All practice questions

Question 7261Question →

The ages, in years, of five participants in a workshop are $2, 4, 5, 7,$ and $12$ . What is the mean deviation of the ages?

2.8

0.0

3.5

6.0

Show answer & explanation

Answer: $2.8$

Answer

2.8

The mean deviation is calculated by taking the average of the absolute differences between each data point and the mean. For the dataset

2, 4, 5, 7,

and

12

, the mean is

6

. The absolute deviations from

6

are

4, 2, 1, 1,

and

6

, which sum to

14

. Dividing

14

by

5

gives

2.8

.

Step-by-Step Solution

1

Calculate the arithmetic mean (

\bar{x}

) of the dataset.

\bar{x} = \frac{2 + 4 + 5 + 7 + 12}{5} = \frac{30}{5} = 6

The mean is needed as the reference point to calculate deviations.

2

Calculate the absolute deviation

|x - \bar{x}|

for each data item.

|2 - 6| = 4, \quad |4 - 6| = 2, \quad |5 - 6| = 1, \quad |7 - 6| = 1, \quad |12 - 6| = 6

Mean deviation measures distance from the mean, so absolute values are taken.

3

Find the average of these absolute deviations.

\text{Mean Deviation} = \frac{4 + 2 + 1 + 1 + 6}{5} = \frac{14}{5} = 2.8

The formula for mean deviation of ungrouped data is

\frac{\sum |x - \bar{x}|}{n}

.

Key Concept

Mean Deviation for Ungrouped Data

Question 7262Question →

The frequency distribution table below shows the mass, in grams, of $20$ harvested oranges recorded during an agricultural experiment:

Mass ( $g$ )	Frequency ( $f$ )
$10 - 19$	$3$
$20 - 29$	$5$
$30 - 39$	$8$
$40 - 49$	$4$

What is the mean mass of the harvested oranges?

26.5\text{ g}

29.5\text{ g}

31.0\text{ g}

35.5\text{ g}

Show answer & explanation

Answer: $31.0\text{ g}$

Answer

The mean mass of the harvested oranges is

31.0\text{ g}

.

The mean of grouped data is given by

\bar{x} = \frac{\sum fx}{\sum f}

. Finding the midpoints (

x

) of the intervals gives

14.5

,

24.5

,

34.5

, and

44.5

. Multiplying each midpoint by its frequency yields

43.5

,

122.5

,

276.0

, and

178.0

. The sum of these products is

620.0

, and dividing by the total frequency (

20

) gives

31.0\text{ g}

.

Step-by-Step Solution

1

Determine the class midpoint (

x

) for each class interval.

Midpoints are:

\frac{10+19}{2} = 14.5

,

\frac{20+29}{2} = 24.5

,

\frac{30+39}{2} = 34.5

, and

\frac{40+49}{2} = 44.5

.

For grouped frequency data, each interval is represented by its central value (class mark or midpoint).

2

Multiply each midpoint (

x

) by its corresponding frequency (

f

) to calculate

fx

.

f_1 x_1 = 3 \times 14.5 = 43.5

,

f_2 x_2 = 5 \times 24.5 = 122.5

,

f_3 x_3 = 8 \times 34.5 = 276.0

,

f_4 x_4 = 4 \times 44.5 = 178.0

.

This computes the total mass contributed by the items in each interval.

3

Calculate the total sum of

fx

and total frequency

\sum f

.

\sum fx = 43.5 + 122.5 + 276.0 + 178.0 = 620.0

and

\sum f = 3 + 5 + 8 + 4 = 20

.

These sums are the numerator and denominator required for the mean formula.

4

Compute the mean using

\bar{x} = \frac{\sum fx}{\sum f}

.

\bar{x} = \frac{620.0}{20} = 31.0\text{ g}

.

Dividing the total weighted value by the total number of items gives the mean.

Key Concept

Calculating Mean for Grouped Frequency Data

Question 7263Question →

A curve has the equation $y = ax^3 + bx^2 + 12x + 1$ , where $a$ and $b$ are constants. If the curve has stationary points at $x = 1$ and $x = 2$ , what is the value of $a$ ?

