All practice questions

Question 7241Question →

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 7242Question →

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 7243Question →

If $y = \ln(\sec 3x + \tan 3x)$ , what is $\frac{dy}{dx}$ ?

\sec 3x

3\sec 3x

3\tan 3x

\frac{3}{\sec 3x + \tan 3x}

Show answer & explanation

Answer: $3\sec 3x$

Answer

3\sec 3x

Using the chain rule for

y = \ln u

where

u = \sec 3x + \tan 3x

, we find

u' = 3\sec 3x\tan 3x + 3\sec^2 3x = 3\sec 3x(\tan 3x + \sec 3x)

. Dividing

u'

by

u

cancels out

(\sec 3x + \tan 3x)

, leaving

3\sec 3x

.

Step-by-Step Solution

1

Apply the chain rule for logarithmic functions

\frac{d}{dx}[\ln(u)] = \frac{1}{u}\frac{du}{dx}

.

Set

u = \sec 3x + \tan 3x

, so

\frac{dy}{dx} = \frac{1}{\sec 3x + \tan 3x} \cdot \frac{d}{dx}(\sec 3x + \tan 3x)

.

The function is a composite function of the form

y = \ln(u(x))

.

2

Differentiate the inner function

u = \sec 3x + \tan 3x

using the chain rule.

\frac{du}{dx} = 3\sec 3x \tan 3x + 3\sec^2 3x.

The derivative of

\sec(ax)

is

a\sec(ax)\tan(ax)

and the derivative of

\tan(ax)

is

a\sec^2(ax)

.

3

Factor out common terms in the numerator and simplify the expression.

\frac{dy}{dx} = \frac{3\sec 3x(\tan 3x + \sec 3x)}{\sec 3x + \tan 3x} = 3\sec 3x.

The term

(\tan 3x + \sec 3x)

in the numerator cancels with the denominator

(\sec 3x + \tan 3x)

.

Key Concept

Differentiation of Logarithmic and Trigonometric Functions via the Chain Rule

Estimated Time:1m 30s

Question 7244Question →

If $y = e^{3x} \sin(2x)$ , what is $\frac{dy}{dx}$ ?

e^{3x}(3\sin 2x + 2\cos 2x)

e^{3x}(\sin 2x + \cos 2x)

e^{3x}(3\sin 2x - 2\cos 2x)

e^{3x}(2\sin 2x + 3\cos 2x)

Show answer & explanation

Answer: $e^{3x}(3\sin 2x + 2\cos 2x)$

Answer

e^{3x}(3\sin 2x + 2\cos 2x)

Using the product rule

\frac{dy}{dx} = u\frac{dv}{dx} + v\frac{du}{dx}

on

y = e^{3x}\sin(2x)

gives

e^{3x} \cdot 2\cos(2x) + \sin(2x) \cdot 3e^{3x}

. Factoring out

e^{3x}

results in

e^{3x}(3\sin 2x + 2\cos 2x)

.

Step-by-Step Solution

1

Identify the function components for the product rule

Let

u = e^{3x}

and

v = \sin(2x)

.

The function

y

is a product of two differentiable functions

u(x)

and

v(x)

.

2

Differentiate each component using the chain rule

\frac{du}{dx} = 3e^{3x}

and

\frac{dv}{dx} = 2\cos(2x)

.

Applying the chain rule gives

\frac{d}{dx}(e^{3x}) = 3e^{3x}

and

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

.

3

Apply the product rule formula

\frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx}

\frac{dy}{dx} = e^{3x}(2\cos 2x) + \sin(2x)(3e^{3x}) = e^{3x}(3\sin 2x + 2\cos 2x)

.

Combining the products and factoring out the common exponential factor

e^{3x}

yields the final derivative.

Key Concept

Differentiation of Transcendental Functions using Product Rule and Chain Rule

Estimated Time:1m 30s

Question 7245Question →

If $y = \frac{4x + 1}{x - 2}$ , calculate the numerical value of $\frac{dy}{dx}$ at $x = 3$ .

Show answer & explanation

Answer: -9

Answer

The numerical value of

\frac{dy}{dx}

at

x = 3

is

-9

.

