All practice questions

Question 801Question →

A trigonometric function is defined as $f(x) = a \sin(b x) + c$ , where $a > 0$ and $b > 0$ . The graph of $y = f(x)$ has a maximum value of $8$ , a minimum value of $-2$ , and a period of $120^\circ$ . What is the value of $a + b + c$ ?

Show answer & explanation

Answer: 11

Answer

The value of

a + b + c

is 11.

Using the maximum (

a + c = 8

) and minimum (

-a + c = -2

), solving simultaneously gives

a = 5

and

c = 3

. Using the period formula

T = \frac{360^\circ}{b} = 120^\circ

, we obtain

b = 3

. Adding these values together yields

5 + 3 + 3 = 11

.

Step-by-Step Solution

1

Set up a system of linear equations for the amplitude and vertical shift from the maximum and minimum bounds

a = 5

and

c = 3

Since

\sin(bx)

ranges from

-1

to

1

, the maximum is

a(1) + c = 8

and minimum is

a(-1) + c = -2

. Solving these simultaneous equations gives

c = 3

and

a = 5

.

2

Calculate the frequency coefficient

b

using the period formula

b = 3

For a sine curve specified in degrees, the period is

T = \frac{360^\circ}{b}

. Substituting

T = 120^\circ

yields

b = 3

.

3

Compute the requested sum

a + b + c

a + b + c = 11

Summing the calculated parameters

5 + 3 + 3 = 11

.

Key Concept

Determining parameters of a trigonometric graph from amplitude, vertical shift, and period.

Question 802Question →

A parallel plate air capacitor of capacitance $8.0\text{ }\mu\text{F}$ is fully charged using a $40.0\text{ V}$ d.c. power supply and then disconnected from the source. A dielectric slab with a relative permittivity of $4.0$ is subsequently inserted to completely fill the region between the plates. Calculate the magnitude of the decrease in electrostatic energy stored in the capacitor, in microjoules ( $\mu\text{J}$ ).

Show answer & explanation

Answer: 4800

Answer

The magnitude of the decrease in stored electrostatic energy is 4800 μJ.

Disconnecting the battery ensures that the charge

Q = C_0 V = 320\text{ }\mu\text{C}

on the plates stays fixed. Inserting the dielectric increases capacitance fourfold to

32.0\text{ }\mu\text{F}

. The energy decreases from

U_i = 6400\text{ }\mu\text{J}

to

U_f = \frac{Q^2}{2C_f} = 1600\text{ }\mu\text{J}

, giving a total decrease of

4800\text{ }\mu\text{J}

.

Step-by-Step Solution

1

Calculate the initial energy stored in the air capacitor before disconnection.

Initial energy

U_i = \frac{1}{2} C_0 V^2 = \frac{1}{2} \times 8.0 \times 10^{-6} \text{ F} \times (40.0 \text{ V})^2 = 6.4 \times 10^{-3} \text{ J} = 6400 \text{ } \mu\text{J}

.

The initial state has known capacitance and potential difference.

2

Calculate the new capacitance with the dielectric present.

Final capacitance

C_f = K C_0 = 4.0 \times 8.0 \text{ } \mu\text{F} = 32.0 \text{ } \mu\text{F}

.

Inserting a dielectric of constant

K

scales the capacitance by

K

.

3

Calculate the final stored energy using charge conservation.

Final energy

U_f = \frac{U_i}{K} = \frac{6400 \text{ } \mu\text{J}}{4.0} = 1600 \text{ } \mu\text{J}

.

Disconnection forces the charge

Q

to remain fixed, so energy scales inversely with capacitance (

U = \frac{Q^2}{2C}

).

4

Find the difference between initial and final energy.

\Delta U = 6400 \text{ } \mu\text{J} - 1600 \text{ } \mu\text{J} = 4800 \text{ } \mu\text{J}.

The loss in electrostatic energy represents the work done by the field pulling the dielectric slab into the plates.

Key Concept

Effect of dielectric insertion on stored electrostatic energy under isolated (constant charge) conditions

Estimated Time:2m 0s

Question 803Question →

Given that $(x - 2)$ is a factor of the polynomial $P(x) = x^3 + kx^2 - 5x + 6$ , find the remainder when $P(x)$ is divided by $(x + 3)$ .