2

6

-2

-6

Show answer & explanation

Answer: $2$

Answer

The value of

a

is

2

.

To find

a

, take the first derivative of the curve, yielding

\frac{dy}{dx} = 3ax^2 + 2bx + 12

. Setting this to zero gives a quadratic equation with roots

x = 1

and

x = 2

. The product of roots for a quadratic equation

Ax^2 + Bx + C = 0

is

\frac{C}{A}

. Therefore,

1 \times 2 = \frac{12}{3a}

, which simplifies to

2 = \frac{4}{a}

, giving

a = 2

.

Step-by-Step Solution

1

Find the derivative of the given curve with respect to

x

.

\frac{dy}{dx} = 3ax^2 + 2bx + 12

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Set the derivative to zero and substitute the stationary point locations

x = 1

and

x = 2

.

The roots of the quadratic equation

3ax^2 + 2bx + 12 = 0

are

x_1 = 1

and

x_2 = 2

.

Since stationary points are given at

x=1

and

x=2

, these values satisfy

\frac{dy}{dx} = 0

.

3

Apply the product of roots formula for a quadratic equation

Ax^2 + Bx + C = 0

, where

x_1 \cdot x_2 = \frac{C}{A}

.

1 \times 2 = \frac{12}{3a} \implies 2 = \frac{4}{a}

Equating the product of roots

1 \times 2 = 2

to

\frac{12}{3a}

isolates parameter

a

.

4

Solve for

a

.

a = 2

Multiplying both sides by

a

gives

2a = 4

, so

a = 2

.

Key Concept

Stationary points occur where

\frac{dy}{dx} = 0

. For a cubic curve, the derivative is a quadratic equation whose roots correspond to the

x

-coordinates of the stationary points.

Estimated Time:2m 0s

Question 7264Question →

A pie chart is used to display the distribution of $720$ candidates registered for a competitive examination across five subjects. If the sector representing Further Mathematics has a central angle of $45^\circ$ , what is the total number of candidates who registered for Further Mathematics?

Show answer & explanation

Answer: 90

Answer

The total number of candidates who registered for Further Mathematics is 90.

To calculate the number of candidates represented by a pie chart sector, multiply the total count by the ratio of the sector's central angle to 360 degrees: (45 / 360) * 720 = 90 candidates.

Step-by-Step Solution

1

Determine the fraction of the total population represented by the Further Mathematics sector.

\frac{45^\circ}{360^\circ} = \frac{1}{8}

A complete pie chart circle corresponds to an angle of 360 degrees.

2

Calculate the actual number of candidates by multiplying the fraction by the total student population.

\frac{1}{8} \times 720 = 90

The number of items in a sector is directly proportional to its central sector angle relative to 360 degrees.

Key Concept

Calculating category frequencies from pie chart sector angles

Question 7265Question →

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 7266Question →

Using differentiation from first principles, evaluate the numerical value of the derivative of the polynomial function $f(x) = 2x^3 - 3x^2 + 4$ at $x = 2$ .

Show answer & explanation

Answer: 12

Answer

The numerical value of the derivative at

x = 2

is 12.

Applying the first principles limit formula

\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

to

f(x) = 2x^3 - 3x^2 + 4

yields

f'(x) = 6x^2 - 6x

. Evaluating this derivative function at

x = 2

yields

6(4) - 12 = 12

.

Step-by-Step Solution

1

Set up the difference quotient using the definition of differentiation from first principles

f'(x) = \lim_{h \to 0} \frac{[2(x+h)^3 - 3(x+h)^2 + 4] - [2x^3 - 3x^2 + 4]}{h}

Differentiation from first principles evaluates the limit of the rate of change as the increment

h

approaches zero.

2

Expand

(x+h)^3

and

(x+h)^2

and subtract

f(x)

f(x+h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 6xh - 3h^2

Expanding the terms allows cancellation of all terms not containing

h

.

3

Divide by

h

and evaluate the limit as

h \to 0

f'(x) = \lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 6x - 3h) = 6x^2 - 6x

Dividing by

h

eliminates the indeterminate form

\frac{0}{0}

, allowing direct substitution of

h = 0

.

