All practice questions

Question 6901Question →

What is the positive value of $y$ that satisfies the simultaneous equations $x - y = 2$ and $x^2 - y^2 = 12$ ?

2

4

6

8

Show answer & explanation

Answer: 2

Answer

The positive value of y is 2.

Factoring

x^2 - y^2

as

(x - y)(x + y)

gives

2(x + y) = 12

, leading to

x + y = 6

. Subtracting

x - y = 2

from

x + y = 6

isolates

2y = 4

, giving

y = 2

.

Step-by-Step Solution

1

Factorize the quadratic difference of squares equation.

x^2 - y^2 = (x - y)(x + y) = 12

Difference of two squares identity allows linear substitution.

2

Substitute

x - y = 2

into the factorized expression.

2(x + y) = 12 \implies x + y = 6

Simplifies the quadratic system into a second linear equation.

3

Solve the system of linear equations

x + y = 6

and

x - y = 2

for

y

.

(x + y) - (x - y) = 6 - 2 \implies 2y = 4 \implies y = 2

Subtracting the two linear equations eliminates

x

and directly isolates

y

.

Key Concept

Simultaneous Linear and Quadratic Equations via Difference of Squares Substitution

Estimated Time:45s

Question 6902Question →

During the industrial processing of coal by destructive distillation in the absence of air, several products are obtained. Which of the following substances is recovered as the solid residue remaining in the retort?

Coke

Coal tar

Ammoniacal liquor

Coal gas

Show answer & explanation

Answer: Coke

Answer

Coke is the solid residue obtained from the destructive distillation of coal.

Destructive distillation involves heating coal in an airtight retort at high temperatures. Volatile compounds escape as vapors and are condensed into liquid coal tar, ammoniacal liquor, and uncondensed coal gas. The non-volatile, porous carbonaceous residue that remains inside the retort is coke.

Step-by-Step Solution

1

Analyze the conditions of destructive distillation of coal.

Coal is heated strongly in the absence of air to prevent combustion and break down complex organic compounds into volatile and non-volatile fractions.

Thermal decomposition drives off gaseous and liquid vapors while leaving fixed carbon behind.

2

Classify the resulting products by their physical states.

Coal gas is gaseous, coal tar and ammoniacal liquor condense into liquids, and coke remains as a solid in the retort.

Non-volatile elemental carbon stays behind as coke, whereas volatile components vaporize.

Key Concept

Products of coal destructive distillation

Question 6903Question →

A radioactive isotope has a half-life of $5\text{ days}$ . If the initial activity of a sample of this isotope is $400\text{ Bq}$ , what is the activity of the portion of the sample that has decayed after $15\text{ days}$ ?

350\text{ Bq}

50\text{ Bq}

100\text{ Bq}

300\text{ Bq}

Show answer & explanation

Answer: $350\text{ Bq}$

Answer

350\text{ Bq}

After 3 half-lives (15 days), one-eighth of the original activity remains, which equals 50 Bq. Therefore, seven-eighths of the original activity has decayed, giving 400 Bq - 50 Bq = 350 Bq.

Step-by-Step Solution

1

Calculate the number of half-lives that elapse in 15 days

Number of half-lives,

n = \frac{15\text{ days}}{5\text{ days}} = 3

Dividing the total time elapsed by the half-life period yields the number of decay half-lives.

2

Calculate the remaining activity of the sample

Remaining activity

A = A_0 \left(\frac{1}{2}\right)^n = 400 \times \left(\frac{1}{2}\right)^3 = 400 \times \frac{1}{8} = 50\text{ Bq}

The remaining fraction after

n

half-lives is

(1/2)^n

of the initial activity.

3

Subtract the remaining activity from the initial activity to find the decayed activity

Decayed activity

A_{\text{decayed}} = A_0 - A = 400\text{ Bq} - 50\text{ Bq} = 350\text{ Bq}

The portion that has decayed is equal to the total initial activity minus the activity that remains.

Key Concept

Distinction between remaining activity and decayed activity in radioactive decay calculations

Question 6904Question →

Using differentiation from first principles, evaluate the value of the derivative $\frac{dy}{dx} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ for the cubic function $f(x) = 2x^3 - 9x^2 + 12x - 5$ at the point $x = 3$ .

Show answer & explanation

Answer: 12

Answer

The derivative evaluated at

x = 3

is equal to 12.

Evaluating the definition of the derivative from first principles for

f(x) = 2x^3 - 9x^2 + 12x - 5

yields

\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = 6x^2 - 18x + 12

. Substituting

x = 3

gives

6(3)^2 - 18(3) + 12 = 54 - 54 + 12 = 12

.

