Algebra Practice Questions — JAMB UTME — Page 6

Question 101Question →

The line $y = 2x - 1$ intersects the curve $y = x^2 - 4x + 4$ at two distinct points, $(x_1, y_1)$ and $(x_2, y_2)$ . Calculate the sum of the $y$ -coordinates of these two points of intersection, $y_1 + y_2$ .

Show answer & explanation

Answer: 10

Answer

The sum of the

y

-coordinates of the points of intersection is 10.

Equating

y = 2x - 1

and

y = x^2 - 4x + 4

yields

x^2 - 6x + 5 = 0

, whose solutions are

x = 1

and

x = 5

. Substituting these values into

y = 2x - 1

gives

y_1 = 1

and

y_2 = 9

. The sum

y_1 + y_2 = 1 + 9 = 10

.

Step-by-Step Solution

1

Equate the linear and quadratic equations to eliminate

y

.

x^2 - 6x + 5 = 0

Setting

2x - 1 = x^2 - 4x + 4

allows finding the

x

-coordinates of the intersection points.

2

Solve the quadratic equation for

x

.

x_1 = 1

and

x_2 = 5

Factoring

(x - 1)(x - 5) = 0

yields the two

x

-values.

3

Determine the corresponding

y

-values using

y = 2x - 1

.

y_1 = 1

and

y_2 = 9

Substituting

x = 1

yields

y = 1

, and substituting

x = 5

yields

y = 9

.

4

Calculate the sum of the

y

-coordinates.

10

Adding

y_1 + y_2 = 1 + 9 = 10

.

Key Concept

Solving simultaneous linear and quadratic equations to find coordinates of intersection

Question 102Question →

The $4^{\text{th}}$ term of an arithmetic progression (A.P.) is $15$ and the $9^{\text{th}}$ term is $35$ . Calculate the sum of the first $10$ terms of the progression.

Show answer & explanation

Answer: 210

Answer

The sum of the first 10 terms of the progression is 210.

Using the

n^{\text{th}}

term formula

T_n = a + (n-1)d

, the equations

a + 3d = 15

and

a + 8d = 35

yield common difference

d = 4

and first term

a = 3

. Substituting these into the sum formula

S_n = \frac{n}{2}[2a + (n-1)d]

for

n = 10

yields

S_{10} = 5[2(3) + 9(4)] = 210

.

Step-by-Step Solution

1

Set up equations for the given terms

a + 3d = 15

and

a + 8d = 35

The

n^{\text{th}}

term of an A.P. is given by

T_n = a + (n-1)d

.

2

Solve for the common difference

d

d = 4

Subtracting

(a + 3d = 15)

from

(a + 8d = 35)

gives

5d = 20

, so

d = 4

.

3

Solve for the first term

a

a = 3

Substituting

d = 4

into

a + 3(4) = 15

yields

a = 15 - 12 = 3

.

4

Calculate the sum of the first 10 terms

S_{10} = 210

Applying

S_{10} = \frac{10}{2}[2(3) + 9(4)] = 5(6 + 36) = 210

.

Key Concept

Finding terms and sums of an Arithmetic Progression using simultaneous linear equations

Estimated Time:1m 30s

Question 103Question →

What are the values of $y$ that satisfy the simultaneous equations $3x + y = 10$ and $x^2 + y = 14$ ?

-1

or

4

-13

or

2

-2

or

13

7

or

22

Show answer & explanation

Answer: $-2$ or $13$

Answer

The values of

y

that satisfy the simultaneous equations are

-2

or

13

.

Rearranging the linear equation gives

y = 10 - 3x

. Substituting this into

x^2 + y = 14

yields

x^2 + 10 - 3x = 14

, which simplifies to

x^2 - 3x - 4 = 0

. Factoring gives

(x - 4)(x + 1) = 0

, so

x = 4

or

x = -1

. Substituting

x = 4

into

y = 10 - 3x

gives

y = -2

, and substituting

x = -1

gives

y = 13

. Thus, the required values of

y

are

-2

or

13

.

Step-by-Step Solution

1

Express

y

in terms of

x

from the linear equation

y = 10 - 3x

Isolation of variable

y

allows straightforward substitution into the quadratic equation.

2

Substitute

y = 10 - 3x

into the quadratic equation

x^2 + y = 14

x^2 + (10 - 3x) = 14 \implies x^2 - 3x - 4 = 0

Creates a quadratic equation in terms of a single variable

x

.

3

Solve the quadratic equation

x^2 - 3x - 4 = 0

for

x

(x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1

Factoring determines the

x

-coordinates of the solution points.