Show answer & explanation

Answer: -15

Answer

The remainder when

P(x)

is divided by

(x + 3)

is

-15

.

According to the Factor Theorem, since

(x - 2)

is a factor of

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

, setting

x = 2

yields

P(2) = 0

. This gives

8 + 4k - 10 + 6 = 0

, which simplifies to

4k + 4 = 0

, giving

k = -1

. The polynomial is therefore

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

. By the Remainder Theorem, dividing

P(x)

by

(x + 3)

produces a remainder of

P(-3)

. Evaluating

P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15

.

Step-by-Step Solution

1

Apply the Factor Theorem to determine the unknown constant

k

.

k = -1

If

(x - 2)

is a factor of

P(x)

, then

P(2) = 0

. Substituting

x = 2

gives

2^3 + k(2)^2 - 5(2) + 6 = 0 \implies 4k + 4 = 0 \implies k = -1

.

2

Substitute

k = -1

into the original polynomial to get the full expression.

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

Replacing

k

with

-1

defines

P(x)

completely.

3

Apply the Remainder Theorem to find the remainder when

P(x)

is divided by

(x + 3)

.

Remainder is

-15

By the Remainder Theorem, dividing

P(x)

by

(x + 3)

leaves a remainder equal to

P(-3)

. Calculating

P(-3) = (-3)^3 - (-3)^2 - 5(-3) + 6 = -27 - 9 + 15 + 6 = -15

.

Key Concept

Factor and Remainder Theorems for Polynomials

Estimated Time:1m 30s

Question 804Question →

A curve has a gradient function defined by $\frac{dy}{dx} = 6x^2 + 8\sin(4x) + 3$ . If the curve passes through the point $(0, 10)$ , what is the value of the constant of integration $C$ ?

Show answer & explanation

Answer: 12

Answer

The constant of integration

C

is

12

.

Integrating

6x^2 + 8\sin(4x) + 3

with respect to

x

gives

y = 2x^3 - 2\cos(4x) + 3x + C

. Substituting

(0, 10)

into the integrated equation gives

10 = 2(0) - 2(1) + 3(0) + C

, which leads directly to

10 = -2 + C

, so

C = 12

.

Step-by-Step Solution

1

Integrate the gradient function to obtain the general equation of the curve.

y = 2x^3 - 2\cos(4x) + 3x + C

The antiderivative of

\sin(kx)

is

-\frac{1}{k}\cos(kx)

and the antiderivative of

x^n

is

\frac{x^{n+1}}{n+1}

.

2

Apply the given initial condition

(x, y) = (0, 10)

to solve for

C

.

10 = 2(0)^3 - 2\cos(0) + 3(0) + C \implies 10 = -2 + C

Evaluating at

x = 0

requires evaluating

\cos(0) = 1

, which leaves

-2

from the trigonometric term.

3

Solve the linear equation for

C

.

C = 12

Adding

2

to both sides of

10 = -2 + C

gives

C = 12

.

Key Concept

Indefinite Integration with Boundary Conditions

Question 805Question →

A curve $y = F(x)$ has a gradient function given by $\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x)$ . Given that $y(0) = 7$ , determine the value of the constant of integration $C$ when the antiderivative is expressed in the standard form $y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C$ .

Show answer & explanation

Answer: 5

Answer

The value of the constant of integration

C

is 5.

Integrating

\frac{dy}{dx} = 9x^2 - 8x + 6\cos(3x) - 4\sin(2x)

yields

y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C

. Substituting

x = 0

into the expression gives

y(0) = 0 - 0 + 0 + 2(1) + C = 2 + C

. Equating to

y(0) = 7

gives

2 + C = 7

, which solves to

C = 5

.

Step-by-Step Solution

1

Integrate the gradient function term-by-term with respect to

x

y = 3x^3 - 4x^2 + 2\sin(3x) + 2\cos(2x) + C

The antiderivative of

9x^2

is

3x^3

, of

-8x

is

-4x^2

, of

6\cos(3x)

is

2\sin(3x)

, and of

-4\sin(2x)

is

2\cos(2x)

.

2

Apply the initial boundary condition

y(0) = 7

3(0)^3 - 4(0)^2 + 2\sin(0) + 2\cos(0) + C = 7 \implies 2 + C = 7

At

x = 0

,

\sin(0) = 0

and

\cos(0) = 1

, making the non-zero constant contribution equal to

2(1) = 2

.