Step-by-Step Solution

1

Set up the difference quotient definition from first principles

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

Differentiation from first principles requires finding the limit of the average rate of change as the increment

h

approaches zero.

2

Substitute

(x+h)

into

f(x) = 2x^3 - 9x^2 + 12x - 5

and expand

f(x+h) = 2x^3 + 6x^2h + 6xh^2 + 2h^3 - 9x^2 - 18xh - 9h^2 + 12x + 12h - 5

Expanding binomial terms

(x+h)^3

and

(x+h)^2

reveals all components involving

h

.

3

Calculate

f(x+h) - f(x)

and factor out

h

f(x+h) - f(x) = h(6x^2 + 6xh + 2h^2 - 18x - 9h + 12)

Terms independent of

h

cancel out completely, isolating

h

as a common factor.

4

Divide by

h

and evaluate the limit as

h \to 0

f'(x) = 6x^2 - 18x + 12

Canceling

h

resolves the

\frac{0}{0}

indeterminate form, allowing direct substitution of

h=0

.

5

Substitute

x = 3

into

f'(x)

f'(3) = 6(3)^2 - 18(3) + 12 = 12

Evaluating at

x = 3

gives the numerical value of the instantaneous rate of change at that specific point.

Key Concept

Differentiation from First Principles

Question 6905Question →

Given the function $y = \frac{(x^2 + 1)^3}{2x - 3}$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 2$ ?

Show answer & explanation

Answer: 50

Answer

The numerical value of

\frac{dy}{dx}

at

x = 2

is 50.

Applying the Quotient Rule

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

along with the Chain Rule to find

u'(x) = 6x(x^2 + 1)^2

, and evaluating all components at

x = 2

gives

u(2) = 125

,

u'(2) = 300

,

v(2) = 1

, and

v'(2) = 2

. Substituting these values yields

\frac{(300)(1) - (125)(2)}{1^2} = 50

.

Step-by-Step Solution

1

Set up the Quotient Rule components

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

and

v(x) = 2x - 3

The given expression is a quotient of two functions requiring

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

.

2

Differentiate

u(x)

using the Chain Rule and

v(x)

using basic power rules

u'(x) = 6x(x^2 + 1)^2

and

v'(x) = 2

Differentiating the outer power 3 gives

3(x^2 + 1)^2

, and multiplying by the derivative of the inner function

(2x)

gives

6x(x^2 + 1)^2

.

3

Evaluate

u(2)

,

u'(2)

,

v(2)

, and

v'(2)

u(2) = 125

,

u'(2) = 300

,

v(2) = 1

, and

v'(2) = 2

Substituting

x = 2

into each function and derivative simplifies calculation of the overall derivative.

4

Substitute values into the Quotient Rule formula

\left.\frac{dy}{dx}\right|_{x=2} = \frac{(300)(1) - (125)(2)}{(1)^2} = 50

Evaluating

\frac{u'(2)v(2) - u(2)v'(2)}{[v(2)]^2}

yields the exact numerical result.

Key Concept

Combined Application of Quotient Rule and Chain Rule

Question 6906Question →

What is the indefinite integral $\int \left( (3 - 2x)(4x + 1) + 8\cos(4x) \right) dx$ ?

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x)

-\frac{8}{3}x^3 + 5x^2 + 3x - 2\sin(4x) + C

-\frac{8}{3}x^3 + 5x^2 + 3x + 32\sin(4x) + C

Show answer & explanation

Answer: $-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C$

Answer

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Expanding the product

(3 - 2x)(4x + 1)

gives

-8x^2 + 10x + 3

. Integrating term-by-term yields

\int -8x^2 dx = -\frac{8}{3}x^3

,

\int 10x dx = 5x^2

,

\int 3 dx = 3x

, and

\int 8\cos(4x) dx = \frac{8}{4}\sin(4x) = 2\sin(4x)

. Adding the integration constant

C

gives the complete result

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

.

Step-by-Step Solution

1

Expand the polynomial product inside the integrand

(3 - 2x)(4x + 1) = 12x + 3 - 8x^2 - 2x = -8x^2 + 10x + 3

Expanding the expression allows term-by-term integration using standard rules.

2

Rewrite the full integrand

\int \left( -8x^2 + 10x + 3 + 8\cos(4x) \right) dx

Substitute the expanded polynomial back into the integral expression.