4

Substitute each

x

-value back into

y = 10 - 3x

to find the corresponding

y

-values

For

x = 4

:

y = 10 - 3(4) = -2

; for

x = -1

:

y = 10 - 3(-1) = 13

Evaluates the exact values of

y

requested by the question.

Key Concept

Simultaneous Linear and Quadratic Equations

Question 104Question →

When the polynomial $P(x) = x^3 - 2x^2 + ax + 8$ is divided by $(x - 3)$ , the remainder is $14$ . What is the value of the constant $a$ ?

Show answer & explanation

Answer: -1

Answer

The value of the constant

a

is

-1

.

According to the Remainder Theorem, dividing

P(x)

by

(x - 3)

means the remainder is

P(3)

. Evaluating

P(3) = 3^3 - 2(3)^2 + 3a + 8 = 27 - 18 + 3a + 8 = 17 + 3a

. Setting this equal to the remainder

14

gives

17 + 3a = 14

, which simplifies to

3a = -3

and yields

a = -1

.

Step-by-Step Solution

1

Apply the Remainder Theorem

The remainder when

P(x)

is divided by

(x - 3)

is equal to

P(3)

.

The Remainder Theorem states that dividing a polynomial

P(x)

by

(x - c)

yields a remainder equal to

P(c)

.

2

Substitute

x = 3

into the polynomial and set equal to the given remainder

3^3 - 2(3)^2 + a(3) + 8 = 14

Setting the value of

P(3)

equal to

14

allows us to form a linear equation for the unknown constant

a

.

3

Simplify numerical terms in the equation

27 - 18 + 3a + 8 = 14 \implies 17 + 3a = 14

Evaluate exponents and multiplication to isolate the term containing

a

.

4

Solve for

a

3a = -3 \implies a = -1

Subtract

17

from both sides and divide by

3

.

Key Concept

Polynomial Remainder Theorem

Question 105Question →

If $\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}$ is expressed in the form $a + b\sqrt{3}$ , where $a$ and $b$ are integers, what is the value of $a + b$ ?

3

2

5

1

Show answer & explanation

Answer: 3

Answer

The value of

a + b

is 3.

Multiplying both numerator and denominator by the conjugate

(\sqrt{6} + \sqrt{2})

yields

\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}

. Comparing this to

a + b\sqrt{3}

gives

a = 2

and

b = 1

, so

a + b = 3

.

Step-by-Step Solution

1

Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator,

(\sqrt{6} + \sqrt{2})

.

\frac{(\sqrt{6} + \sqrt{2})(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}

Multiplying by the conjugate eliminates radicals from the denominator using the difference of squares.

2

Expand the numerator

(\sqrt{6} + \sqrt{2})^2

using

(x+y)^2 = x^2 + 2xy + y^2

.

(\sqrt{6})^2 + 2\sqrt{12} + (\sqrt{2})^2 = 6 + 2(2\sqrt{3}) + 2 = 8 + 4\sqrt{3}

Simplifying

\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}

converts compound surds to standard form.

3

Expand the denominator using the difference of two squares

(x-y)(x+y) = x^2 - y^2

.

(\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4

Squaring each square root leaves rational integers in the denominator.

4

Divide the expanded numerator by the denominator to find

a

and

b

.

\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}

Dividing each term by 4 gives

a = 2

and

b = 1

.

5

Compute the sum

a + b

.

2 + 1 = 3

Adding the coefficients

a

and

b

yields the required value.

Key Concept

Rationalisation of Binomial Denominators

Question 106Question →

Let the universal set $\mathcal{U} = \{x \in \mathbb{Z}^+ : x \le 60\}$ . Subsets $A$ , $B$ , and $C$ of $\mathcal{U}$ are defined as:

$A = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$
$B = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}$
$C = \{x \in \mathcal{U} : x \text{ is a factor of } 60\}$

What is the cardinality of the set $(A \Delta B) \setminus C'$ , where $A \Delta B$ represents the symmetric difference of $A$ and $B$ , and $C'$ is the complement of $C$ in $\mathcal{U}$ ?

6

8

12

19

Show answer & explanation

Answer: 6

Answer

6

Using the set identity

X \setminus Y' = X \cap Y

, the expression

(A \Delta B) \setminus C'

simplifies directly to

(A \Delta B) \cap C

. The set

C

consists of all factors of 60:

\{1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\}

. The symmetric difference

A \Delta B

consists of numbers that are divisible by 3 or 4, but not both (excluding multiples of 12). Checking the 12 factors of 60 against this rule, the numbers 3, 6, 15, and 30 are multiples of 3 but not 4, while 4 and 20 are multiples of 4 but not 3. The numbers 12 and 60 are multiples of both 3 and 4, so they are excluded. Thus, the resulting set is

\{3, 4, 6, 15, 20, 30\}

, which contains exactly 6 elements.

