its minor axis (b) = 12cm

Akọwa Nkọwa

To solve the problem, we need to use the concept of proportionality. We can set up a proportion between the return and the investment amount: Return/Investment = X/₦450,000 We can rearrange this to solve for X: X = (Return * ₦450,000) / Investment Now, we can use this equation to find the return Y on an investment of ₦700,000: Y = (X * ₦700,000) / ₦450,000 Substituting the given values, we get: X = (₦25,000 * ₦450,000) / ₦450,000 = ₦25,000 Y = (₦25,000 * ₦700,000) / ₦450,000 = ₦38,888.89 (rounded to ₦38,888.90K) Therefore, the answer is option D: ₦38,888.90K.

Akọwa Nkọwa

tan x = $\frac{2}{3}$(given), is illustrated in a right-angled $\Delta$
thus m2 = 22 + 32
= 4 + 9 = 13
m = $\sqrt{13}$
Hence, 2sin x = 2 x $\frac{2}{m}$
2 x$\frac{2}{\sqrt{13}}$
= $\frac{4}{\sqrt{13}}$
= $\frac{4}{\sqrt{13}}=\frac{\sqrt{13}}{\sqrt{13}}$
= $\frac{4\sqrt{13}}{13}$

Akọwa Nkọwa

The formula for finding the simple interest on a principal amount P for a period of t years at an interest rate of r per annum is:

simple interest = P * r * t / 100

In this case, P = 325, t = 5, and r = 3, so we can plug in the values into the formula:

simple interest = 325 * 3 * 5 / 100
simple interest = 48.75

So, the simple interest on ₦325 in 5 years at 3% per annum is ₦48.75.

Akọwa Nkọwa

The given inequalities are: 5 + x ≤ 8 13 + x ≥ 7 Simplifying them, we get: x ≤ 3 x ≥ -6 So the range of values of x that satisfy both the inequalities is: -6 ≤ x ≤ 3 Therefore, option (C) is the correct answer. To arrive at this answer, we first solved each inequality separately to get an idea of the possible range of values for x. Then we looked for the overlap between these ranges, which gives us the range of values that satisfy both inequalities.