3

Integrate each term individually

\int -8x^2 dx = -\frac{8}{3}x^3

,

\int 10x dx = 5x^2

,

\int 3 dx = 3x

, and

\int 8\cos(4x) dx = 8 \cdot \frac{\sin(4x)}{4} = 2\sin(4x)

Apply the power rule

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

and linear trigonometric rule

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

.

4

Combine terms and append the constant of integration

-\frac{8}{3}x^3 + 5x^2 + 3x + 2\sin(4x) + C

Indefinite integration requires adding an arbitrary constant

C

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Estimated Time:2m 0s

Question 6907Question →

If $43_x + 56_x = 121_x$ , where $x$ represents a positive integer base, find the value of $x$ .

Show answer & explanation

Answer: 8

Answer

The value of the base

x

is 8.

Expanding each base

x

number into polynomial form gives

(4x + 3) + (5x + 6) = x^2 + 2x + 1

. Simplifying yields the quadratic equation

x^2 - 7x - 8 = 0

, which factors as

(x - 8)(x + 1) = 0

. Since a number base must be a positive integer greater than any digit present in the problem (

x > 6

),

x = 8

.

Step-by-Step Solution

1

Convert all base

x

numbers into base 10 algebraic expressions.

43_x = 4x + 3

,

56_x = 5x + 6

, and

121_x = x^2 + 2x + 1

.

Place-value expansion expresses numbers in base

x

as polynomials in

x

.

2

Set up the algebraic equation corresponding to the addition.

(4x + 3) + (5x + 6) = x^2 + 2x + 1 \implies 9x + 9 = x^2 + 2x + 1

.

The sum of the left-hand terms equals the right-hand term.

3

Rearrange into standard quadratic form and factor.

x^2 - 7x - 8 = 0 \implies (x - 8)(x + 1) = 0

.

Moving all terms to one side allows solving for the roots of the quadratic equation.

4

Determine the valid base value.

x = 8

.

Number bases must be positive integers greater than all individual digits present in the expression (

x > 6

).

Key Concept

Unknown base equations and expansion

Question 6908Question →

What is the equation of the normal line to the curve $y = x^3 - 3x + 2$ at the point where $x = 2$ ?

x + 9y - 38 = 0

9x - y - 14 = 0

x - 9y + 34 = 0

x + 9y - 34 = 0

Show answer & explanation

Answer: $x + 9y - 38 = 0$

Answer

The equation of the normal line is

x + 9y - 38 = 0

.

Evaluating

y

at

x = 2

gives the point

(2, 4)

. Differentiating

y = x^3 - 3x + 2

yields

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

. At

x = 2

, the tangent slope is

m_t = 9

, making the normal slope

m_n = -1/9

. Using the point-slope formula

y - 4 = -1/9(x - 2)

simplifies to

x + 9y - 38 = 0

.

Step-by-Step Solution

1

Find the

y

-coordinate at

x = 2

y = (2)^3 - 3(2) + 2 = 8 - 6 + 2 = 4

. The point on the curve is

(2, 4)

.

The point of tangency/normal line intersection must be determined on the curve.

2

Calculate the derivative

\frac{dy}{dx}

to find the slope of the tangent line

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

. At

x = 2

,

\frac{dy}{dx} = 3(2)^2 - 3 = 12 - 3 = 9

.

The derivative evaluated at

x = 2

gives the gradient of the tangent line.

3

Find the slope of the normal line

m_n = -\frac{1}{m_t} = -\frac{1}{9}

.

The normal line is perpendicular to the tangent line, so its slope is the negative reciprocal of the tangent slope.

4

Use the point-slope form to find the equation of the normal line

y - 4 = -\frac{1}{9}(x - 2) \implies 9(y - 4) = -(x - 2) \implies 9y - 36 = -x + 2 \implies x + 9y - 38 = 0

.

Substituting point

(2, 4)

and slope

-1/9

into

y - y_1 = m_n(x - x_1)

yields the standard line equation.

Key Concept

The slope of the normal line to a curve

y = f(x)

at

(x_1, y_1)

is

m_n = -\frac{1}{f'(x_1)}

.

Question 6909Question →

If the surd expression $\sqrt{75} - \sqrt{12} + \frac{6}{\sqrt{3}}$ is simplified to the form $k\sqrt{3}$ , what is the value of $k$ ?

Show answer & explanation

Answer: 5

Answer

The value of

k

is 5.