4

Substitute

x = 2

into the derivative function

f'(x)

f'(2) = 6(2)^2 - 6(2) = 24 - 12 = 12

Evaluating at the given point gives the slope of the tangent line at

x = 2

.

Key Concept

Differentiation from first principles using limit of difference quotient

Question 7267Question →

In a convex polygon of $n$ sides, three of its interior angles are right angles, and each of the remaining interior angles is equal to $150^\circ$ . What is the total number of diagonals in this polygon?

9

15

6

18

Show answer & explanation

Answer: 9

Answer

9 diagonals

The total sum of interior angles of an

n

-sided polygon is

(n - 2) \times 180^\circ

. Setting this equal to

3(90^\circ) + (n - 3)(150^\circ)

simplifies to

180n - 360 = 150n - 180

, which solves to

n = 6

. Substituting

n = 6

into the diagonal formula

D = \frac{n(n - 3)}{2}

yields

\frac{6 \times 3}{2} = 9

diagonals.

Step-by-Step Solution

1

Formulate the interior angle sum equation for the

n

-sided convex polygon.

The total interior angle sum is

(n - 2) \times 180^\circ

. Summing the individual angles gives

3 \times 90^\circ + (n - 3) \times 150^\circ

. Thus:

(n - 2) \times 180^\circ = 270^\circ + (n - 3) \times 150^\circ

.

The sum of all interior angles of any

n

-sided polygon is given by

(n - 2) \times 180^\circ

.

2

Solve the equation for the number of sides

n

.

180n - 360 = 270 + 150n - 450 \implies 180n - 360 = 150n - 180 \implies 30n = 180 \implies n = 6

.

Expanding terms and isolating

n

reveals that the polygon has 6 sides (a hexagon).

3

Calculate the total number of diagonals using the formula

D = \frac{n(n - 3)}{2}

.

D = \frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9

.

Each of the

n

vertices can connect to

n - 3

non-adjacent vertices, and dividing by 2 avoids double-counting each diagonal line segment.

Key Concept

Polygon interior angle sum theorem and diagonal calculation

Question 7268Question →

What is the derivative of the function $f(x) = \frac{\sin(2x)}{(1 + x^2)^2}$ with respect to $x$ ?

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

\frac{(1 + x^2)\cos(2x) - 4x\sin(2x)}{(1 + x^2)^3}

\frac{2[(1 + x^2)\cos(2x) - \sin(2x)]}{(1 + x^2)^3}

\frac{2[(1 + x^2)\cos(2x) + 2x\sin(2x)]}{(1 + x^2)^3}

Show answer & explanation

Answer: $\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}$

Answer

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Applying the quotient rule

\frac{u'v - uv'}{v^2}

with

u = \sin(2x)

and

v = (1 + x^2)^2

yields

u' = 2\cos(2x)

and

v' = 4x(1 + x^2)

. Substituting these into the formula gives

\frac{2\cos(2x)(1+x^2)^2 - 4x\sin(2x)(1+x^2)}{(1+x^2)^4}

. Factoring out

(1+x^2)

from numerator and denominator reduces the expression to

\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

.

Step-by-Step Solution

1

Identify the numerator

u(x)

and denominator

v(x)

for the quotient rule.

u(x) = \sin(2x)

and

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

The quotient rule states that

\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}

.

2

Find the derivatives

u'(x)

and

v'(x)

using the chain rule.

u'(x) = 2\cos(2x)

and

v'(x) = 2(1 + x^2) \cdot (2x) = 4x(1 + x^2)

By the chain rule,

\frac{d}{dx}\sin(2x) = \cos(2x) \cdot 2

and

\frac{d}{dx}(1 + x^2)^2 = 2(1 + x^2) \cdot 2x

.

3

Substitute

u, u', v, v'

into the quotient rule formula.

f'(x) = \frac{2\cos(2x)(1 + x^2)^2 - \sin(2x) \cdot 4x(1 + x^2)}{((1 + x^2)^2)^2}

Applying

\frac{u'v - uv'}{v^2}

directly.