Step-by-Step Solution

1

Simplify the set expression using complement and relative difference identities.

(A \Delta B) \setminus C' = (A \Delta B) \cap (C')' = (A \Delta B) \cap C

By definition of set difference,

X \setminus Y = X \cap Y'

. Replacing

Y

with

C'

gives

X \cap (C')' = X \cap C

.

2

List all elements in set

C

(factors of 60).

C = \{1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\}

60 prime factorizes as

2^2 \times 3^1 \times 5^1

, giving

(2+1)(1+1)(1+1) = 12

positive factors.

3

Recall the definition of symmetric difference

A \Delta B

.

A \Delta B = (A \cup B) \setminus (A \cap B)

An element belongs to

A \Delta B

if it is a multiple of 3 or a multiple of 4, but NOT a multiple of both 3 and 4 (i.e. not a multiple of 12).

4

Evaluate which elements of

C

belong to

A \Delta B

.

The elements in

C

that are multiples of 3 only (not 12) are

\{3, 6, 15, 30\}

. The elements in

C

that are multiples of 4 only (not 12) are

\{4, 20\}

. Elements

\{12, 60\}

are multiples of 12, so they are excluded.

Testing each factor of 60 against the criteria for symmetric difference.

5

Form the set

(A \Delta B) \cap C

and find its cardinality.

(A \Delta B) \cap C = \{3, 4, 6, 15, 20, 30\}

, so

n((A \Delta B) \cap C) = 6

.

Counting the elements in the intersection.

Key Concept

Symmetric Difference and Set Complement Identities

Estimated Time:2m 30s

Question 107Question →

The time, $t$ hours, required to complete a road maintenance project varies inversely as the number of workers, $w$ , assigned to the project. If 8 workers can finish the project in 15 hours, calculate the time, in hours, required for 12 workers to finish the same project.

Show answer & explanation

Answer: 10

Answer

10 hours

Because the time

t

varies inversely as the number of workers

w

, the total worker-hours required for the project is constant:

k = 8 \times 15 = 120

worker-hours. Dividing this total work by 12 workers gives

\frac{120}{12} = 10

hours.

Step-by-Step Solution

1

Set up the inverse variation equation

t = \frac{k}{w}

, where

k

is the constant of variation.

Inverse variation implies that as the number of workers increases, the time required decreases proportionally.

2

Determine the value of the constant of variation

k

k = t \times w = 15 \times 8 = 120

.

Substitute the known pair of values (

w = 8, t = 15

) into the equation.

3

Compute the new value of

t

for 12 workers

t = \frac{120}{12} = 10

hours.

Substitute

w = 12

and

k = 120

into

t = \frac{k}{w}

.

Key Concept

Inverse Variation

Question 108Question →

The total energy loss $E$ in Joules per minute in a magnetic core circuit is partly constant and partly varies directly as the square of the frequency $f$ in Hz of the alternating current. Given that $E = 120\text{ J}$ when $f = 10\text{ Hz}$ , and $E = 360\text{ J}$ when $f = 20\text{ Hz}$ , what is the value of $E$ in Joules when $f = 15\text{ Hz}$ ?

Show answer & explanation

Answer: 220

Answer

220

The relation describes a partial variation model

E = c + k f^2

. Substituting the two given conditions (

f=10, E=120

and

f=20, E=360

) yields the system of equations

c + 100k = 120

and

c + 400k = 360

. Solving this system gives

k = 0.8

and

c = 40

. Evaluating

E = 40 + 0.8(15)^2

results in

220\text{ J}

.

Step-by-Step Solution

1

Write the general equation for partial variation involving a constant term and a term proportional to

f^2

E = c + k f^2

Partial variation consists of a sum of a constant component and a variable component.

2

Substitute the known conditions into the variation equation to create a system of linear equations

Equation 1:

c + 100k = 120

; Equation 2:

c + 400k = 360

Two pairs of values are provided to determine the two unknown constants

c

and

k

.

3

Solve the system of simultaneous equations for

k

and

c

k = 0.8

and

c = 40

Subtracting Equation 1 from Equation 2 eliminates

c

, allowing direct calculation of

k

, after which

c

is found by substitution.

4

Calculate the required value of

E

when

f = 15

E = 40 + 0.8(15^2) = 40 + 0.8(225) = 220

Applying the discovered constants to the target frequency yields the final energy loss value.