3

Solve for the constant of integration

C

C = 5

Subtracting 2 from both sides of

2 + C = 7

yields

C = 5

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions with Boundary Conditions

Question 806Question →

Determine the number of distinct solutions to the trigonometric equation $2\cos^2 \theta + \sin \theta - 1 = 0$ within the interval $0^\circ \le \theta \le 360^\circ$ .

Show answer & explanation

Answer: 3

Answer

The total number of distinct solutions in the given interval is 3.

Substituting

\cos^2 \theta = 1 - \sin^2 \theta

yields the quadratic equation

2\sin^2 \theta - \sin \theta - 1 = 0

. Factoring gives

\sin \theta = 1

and

\sin \theta = -\frac{1}{2}

. Within

0^\circ \le \theta \le 360^\circ

,

\sin \theta = 1

gives one solution (

90^\circ

), while

\sin \theta = -\frac{1}{2}

gives two solutions (

210^\circ

and

330^\circ

). In total, there are 3 distinct solutions.

Step-by-Step Solution

1

Substitute the identity

\cos^2 \theta = 1 - \sin^2 \theta

into the original equation

2(1 - \sin^2 \theta) + \sin \theta - 1 = 0

Converting the equation into a single trigonometric ratio allows for algebraic solving.

2

Simplify and arrange into quadratic form

2\sin^2 \theta - \sin \theta - 1 = 0

This puts the expression into standard quadratic form

au^2 + bu + c = 0

where

u = \sin \theta

.

3

Factor the quadratic equation

(2\sin \theta + 1)(\sin \theta - 1) = 0

Factoring determines the roots for

\sin \theta

.

4

Solve for possible values of

\sin \theta

\sin \theta = 1

or

\sin \theta = -\frac{1}{2}

By the zero-product property, at least one factor must equal zero.

5

Find the angles for each ratio in the interval

0^\circ \le \theta \le 360^\circ

\theta = 90^\circ, 210^\circ, 330^\circ

\sin \theta = 1

has one solution (

90^\circ

) and

\sin \theta = -0.5

has two solutions in the 3rd and 4th quadrants (

210^\circ

and

330^\circ

).

6

Count the solutions

3

There are 3 distinct values of

\theta

satisfying the condition.

Key Concept

Solving quadratic trigonometric equations using basic identities and quadrant analysis

Estimated Time:1m 30s

Question 807Question →

Find the acute angle $\theta$ , in degrees, that satisfies the trigonometric equation $\sqrt{3}\tan \theta - 3 = 0$ .

Show answer & explanation

Answer: 60

Answer

The acute angle

\theta

is

60^\circ

.

Rearranging the equation

\sqrt{3}\tan \theta - 3 = 0

gives

\sqrt{3}\tan \theta = 3

, so

\tan \theta = \frac{3}{\sqrt{3}} = \sqrt{3}

. For an acute angle (

0^\circ < \theta < 90^\circ

), the angle with a tangent equal to

\sqrt{3}

is

60^\circ

.

Step-by-Step Solution

1

Isolate the trigonometric ratio

\tan \theta

\tan \theta = \sqrt{3}

Add

3

to both sides and divide by

\sqrt{3}

, giving

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

.

2

Determine the value of the acute angle

\theta

\theta = 60^\circ

From special angle exact values,

\tan(60^\circ) = \sqrt{3}

.

Key Concept

Solving Simple Trigonometric Equations

Question 808Question →

The couple per unit twist $C$ (torque per unit angle of twist) of a solid wire of length $L$ , radius $r$ , and shear modulus $\eta$ is modeled by the equation:

C = \frac{\pi \eta r^x}{2 L}

Using dimensional analysis, determine the numerical value of the exponent $x$ .

Show answer & explanation

Answer: 4

Answer

The numerical value of the exponent x is 4.

Applying the principle of dimensional homogeneity requires the dimensions of couple per unit twist

[\text{M L}^2 \text{T}^{-2}]

to equal the dimensions of

\frac{\eta r^x}{L}

, which simplifies to

[\text{M L}^{x-2} \text{T}^{-2}]

. Equating exponents of length gives

2 = x - 2

, yielding

x = 4

.

Step-by-Step Solution

1

Determine the dimensional formula of couple per unit twist

C

.