Using the quotient rule

\frac{dy}{dx} = \frac{v\frac{du}{dx} - u\frac{dv}{dx}}{v^2}

for

y = \frac{4x + 1}{x - 2}

gives

\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{-9}{(x - 2)^2}

. Substituting

x = 3

produces

\frac{-9}{(3 - 2)^2} = -9

.

Step-by-Step Solution

1

Identify

u(x)

and

v(x)

for the quotient rule formula.

u = 4x + 1

and

v = x - 2

.

The given equation

y = \frac{4x + 1}{x - 2}

is a quotient of two functions of

x

.

2

Calculate the individual derivatives

\frac{du}{dx}

and

\frac{dv}{dx}

.

\frac{du}{dx} = 4

and

\frac{dv}{dx} = 1

.

These derivatives are required components of the quotient rule.

3

Substitute the expressions into the quotient rule formula

\frac{dy}{dx} = \frac{v u' - u v'}{v^2}

and simplify.

\frac{dy}{dx} = \frac{(x - 2)(4) - (4x + 1)(1)}{(x - 2)^2} = \frac{4x - 8 - 4x - 1}{(x - 2)^2} = \frac{-9}{(x - 2)^2}

.

Simplifying the numerator yields the general derivative function.

4

Evaluate the derivative at

x = 3

.

\frac{dy}{dx}\Big|_{x=3} = \frac{-9}{(3 - 2)^2} = \frac{-9}{1} = -9

.

Substituting

x = 3

gives the requested numerical value.

Key Concept

Quotient Rule of Differentiation

Question 7246Question →

Evaluate the limit $\lim_{x \to 3} \frac{x^2 - 5x + 6}{x - 3}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 1

Answer

The numerical value of the limit is 1.

Substituting x = 3 into the original expression produces the indeterminate form 0/0. Factoring the numerator gives (x - 3)(x - 2). Canceling the common factor (x - 3) yields x - 2. Substituting x = 3 into x - 2 gives 3 - 2 = 1.

Step-by-Step Solution

1

Check the expression using direct substitution at x = 3.

The substitution yields \frac{3^2 - 5(3) + 6}{3 - 3} = \frac{0}{0}, an indeterminate form.

Direct substitution results in 0/0, indicating that algebraic factorization is needed.

2

Factorize the quadratic polynomial in the numerator.

x^2 - 5x + 6 = (x - 3)(x - 2)

Finding factors of +6 that sum to -5 allows cancellation of the denominator term.

3

Cancel the factor (x - 3) and evaluate the remaining linear expression at x = 3.

\lim_{x \to 3} (x - 2) = 3 - 2 = 1

The factor causing zero in the denominator is removed, leaving a continuous function.

Key Concept

Evaluating indeterminate limits (0/0) by algebraic factorization

Question 7247Question →

Match each statistical data representation term on the left with its corresponding definition or mathematical property on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Class Boundary matches with the value separating adjacent non-overlapping class intervals; Sector Angle matches with the central angle in a pie chart calculated as

\frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ

; Frequency Density matches with the quotient of class frequency and class width; Ogive matches with a line graph produced by plotting cumulative frequencies against upper class boundaries.

Each data representation term directly corresponds to its core definition: class boundary closes gaps between discrete class intervals, sector angle measures central circle proportion in a pie chart, frequency density scales histogram height when class widths differ, and an ogive graphs cumulative frequency against upper boundaries.

Step-by-Step Solution

1

Define Class Boundary

Class boundary is the continuous point midway between adjacent class limits.

Class boundaries remove gaps in discrete grouped frequency distributions.

2

Define Sector Angle formula for a pie chart

Sector Angle

= \frac{\text{Frequency}}{\text{Total Frequency}} \times 360^\circ

.

The complete circle represents total frequency, so individual sectors scale proportionally with

360^\circ

.

3

Define Frequency Density for histograms

Frequency Density

= \frac{\text{Frequency}}{\text{Class Width}}

.

Histogram area equals frequency; when widths differ, height must represent frequency per unit width.

4

Define Ogive

An Ogive is a cumulative frequency curve plotted against upper boundaries.

Each point on an ogive shows the cumulative frequency up to that class's upper boundary.