Key Concept

Partial Variation and Simultaneous Equations

Estimated Time:2m 0s

Question 109Question →

If $y$ varies directly as $x$ , and $y = 20$ when $x = 4$ , what is the value of $y$ when $x = 10$ ?

50

8

5

2

Show answer & explanation

Answer: 50

Answer

The value of

y

when

x = 10

is 50.

Direct variation means

y = kx

. Substituting

y = 20

and

x = 4

yields

k = 5

. Substituting

k = 5

and

x = 10

gives

y = 50

.

Step-by-Step Solution

1

Set up the equation for direct variation

y = kx

, where

k

is the constant of variation.

Direct variation implies that

y

is directly proportional to

x

.

2

Substitute the given initial values (

y = 20

,

x = 4

) to find

k

20 = k(4) \implies k = \frac{20}{4} = 5

.

Determining the constant of variation allows us to write the specific relationship equation.

3

Calculate

y

when

x = 10

using the constant

k = 5

y = 5 \times 10 = 50

.

Substituting

x = 10

into

y = 5x

yields the required value.

Key Concept

Direct Variation (

y = kx

)

Estimated Time:45s

Question 110Question →

What is the set of real values of $x$ that satisfies the quadratic inequality $2x^2 - 5x - 3 \le 0$ ?

-\frac{1}{2} \le x \le 3

x \le -\frac{1}{2}

or

x \ge 3

-3 \le x \le \frac{1}{2}

x \le -3

or

x \ge \frac{1}{2}

Show answer & explanation

Answer: $-\frac{1}{2} \le x \le 3$

Answer

-\frac{1}{2} \le x \le 3

Factoring

2x^2 - 5x - 3

yields

(2x + 1)(x - 3) \le 0

. Setting the factors to zero gives roots

x = -1/2

and

x = 3

. Because the coefficient of

x^2

is positive, the quadratic curve opens upwards and is less than or equal to zero in the closed interval between the roots, resulting in

-\frac{1}{2} \le x \le 3

.

Step-by-Step Solution

1

Factor the quadratic expression

2x^2 - 5x - 3

(2x + 1)(x - 3) \le 0

Finding the factors helps identify the critical points (roots) of the inequality.

2

Determine the critical points by setting each factor equal to zero

x = -\frac{1}{2}

and

x = 3

The critical points divide the real number line into test intervals.

3

Determine the region where

(2x + 1)(x - 3) \le 0

-\frac{1}{2} \le x \le 3

Since the quadratic coefficient is positive (

2 > 0

), the parabola opens upwards, so the function values are less than or equal to zero between the two roots.

Key Concept

Solving Quadratic Inequalities by Factorisation

Question 111Question →

Find the smallest positive integer $k$ for which the inequality $(k - 2)x^2 + 8x + k + 4 > 0$ holds for all real values of $x$ .

Show answer & explanation

Answer: 5

Answer

5

For the quadratic expression

(k - 2)x^2 + 8x + k + 4

to be positive for all real values of

x

, two conditions must be satisfied simultaneously: the leading coefficient must be positive (

k - 2 > 0 \implies k > 2

) and the discriminant must be strictly negative (

\Delta < 0

). Calculating the discriminant gives

\Delta = 8^2 - 4(k - 2)(k + 4) = 96 - 8k - 4k^2

. Setting

96 - 8k - 4k^2 < 0

and dividing by

-4

(reversing the inequality) yields

k^2 + 2k - 24 > 0

, which factors as

(k + 6)(k - 4) > 0

. This gives

k < -6

or

k > 4

. Intersecting with

k > 2

results in

k > 4

. The smallest integer greater than 4 is 5.

Step-by-Step Solution

1

Determine the conditions for positivity for all real numbers

k - 2 > 0

and

\Delta < 0

A quadratic

Ax^2 + Bx + C

remains above the x-axis for all real

x

if and only if its parabola opens upwards (

A > 0

) and has no real roots (

\Delta < 0

).

2

Set up and solve the discriminant inequality

k^2 + 2k - 24 > 0 \implies (k + 6)(k - 4) > 0

Expanding

8^2 - 4(k - 2)(k + 4) < 0

gives

64 - 4(k^2 + 2k - 8) < 0

, which simplifies to

k^2 + 2k - 24 > 0

after dividing by

-4

and reversing the inequality sign.

3

Intersect solution sets and find the smallest integer

k = 5

The intersection of

k > 2

and

(k < -6 \text{ or } k > 4)

gives

k > 4

. The smallest integer strictly greater than 4 is 5.

