Calculus Practice Questions — JAMB UTME — Page 2

Evaluate the limit $\lim_{x \to 2} \frac{x^2 - 4}{x - 2}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 4

Answer

The value of the limit is

4

.

Direct substitution of

x = 2

produces the indeterminate form

\frac{0}{0}

. Factoring the numerator gives

\frac{(x-2)(x+2)}{x-2}

. Canceling the non-zero factor

(x-2)

simplifies the expression to

x+2

. Evaluating the limit as

x

approaches

2

yields

2 + 2 = 4

.

Step-by-Step Solution

1

Check the form by direct substitution of

x = 2

Obtained the indeterminate form

\frac{0}{0}

Direct substitution results in division by zero, requiring algebraic simplification.

2

Factor the polynomial in the numerator

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

Difference of two squares factorization allows canceling common terms.

3

Cancel the common factor

(x - 2)

\frac{(x - 2)(x + 2)}{x - 2} = x + 2

For

x \neq 2

, division by

(x - 2)

is valid.

4

Evaluate the simplified limit as

x \to 2

2 + 2 = 4

Substitute

x = 2

directly into the continuous polynomial

x + 2

.

Key Concept

Evaluating indeterminate limits of the form

\frac{0}{0}

via factorization

Question 22Question →

A curve is defined by the equation $y = x^3 + px^2 + qx + 5$ , where $p$ and $q$ are constants. If the curve has a stationary point with a local maximum at $x = -1$ and a local minimum at $x = 3$ , what is the value of $p + q$ ?

-12

12

-15

-6

Show answer & explanation

Answer: $-12$

Answer

The value of

p + q

is

-12

.

The derivative of

y = x^3 + px^2 + qx + 5

is

\frac{dy}{dx} = 3x^2 + 2px + q

. Setting

\frac{dy}{dx} = 0

at the stationary points

x = -1

and

x = 3

means

\frac{dy}{dx} = 3(x + 1)(x - 3) = 3x^2 - 6x - 9

. Comparing coefficients gives

2p = -6 \implies p = -3

and

q = -9

. Summing these constants gives

p + q = -12

. Evaluating the second derivative

\frac{d^2y}{dx^2} = 6x + 2p = 6x - 6

confirms a maximum at

x = -1

(

\frac{d^2y}{dx^2} = -12 < 0

) and a minimum at

x = 3

(

\frac{d^2y}{dx^2} = 12 > 0

).

Step-by-Step Solution

1

Find the derivative of the given function.

\frac{dy}{dx} = 3x^2 + 2px + q

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Use the stationary points

x = -1

and

x = 3

to form a quadratic equation for the derivative.

\frac{dy}{dx} = 3(x + 1)(x - 3) = 3(x^2 - 2x - 3) = 3x^2 - 6x - 9

Since

x = -1

and

x = 3

are roots of

\frac{dy}{dx} = 0

, the derivative must factor as

3(x - (-1))(x - 3)

.

3

Equate coefficients of the two derivative expressions to solve for

p

and

q

.

2p = -6 \implies p = -3

and

q = -9

Matching corresponding terms gives

2p = -6

and

q = -9

.

4

Calculate the required sum

p + q

.

p + q = -3 + (-9) = -12

Summing the calculated constants yields

-12

.

Key Concept

Determining parameters of a polynomial function from given stationary points using differentiation and coefficient matching.

Question 23Question →

Using differentiation from first principles, what is the derivative of the function $f(x) = x^2 + 3x$ with respect to $x$ ?

2x + 3

2x + 3 + h

x^2 + 3

2x

Show answer & explanation

Answer: $2x + 3$

Answer

2x + 3

Applying the first principles limit formula

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

gives

\frac{x^2 + 2xh + h^2 + 3x + 3h - (x^2 + 3x)}{h} = \frac{2xh + h^2 + 3h}{h} = 2x + h + 3

. As

h \to 0

, this expression evaluates to

2x + 3

.

Step-by-Step Solution

1

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

\frac{f(x+h) - f(x)}{h} = \frac{[(x+h)^2 + 3(x+h)] - [x^2 + 3x]}{h}

The definition of derivative from first principles requires finding

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

.

2

Expand algebraic terms in the numerator

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

Expanding

(x+h)^2 = x^2 + 2xh + h^2

and

3(x+h) = 3x + 3h

allows terms to be combined.

3

Simplify the numerator by canceling like terms and dividing by

h

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

The terms

x^2 - x^2 = 0

and

3x - 3x = 0

cancel out, leaving terms containing

h

, which can be divided by

h

.

