All practice questions

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 602Question →

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 603Question →

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 604Question →

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 605Question →

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 606Question →

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 607Question →

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 608Question →

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

Question 609Question →

An environmental research station recorded the daily particulate matter concentration (in $\mu\text{g/m}^3$ ) near an urban center over a period of $50$ days. The observations are summarized in the table below:

Particulate Matter ( $\mu\text{g/m}^3$ )	Number of Days ( $f$ )
$20 - 29$	$6$
$30 - 39$	$10$
$40 - 49$	$15$
$50 - 59$	$11$
$60 - 69$	$8$

Find the estimated mean particulate matter concentration, in $\mu\text{g/m}^3$ , for the 50-day period.

Show answer & explanation

Answer: 45.5

Answer

The estimated mean particulate matter concentration is

45.5\text{ }\mu\text{g/m}^3

.

To calculate the mean of grouped data, determine the midpoint (

x

) of each class interval:

24.5

,

34.5

,

44.5

,

54.5

, and

64.5

. Multiply each midpoint by its frequency (

f

) to get the products

147

,

345

,

667.5

,

599.5

, and

516

. The sum of these products (

\sum fx

) is

2275

. Dividing

\sum fx

by the total frequency (

\sum f = 50

) yields the estimated mean of

45.5\text{ }\mu\text{g/m}^3

.

Step-by-Step Solution

1

Calculate the class midpoints (

x

) for each class interval

Midpoints are

24.5

,

34.5

,

44.5

,

54.5

, and

64.5

.

Grouped data uses the midpoint of each interval to represent all values falling within that interval.

2

Compute the product of frequency and midpoint (

fx

) for each class

6 \times 24.5 = 147

,

10 \times 34.5 = 345

,

15 \times 44.5 = 667.5

,

11 \times 54.5 = 599.5

,

8 \times 64.5 = 516

.

This determines the estimated sum of data values within each group.

3

Sum all products

\sum fx

and total frequency

\sum f

\sum fx = 147 + 345 + 667.5 + 599.5 + 516 = 2275

and

\sum f = 50

.

These totals are required for the mean formula.

4

Calculate the mean concentration using

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

\bar{x} = \frac{2275}{50} = 45.5

.

Dividing the total estimated sum by the total number of days gives the estimated mean.

Key Concept

Calculation of Mean for Grouped Frequency Data

Question 610Question →

Given that

x

and

y

are real numbers satisfying the simultaneous exponential equations

3^x \times 9^y = 81

and

8^x \times 4^y = 256

find the value of

x^2 + y^2

.

Show answer & explanation

Answer: 5

Answer

The value of

x^2 + y^2

is 5.

Converting all terms to their prime bases (

3

for the first equation and

2

for the second equation) produces the simultaneous linear equations

x + 2y = 4

and

3x + 2y = 8

. Solving these gives

x = 2

and

y = 1

. Substituting these values into

x^2 + y^2

gives

2^2 + 1^2 = 5

.

Step-by-Step Solution

1

Convert all terms in the first equation to powers of base 3.

x + 2y = 4

Since

9 = 3^2

and

81 = 3^4

, applying the product law of indices

a^m \times a^n = a^{m+n}

gives

3^{x+2y} = 3^4

. Equating the exponents gives

x + 2y = 4

.

2

Convert all terms in the second equation to powers of base 2.

3x + 2y = 8

Since

8 = 2^3

,

4 = 2^2

, and

256 = 2^8

, applying the laws of indices yields

2^{3x} \times 2^{2y} = 2^8 \implies 2^{3x+2y} = 2^8

. Equating exponents gives

3x + 2y = 8

.

3

Solve the system of simultaneous linear equations for

x

and

y

.

x = 2

and

y = 1

Subtracting

x + 2y = 4

from

3x + 2y = 8

yields

2x = 4 \implies x = 2

. Substituting

x = 2

into

x + 2y = 4

gives

2 + 2y = 4 \implies y = 1

.

4

Evaluate the target expression

x^2 + y^2

.

5

Substitute

x = 2

and

y = 1

into

x^2 + y^2

to obtain

2^2 + 1^2 = 4 + 1 = 5

.

Key Concept

Converting exponential terms to common prime bases to reduce exponential equations into linear equations.

Question 611Question →

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

Show answer & explanation

Answer: 24

Answer

The numerical value of the derivative

\frac{dy}{dx}

at

x = 1

is 24.

Evaluating the derivative requires combined application of the quotient rule and chain rule. Differentiating

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

via the chain rule yields

u'(x) = 12x(2x^2 + 1)^2

. At

x = 1

,

u(1) = 27

,

u'(1) = 108

,

v(1) = 3

, and

v'(1) = 4

. Substituting these into the quotient rule

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

produces

\frac{324 - 108}{9} = 24

.