Key Concept

Condition for Positive Definite Quadratic Inequalities

Question 112Question →

If the polynomial $P(x) = x^4 + ax^3 - 7x^2 + bx + 12$ is completely divisible by $x^2 - 2x - 3$ , what is the value of $a - b$ ?

-10

10

-5

5

Show answer & explanation

Answer: $-10$

Answer

The value of

a - b

is

-10

.

Factoring

x^2 - 2x - 3

gives

(x - 3)(x + 1)

. By the Factor Theorem,

P(3) = 0

and

P(-1) = 0

. Substituting these into

P(x)

produces the linear system

9a + b = -10

and

a + b = 6

. Solving this system yields

a = -2

and

b = 8

. Subtracting gives

a - b = -2 - 8 = -10

.

Step-by-Step Solution

1

Factor the quadratic divisor to find the roots.

x^2 - 2x - 3 = (x - 3)(x + 1)

, so the roots are

x = 3

and

x = -1

.

By the Factor Theorem, if a polynomial is divisible by a quadratic expression,

P(x)

must evaluate to zero at each root of the divisor.

2

Set up equations by evaluating

P(3) = 0

and

P(-1) = 0

.

For

x = 3

:

3^4 + a(3)^3 - 7(3)^2 + b(3) + 12 = 0 \implies 81 + 27a - 63 + 3b + 12 = 0 \implies 9a + b = -10

.
For

x = -1

:

(-1)^4 + a(-1)^3 - 7(-1)^2 + b(-1) + 12 = 0 \implies 1 - a - 7 - b + 12 = 0 \implies a + b = 6

.

Evaluating the polynomial at each root yields a system of two linear equations in variables

a

and

b

.

3

Solve the simultaneous equations for

a

and

b

.

Subtracting

(a + b = 6)

from

(9a + b = -10)

gives

8a = -16 \implies a = -2

.
Substituting

a = -2

into

a + b = 6

gives

-2 + b = 6 \implies b = 8

.

Elimination isolates

a

, allowing both

a

and

b

to be uniquely determined.

4

Calculate

a - b

.

a - b = -2 - 8 = -10

.

This computes the required expression value.

Key Concept

Factor Theorem for quadratic divisors

Estimated Time:2m 0s

Question 113Question →

Let the universal set $\mathcal{U} = \{x \in \mathbb{Z} : 10 \le x \le 99\}$ . Consider two subsets $P$ and $Q$ of $\mathcal{U}$ , where $P = \{10a + b \in \mathcal{U} : a > b\}$ and $Q = \{10a + b \in \mathcal{U} : a \cdot b \text{ is odd}\}$ , with $a$ and $b$ representing the tens and units digits of $x$ respectively. What is the cardinality of the set $P \setminus Q$ ?

Show answer & explanation

Answer: 35; 35 elements

Answer

35

To evaluate

n(P \setminus Q)

, we calculate

n(P) - n(P \cap Q)

. Set

P

consists of all two-digit numbers where the tens digit exceeds the units digit, giving

1 + 2 + \dots + 9 = 45

numbers. Set

Q

consists of numbers with two odd digits. The intersection

P \cap Q

contains pairs of distinct odd digits

(a,b)

where

a > b

, which corresponds to

\binom{5}{2} = 10

pairs. Thus,

n(P \setminus Q) = 45 - 10 = 35

.

Step-by-Step Solution

1

Calculate the cardinality of set

P

, denoted as

n(P)

.

n(P) = 45

For a two-digit integer

10a + b

,

a \in \{1, 2, \dots, 9\}

and

b \in \{0, 1, \dots, 9\}

. The condition

a > b

yields

0

valid values of

b

for

a=1

(1 choice: 10), 2 choices for

a=2

(20, 21), up to 9 choices for

a=9

(90 through 98). The total is

\sum_{a=1}^9 a = \frac{9 \times 10}{2} = 45

.

2

Calculate the cardinality of the intersection

P \cap Q

, denoted as

n(P \cap Q)

.

n(P \cap Q) = 10

Set

Q

contains numbers where the product

a \cdot b

is odd, which requires both

a

and

b

to be odd digits chosen from

\{1, 3, 5, 7, 9\}

. For

P \cap Q

, we need

a > b

with distinct odd digits. The number of such pairs

(a, b)

is given by choosing 2 distinct odd digits out of 5:

\binom{5}{2} = 10

.

3

Calculate the cardinality of the set difference

P \setminus Q

.

n(P \setminus Q) = 35

Using the identity

n(P \setminus Q) = n(P) - n(P \cap Q)

, we get

45 - 10 = 35

.