Key Concept

Data Representation Terms and Formulas

Estimated Time:1m 30s

Question 7248Question →

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 7249Question →

A rhombus has diagonals of lengths $12\text{ cm}$ and $16\text{ cm}$ . What is the perimeter of the rhombus in $\text{cm}$ ?

Show answer & explanation

Answer: 40

Answer

The perimeter of the rhombus is

40\text{ cm}

.

The diagonals of a rhombus bisect each other at right angles, dividing the rhombus into four congruent right-angled triangles. Each triangle has legs measuring

6\text{ cm}

and

8\text{ cm}

. Applying the Pythagorean theorem, the hypotenuse (which is the side length of the rhombus) is

\sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}

. Since all four sides of a rhombus are equal, the perimeter is

4 \times 10 = 40\text{ cm}

.

Step-by-Step Solution

1

Calculate the lengths of the semi-diagonals

The semi-diagonals are

6\text{ cm}

and

8\text{ cm}

.

The diagonals of a rhombus bisect each other perpendicularly.

2

Determine the side length of the rhombus using the Pythagorean theorem

Side length

s = \sqrt{6^2 + 8^2} = \sqrt{100} = 10\text{ cm}

.

Each side of the rhombus forms the hypotenuse of a right-angled triangle formed by the semi-diagonals.

3

Calculate the total perimeter

Perimeter

P = 4 \times 10 = 40\text{ cm}

.

All four sides of a rhombus are equal in length.

Key Concept

Perimeter of a rhombus derived from diagonal lengths using right-triangle properties

Estimated Time:1m 0s

Question 7250Question →

A piecewise function

f(x)

is defined by

f(x) = \begin{cases} \frac{\sin(3x) + \tan(5x)}{2x}, & x \neq 0 \\ a^2 - 5, & x = 0 \end{cases}

If

f(x)

is continuous at

x = 0

, where

a > 0

, determine the numerical value of

a

.

Show answer & explanation

Answer: 3

Answer

The numerical value of

a

is 3.

For the function to be continuous at

x = 0

, the limit as

x \to 0

must equal the function value

f(0)

. Splitting the trigonometric limit gives

\frac{3}{2} + \frac{5}{2} = 4

. Setting

a^2 - 5 = 4

yields

a^2 = 9

. Because

a > 0

, taking the positive square root gives

a = 3

.

Step-by-Step Solution

1

Evaluate the limit of the trigonometric expression as

x

approaches 0.

\lim_{x \to 0} \frac{\sin(3x) + \tan(5x)}{2x} = 4

Using standard trigonometric limit principles: \lim_{x \to 0} \frac{\sin(kx)}{x} = k and \lim_{x \to 0} \frac{\tan(kx)}{x} = k.

2

Equate the limit value to

f(0)

to ensure continuity at

x = 0

.

a^2 - 5 = 4

By definition, a function is continuous at x = c if and only if \lim_{x \to c} f(x) = f(c).

3

Solve the resulting quadratic equation for the positive constant

a

.

a = 3

Adding 5 to both sides gives a^2 = 9; taking the principal square root gives a = 3 since a > 0.

Key Concept

Limits and Continuity of Functions

Question 7251Question →

A binary operation $\Delta$ defined on the set of integers modulo $11$ is given by $a \Delta b \equiv (2a^2 - 3ab + b^2) \pmod{11}$ . Find the smallest non-negative integer $x$ that satisfies the equation $4 \Delta x \equiv 5 \pmod{11}$ .

Show answer & explanation

Answer: 3

Answer

The smallest non-negative integer

x

that satisfies

4 \Delta x \equiv 5 \pmod{11}

is 3.

Evaluating

4 \Delta x

gives

32 - 12x + x^2 \equiv x^2 - x + 10 \pmod{11}

. Setting this congruent to

5 \pmod{11}

yields

x^2 - x + 5 \equiv 0 \pmod{11}

, which converts to

x^2 - x - 6 \equiv 0 \pmod{11}

. Factoring gives

(x - 3)(x + 2) \equiv 0 \pmod{11}

, which yields solutions

x \equiv 3 \pmod{11}

and

x \equiv 9 \pmod{11}

. The smallest non-negative integer among these solutions is

3

.