4

Evaluate the limit as

h \to 0

\lim_{h \to 0} (2x + h + 3) = 2x + 3

Setting

h = 0

in the simplified quotient yields the final derivative

f'(x) = 2x + 3

.

Key Concept

Differentiation from first principles

Question 24Question →

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

Show answer & explanation

Answer: 54

Answer

54

Applying the chain rule gives

\frac{dy}{dx} = 3(2x + 1)^2 \cdot 2 = 6(2x + 1)^2

. Evaluating at

x = 1

gives

6(3)^2 = 54

.

Step-by-Step Solution

1

Differentiate

y = (2x + 1)^3

using the chain rule.

\frac{dy}{dx} = 6(2x + 1)^2

According to the chain rule,

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

, where

u = 2x + 1

and

\frac{du}{dx} = 2

.

2

Evaluate the derivative at

x = 1

.

\frac{dy}{dx}\Big|_{x=1} = 54

Substituting

x = 1

into

6(2x + 1)^2

yields

6(3)^2 = 54

.

Key Concept

Chain Rule for Differentiation

Question 25Question →

If $y = e^{4x} - \sin x$ , what is $\frac{dy}{dx}$ ?

4e^{4x} - \cos x

4e^{4x} + \cos x

e^{4x} - \cos x

e^{4x} + \cos x

Show answer & explanation

Answer: $4e^{4x} - \cos x$

Answer

4e^{4x} - \cos x

Differentiating

e^{4x}

gives

4e^{4x}

by applying the chain rule, and differentiating

-\sin x

yields

-\cos x

. Combining these terms gives the correct derivative

4e^{4x} - \cos x

.

Step-by-Step Solution

1

Apply the sum/difference rule of differentiation.

\frac{dy}{dx} = \frac{d}{dx}(e^{4x}) - \frac{d}{dx}(\sin x)

The derivative of a difference of two terms is the difference of their individual derivatives.

2

Differentiate the exponential term

e^{4x}

using the chain rule.

\frac{d}{dx}(e^{4x}) = 4e^{4x}

By the chain rule, \frac{d}{dx}(e^{k x}) = k e^{k x}.

3

Differentiate the trigonometric term

\sin x

.

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

The standard derivative of

\sin x

with respect to

x

is

\cos x

.

4

Combine the results.

\frac{dy}{dx} = 4e^{4x} - \cos x

Subtracting the derivative of

\sin x

from the derivative of

e^{4x}

gives the final answer.

Key Concept

Differentiation of Exponential and Trigonometric Functions

Question 26Question →

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

112

56

80

104

Show answer & explanation

Answer: $112$

Answer

112

Applying the product rule

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

along with the chain rule gives

u'(1) = 6

,

v(1) = 16

,

u(1) = 1

, and

v'(1) = 16

. Evaluating

u'(1)v(1) + u(1)v'(1)

yields

6 \times 16 + 1 \times 16 = 96 + 16 = 112

.

Step-by-Step Solution

1

Decompose the function into two components for the Product Rule

Let

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

and

v(x) = (x^2 + 3)^2

, so

y = u(x) v(x)

.

The function is a product of two composite expressions.

2

Differentiate each term using the Chain Rule

u'(x) = 3(2x - 1)^2 \cdot \frac{d}{dx}(2x - 1) = 6(2x - 1)^2

v'(x) = 2(x^2 + 3) \cdot \frac{d}{dx}(x^2 + 3) = 4x(x^2 + 3)

Applying the Chain Rule requires differentiating the outer function and multiplying by the derivative of the inner function.

3

Apply the Product Rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x) = 6(2x - 1)^2 (x^2 + 3)^2 + 4x(2x - 1)^3 (x^2 + 3)

The derivative of a product

u \cdot v

is

u'v + uv'

.

4

Evaluate the derivative at

x = 1

u(1) = (2(1) - 1)^3 = 1

u'(1) = 6(2(1) - 1)^2 = 6

v(1) = (1^2 + 3)^2 = 16

v'(1) = 4(1)(1^2 + 3) = 16

\left.\frac{dy}{dx}\right|_{x=1} = (6)(16) + (1)(16) = 96 + 16 = 112

Substitute

x = 1

into all expressions to find the numerical derivative.

Key Concept

Combined Product Rule and Chain Rule of Differentiation

Estimated Time:2m 0s

Question 27Question →

What is the gradient of the normal to the curve $y = x^2 + 2x - 1$ at the point where $x = 1$ ?