4

Factor out

(1 + x^2)

from the numerator and simplify the denominator.

f'(x) = \frac{(1 + x^2)[2\cos(2x)(1 + x^2) - 4x\sin(2x)]}{(1 + x^2)^4} = \frac{2(1 + x^2)\cos(2x) - 4x\sin(2x)}{(1 + x^2)^3}

Canceling the common factor

(1 + x^2)

from numerator and denominator.

5

Factor out the constant

2

from the numerator.

f'(x) = \frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Putting the derivative into simplified factored form.

Key Concept

Combining Quotient Rule and Chain Rule for Composite Functions

Question 7269Question →

What is the gradient of the tangent line to the curve $y = 4x^2 - 7x + 2$ at the point where $x = 3$ ?

Show answer & explanation

Answer: 17

Answer

The gradient of the tangent line to the curve at

x = 3

is 17.

Differentiating

y = 4x^2 - 7x + 2

gives

\frac{dy}{dx} = 8x - 7

. Substituting

x = 3

yields

8(3) - 7 = 17

.

Step-by-Step Solution

1

Find the derivative of the curve's equation

\frac{dy}{dx} = 8x - 7

The gradient of the curve at any point is given by its first derivative with respect to x.

2

Evaluate the derivative at x = 3

m = 8(3) - 7 = 17

Substituting the given x-coordinate into the derivative gives the specific slope of the tangent line at that point.

Key Concept

The gradient of the tangent to a curve

y = f(x)

at

x = a

is the value of the first derivative

f'(a)

.

Question 7270Question →

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 7271Question →

What is the value of $\log_2 81 \cdot \log_3 32$ ?

Show answer & explanation

Answer: 20

Answer

The value of the logarithmic expression is 20.

Rewriting

81

as

3^4

and

32

as

2^5

allows extraction of coefficients

4

and

5

. Utilizing the change of base property

\log_2 3 \cdot \log_3 2 = 1

reduces the expression to

4 \times 5 = 20

.

Step-by-Step Solution

1

Rewrite arguments as prime powers

\log_2 81 = \log_2(3^4)

and

\log_3 32 = \log_3(2^5)

Simplifies terms to prime base components

2

Apply the power law of logarithms

4\log_2 3 \cdot 5\log_3 2

Brings powers out as numerical coefficients

3

Apply the change of base identity

\log_b a \cdot \log_a b = 1

4 \cdot 5 \cdot (\log_2 3 \cdot \log_3 2) = 20 \cdot 1 = 20

Reciprocal logarithm bases multiply to give 1

Key Concept

Logarithms and Change of Base

Estimated Time:45s

Question 7272Question →

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 7273Question →

If $\theta$ is an acute angle such that $\tan \theta = 2$ , calculate the exact numerical value of $\frac{3\sin \theta + \cos \theta}{\sin \theta + 2\cos \theta}$ .

Show answer & explanation

Answer: 1.75

Answer

The exact numerical value of the trigonometric expression is 1.75.

Dividing every term in the expression by

\cos \theta

converts

\sin \theta

terms into

\tan \theta

. The quotient becomes

\frac{3\tan \theta + 1}{\tan \theta + 2}

. Substituting

\tan \theta = 2

gives

\frac{3(2) + 1}{2 + 2} = \frac{7}{4} = 1.75

.

Step-by-Step Solution

1

Divide numerator and denominator by

\cos \theta

The expression becomes

\frac{3\left(\frac{\sin \theta}{\cos \theta}\right) + 1}{\frac{\sin \theta}{\cos \theta} + 2} = \frac{3\tan \theta + 1}{\tan \theta + 2}

.

Using the trigonometric identity

\tan \theta = \frac{\sin \theta}{\cos \theta}

simplifies the calculation without needing to evaluate the hypotenuse.

2

Substitute

\tan \theta = 2

\frac{3(2) + 1}{2 + 2} = \frac{7}{4}

Replace

\tan \theta

with its given numerical value of 2.

3

Convert fraction to decimal

1.75

Decimal representation of the exact fraction

\frac{7}{4}

.

Key Concept

Basic Trigonometric Ratios and Quotient Identity

Question 7274Question →

If $y = (x^2 - 2x + 2)^3$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 2$ ?