Step-by-Step Solution

1

Substitute the given value

a = 4

into the operation definition.

4 \Delta x = 2(4)^2 - 3(4)x + x^2 = 32 - 12x + x^2

This establishes the explicit algebraic polynomial in terms of

x

.

2

Reduce coefficients modulo

11

.

32 \equiv 10 \pmod{11}

and

-12x \equiv -x \pmod{11}

, giving

x^2 - x + 10 \pmod{11}

.

Simplifying coefficients reduces computational complexity during equation solving.

3

Form the modular quadratic equation and set it to zero.

x^2 - x + 10 \equiv 5 \pmod{11} \implies x^2 - x + 5 \equiv 0 \pmod{11} \implies x^2 - x - 6 \equiv 0 \pmod{11}

.

Expressing

5 \equiv -6 \pmod{11}

allows standard integer factorisation.

4

Factor the quadratic polynomial and solve for

x

.

(x - 3)(x + 2) \equiv 0 \pmod{11} \implies x \equiv 3 \text{ or } x \equiv -2 \equiv 9 \pmod{11}

.

Since

11

is prime, a product congruent to

0 \pmod{11}

implies at least one factor is congruent to

0 \pmod{11}

.

5

Select the smallest non-negative integer from the valid solution set

\{3, 9\}

.

x = 3

3

is non-negative and strictly smaller than

9

.

Key Concept

Modular Arithmetic Binary Operations and Quadratic Congruences

Estimated Time:3m 0s

Question 7252Question →

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 7253Question →

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 7254Question →

If $y = e^{\sin 2x} \ln(\tan x)$ , what is the value of $\frac{dy}{dx}$ at $x = \frac{\pi}{4}$ ?

2e

e

-2e

0

Show answer & explanation

Answer: $2e$

Answer

The derivative evaluated at

x = \frac{\pi}{4}

is

2e

.

Applying the product rule

\frac{dy}{dx} = u'v + uv'

to

y = e^{\sin 2x} \ln(\tan x)

gives

u' = 2\cos(2x)e^{\sin 2x}

and

v' = 2\csc(2x)

. At

x = \frac{\pi}{4}

,

\cos(2x) = \cos(\pi/2) = 0

, which vanishes the first term. The second term evaluates to

e^{\sin(\pi/2)} \cdot 2\csc(\pi/2) = e^1 \cdot 2(1) = 2e

.

Step-by-Step Solution

1

Identify the product rule structure for

y = u \cdot v

, where

u = e^{\sin 2x}

and

v = \ln(\tan x)

.

The product rule states

\frac{dy}{dx} = u'v + uv'

.

The given function is a product of an exponential function and a logarithmic function.

2

Differentiate

u = e^{\sin 2x}

using the chain rule.

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

.

The inner function is

\sin 2x

, whose derivative is

2\cos 2x

.

3

Differentiate

v = \ln(\tan x)

using the chain rule.

\frac{dv}{dx} = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = 2\csc(2x)

.

Simplifying

\frac{\sec^2 x}{\tan x}

gives

\frac{2}{\sin 2x} = 2\csc(2x)

.

4

Combine terms into the product rule expression.

\frac{dy}{dx} = [2\cos(2x) e^{\sin 2x} \ln(\tan x)] + [e^{\sin 2x} \cdot 2\csc(2x)]

.

Substitute

u, u', v, v'

into

\frac{dy}{dx} = u'v + uv'

.

5

Evaluate the derivative at

x = \frac{\pi}{4}

.

At

x = \frac{\pi}{4}

,

2x = \frac{\pi}{2}

. Thus,

\cos(\pi/2) = 0

,

\tan(\pi/4) = 1 \implies \ln(1) = 0

,

\sin(\pi/2) = 1

, and

\csc(\pi/2) = 1

. The first term becomes

0

, and the second term becomes

e^{1} \cdot 2(1) = 2e

. Thus,

\frac{dy}{dx} = 2e

.

Substitute special angle values into the expression to compute the numerical result.