[C] = \text{M L}^2 \text{T}^{-2}

Couple (torque) is force multiplied by perpendicular distance, which has dimensions

[\text{M L T}^{-2}][\text{L}] = [\text{M L}^2 \text{T}^{-2}]

. The angle of twist (in radians) is dimensionless.

2

Determine the dimensional formula of shear modulus

\eta

.

[\eta] = \text{M L}^{-1} \text{T}^{-2}

Shear modulus is defined as shear stress divided by shear strain. Stress has dimensions of force per unit area

[\text{M L T}^{-2}]/[\text{L}^2] = [\text{M L}^{-1} \text{T}^{-2}]

, while strain is dimensionless.

3

Set up the dimensional equation for the relation

C = \frac{\pi \eta r^x}{2 L}

.

[\text{M L}^2 \text{T}^{-2}] = \frac{[\text{M L}^{-1} \text{T}^{-2}][\text{L}]^x}{[\text{L}]} = [\text{M L}^{x-2} \text{T}^{-2}]

Pure numerical constants such as

\pi

and

2

are dimensionless. Length

L

and radius

r

both have dimension

[\text{L}]

.

4

Equate the exponents of length

\text{L}

on both sides of the dimensional equation.

x = 4

Comparing powers of

\text{L}

on both sides gives

2 = x - 2

, which solves to

x = 4

.

Key Concept

Dimensional Homogeneity in Mechanics

Question 809Question →

The length of the mercury column in an uncalibrated thermometer is $4.0\,\text{cm}$ at the ice point ( $0^\circ\text{C}$ ) and $24.0\,\text{cm}$ at the steam point ( $100^\circ\text{C}$ ). What is the temperature in degrees Celsius when the length of the mercury column is $19.0\,\text{cm}$ ?

Show answer & explanation

Answer: 75

Answer

The temperature corresponding to a mercury column length of

19.0\,\text{cm}

is

75^\circ\text{C}

.

The temperature on the Celsius scale is determined by the ratio of the length change above the ice point to the total length change between the ice and steam points:

T = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

. Substituting

L_0 = 4.0\,\text{cm}

,

L_{100} = 24.0\,\text{cm}

, and

L_T = 19.0\,\text{cm}

gives

T = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}

.

Step-by-Step Solution

1

Identify given thermometric length values at fixed points and at the unknown temperature

L_0 = 4.0\,\text{cm}

,

L_{100} = 24.0\,\text{cm}

, and

L_T = 19.0\,\text{cm}

These represent the length at the lower fixed point (

0^\circ\text{C}

), upper fixed point (

100^\circ\text{C}

), and intermediate temperature

T

respectively.

2

Set up the linear interpolation equation on the Celsius scale

T = \frac{L_T - L_0}{L_{100} - L_0} \times 100^\circ\text{C}

Thermometric expansion is assumed to vary linearly with temperature over the operational range.

3

Substitute the given values and perform arithmetic calculation

T = \frac{19.0 - 4.0}{24.0 - 4.0} \times 100 = \frac{15.0}{20.0} \times 100 = 75^\circ\text{C}

Simplifying

\frac{15.0}{20.0}

gives

0.75

, which multiplied by

100

equals

75

.

Key Concept

Temperature measurement using linear variation of thermometric properties

Question 810Question →

The mean of a set of $7$ numbers arranged in ascending order is $16$ . If the median of the set is $16$ and the mean of the smallest $3$ numbers is $11$ , what is the mean of the largest $3$ numbers?

Show answer & explanation

Answer: 21

Answer

The mean of the largest 3 numbers is 21.

For a set of 7 ordered numbers, the median is the 4th number, which is given as 16. The total sum of all 7 numbers is 7 × 16 = 112. The sum of the 3 smallest numbers is 3 × 11 = 33. Subtracting the 3 smallest numbers and the median from the total sum leaves the sum of the 3 largest numbers: 112 - 33 - 16 = 63. Dividing 63 by 3 gives a mean of 21.

Step-by-Step Solution

1

Find the total sum of the set of 7 numbers

Sum = 7 × 16 = 112

The mean of a set is the total sum divided by the number of elements.

2

Determine the median value and the sum of the 3 smallest numbers

Median (4th number) = 16, Sum of 3 smallest = 3 × 11 = 33

For an ordered set of 7 numbers, the 4th number is the median, and the mean of the first 3 numbers gives their sum when multiplied by 3.