Key Concept

Cardinality of set difference and set-builder operations

Estimated Time:2m 30s

Question 114Question →

Simplify the expression $\frac{\sqrt{3}}{\sqrt{3} - 1} - \frac{\sqrt{3}}{\sqrt{3} + 1}$ .

\sqrt{3}

2\sqrt{3}

3

0

Show answer & explanation

Answer: $\sqrt{3}$

Answer

\sqrt{3}

Combining the fractions using the common denominator

(\sqrt{3}-1)(\sqrt{3}+1) = 2

gives a numerator of

\sqrt{3}(\sqrt{3}+1) - \sqrt{3}(\sqrt{3}-1) = 3 + \sqrt{3} - 3 + \sqrt{3} = 2\sqrt{3}

. Dividing

2\sqrt{3}

by

2

yields the simplified answer

\sqrt{3}

.

Step-by-Step Solution

1

Find a common denominator for the two fractions

The common denominator is

(\sqrt{3} - 1)(\sqrt{3} + 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2

The denominators are conjugate surds, so their product simplifies to a rational number using the difference of two squares.

2

Express the numerator over the common denominator

Numerator

= \sqrt{3}(\sqrt{3} + 1) - \sqrt{3}(\sqrt{3} - 1)

Multiply each numerator by the missing factor of the common denominator.

3

Expand and simplify the numerator

Numerator

= (3 + \sqrt{3}) - (3 - \sqrt{3}) = 3 + \sqrt{3} - 3 + \sqrt{3} = 2\sqrt{3}

Distribute

\sqrt{3}

and handle the subtraction sign carefully.

4

Divide the simplified numerator by the common denominator

\frac{2\sqrt{3}}{2} = \sqrt{3}

Cancel the common factor of 2.

Key Concept

Rationalisation of denominators and algebraic manipulation of surd fractions

Estimated Time:1m 15s

Question 115Question →

If $(x + 3)$ is a factor of the polynomial $P(x) = x^3 + 4x^2 + mx - 6$ , what is the value of $m$ ?

Show answer & explanation

Answer: 1

Answer

The value of

m

is 1.

According to the Factor Theorem,

(x + 3)

is a factor of

P(x)

if

P(-3) = 0

. Substituting

x = -3

into

P(x) = x^3 + 4x^2 + mx - 6

yields

(-3)^3 + 4(-3)^2 + m(-3) - 6 = 0

, which simplifies to

-27 + 36 - 3m - 6 = 0

, giving

3 - 3m = 0

and thus

m = 1

.

Step-by-Step Solution

1

Apply the Factor Theorem

P(-3) = 0

By the Factor Theorem, for a linear divisor

(x - a)

to be a factor of

P(x)

,

P(a)

must equal zero. Here

x + 3 = 0 \implies x = -3

.

2

Substitute

x = -3

into

P(x) = x^3 + 4x^2 + mx - 6

(-3)^3 + 4(-3)^2 + m(-3) - 6 = 0

Evaluating

P(-3)

sets up an equation to find the unknown coefficient

m

.

3

Simplify and solve for

m

-27 + 36 - 3m - 6 = 0 \implies 3 - 3m = 0 \implies m = 1

Combine the constant terms

-27 + 36 - 6 = 3

and solve the linear equation in terms of

m

.

Key Concept

Factor Theorem

Question 116Question →

For what set of real values of $k$ does the quadratic inequality $(k + 2)x^2 - 2(k - 1)x + (k + 5) > 0$ hold for all real values of $x$ ?

k > -1

-2 < k < -1

k < -1

k > -2

Show answer & explanation

Answer: $k > -1$

Answer

The condition holds for

k > -1

.

A quadratic function is strictly positive for all real values of

x

if and only if its parabola opens upwards (

a > 0

) and it has no real x-intercepts (

\Delta < 0

). Here

a = k + 2 > 0

gives

k > -2

. The discriminant

\Delta = [-2(k-1)]^2 - 4(k+2)(k+5) = -36(k+1)

. Setting

-36(k+1) < 0

requires dividing by

-36

and reversing the inequality sign, giving

k + 1 > 0 \implies k > -1

. The intersection of

k > -2

and

k > -1

is

k > -1

.

Step-by-Step Solution

1

Identify the conditions required for a quadratic expression

ax^2 + bx + c

to be strictly positive for all real

x

.

The coefficient of

x^2

must be positive (

a > 0

) and the discriminant must be strictly negative (

\Delta < 0

).