Show answer & explanation

Answer: 24

Answer

24

Applying the chain rule gives

\frac{dy}{dx} = 3(x^2 - 2x + 2)^2(2x - 2)

. Evaluating this derivative at

x = 2

yields

3(2)^2(2) = 24

.

Step-by-Step Solution

1

Identify the composite structure and state the chain rule formula.

Let

u = x^2 - 2x + 2

, so

y = u^3

. By the chain rule,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

The function is an expression raised to a power, requiring the chain rule.

2

Differentiate the outer function with respect to

u

and the inner function with respect to

x

.

\frac{dy}{du} = 3(x^2 - 2x + 2)^2 and \frac{du}{dx} = 2x - 2.

Apply the power rule to both the outer expression and inner polynomial.

3

Multiply the two derivatives to express

\frac{dy}{dx}

.

\frac{dy}{dx} = 3(x^2 - 2x + 2)^2(2x - 2).

Product of outer and inner derivatives per the chain rule.

4

Evaluate the expression at

x = 2

.

\frac{dy}{dx}\Big|_{x=2} = 3(4 - 4 + 2)^2(4 - 2) = 3(2)^2(2) = 24.

Substitute

x = 2

into the derivative expression and simplify.

Key Concept

Chain Rule of Differentiation

Question 7275Question →

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 7276Question →

The table below shows the cumulative frequency distribution of the masses (in grams) of $100$ cocoa beans sampled from an agricultural yield:

Mass Class Interval (g)	Cumulative Frequency
$10 - 19$	$10$
$20 - 29$	$30$
$30 - 39$	$65$
$40 - 49$	$90$
$50 - 59$	$100$

Using linear interpolation from the cumulative frequency distribution, calculate the 75th percentile ( $Q_3$ ) mass of the cocoa beans in grams.

Show answer & explanation

Answer: 43.5

Answer

The 75th percentile (

Q_3

) mass of the cocoa beans is

43.5\text{ g}

.

To find the 75th percentile (

Q_3

) from cumulative frequency data, calculate the rank

\frac{75}{100} \times 100 = 75

. This falls into the

40 - 49

class interval (boundaries

39.5 - 49.5

). Applying

Q_3 = L + \left(\frac{75 - F}{f}\right)c

yields

39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 43.5\text{ g}

.

Step-by-Step Solution

1

Calculate the percentile position rank

Rank position is

75

The 75th percentile corresponds to

75\%

of the total sample size

N = 100

, giving

\frac{75}{100} \times 100 = 75

.

2

Locate the 75th percentile class interval and its boundaries

Class interval is

40 - 49

, with lower boundary

L = 39.5

and upper boundary

49.5

Cumulative frequency before

40 - 49

is

65

, and up to

40 - 49

is

90

. Since

65 < 75 \leq 90

, the 75th item falls in this interval.

3

Identify class parameters for interpolation

L = 39.5

,

F = 65

,

f = 25

,

c = 10

Lower boundary

L = 39.5

, previous cumulative frequency

F = 65

, class frequency

f = 90 - 65 = 25

, class width

c = 49.5 - 39.5 = 10

.

4

Compute

Q_3

using the linear interpolation formula

Q_3 = 43.5\text{ g}

Q_3 = 39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 39.5 + 4 = 43.5

.

Key Concept

Linear Interpolation of Percentiles from Cumulative Frequency Data

Question 7277Question →

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 7278Question →

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

Question 7279Question →

A point $P(x, y)$ moves in a Cartesian plane such that the sum of the squares of its distances from two fixed points $A(0, 0)$ and $B(8, 0)$ is equal to $82$ , defining a locus $L_1$ . A second locus $L_2$ is the set of all points equidistant from the parallel lines $y = -1$ and $y = 7$ . Given that $L_1$ and $L_2$ intersect at two distinct points $M$ and $N$ , what is the length of the line segment $MN$ ?

Show answer & explanation

Answer: 8

Answer

The length of the line segment

MN

is 8 units.