Step-by-Step Solution

1

Decompose the function for the Quotient Rule

Let

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

and

v(x) = 4x - 1

.

The given expression is a quotient of two differentiable functions.

2

Differentiate the numerator using the Chain Rule

u'(x) = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2

The chain rule states that

\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)

.

3

Differentiate the denominator and evaluate all components at

x = 1

v'(1) = 4

,

v(1) = 3

,

u(1) = 27

, and

u'(1) = 108

Evaluating each term individually reduces algebraic complexity prior to applying the quotient rule.

4

Apply the Quotient Rule formula and evaluate

\frac{dy}{dx}\Big|_{x=1} = \frac{(108)(3) - (27)(4)}{3^2} = \frac{324 - 108}{9} = 24

The quotient rule formula is

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

.

Key Concept

Combination of Quotient Rule and Chain Rule

Estimated Time:1m 30s

Question 612Question →

Using differentiation from first principles, evaluate the value of the derivative of the function $f(x) = 2x^2 + 3x - 1$ at the point where $x = 1$ .

Show answer & explanation

Answer: 7

Answer

The value of the derivative at

x = 1

is 7.

Using first principles, the derivative is defined as

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

. Evaluating at

x = 1

, we find

f(1) = 4

and

f(1+h) = 4 + 7h + 2h^2

. Subtracting

f(1)

leaves

7h + 2h^2

, and dividing by

h

yields

7 + 2h

. Taking the limit as

h \to 0

yields the final result of 7.

Step-by-Step Solution

1

Calculate the value of the function at

x = 1

f(1) = 2(1)^2 + 3(1) - 1 = 4

This establishes the base value needed for the difference quotient.

2

Expand and simplify

f(1+h)

f(1+h) = 2(1+h)^2 + 3(1+h) - 1 = 4 + 7h + 2h^2

This gives the value of the function at the incremented point

x + h

.

3

Form and simplify the difference quotient

\frac{f(1+h) - f(1)}{h}

\frac{4 + 7h + 2h^2 - 4}{h} = 7 + 2h

Dividing by

h

eliminates the indeterminate form before taking the limit.

4

Evaluate the limit as

h \to 0

f'(1) = \lim_{h \to 0} (7 + 2h) = 7

Taking

h = 0

in the simplified quotient gives the exact rate of change at

x = 1

.

Key Concept

Differentiation from First Principles

Estimated Time:1m 0s

Question 613Question →

How many positive integer values of $x$ satisfy the quadratic inequality $x^2 - 4x - 5 < 0$ ?

Show answer & explanation

Answer: 4

Answer

There are 4 positive integer values of

x

that satisfy the inequality.

Factoring

x^2 - 4x - 5 < 0

yields

(x - 5)(x + 1) < 0

, giving the real solution interval

-1 < x < 5

. Filtering for positive integers (

x \in \mathbb{Z}^+

) yields the set

\{1, 2, 3, 4\}

, which contains exactly 4 values.

Step-by-Step Solution

1

Factor the quadratic inequality

(x - 5)(x + 1) < 0

Factoring helps find the boundary roots of the quadratic equation.

2

Determine the solution set interval

-1 < x < 5

For a quadratic inequality of the form

(x - a)(x - b) < 0

with

a < b

, the solution interval is

a < x < b

.

3

List the positive integer solutions in the range

-1 < x < 5

x \in \{1, 2, 3, 4\}

Positive integers are whole numbers strictly greater than 0.

4

Count the number of positive integers

4

Counting the elements in the set

\{1, 2, 3, 4\}

gives a total of 4.

Key Concept

Solving quadratic inequalities and finding valid integer solutions within a target domain.

Estimated Time:45s

Question 614Question →

If $y = (3x - 1)^4$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

Show answer & explanation

Answer: 96

Answer

96

Applying the chain rule

\frac{d}{dx}[u^n] = n u^{n-1} \cdot \frac{du}{dx}

to

y = (3x - 1)^4

gives

\frac{dy}{dx} = 4(3x - 1)^3 \cdot 3 = 12(3x - 1)^3

. Evaluating this expression at

x = 1

yields

12(3(1) - 1)^3 = 12(2)^3 = 96

.

Step-by-Step Solution

1

Differentiate

y = (3x - 1)^4

with respect to

x

using the chain rule.

\frac{dy}{dx} = 12(3x - 1)^3

Differentiate the outer power function to get

4(3x - 1)^3

and multiply by the derivative of the inner linear function

3x - 1

, which is 3.