Simplifying

\sqrt{75}

yields

5\sqrt{3}

, simplifying

\sqrt{12}

yields

2\sqrt{3}

, and rationalizing

\frac{6}{\sqrt{3}}

gives

2\sqrt{3}

. Adding these together gives

5\sqrt{3} - 2\sqrt{3} + 2\sqrt{3} = 5\sqrt{3}

. Equating

5\sqrt{3}

to

k\sqrt{3}

yields

k = 5

.

Step-by-Step Solution

1

Simplify the individual square roots.

\sqrt{75} = \sqrt{25 \times 3} = 5\sqrt{3}

and

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

.

Factoring out perfect square numbers allows surds to be written in basic radical form.

2

Rationalize the fractional surd term

\frac{6}{\sqrt{3}}

.

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

.

Multiplying numerator and denominator by

\sqrt{3}

eliminates the radical from the denominator.

3

Combine like surd terms.

5\sqrt{3} - 2\sqrt{3} + 2\sqrt{3} = 5\sqrt{3}

.

Like surds share the same radical factor and can be added or subtracted algebraically.

4

Equate the simplified expression to

k\sqrt{3}

.

k = 5

.

Comparing coefficients of

\sqrt{3}

reveals the value of

k

.

Key Concept

Surd Simplification and Rationalization of Denominators

Estimated Time:1m 30s

Question 6910Question →

What is the value of $\log_3 16 \times \log_2 27$ ?

12

7

432

6

Show answer & explanation

Answer: $12$

Answer

The value of

\log_3 16 \times \log_2 27

is

12

.

By applying the power law of logarithms,

\log_3 16 = 4\log_3 2

and

\log_2 27 = 3\log_2 3

. Multiplying these gives

(4 \times 3)(\log_3 2 \times \log_2 3)

. Using the change of base identity

\log_3 2 \times \log_2 3 = 1

, the product simplifies to

12

.

Step-by-Step Solution

1

Rewrite

16

and

27

as prime powers

\log_3 16 = \log_3 (2^4)

and

\log_2 27 = \log_2 (3^3)

Express numbers in terms of base prime factors to simplify logarithmic powers.

2

Apply the power law of logarithms

\log_b (a^k) = k \log_b a

\log_3 (2^4) = 4 \log_3 2

and

\log_2 (3^3) = 3 \log_2 3

Bring the exponents out as multipliers.

3

Multiply the expressions and apply the change of base reciprocal property

\log_a b \times \log_b a = 1

(4 \log_3 2) \times (3 \log_2 3) = 12 \times (\log_3 2 \times \log_2 3) = 12 \times 1 = 12

The logarithmic terms are reciprocals of each other, simplifying their product to 1.

Key Concept

Change of Base and Power Laws of Logarithms

Estimated Time:45s

Question 6911Question →

What is the numerical value of the limit

\lim_{x \to 4} \frac{x - 4}{\sqrt{x + 5} - 3}

?

6

3

0

\frac{1}{6}

Show answer & explanation

Answer: $6$

Answer

The value of the limit is

6

.

Multiplying both the numerator and the denominator by the conjugate of the denominator

(\sqrt{x + 5} + 3)

allows the factor

(x - 4)

to cancel out, leaving

\sqrt{x + 5} + 3

. Evaluating this expression as

x \to 4

yields

3 + 3 = 6

.

Step-by-Step Solution

1

Check for direct substitution.

Substituting

x = 4

yields

\frac{4 - 4}{\sqrt{4 + 5} - 3} = \frac{0}{3 - 3} = \frac{0}{0}

, which is an indeterminate form.

Direct substitution gives

\frac{0}{0}

, requiring algebraic simplification such as rationalization.

2

Rationalize the denominator by multiplying the numerator and denominator by the conjugate

(\sqrt{x + 5} + 3)

.

\lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{(\sqrt{x + 5} - 3)(\sqrt{x + 5} + 3)} = \lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{(x + 5) - 9}

Using the difference of squares formula

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

eliminates the square root in the denominator.

3

Simplify the denominator and cancel out the common factor

(x - 4)

.

\lim_{x \to 4} \frac{(x - 4)(\sqrt{x + 5} + 3)}{x - 4} = \lim_{x \to 4} (\sqrt{x + 5} + 3)

Since

x \neq 4

when evaluating the limit, the indeterminate factor

(x - 4)

cancels out.

4

Substitute

x = 4

into the simplified expression.

\sqrt{4 + 5} + 3 = \sqrt{9} + 3 = 3 + 3 = 6

Evaluates the limit after removing the zero-denominator condition.

Key Concept

Limits of indeterminate forms

\frac{0}{0}

involving radicals (Rationalization Technique)

Estimated Time:1m 30s

Question 6912Question →

If $4^{2x + 1} \times 8^{1 - x} = 32^{x - 1}$ , find the value of $x$ .