The locus

L_1

simplifies to the circle

(x - 4)^2 + y^2 = 25

with center

(4,0)

and radius

5

. The locus

L_2

is the horizontal line

y = 3

. Substituting

y = 3

into the circle equation yields

(x - 4)^2 = 16

, giving intersection points at

(0,3)

and

(8,3)

. The distance between these points is

8

units.

Step-by-Step Solution

1

Formulate the algebraic expression for locus

L_1

The distance squared

PA^2 = x^2 + y^2

and

PB^2 = (x - 8)^2 + y^2

. Setting

PA^2 + PB^2 = 82

gives

x^2 + y^2 + (x - 8)^2 + y^2 = 82

.

By the distance formula, the sum of squared distances from

P(x, y)

to

A(0,0)

and

B(8,0)

equals 82.

2

Simplify and complete the square for locus

L_1

Expanding yields

2x^2 - 16x + 64 + 2y^2 = 82 \implies 2x^2 - 16x + 2y^2 = 18 \implies x^2 - 8x + y^2 = 9

. Completing the square gives

(x - 4)^2 + y^2 = 25

.

This shows

L_1

is a circle with center

(4, 0)

and radius

R = 5

.

3

Find the equation for locus

L_2

The locus of points equidistant from two parallel lines

y = -1

and

y = 7

is the line

y = \frac{-1 + 7}{2} = 3

.

The locus equidistant from two parallel lines is a parallel line running midway between them.

4

Find the intersection points

M

and

N

Substitute

y = 3

into

(x - 4)^2 + y^2 = 25

:

(x - 4)^2 + 3^2 = 25 \implies (x - 4)^2 = 16 \implies x - 4 = \pm 4

, so

x = 8

or

x = 0

. Thus,

M = (0, 3)

and

N = (8, 3)

.

Solving the circle and line equations simultaneously gives the

x

-coordinates of the intersection points.

5

Compute the distance between

M

and

N

Since both points share the ordinate

y = 3

, distance

MN = 8 - 0 = 8

.

The length of a horizontal segment is the difference between its endpoints'

x

-coordinates.

Key Concept

Intersection of loci involving circles and parallel lines

Question 7280Question →

A point $P(x, y)$ moves such that it is equidistant from two parallel lines $3x - 4y + 11 = 0$ and $3x - 4y - 1 = 0$ , defining locus $L_1$ . A second locus $L_2$ consists of all points that are at a constant distance of $5$ units from the fixed point $(1, 7)$ . Calculate the distance between the two points of intersection of locus $L_1$ and locus $L_2$ .

Show answer & explanation

Answer: 6

Answer

The distance between the two points of intersection of locus L1 and locus L2 is 6 units.

Locus L1 is the line midway between the two given parallel lines, having the equation 3x - 4y + 5 = 0. Locus L2 is a circle centered at (1, 7) with radius 5. The perpendicular distance from the center (1, 7) to line L1 is 4 units. Applying Pythagoras' theorem to the right triangle formed by the radius, perpendicular distance, and half-chord gives a half-chord length of sqrt(5^2 - 4^2) = 3. Therefore, the distance between the two intersection points (the full chord length) is 2 * 3 = 6 units.

Step-by-Step Solution

1

Determine the equation of locus L1

3x - 4y + 5 = 0

The locus of points equidistant from two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is a line parallel to both and midway between them, given by ax + by + (c1 + c2)/2 = 0.

2

Identify the geometric representation of locus L2

A circle centered at (1, 7) with radius r = 5

The locus of points at a constant distance r from a fixed point (x0, y0) is a circle.

3

Calculate the perpendicular distance d from the circle's center (1, 7) to line L1

d = 4 units

Using the perpendicular distance formula d = |ax0 + by0 + c| / sqrt(a^2 + b^2), we get d = |3(1) - 4(7) + 5| / sqrt(3^2 + (-4)^2) = |-20| / 5 = 4.

4

Compute the total distance between the two intersection points

6 units

The line intersects the circle along a chord. Using Pythagoras' theorem, half the chord length is sqrt(r^2 - d^2) = sqrt(5^2 - 4^2) = 3. Multiplying by 2 gives the full distance of 6 units.

Key Concept

Intersection of Loci (Parallel Line Bisector and Circle)

Estimated Time:2m 30s

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