Key Concept

Product Rule and Chain Rule for Transcendental Functions

Question 7255Question →

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 7256Question →

Line $AB$ is parallel to line $CD$ . A transversal line $EF$ intersects line $AB$ at point $P$ and line $CD$ at point $Q$ . If $\angle APQ = (4x - 10)^\circ$ and $\angle PQD = (2x + 30)^\circ$ are alternate interior angles, what is the measure of $\angle BPQ$ ?

110^\circ

70^\circ

20^\circ

100^\circ

Show answer & explanation

Answer: $110^\circ$

Answer

110^\circ

Since alternate interior angles are equal,

4x - 10 = 2x + 30

, yielding

x = 20

. Thus,

\angle APQ = 70^\circ

. Because

\angle APQ

and

\angle BPQ

are adjacent angles on straight line

AB

, their sum is

180^\circ

, giving

\angle BPQ = 110^\circ

.

Step-by-Step Solution

1

Set up the equation for alternate interior angles.

4x - 10 = 2x + 30

Alternate interior angles formed by a transversal cutting parallel lines are equal.

2

Solve the linear equation for

x

.

2x = 40 \implies x = 20

Subtract

2x

and add

10

to both sides.

3

Calculate the measure of

\angle APQ

.

\angle APQ = 4(20) - 10 = 70^\circ$

Substitute

x = 20

into the expression

(4x - 10)^\circ

.

4

Determine

\angle BPQ

using the straight line angle property.

\angle BPQ = 180^\circ - 70^\circ = 110^\circ$

Angles

\angle APQ

and

\angle BPQ

form a linear pair on straight line

AB

, summing to

180^\circ

.

Key Concept

Alternate interior angles of parallel lines and angles on a straight line

Question 7257Question →

When evaluating the derivative of the cubic function $f(x) = 2x^3 - 5x$ from first principles, which of the following expressions represents the fully simplified difference quotient $\frac{f(x+h) - f(x)}{h}$ prior to taking the limit as $h \to 0$ ?

6x^2 + 6xh + 2h^2 - 5

6x^2 + 6xh + 2h^2 + 5

2x^2 + 2xh + 2h^2 - 5

6x^2 - 5

Show answer & explanation

Answer: $6x^2 + 6xh + 2h^2 - 5$

Answer

The simplified difference quotient before evaluating the limit is

6x^2 + 6xh + 2h^2 - 5

.

Expanding

f(x+h) = 2(x+h)^3 - 5(x+h)

using the binomial expansion yields

2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h

. Subtracting

f(x) = 2x^3 - 5x

leaves

6x^2h + 6xh^2 + 2h^3 - 5h

. Dividing each term by

h

gives the correct simplified difference quotient

6x^2 + 6xh + 2h^2 - 5

.

Step-by-Step Solution

1

Express

f(x+h)

for the function

f(x) = 2x^3 - 5x

f(x+h) = 2(x+h)^3 - 5(x+h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) - 5x - 5h = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h

Substitute

(x+h)

into

f(x)

and expand the binomial terms completely.

2

Subtract

f(x)

from

f(x+h)

to find the numerator

f(x+h) - f(x)

f(x+h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 - 5x - 5h) - (2x^3 - 5x) = 6x^2h + 6xh^2 + 2h^3 - 5h

Cancel out the common terms

2x^3

and

-5x

.

3

Divide the numerator by

h

to form the difference quotient

\frac{6x^2h + 6xh^2 + 2h^3 - 5h}{h} = 6x^2 + 6xh + 2h^2 - 5

Factor out

h

from each term in the numerator and divide by

h

.

Key Concept

Differentiation from First Principles

Question 7258Question →

A rectangular garden measuring $12\text{ m}$ by $8\text{ m}$ has a paved path of uniform width $x\text{ m}$ constructed inside it along its perimeter. If the area of the remaining inner garden is $60\text{ m}^2$ , what is the width $x$ of the path?

1\text{ m}

2\text{ m}

1.5\text{ m}

3\text{ m}

Show answer & explanation

Answer: $1\text{ m}$

Answer

The width of the path is

1\text{ m}

.

The total outer garden area is

12 \times 8 = 96\text{ m}^2

. Subtracting the path width

x

from both ends gives inner dimensions

(12 - 2x)

and

(8 - 2x)

. Equating the inner area

(12 - 2x)(8 - 2x) = 60

yields

x^2 - 10x + 9 = 0

. The realistic physical solution is

x = 1\text{ m}

.