3

Calculate the sum of the 3 largest numbers

Sum of 3 largest = 112 - 33 - 16 = 63

The total sum is the sum of the 3 smallest numbers, the median (4th number), and the 3 largest numbers.

4

Compute the mean of the 3 largest numbers

Mean = 63 / 3 = 21

Dividing the sum of the 3 largest numbers by 3 gives their arithmetic mean.

Key Concept

Arithmetic Mean and Median of Ungrouped Data Subgroups

Question 811Question →

An aluminium rod of initial length $2.0\text{ m}$ at $20^\circ\text{C}$ expands by $0.96\text{ mm}$ when heated. If the linear expansivity of aluminium is $2.4 \times 10^{-5}\text{ K}^{-1}$ , what is the rise in temperature of the rod?

Show answer & explanation

Answer: 20

Answer

The rise in temperature of the aluminium rod is

20\text{ K}

.

The fractional change in length depends on linear expansivity and temperature change through

\Delta L = L_0 \alpha \Delta T

. Substituting the converted expansion

\Delta L = 9.6 \times 10^{-4}\text{ m}

, initial length

L_0 = 2.0\text{ m}

, and linear expansivity

\alpha = 2.4 \times 10^{-5}\text{ K}^{-1}

gives

\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times 2.4 \times 10^{-5}} = 20\text{ K}

.

Step-by-Step Solution

1

Convert change in length from millimeters to meters

\Delta L = 9.6 \times 10^{-4}\text{ m}

Units must be consistent with initial length in meters.

2

Rearrange the linear thermal expansion formula \Delta L = L_0 \alpha \Delta T for temperature change \Delta T

\Delta T = \frac{\Delta L}{L_0 \alpha}

To isolate the unknown quantity \Delta T.

3

Substitute values into the rearranged formula and compute \Delta T

\Delta T = \frac{9.6 \times 10^{-4}}{2.0 \times (2.4 \times 10^{-5})} = 20\text{ K}

Evaluating the mathematical expression yields the required temperature rise.

Key Concept

Linear Expansivity and Thermal Expansion of Solids

Question 812Question →

Given the matrix $A = \begin{pmatrix} 1 & 2 & 0 \\ 3 & 4 & 1 \\ 2 & 0 & k \end{pmatrix}$ , if $\det(A) = 10$ , what is the value of $k$ ?

Show answer & explanation

Answer: -3

Answer

The value of

k

is

-3

.

Expanding the determinant along the first row gives

1(4k) - 2(3k - 2) + 0 = -2k + 4

. Setting

-2k + 4 = 10

leads directly to

-2k = 6

, giving

k = -3

.

Step-by-Step Solution

1

Expand the determinant of matrix

A

along the first row.

\det(A) = 1(4k - 0) - 2(3k - 2) + 0 = -2k + 4

Using cofactor expansion along the top row to find the expression for the determinant.

2

Set the calculated determinant equal to the given value and solve for

k

.

-2k + 4 = 10 \implies -2k = 6 \implies k = -3

Equating the determinant algebraic expression to 10.

Key Concept

Determinant of a 3x3 Matrix

Question 813Question →

The area of the region bounded by the curve $y = 3x^2 - 4x + 3$ , the $x$ -axis, and the vertical lines $x = 0$ and $x = k$ (where $k > 0$ ) is $18$ square units. What is the value of $k$ ?

Show answer & explanation

Answer: 3

Answer

The value of

k

is

3

.

The area under the non-negative curve

y = 3x^2 - 4x + 3

from

x = 0

to

x = k

is found by calculating

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = k^3 - 2k^2 + 3k

. Setting this expression equal to

18

gives

k^3 - 2k^2 + 3k - 18 = 0

. By the factor theorem, testing

k = 3

yields

3^3 - 2(3)^2 + 3(3) - 18 = 0

. Factoring as

(k - 3)(k^2 + k + 6) = 0

confirms

k = 3

as the only real solution.

Step-by-Step Solution

1

Set up the definite integral for the area bounded by the curve and the x-axis

\int_{0}^{k} (3x^2 - 4x + 3) \, dx = 18

The curve

y = 3x^2 - 4x + 3

lies entirely above the x-axis for all real

x

because its leading coefficient is positive and its discriminant

(-4)^2 - 4(3)(3) = -20 < 0

.