If

a < 0

, the parabola opens downwards and yields negative values. If

\Delta \ge 0

, the quadratic has real roots or a double root, taking non-positive values.

2

Apply the first condition

a > 0

to the coefficient of

x^2

.

k + 2 > 0 \implies k > -2

.

Ensures the parabola opens upwards.

3

Calculate the discriminant

\Delta = b^2 - 4ac

and set

\Delta < 0

.

\Delta = [-2(k - 1)]^2 - 4(k + 2)(k + 5) = 4(k^2 - 2k + 1) - 4(k^2 + 7k + 10) = 4(-9k - 9) = -36(k + 1)

. Setting

-36(k + 1) < 0

.

Ensures the quadratic equation has no real roots and does not cross or touch the x-axis.

4

Solve the linear inequality

-36(k + 1) < 0

and combine with the first condition.

Dividing by

-36

reverses the inequality:

k + 1 > 0 \implies k > -1

. Taking the intersection of

k > -2

and

k > -1

yields

k > -1

.

Dividing an inequality by a negative number flips the inequality symbol.

Key Concept

Conditions for positive definiteness of quadratic expressions

Question 117Question →

Let the universal set be $\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 36\}$ . Subsets $A$ , $B$ , and $C$ of $\mathcal{U}$ are defined as follows:
$A = \{x \in \mathcal{U} : x \text{ is a perfect square}\}$ ,
$B = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}$ , and
$C = \{x \in \mathcal{U} : x \text{ is an even number}\}$ .
What is the cardinality of the set $(A \cup B)' \cap C$ ?

8

9

10

12

Show answer & explanation

Answer: 10

Answer

The cardinality of the set

(A \cup B)' \cap C

is 10.

The set

(A \cup B)' \cap C

represents elements in

C

that are neither in

A

nor in

B

. Set

C

contains 18 even integers up to 36. Among these 18 numbers, three are perfect squares (4, 16, 36) and six are multiples of 3 (6, 12, 18, 24, 30, 36). Since 36 satisfies both conditions, the total number of even numbers in

A \cup B

is

3 + 6 - 1 = 8

. Subtracting these 8 elements from 18 leaves 10 elements.

Step-by-Step Solution

1

Identify the elements of set C (even numbers in the universal set)

C = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36\}

, so

n(C) = 18

.

The complement set operation is restricted to elements within set C.

2

Find the elements of set C that belong to set A (even perfect squares)

C \cap A = \{4, 16, 36\}

, so

n(C \cap A) = 3

.

Identify which elements in C must be excluded due to being perfect squares.

3

Find the elements of set C that belong to set B (even multiples of 3, i.e., multiples of 6)

C \cap B = \{6, 12, 18, 24, 30, 36\}

, so

n(C \cap B) = 6

.

Identify which elements in C must be excluded due to being multiples of 3.

4

Find the overlap between the excluded sets within C

C \cap A \cap B = \{36\}

, so

n(C \cap A \cap B) = 1

.

Avoid double-counting 36 when subtracting excluded elements.

5

Calculate

n(C \cap (A \cup B))

using inclusion-exclusion and find the remaining complement elements in C

n(C \cap (A \cup B)) = 3 + 6 - 1 = 8

. Therefore,

n((A \cup B)' \cap C) = 18 - 8 = 10

.

Subtracting the 8 excluded even numbers from the total 18 even numbers yields the final cardinality.

Key Concept

Set Complement and Relative Difference with Inclusion-Exclusion

Estimated Time:2m 0s

Question 118Question →

If $\frac{7}{3\sqrt{2} - \sqrt{11}} = a\sqrt{2} + b\sqrt{11}$ , where $a$ and $b$ are rational numbers, what is the value of $a + b$ ?

4

2

12

8

Show answer & explanation

Answer: 4

Answer

4

Multiplying both the top and bottom by the conjugate

(3\sqrt{2} + \sqrt{11})

transforms the denominator into

(3\sqrt{2})^2 - (\sqrt{11})^2 = 18 - 11 = 7

. Dividing the numerator

7(3\sqrt{2} + \sqrt{11})

by

7

simplifies to

3\sqrt{2} + \sqrt{11}

. Matching coefficients yields

a = 3

and

b = 1

, giving

a + b = 4

.

Step-by-Step Solution

1

Multiply the numerator and denominator by the conjugate of the denominator,

(3\sqrt{2} + \sqrt{11})

.

\frac{7(3\sqrt{2} + \sqrt{11})}{(3\sqrt{2} - \sqrt{11})(3\sqrt{2} + \sqrt{11})}

Rationalising the denominator eliminates radicals from the bottom of the fraction.