Step-by-Step Solution

1

Express the inner dimensions in terms of path width

x

Length =

(12 - 2x)\text{ m}

, Width =

(8 - 2x)\text{ m}

The path reduces each side dimension by

x

at both ends.

2

Set up the area equation for the inner rectangular garden

(12 - 2x)(8 - 2x) = 60

The area of a rectangle is length multiplied by width.

3

Expand and simplify the quadratic equation

96 - 40x + 4x^2 = 60 \implies 4x^2 - 40x + 36 = 0 \implies x^2 - 10x + 9 = 0

Divide the whole equation by 4 to simplify quadratic terms.

4

Solve for

x

by factoring

(x - 1)(x - 9) = 0 \implies x = 1

or

x = 9

Since the width

x

cannot exceed half of the smaller side (

x < 4\text{ m}

),

x = 9

is extraneous.

Key Concept

Perimeter and Area of Rectangles with Uniform Borders

Estimated Time:1m 30s

Question 7259Question →

What is the exact simplified value of $\sin 60^\circ + \cos 45^\circ$ ?

\frac{\sqrt{3} + \sqrt{2}}{2}

\frac{\sqrt{5}}{2}

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

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

Show answer & explanation

Answer: $\frac{\sqrt{3} + \sqrt{2}}{2}$

Answer

\frac{\sqrt{3} + \sqrt{2}}{2}

Substituting the exact values

\sin 60^\circ = \frac{\sqrt{3}}{2}

and

\cos 45^\circ = \frac{\sqrt{2}}{2}

gives

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

.

Step-by-Step Solution

1

Identify the exact trigonometric values for the special angles.

\sin 60^\circ = \frac{\sqrt{3}}{2}

and

\cos 45^\circ = \frac{\sqrt{2}}{2}

Standard special angle values in trigonometry.

2

Substitute the values into the given expression.

\frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2}

Direct substitution of known ratio values.

3

Combine the fractions over the common denominator of

2

.

\frac{\sqrt{3} + \sqrt{2}}{2}

Adding numerators over a shared common denominator.

Key Concept

Special Angles and Exact Trigonometric Values

Question 7260Question →

A normal line is drawn to the curve $y = \frac{x^2 - 5}{x - 3}$ at the point where $x = 2$ . What is the $x$ -intercept of this normal line?

Show answer & explanation

Answer: -1

Answer

The

x

-intercept of the normal line is

-1

.

Substituting

x = 2

into the curve yields the point

(2, 1)

. Differentiation via the quotient rule gives

\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}

. At

x = 2

, the tangent slope is

-3

, making the normal slope

\frac{1}{3}

. The line equation

y - 1 = \frac{1}{3}(x - 2)

simplifies to

x - 3y + 1 = 0

. Setting

y = 0

gives

x = -1

.

Step-by-Step Solution

1

Calculate the y-coordinate of the point of tangency

y = 1

Substitute

x = 2

into the curve equation

y = \frac{x^2 - 5}{x - 3}

to get the point

(2, 1)

.

2

Differentiate the function using the quotient rule

\frac{dy}{dx} = \frac{x^2 - 6x + 5}{(x - 3)^2}

Applying

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

to

u = x^2 - 5

and

v = x - 3

.

3

Find the tangent gradient at

x = 2

m_t = -3

Evaluate

\frac{dy}{dx}

at

x = 2

to obtain

m_t = \frac{2^2 - 6(2) + 5}{(2-3)^2} = -3

.

4

Calculate the gradient of the normal line

m_n = \frac{1}{3}

The normal line is perpendicular to the tangent line, so

m_n = -\frac{1}{m_t} = \frac{1}{3}

.

5

Derive the line equation for the normal

x - 3y + 1 = 0

Use point-slope form

y - y_1 = m_n(x - x_1)

with

(2, 1)

and

m_n = \frac{1}{3}

.

6

Find the x-intercept

x = -1

Set

y = 0

in the normal equation

x - 3(0) + 1 = 0

, giving

x = -1

.

Key Concept

Equation of normal line to a rational curve and finding its axis intercepts

Answer

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Common Mistakes

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