2

Substitute

x = 1

into the derivative function.

96

Evaluating

12(3(1) - 1)^3

gives

12(2)^3 = 12 \times 8 = 96

.

Key Concept

Chain Rule of Differentiation

Question 615Question →

If $27^{x - 1} = 9^{x + 1}$ , determine the value of $x$ .

Show answer & explanation

Answer: 5

Answer

The value of

x

is 5.

Rewriting 27 as

3^3

and 9 as

3^2

transforms the given equation into

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

. Equating exponents gives

3x - 3 = 2x + 2

, which simplifies directly to

x = 5

.

Step-by-Step Solution

1

Express numbers in terms of a common base

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

Both 27 and 9 are powers of 3, allowing reduction to a single base.

2

Apply power of a power index law

3^{3x - 3} = 3^{2x + 2}

Multiply the base power by the expression in the exponent:

3 \times (x - 1) = 3x - 3

and

2 \times (x + 1) = 2x + 2

.

3

Equate exponents and solve for

x

3x - 3 = 2x + 2 \implies x = 5

Equal bases imply that the index powers must be equal.

Key Concept

Equating exponential expressions using a common base

Question 616Question →

Find the $x$ -intercept of the normal line to the curve $y = x^3 - 3x^2 + 4x - 1$ at the point where $x = 1$ .

Show answer & explanation

Answer: 2

Answer

The

x

-intercept of the normal line is

2

.

At

x = 1

, the point on the curve

y = x^3 - 3x^2 + 4x - 1

is

(1, 1)

. The derivative is

\frac{dy}{dx} = 3x^2 - 6x + 4

, which equals

1

at

x = 1

. The normal gradient is therefore

-1

. The normal line equation is

y - 1 = -1(x - 1)

, which simplifies to

y = -x + 2

. Setting

y = 0

gives

x = 2

.

Step-by-Step Solution

1

Find the y-coordinate at x = 1

y = 1, so the point on the curve is (1, 1)

The point of contact is needed to construct the equation of the normal line.

2

Differentiate the function with respect to x

dy/dx = 3x^2 - 6x + 4

The derivative gives the expression for the gradient of the tangent to the curve.

3

Find the gradient of the tangent and normal at x = 1

Gradient of tangent m_t = 1; gradient of normal m_n = -1

The normal line is perpendicular to the tangent line, so m_n = -1 / m_t.

4

Determine the equation of the normal line

y - 1 = -1(x - 1) => y = -x + 2

Applying the straight-line equation y - y_1 = m(x - x_1).

5

Find the x-intercept of the normal line

x = 2

Setting y = 0 in the normal line equation yields the x-intercept.

Key Concept

Equation and axis intercepts of a normal line to a curve

Question 617Question →

A solid right circular cylinder has a height of $10\text{ cm}$ and a total surface area of $192\pi\text{ cm}^2$ . What is the volume of the cylinder in $\text{cm}^3$ expressed in terms of $\pi$ ?

Show answer & explanation

Answer: 360

Answer

360

The total surface area of a closed cylinder is

A = 2\pi r^2 + 2\pi r h

. Given

A = 192\pi

and

h = 10\text{ cm}

, we have

2\pi r^2 + 20\pi r = 192\pi

. Dividing by

2\pi

yields

r^2 + 10r - 96 = 0

. Factoring the quadratic gives

(r + 16)(r - 6) = 0

, so

r = 6\text{ cm}

(since radius must be positive). Substituting

r = 6\text{ cm}

and

h = 10\text{ cm}

into the volume formula

V = \pi r^2 h

yields

V = \pi (6^2)(10) = 360\pi\text{ cm}^3

. The numerical coefficient of

\pi

is

360

.

Step-by-Step Solution

1

Set up the equation for the total surface area of the cylinder

2\pi r^2 + 2\pi r(10) = 192\pi

The total surface area of a solid cylinder comprises the areas of the top and bottom circular bases (

2\pi r^2

) plus the curved surface area (

2\pi r h

).

2

Simplify and solve the resulting quadratic equation for radius

r

r^2 + 10r - 96 = 0 \implies (r + 16)(r - 6) = 0 \implies r = 6\text{ cm}

Dividing the equation by

2\pi

yields

r^2 + 10r - 96 = 0

. Since radius must be positive,

r = 6\text{ cm}

.

3

Calculate the volume of the cylinder using the radius and height

V = \pi r^2 h = \pi (6)^2 (10) = 360\pi\text{ cm}^3

The volume of a cylinder is calculated using the formula

V = \pi r^2 h

.