2

Expand the denominator using the difference of squares formula

(x - y)(x + y) = x^2 - y^2

.

(3\sqrt{2})^2 - (\sqrt{11})^2 = (9 \times 2) - 11 = 18 - 11 = 7

Squaring each term simplifies the denominator into an integer.

3

Simplify the overall rational fraction by cancelling common factors.

\frac{7(3\sqrt{2} + \sqrt{11})}{7} = 3\sqrt{2} + \sqrt{11}

The factor of 7 in the numerator and denominator cancels out.

4

Equate

3\sqrt{2} + 1\sqrt{11}

with

a\sqrt{2} + b\sqrt{11}

to determine the values of

a

and

b

, then sum them.

a = 3, b = 1 \implies a + b = 3 + 1 = 4

Comparing coefficients of corresponding surd terms gives the required values.

Key Concept

Rationalisation of binomial surd denominators using conjugates

Question 119Question →

Find the value of $x$ that satisfies the logarithmic equation $\log_2(x^2 - 1) - \log_2(x - 1) = 3$ .

Show answer & explanation

Answer: 7

Answer

The value of

x

is

7

.

Applying the logarithm quotient law reduces

\log_2(x^2 - 1) - \log_2(x - 1)

to

\log_2\left(\frac{x^2 - 1}{x - 1}\right)

. Factoring

x^2 - 1

into

(x - 1)(x + 1)

and canceling

(x - 1)

leaves

\log_2(x + 1) = 3

. Converting this to exponential form gives

x + 1 = 2^3 = 8

, which yields

x = 7

.

Step-by-Step Solution

1

Apply the quotient rule for logarithms to combine the left-hand side into a single logarithmic expression.

\log_2\left(\frac{x^2 - 1}{x - 1}\right) = 3

According to logarithmic rules, the difference of two logarithms of the same base is equal to the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

2

Factor the numerator as a difference of two squares and simplify the rational expression.

\log_2(x + 1) = 3

Since

x^2 - 1 = (x - 1)(x + 1)

, dividing by

(x - 1)

leaves

(x + 1)

, given

x \neq 1

.

3

Rewrite the logarithmic equation in exponential form.

x + 1 = 2^3 = 8

By definition of logarithms,

\log_b y = z \iff b^z = y

.

4

Solve the resulting linear equation for

x

.

x = 7

Subtracting

1

from both sides gives

x = 7

.

Key Concept

Quotient Law of Logarithms and Logarithmic Equations

Estimated Time:1m 30s

Question 120Question →

The power consumption $P$ (in watts) of a variable-speed motor is partly constant and partly varies directly as the square of its operational speed $v$ (in revolutions per second). If $P = 250\text{ W}$ when $v = 10\text{ rev/s}$ and $P = 700\text{ W}$ when $v = 20\text{ rev/s}$ , calculate the value of $P$ (in watts) when $v = 15\text{ rev/s}$ .

Show answer & explanation

Answer: 437.5

Answer

437.5 W

The partial variation equation is

P = k_1 + k_2 v^2

. Setting up simultaneous equations

k_1 + 100 k_2 = 250

and

k_1 + 400 k_2 = 700

yields

k_1 = 100

and

k_2 = 1.5

. Substituting

v = 15

into

P = 100 + 1.5(15^2)

gives

P = 437.5\text{ W}

.

Step-by-Step Solution

1

Express the partial variation mathematically.

P = k_1 + k_2 v^2

, where

k_1

and

k_2

are constants.

The total power consumption is the sum of a fixed baseline constant

k_1

and a variable component proportional to

v^2

.

2

Form simultaneous linear equations using the provided data points.

k_1 + 100 k_2 = 250

and

k_1 + 400 k_2 = 700

.

Substituting

v = 10

gives

10^2 = 100

, and substituting

v = 20

gives

20^2 = 400

.

3

Solve for the constants

k_1

and

k_2

.

k_2 = 1.5

and

k_1 = 100

.

Subtracting the two equations eliminates

k_1

, yielding

300 k_2 = 450 \implies k_2 = 1.5

. Substituting

k_2 = 1.5

back into

k_1 + 100 k_2 = 250

gives

k_1 = 100

.

4

Calculate the value of

P

at

v = 15\text{ rev/s}

.

P = 437.5\text{ W}

.

Substitute

v = 15

,

k_1 = 100

, and

k_2 = 1.5

into the governing formula

P = 100 + 1.5(15^2) = 100 + 337.5 = 437.5

.

Key Concept

Partial Variation and Simultaneous Linear Equations

Estimated Time:1m 30s

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