\frac{5}{2}

\frac{9}{4}

\frac{3}{2}

3

Show answer & explanation

Answer: $\frac{5}{2}$

Answer

The value of

x

is

\frac{5}{2}

.

Converting all terms to base

2

yields

4^{2x+1} = 2^{4x+2}

,

8^{1-x} = 2^{3-3x}

, and

32^{x-1} = 2^{5x-5}

. Applying the law

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

gives

2^{(4x+2)+(3-3x)} = 2^{x+5}

. Equating exponents gives

x+5 = 5x-5

, which solves cleanly to

x = \frac{5}{2}

.

Step-by-Step Solution

1

Express all terms with a common prime base (base 2)

4 = 2^2

,

8 = 2^3

, and

32 = 2^5

To apply the laws of indices, all terms must share the same base.

2

Apply the power of a power law

(a^m)^n = a^{mn}

to each term

(2^2)^{2x+1} \times (2^3)^{1-x} = (2^5)^{x-1} \implies 2^{2(2x+1)} \times 2^{3(1-x)} = 2^{5(x-1)} \implies 2^{4x+2} \times 2^{3-3x} = 2^{5x-5}

Multiplying the inner exponent by the outer exponent simplifies nested powers.

3

Apply the multiplication law

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

on the left-hand side

2^{(4x+2) + (3-3x)} = 2^{5x-5} \implies 2^{x+5} = 2^{5x-5}

Powers with the same base being multiplied require adding their exponents.

4

Equate the exponents and solve for

x

x + 5 = 5x - 5 \implies 5 + 5 = 5x - x \implies 10 = 4x \implies x = \frac{10}{4} = \frac{5}{2}

Since the bases are equal and non-zero, their exponents must be equal.

Key Concept

Solving exponential equations by expressing numbers in terms of a common prime base and applying laws of indices.

Question 6913Question →

Find the product of all real solutions to the exponential equation $9^{x+1} - 28 \cdot 3^x + 3 = 0$ .

Show answer & explanation

Answer: -2

Answer

The product of all real solutions is -2.

Using the index laws

9^{x+1} = 9 \cdot (3^x)^2

, let

u = 3^x

. The equation reduces to the quadratic

9u^2 - 28u + 3 = 0

, which factors as

(9u - 1)(u - 3) = 0

. This gives

u = 1/9

or

u = 3

. Solving

3^x = 1/9

yields

x = -2

, and solving

3^x = 3

yields

x = 1

. The product of these solutions is

(-2) \times 1 = -2

.

Step-by-Step Solution

1

Express

9^{x+1}

in terms of

3^x

9^{x+1} = 9^1 \cdot 9^x = 9 \cdot (3^2)^x = 9 \cdot (3^x)^2

Apply index laws

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

and

(a^m)^n = (a^n)^m

to establish a common base of 3.

2

Substitute

u = 3^x

to form a quadratic equation

9u^2 - 28u + 3 = 0

Transform the exponential equation into a standard quadratic algebraic equation.

3

Solve the quadratic equation for

u

(9u - 1)(u - 3) = 0 \implies u = \frac{1}{9}

or

u = 3

Factorize the quadratic expression to determine its roots.

4

Substitute back

u = 3^x

to solve for

x

3^x = 3^{-2} \implies x = -2

, and

3^x = 3^1 \implies x = 1

Equate exponents with matching bases to find all valid real solutions for

x

.

5

Find the product of the two solutions

(-2) \times 1 = -2

Calculate the required mathematical product of the solutions.

Key Concept

Solving exponential equations reducible to quadratic form using laws of indices

Question 6914Question →

In a medical study of $120$ hospital outpatients, $68$ are prescribed Green tea extract ( $G$ ), $52$ are prescribed Herbal infusion ( $H$ ), and $44$ are prescribed Chamomile syrup ( $C$ ). Furthermore, $24$ patients receive both Green tea extract and Herbal infusion, $18$ receive both Green tea extract and Chamomile syrup, $16$ receive both Herbal infusion and Chamomile syrup, while $8$ receive all three prescriptions. How many patients receive at most one of these three prescriptions?

72

78

54

112

Show answer & explanation

Answer: $78$

Answer

The number of patients who receive at most one of the three prescriptions is

78

.