Key Concept

Surface Area and Volume of 3D Solids

Question 618Question →

A piece of wire of length $88\text{ cm}$ is bent to form the perimeter of a sector of a circle of radius $21\text{ cm}$ . What is the area of the sector in $\text{cm}^2$ ?

Show answer & explanation

Answer: 483

Answer

The area of the sector is

483\text{ cm}^2

.

The total length of the wire forms the perimeter of the sector, which consists of two radii and the arc length (

P = 2r + l

). With

P = 88\text{ cm}

and

r = 21\text{ cm}

, the arc length is

l = 88 - 2(21) = 46\text{ cm}

. Using the sector area formula

A = \frac{1}{2} r l

, the area is

\frac{1}{2} \times 21 \times 46 = 483\text{ cm}^2

.

Step-by-Step Solution

1

Determine the arc length of the sector from the given perimeter

l = 46\text{ cm}

The total perimeter of a sector includes two straight radii and the curved arc length:

P = 2r + l

. Subtracting twice the radius (

2 \times 21 = 42\text{ cm}

) from the total wire length of

88\text{ cm}

gives an arc length of

46\text{ cm}

.

2

Calculate the area of the sector

A = 483\text{ cm}^2

The area of a sector can be computed using the formula

A = \frac{1}{2} r l

. Substituting

r = 21\text{ cm}

and

l = 46\text{ cm}

yields

A = \frac{1}{2} \times 21 \times 46 = 483\text{ cm}^2

.

Key Concept

Perimeter and Area of a Sector of a Circle

Question 619Question →

Evaluate the limit:

\lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x + 2} - 2}

What is the numerical value of this limit?

Show answer & explanation

Answer: 48

Answer

The numerical value of the limit is 48.

Evaluating the limit of

\frac{x^3 - 8}{\sqrt{x + 2} - 2}

as

x \to 2

gives an indeterminate form

\frac{0}{0}

. Factorizing the numerator gives

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

, and rationalizing the denominator by multiplying numerator and denominator by

(\sqrt{x + 2} + 2)

converts the denominator to

x - 2

. Canceling

(x - 2)

leaves

(x^2 + 2x + 4)(\sqrt{x + 2} + 2)

. Evaluating at

x = 2

gives

(4 + 4 + 4)(\sqrt{4} + 2) = 12 \times 4 = 48

.

Step-by-Step Solution

1

Identify the limit form via direct substitution

Substituting

x = 2

yields

\frac{0}{0}

.

Direct evaluation results in an indeterminate form, requiring algebraic manipulation to eliminate the zero factor.

2

Factorize the numerator using the difference of cubes formula

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Exposing the factor

(x - 2)

is essential to resolving the zero denominator.

3

Rationalize the denominator using its algebraic conjugate

Multiply top and bottom by

(\sqrt{x + 2} + 2)

to get denominator

(x + 2) - 4 = x - 2

.

Applying

(a-b)(a+b) = a^2 - b^2

eliminates the square root from the denominator.

4

Cancel the common factor and compute the final value

\lim_{x \to 2} (x^2 + 2x + 4)(\sqrt{x + 2} + 2) = (12)(4) = 48.

With

(x - 2)

cancelled for

x \neq 2

, direct substitution now yields a defined real number.

Key Concept

Limits of Indeterminate Forms using Difference of Cubes and Surd Rationalization

Estimated Time:2m 30s

Question 620Question →

If $y = 3e^{2x} + \ln(\cos x)$ , calculate the value of $\frac{dy}{dx}$ at $x = 0$ .

Show answer & explanation

Answer: 6

Answer

The value of the derivative of

y = 3e^{2x} + \ln(\cos x)

at

x = 0

is 6.

Differentiating each transcendental term individually using the chain rule yields

\frac{dy}{dx} = 6e^{2x} - \tan x

. Substituting

x = 0

gives

6e^0 - \tan 0 = 6 - 0 = 6

.

Step-by-Step Solution

1

Differentiate the exponential term

3e^{2x}

6e^{2x}

Applying the derivative rule for exponential functions

\frac{d}{dx}[a e^{kx}] = a k e^{kx}

.

2

Differentiate the logarithmic term

\ln(\cos x)

-\tan x

Applying the chain rule

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

where

u = \cos x

gives

\frac{-\sin x}{\cos x} = -\tan x

.

3

Combine terms and evaluate at

x = 0

6

Substituting

x = 0

into

\frac{dy}{dx} = 6e^{2x} - \tan x

yields

6e^0 - \tan 0 = 6(1) - 0 = 6

.

Key Concept

Differentiation of Trigonometric, Exponential, and Logarithmic Functions

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