2

Evaluate the definite integral in terms of

k

\left[ x^3 - 2x^2 + 3x \right]_0^k = (k^3 - 2k^2 + 3k) - 0 = k^3 - 2k^2 + 3k

Applying the fundamental theorem of calculus by integrating term by term.

3

Form and simplify the polynomial equation

k^3 - 2k^2 + 3k - 18 = 0

Equating the definite integral expression to the given area value of 18.

4

Solve for real values of

k > 0

k = 3

Using the factor theorem on

k^3 - 2k^2 + 3k - 18 = 0

,

k = 3

yields zero (

27 - 18 + 9 - 18 = 0

). Factoring out

(k - 3)

gives

(k - 3)(k^2 + k + 6) = 0

, where

k^2 + k + 6 = 0

has complex roots.

Key Concept

Determining an unknown boundary limit of a definite integral representing area under a curve

Question 814Question →

Calculate the gradient of the normal line to the curve $y = \frac{6}{x}$ at the point where $x = 3$ .

Show answer & explanation

Answer: 1.5

Answer

The gradient of the normal line is 1.5.

Differentiating

y = 6x^{-1}

yields

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

. Evaluating this derivative at

x = 3

gives the tangent gradient

m_t = -\frac{6}{9} = -\frac{2}{3}

. Because the normal line is perpendicular to the tangent line, its gradient is the negative reciprocal

m_n = -\frac{1}{m_t} = \frac{3}{2} = 1.5

.

Step-by-Step Solution

1

Differentiate the function

y = 6x^{-1}

with respect to

x

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

The first derivative represents the formula for the tangent gradient to the curve at any given point.

2

Evaluate the derivative at

x = 3

to find the tangent slope (

m_t

)

m_t = -\frac{6}{3^2} = -\frac{6}{9} = -\frac{2}{3}

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

3

Calculate the normal slope (

m_n

) as the negative reciprocal of

m_t

m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{2}{3}} = \frac{3}{2} = 1.5

The normal line is perpendicular to the tangent line, meaning

m_t \cdot m_n = -1

.

Key Concept

The gradient of the normal to a curve

y = f(x)

at

x = a

is the negative reciprocal of the derivative evaluated at that point:

m_n = -\frac{1}{f'(a)}

.

Question 815Question →

A pie chart illustrates the distribution of undergraduate students enrolled across four faculties at a university: Arts, Science, Law, and Medicine. The central angles for the sectors representing Arts, Science, and Law are $120^\circ$ , $90^\circ$ , and $75^\circ$ , respectively. If $300$ students are enrolled in the Faculty of Medicine, what is the total number of students enrolled in the university?

Show answer & explanation

Answer: 1440

Answer

The total number of students enrolled in the university is 1440.

The total sum of central angles in any pie chart is

360^\circ

. Subtracting the given angles for Arts (

120^\circ

), Science (

90^\circ

), and Law (

75^\circ

) from

360^\circ

gives the sector angle for Medicine as

75^\circ

. Since

75^\circ

represents

300

students, each degree represents

\frac{300}{75} = 4

students. Multiplying

4

students per degree by the total

360^\circ

yields

1440

total students in the university.

Step-by-Step Solution

1

Determine the sector angle for Medicine.

Sector angle for Medicine =

75^\circ

The sum of central angles in a pie chart is

360^\circ

. Subtracting

120^\circ + 90^\circ + 75^\circ = 285^\circ

from

360^\circ

gives

75^\circ

.

2

Formulate the proportion relating sector angle to frequency.

\frac{75^\circ}{360^\circ} \times N = 300

, where

N

represents total students.

The fractional portion of the angle (

75^\circ

out of

360^\circ

) equals the fractional portion of the total student count (

300

out of

N

).

3

Calculate the total student population

N

.

N = \frac{300 \times 360}{75} = 1440

Dividing

300

by

75

yields

4

students per degree. Multiplying

4

by

360

gives

1440

students.

Key Concept

Pie Chart Sector Angle and Total Population Calculation

Question 816Question →

Given two matrices $A = \begin{pmatrix} x & 2 \\ 2 & 3 \end{pmatrix}$ and $B = \begin{pmatrix} x & 1 \\ -1 & 1 \end{pmatrix}$ , where $x > 0$ . If the determinant of the product matrix $AB$ is $20$ , find the value of $x$ .