To find the number of patients who receive at most one prescription, we must calculate the number of patients who receive exactly one prescription as well as those who receive none (the complement of the union relative to the universal set of 120 outpatients). The number receiving only Green tea is

34

, only Herbal infusion is

20

, and only Chamomile syrup is

18

, giving

72

patients taking exactly one prescription. The total taking at least one prescription is

114

, leaving

120 - 114 = 6

patients receiving none. Adding these together yields

72 + 6 = 78

.

Step-by-Step Solution

1

Calculate the number of patients in each exclusive intersection region.

Patients receiving all three

n(G \cap H \cap C) = 8

. Patients receiving only

G \cap H = 24 - 8 = 16

. Patients receiving only

G \cap C = 18 - 8 = 10

. Patients receiving only

H \cap C = 16 - 8 = 8

.

Subtracting the triple intersection isolates the regions representing patients taking exactly two prescriptions.

2

Calculate the number of patients receiving exactly one prescription.

n(G \text{ only}) = 68 - (16 + 10 + 8) = 34

.

n(H \text{ only}) = 52 - (16 + 8 + 8) = 20

.

n(C \text{ only}) = 44 - (10 + 8 + 8) = 18

. Total receiving exactly one

= 34 + 20 + 18 = 72

.

Each set total is reduced by its overlapping intersection regions to yield the single-prescription regions.

3

Determine the number of patients receiving at least one prescription using the inclusion-exclusion principle.

n(G \cup H \cup C) = (68 + 52 + 44) - (24 + 18 + 16) + 8 = 164 - 58 + 8 = 114

.

The principle of inclusion-exclusion accounts for double-counted pairwise intersections and triple-counted intersections.

4

Calculate the number of patients receiving none of the prescriptions.

n((G \cup H \cup C)') = 120 - 114 = 6

.

Subtracting the total taking at least one prescription from the universal set gives the complement count.

5

Sum the patients receiving exactly one prescription and those receiving zero prescriptions.

Total taking at most one

= 72 + 6 = 78

.

'At most one' encompasses both 'zero prescriptions' and 'exactly one prescription'.

Key Concept

Three-set principle of inclusion-exclusion and complementary sets within a universal set

Question 6915Question →

Two point charges, $q_1 = +1.6 \times 10^{-8}\text{ C}$ and $q_2 = +6.4 \times 10^{-8}\text{ C}$ , are fixed in a vacuum at a distance of $0.60\text{ m}$ apart. At what distance from $q_1$ along the line joining the two charges is the net electric field intensity equal to zero?

0.20\text{ m}

0.40\text{ m}

0.12\text{ m}

0.30\text{ m}

Show answer & explanation

Answer: $0.20\text{ m}$

Answer

The distance from

q_1

where the net electric field intensity is zero is

0.20\text{ m}

.

The net electric field is zero where the magnitudes of the electric fields produced by both charges are equal (

E_1 = E_2

). Setting up

\frac{k q_1}{x^2} = \frac{k q_2}{(0.60 - x)^2}

with

q_2 = 4 q_1

yields

\frac{1}{x^2} = \frac{4}{(0.60 - x)^2}

. Taking the square root gives

\frac{1}{x} = \frac{2}{0.60 - x}

, which yields

x = 0.20\text{ m}

from

q_1

.

Step-by-Step Solution

1

Set up the condition for zero net electric field intensity.

The electric field magnitudes produced by

q_1

and

q_2

at distance

x

from

q_1

must be equal in magnitude and opposite in direction:

E_1 = E_2

.

Since both charges are positive, the point of zero net electric field must lie on the line segment connecting them.

2

Substitute the electric field formula into the equilibrium equation.

\frac{k q_1}{x^2} = \frac{k q_2}{(d - x)^2}

, where

d = 0.60\text{ m}

.

Electric field intensity due to a point charge is given by

E = \frac{k q}{r^2}

.

3

Simplify the equation by canceling common terms and substituting known charge values.

\frac{1.6 \times 10^{-8}}{x^2} = \frac{6.4 \times 10^{-8}}{(0.60 - x)^2} \implies \frac{1}{x^2} = \frac{4}{(0.60 - x)^2}

.

Dividing both sides by

k \times 1.6 \times 10^{-8}

reduces the numerical coefficients to simple integers.

4

Take the square root of both sides and solve for

x

.

\frac{1}{x} = \frac{2}{0.60 - x} \implies 0.60 - x = 2x \implies 3x = 0.60 \implies x = 0.20\text{ m}

.

Taking the square root removes the quadratic terms and gives a linear relation for the distance

x

from

q_1

.