-\frac{1}{4}

4

\frac{1}{4}

-4

Show answer & explanation

Answer: $-\frac{1}{4}$

Answer

The gradient of the normal to the curve at

x = 1

is

-\frac{1}{4}

.

Differentiating

y = x^2 + 2x - 1

gives

\frac{dy}{dx} = 2x + 2

. Substituting

x = 1

yields a tangent slope of

4

. Taking the negative reciprocal gives

-\frac{1}{4}

, which is the correct gradient of the normal.

Step-by-Step Solution

1

Find the derivative of the curve equation to determine the general tangent gradient function.

\frac{dy}{dx} = 2x + 2

The derivative of a function gives the slope of the tangent line at any point

x

.

2

Evaluate the tangent gradient at the specific point

x = 1

.

m = 2(1) + 2 = 4

Substituting

x = 1

into the derivative gives the slope of the tangent at that specific x-coordinate.

3

Calculate the gradient of the normal line using the perpendicular condition.

m_n = -\frac{1}{m} = -\frac{1}{4}

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

Key Concept

Relationship between tangent and normal gradients (

m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}

)

Estimated Time:1m 0s

Question 28Question →

If $y = 5\sin x + e^{3x}$ , what is the value of $\frac{dy}{dx}$ at $x = 0$ ?

Show answer & explanation

Answer: 8

Answer

The value of the derivative at

x = 0

is 8.

Differentiating

y = 5\sin x + e^{3x}

yields

\frac{dy}{dx} = 5\cos x + 3e^{3x}

. Substituting

x = 0

gives

5\cos(0) + 3e^{0} = 5(1) + 3(1) = 8

.

Step-by-Step Solution

1

Differentiate each term of the function

y = 5\sin x + e^{3x}

with respect to

x

.

\frac{dy}{dx} = 5\cos x + 3e^{3x}

The derivative of

\sin x

is

\cos x

, and applying the chain rule to

e^{3x}

gives

3e^{3x}

.

2

Evaluate the derivative

\frac{dy}{dx}

at

x = 0

.

5\cos(0) + 3e^0 = 5(1) + 3(1) = 8

Evaluating trigonometric and exponential functions at zero gives

\cos(0) = 1

and

e^0 = 1

.

Key Concept

Differentiation of trigonometric and exponential functions and evaluation of derivatives at specific points.

Question 29Question →

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

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

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

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

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

Show answer & explanation

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

Answer

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

Applying the product rule to

u = e^{2x}

and

v = \cos 3x

yields

\frac{du}{dx} = 2e^{2x}

and

\frac{dv}{dx} = -3\sin 3x

. Substituting these into

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

gives

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

.

Step-by-Step Solution

1

Identify the components for the product rule

Let

u = e^{2x}

and

v = \cos 3x

.

The function

y = e^{2x} \cos 3x

is a product of two differentiable functions.

2

Differentiate

u = e^{2x}

with respect to

x

\frac{du}{dx} = 2e^{2x}

By the chain rule,

\frac{d}{dx}(e^{kx}) = k e^{kx}

.

3

Differentiate

v = \cos 3x

with respect to

x

\frac{dv}{dx} = -3\sin 3x

By the chain rule,

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

.

4

Apply the product rule formula

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

and factor out

e^{2x}

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

Combining the products and factoring out the common exponential factor simplifies the expression.

Key Concept

Product Rule and Chain Rule for Exponential and Trigonometric Functions

Question 30Question →

A curve is defined by the equation $y = x^2 - 6x + 11$ . What is the minimum value of $y$ on this curve?

Show answer & explanation

Answer: 2

Answer

The minimum value of

y

on the curve is 2.

Differentiating

y = x^2 - 6x + 11

gives

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

. Setting this derivative to zero yields

2x - 6 = 0

, so

x = 3

. Substituting

x = 3

into the original function gives

y = (3)^2 - 6(3) + 11 = 2

. Since

\frac{d^2y}{dx^2} = 2 > 0

, the point at

x = 3

is a local minimum, making 2 the minimum value of

y

.

Step-by-Step Solution

1

Find the first derivative of the curve function.

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

Stationary points occur where the derivative is equal to zero.

2

Solve for the

x

-coordinate at the stationary point.

2x - 6 = 0 \implies x = 3

Setting the derivative to zero determines the input value where the slope is horizontal.

3

Calculate the corresponding

y

-value at

x = 3

.

y = (3)^2 - 6(3) + 11 = 2

Evaluating the original equation at

x = 3

yields the minimum value of

y

.