Show answer & explanation

Answer: 3

Answer

The positive value of

x

is 3.

Using the property

\det(AB) = \det(A) \cdot \det(B)

, we compute

\det(A) = 3x - 4

and

\det(B) = x + 1

. Equating their product to 20 gives

(3x - 4)(x + 1) = 20

, which simplifies to the quadratic

3x^2 - x - 24 = 0

. Factorizing yields

(3x + 8)(x - 3) = 0

. Since

x

must be a positive number, the correct value is

x = 3

.

Step-by-Step Solution

1

Calculate the determinant of matrix A

\det(A) = 3x - 4

The determinant of a 2x2 matrix \begin{pmatrix} a & b \\ c & d \end{pmatrix} is ad - bc.

2

Calculate the determinant of matrix B

\det(B) = x + 1

Applying ad - bc gives (x)(1) - (1)(-1) = x + 1.

3

Use the determinant product property det(AB) = det(A) * det(B)

(3x - 4)(x + 1) = 20

The determinant of the product of two square matrices equals the product of their individual determinants.

4

Form and solve the quadratic equation

3x^2 - x - 24 = 0, which factorizes into (3x + 8)(x - 3) = 0

Expanding (3x - 4)(x + 1) gives 3x^2 - x - 4. Subtracting 20 yields 3x^2 - x - 24 = 0.

5

Determine the positive solution for x

x = 3

Solving the factors gives x = -8/3 or x = 3. Since x must be positive (x > 0), x = 3.

Key Concept

Determinant of Matrix Product and 2x2 Determinants

Estimated Time:2m 0s

Question 817Question →

If $\int (8x^3 - 9\sin(3x) + 2) \, dx = ax^4 + b\cos(3x) + cx + C$ , where $a$ , $b$ , and $c$ are constant coefficients and $C$ is the constant of integration, what is the value of $a + b + c$ ?

Show answer & explanation

Answer: 7

Answer

The value of

a + b + c

is 7.

Integrating term-by-term yields

\int 8x^3 dx = 2x^4

,

\int -9\sin(3x) dx = 3\cos(3x)

, and

\int 2 dx = 2x

. Equating coefficients with

ax^4 + b\cos(3x) + cx

gives

a = 2

,

b = 3

, and

c = 2

. Therefore,

a + b + c = 2 + 3 + 2 = 7

.

Step-by-Step Solution

1

Integrate the polynomial term

8x^3

\int 8x^3 \, dx = 2x^4

, identifying

a = 2

Apply the power rule of integration:

\int x^n \, dx = \frac{x^{n+1}}{n+1}

.

2

Integrate the trigonometric term

-9\sin(3x)

\int -9\sin(3x) \, dx = 3\cos(3x)

, identifying

b = 3

Apply the standard trigonometric integral formula:

\int \sin(kx) \, dx = -\frac{1}{k}\cos(kx)

.

3

Integrate the constant term

2

\int 2 \, dx = 2x

, identifying

c = 2

The integral of a constant

k

with respect to

x

is

kx

.

4

Calculate the requested sum

a + b + c

a + b + c = 2 + 3 + 2 = 7

Summing the coefficients derived from each term's antiderivative.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 818Question →

A progressive transverse wave traveling along a taut string is governed by the displacement equation $y(x,t) = 0.05 \sin(200\pi t - 10\pi x)$ , where $x$ and $y$ are measured in meters and $t$ in seconds. Calculate the distance, in meters, traveled by the wave front during the time taken for a single particle on the string to complete $15$ full oscillations.

Show answer & explanation

Answer: 3

Answer

The distance traveled by the wave front during 15 full particle oscillations is

3.0\text{ m}

.

Comparing

y(x,t) = 0.05 \sin(200\pi t - 10\pi x)

to the standard wave equation

y = A \sin(\omega t - kx)

gives

\omega = 200\pi\text{ rad/s}

and

k = 10\pi\text{ rad/m}

. The wave speed

v = \frac{\omega}{k} = 20\text{ m/s}

. The time for one full oscillation is

T = \frac{2\pi}{\omega} = 0.01\text{ s}

, so 15 full oscillations take

t = 15 \times 0.01\text{ s} = 0.15\text{ s}

. The distance traveled by the wave front is

d = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}

. Alternatively, because a wave travels a distance of one wavelength

\lambda = \frac{2\pi}{k} = 0.2\text{ m}

during each period (1 full oscillation), in 15 full oscillations the wave travels

15 \times \lambda = 15 \times 0.2\text{ m} = 3.0\text{ m}

.