Key Concept

Electric field superposition and point of zero field intensity between like point charges

Question 6916Question →

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

3 + \tan x

3 - \tan x

3 - \cot x

3 + \cot x

Show answer & explanation

Answer: $3 - \tan x$

Answer

3 - \tan x

Applying the product rule for logarithms converts

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

into

y = 3x + \ln(\cos x)

. Differentiating

3x

yields

3

, and differentiating

\ln(\cos x)

by chain rule yields

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

. Combining these gives

3 - \tan x

.

Step-by-Step Solution

1

Simplify the logarithmic function using log laws

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

Splitting the logarithm of a product simplifies the differentiation process.

2

Differentiate each term with respect to

x

\frac{dy}{dx} = \frac{d}{dx}(3x) + \frac{d}{dx}(\ln(\cos x)) = 3 + \frac{1}{\cos x} \cdot (-\sin x)

The derivative of

3x

is

3

, and applying the chain rule to

\ln(u)

gives

\frac{1}{u} \cdot \frac{du}{dx}

where

u = \cos x

.

3

Simplify the trigonometric ratio

\frac{dy}{dx} = 3 - \frac{\sin x}{\cos x} = 3 - \tan x

Replacing

\frac{\sin x}{\cos x}

with

\tan x

yields the simplified final derivative.

Key Concept

Differentiation of Logarithmic and Trigonometric Functions via Chain Rule and Log Laws

Question 6917Question →

A simple pendulum of length $L$ with a bob of mass $m$ has a period of oscillation of $2.0\text{ s}$ . If the mass of the bob is increased to $4m$ and the length of the pendulum string is increased to $4L$ , what is the new period of oscillation?

2.0\text{ s}

4.0\text{ s}

8.0\text{ s}

16.0\text{ s}

Show answer & explanation

Answer: $4.0\text{ s}$

Answer

The new period of oscillation is

4.0\text{ s}

.

The period of a simple pendulum undergoing simple harmonic motion is given by

T = 2\pi \sqrt{\frac{L}{g}}

. The mass of the bob does not enter the period formula, meaning changes to bob mass have zero effect on the period. When the length of the pendulum is quadrupled (

L' = 4L

), the new period becomes

T' = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

. Since the original period was

2.0\text{ s}

, the new period is

2 \times 2.0\text{ s} = 4.0\text{ s}

.

Step-by-Step Solution

1

Identify the formula for the period of a simple pendulum.

The period is given by

T = 2\pi \sqrt{\frac{L}{g}}

.

The period of simple harmonic motion for a simple pendulum depends only on the length of the string

L

and the acceleration due to gravity

g

, and is independent of the mass of the bob

m

.

2

Substitute the scaled values into the formula to find the new period

T'

.

T' = 2\pi \sqrt{\frac{4L}{g}} = 2 \times \left(2\pi \sqrt{\frac{L}{g}}\right) = 2T

.

Taking the square root of

4L

factors out a multiplier of

\sqrt{4} = 2

, while the change in mass from

m

to

4m

has no effect on the period.

3

Calculate the numerical value of the new period.

T' = 2 \times 2.0\text{ s} = 4.0\text{ s}

.

Multiplying the initial period of

2.0\text{ s}

by

2

yields

4.0\text{ s}

.

Key Concept

Mass Independence and Length Relationship of a Simple Pendulum

Question 6918Question →

Calcium hydride ( $\text{CaH}_2$ ) reacts vigorously with water to produce calcium hydroxide and hydrogen gas. What volume of dry hydrogen gas, in $\text{dm}^3$ , measured at standard temperature and pressure (s.t.p.), is liberated when $10.5\text{ g}$ of pure calcium hydride reacts completely with excess water?

[Relative atomic masses: $\text{Ca} = 40$ , $\text{H} = 1$ ; Molar volume of gas at s.t.p. = $22.4\text{ dm}^3\text{ mol}^{-1}$ ]

Show answer & explanation

Answer: 11.2

Answer

The volume of dry hydrogen gas liberated at s.t.p. is 11.2 dm³.

Calcium hydride reacts with water according to the reaction CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂. Given 10.5 g of CaH₂ (molar mass 42 g/mol), there are 0.25 moles of CaH₂. Based on the 1:2 stoichiometric ratio, 0.50 moles of H₂ gas are generated. Multiplying by the molar volume at s.t.p. (22.4 dm³/mol) yields 11.2 dm³.

Step-by-Step Solution

1

Write the balanced chemical equation for the reaction of calcium hydride with water

CaH₂ + 2H₂O → Ca(OH)₂ + 2H₂

Establishing the stoichiometric mole ratio between the reactant CaH₂ and the product H₂ gas.