Step-by-Step Solution

1

Extract wave parameters from the progressive wave equation

Angular frequency

\omega = 200\pi\text{ rad/s}

and wave number

k = 10\pi\text{ rad/m}

.

Matching the given equation

y = 0.05 \sin(200\pi t - 10\pi x)

with the standard form

y = A \sin(\omega t - kx)

identifies

\omega

and

k

.

2

Calculate the wave propagation velocity

v = 20\text{ m/s}

.

Wave speed is given by the relation

v = \frac{\omega}{k} = \frac{200\pi}{10\pi} = 20\text{ m/s}

.

3

Find the period of oscillation and total elapsed time

Period

T = 0.01\text{ s}

, total time

t = 0.15\text{ s}

.

The period

T = \frac{2\pi}{\omega} = \frac{2\pi}{200\pi} = 0.01\text{ s}

. For 15 complete oscillations,

t = 15 \times 0.01\text{ s} = 0.15\text{ s}

.

4

Compute the total distance traveled by the wave

Distance

d = 3.0\text{ m}

.

Using linear motion at constant wave speed,

d = v \times t = 20\text{ m/s} \times 0.15\text{ s} = 3.0\text{ m}

.

Key Concept

Wave equation parameters, particle oscillation period, and wave propagation distance

Estimated Time:2m 0s

Question 819Question →

What is the maximum value of the trigonometric function $y = 3\sin x + 2$ ?

Show answer & explanation

Answer: 5

Answer

The maximum value of the function is 5.

The basic sine function

\sin x

reaches a maximum value of

1

. Substituting

\sin x = 1

into

y = 3\sin x + 2

yields

y = 3(1) + 2 = 5

.

Step-by-Step Solution

1

Identify the maximum value of the sine term

The range of

\sin x

is

[-1, 1]

, so its maximum value is

1

.

The sine function oscillates between a minimum of

-1

and a maximum of

1

for all real numbers

x

.

2

Calculate the maximum value of the transformed function

y_{\text{max}} = 3(1) + 2 = 5

.

Multiplying by the positive amplitude coefficient

3

scales the peak to

3

, and adding the vertical shift of

2

raises the peak to

5

.

Key Concept

Maximum and Minimum Values of Trigonometric Functions

Question 820Question →

Evaluate the numerical expression $\frac{0.00054 \times 0.002}{0.00036}$ and state the final result as a decimal.

Show answer & explanation

Answer: 0.003

Answer

The correct value as a decimal is 0.003.

Converting all terms to scientific notation yields

\frac{(5.4 \times 10^{-4}) \times (2 \times 10^{-3})}{3.6 \times 10^{-4}} = \frac{10.8 \times 10^{-7}}{3.6 \times 10^{-4}} = 3 \times 10^{-3} = 0.003

.

Step-by-Step Solution

1

Convert each decimal in the expression to scientific notation (standard form)

0.00054 = 5.4 \times 10^{-4}

,

0.002 = 2 \times 10^{-3}

,

0.00036 = 3.6 \times 10^{-4}

Converting decimals with leading zeros to powers of 10 prevents errors in decimal point alignment during multiplication and division.

2

Simplify the numerator by multiplying coefficients and adding exponents

(5.4 \times 10^{-4}) \times (2 \times 10^{-3}) = 10.8 \times 10^{-7}

According to the laws of indices,

10^a \times 10^b = 10^{a+b}

, so

-4 + (-3) = -7

.

3

Divide the simplified numerator by the denominator

\frac{10.8 \times 10^{-7}}{3.6 \times 10^{-4}} = \left(\frac{10.8}{3.6}\right) \times 10^{-7 - (-4)} = 3.0 \times 10^{-3}

Dividing the coefficients gives

10.8 \div 3.6 = 3

, and subtracting the exponents gives

-7 - (-4) = -3

.

4

Express the result in standard decimal form

3.0 \times 10^{-3} = 0.003

Shifting the decimal point 3 positions to the left converts

10^{-3}

to standard decimal representation.

Key Concept

Simplifying Decimal Expressions using Standard Form and Laws of Indices

Answer

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