2

Calculate the molar mass of CaH₂

42 g/mol

Molar mass is required to convert the given mass of CaH₂ into moles.

3

Determine the amount of CaH₂ in moles

0.25 mol

Moles = Mass / Molar mass = 10.5 g / 42 g/mol.

4

Calculate the moles of H₂ gas liberated using the 1:2 stoichiometric ratio

0.50 mol

1 mole of CaH₂ produces 2 moles of H₂ gas.

5

Calculate the volume of H₂ gas produced at standard temperature and pressure (s.t.p.)

11.2 dm³

Volume at s.t.p. = Moles × Molar volume at s.t.p. = 0.50 mol × 22.4 dm³/mol.

Key Concept

Laboratory and industrial preparation of hydrogen using metal hydrides and mole-volume stoichiometric calculations at s.t.p.

Estimated Time:2m 0s

Question 6919Question →

A radioactive sample has an initial activity of $80\text{ Bq}$ and a half-life of $4\text{ days}$ . What is the activity of the sample that has decayed after $12\text{ days}$ ?

70\text{ Bq}

10\text{ Bq}

40\text{ Bq}

20\text{ Bq}

Show answer & explanation

Answer: $70\text{ Bq}$

Answer

The activity that has decayed after 12 days is

70\text{ Bq}

.

After 12 days, which equals 3 half-lives (

12 / 4 = 3

), the remaining activity of the sample is

80 \times (1/2)^3 = 10\text{ Bq}

. Consequently, the activity that has decayed is the initial activity minus the remaining activity:

80\text{ Bq} - 10\text{ Bq} = 70\text{ Bq}

.

Step-by-Step Solution

1

Determine the number of elapsed half-lives

n = \frac{t}{T_{1/2}} = \frac{12\text{ days}}{4\text{ days}} = 3\text{ half-lives}

Dividing the total time elapsed by the half-life period gives the number of decay cycles.

2

Calculate the remaining activity

A_{\text{remaining}} = A_0 \left(\frac{1}{2}\right)^n = 80 \times \left(\frac{1}{2}\right)^3 = 80 \times \frac{1}{8} = 10\text{ Bq}

The remaining quantity decreases by half for each half-life cycle.

3

Calculate the decayed activity

A_{\text{decayed}} = A_0 - A_{\text{remaining}} = 80\text{ Bq} - 10\text{ Bq} = 70\text{ Bq}

Subtracting the undecayed remaining activity from the initial activity gives the total decayed activity.

Key Concept

Radioactive Decay Law and Half-life

Estimated Time:45s

Question 6920Question →

The atomic mass of a lithium nucleus $^{7}_{3}\text{Li}$ is $7.0160\text{ u}$ . Given that the mass of a proton is $1.0073\text{ u}$ and the mass of a neutron is $1.0087\text{ u}$ , what is the total binding energy of the lithium nucleus? ( $1\text{ u} = 931\text{ MeV}$ )

37.89\text{ MeV}

5.41\text{ MeV}

36.59\text{ MeV}

0.041\text{ MeV}

Show answer & explanation

Answer: $37.89\text{ MeV}$

Answer

The total binding energy of the lithium nucleus is

37.89\text{ MeV}

.

The correct answer is derived by finding the mass defect

\Delta m = [3(1.0073) + 4(1.0087)] - 7.0160 = 0.0407\text{ u}

and multiplying by

931\text{ MeV/u}

to obtain

37.89\text{ MeV}

.

Step-by-Step Solution

1

Determine the number of protons and neutrons in

^{7}_{3}\text{Li}

Z = 3

protons,

N = A - Z = 7 - 3 = 4

neutrons

The mass number

A=7

and atomic number

Z=3

define the nuclear composition.

2

Calculate total mass of individual nucleons

Mass of nucleons

= 3(1.0073\text{ u}) + 4(1.0087\text{ u}) = 3.0219\text{ u} + 4.0348\text{ u} = 7.0567\text{ u}

Summing the masses of constituent protons and neutrons.

3

Calculate mass defect

(\Delta m)

\Delta m = 7.0567\text{ u} - 7.0160\text{ u} = 0.0407\text{ u}

Mass defect is the difference between total constituent mass and nuclear mass.

4

Convert mass defect to energy

E_b = 0.0407\text{ u} \times 931\text{ MeV/u} = 37.8917\text{ MeV} \approx 37.89\text{ MeV}

Using the equivalence

1\text{ u} = 931\text{ MeV}

.

Key Concept

Mass Defect and Nuclear Binding Energy

Answer

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