Key Concept

Finding the minimum value of a function using differentiation

Question 31Question →

What is the indefinite integral $\int (3x^2 + 4\cos x) \, dx$ ?

x^3 + 4\sin x + C

x^3 - 4\sin x + C

x^3 + 4\sin x

6x + 4\sin x + C

Show answer & explanation

Answer: $x^3 + 4\sin x + C$

Answer

x^3 + 4\sin x + C

Integrating

3x^2

gives

x^3

via the power rule, and integrating

4\cos x

gives

4\sin x

. Combining these terms along with the required constant of integration

C

results in

x^3 + 4\sin x + C

.

Step-by-Step Solution

1

Integrate the polynomial term

3x^2

using the power rule

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

.

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

The power rule increases the exponent by 1 and divides by the new exponent.

2

Integrate the trigonometric term

4\cos x

using the standard integral

\int \cos x \, dx = \sin x

.

\int 4\cos x \, dx = 4\sin x

The antiderivative of cosine is positive sine.

3

Combine the results and append the constant of integration

C

.

x^3 + 4\sin x + C

All indefinite integrals must include an arbitrary constant

C

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 32Question →

An open rectangular box with a square base of side length $x\text{ cm}$ is to be constructed such that its total surface area is $108\text{ cm}^2$ . What is the maximum volume of the box in $\text{cm}^3$ ?

54\text{ cm}^3

72\text{ cm}^3

108\text{ cm}^3

432\text{ cm}^3

Show answer & explanation

Answer: $108\text{ cm}^3$

Answer

108\text{ cm}^3

For an open box with a square base of side

x

and height

h

, total surface area is

A = x^2 + 4xh = 108\text{ cm}^2

. Solving for

h

gives

h = \frac{108 - x^2}{4x}

. Substituting

h

into the volume formula gives

V(x) = x^2 h = 27x - \frac{1}{4}x^3

. Differentiating gives

\frac{dV}{dx} = 27 - \frac{3}{4}x^2

. Setting

\frac{dV}{dx} = 0

yields

x = 6

. The second derivative

\frac{d^2V}{dx^2} = -\frac{3}{2}x

evaluated at

x = 6

is

-9 < 0

, confirming a local maximum. Evaluating

V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3

yields the correct maximum volume.

Step-by-Step Solution

1

Set up the surface area equation for the open box and express height

h

in terms of

x

.

Surface area

A = x^2 + 4xh = 108 \implies h = \frac{108 - x^2}{4x}

.

An open box with a square base has 1 base face (

x^2

) and 4 vertical side faces (

xh

).

2

Formulate the volume function

V(x)

in terms of

x

.

V(x) = x^2 h = x^2 \left(\frac{108 - x^2}{4x}\right) = \frac{1}{4}(108x - x^3) = 27x - \frac{1}{4}x^3

.

Substitute the expression for

h

into the volume formula

V = x^2 h

.

3

Find the critical point by differentiating

V(x)

with respect to

x

and setting

\frac{dV}{dx} = 0

.

\frac{dV}{dx} = 27 - \frac{3}{4}x^2 = 0 \implies \frac{3}{4}x^2 = 27 \implies x^2 = 36 \implies x = 6\text{ cm}

.

Stationary points occur where the first derivative of the volume function equals zero.

4

Verify that

x = 6

gives a maximum volume and calculate

V(6)

.

\frac{d^2V}{dx^2} = -\frac{3}{2}x \implies \left.\frac{d^2V}{dx^2}\right|_{x=6} = -9 < 0

(maximum). Volume

V(6) = 27(6) - \frac{1}{4}(6^3) = 162 - 54 = 108\text{ cm}^3

.

The negative second derivative confirms a maximum turning point.

Key Concept

Optimization of physical quantities using the first and second derivative tests.

Estimated Time:2m 0s

Question 33Question →

Given that $F(x) = \int (3x^2 - 2\sin x) \, dx$ and $F(0) = 6$ , what is the value of the constant of integration $C$ ?

Show answer & explanation

Answer: 4

Answer

The value of the constant of integration

C

is

4

.

Integrating

3x^2 - 2\sin x

yields

F(x) = x^3 + 2\cos x + C

. Substituting

x = 0

gives

F(0) = 2(1) + C = 2 + C

. Since

F(0) = 6

, setting

2 + C = 6

yields

C = 4

.

Step-by-Step Solution

1

Integrate the function

f(x) = 3x^2 - 2\sin x

with respect to

x

F(x) = x^3 + 2\cos x + C

The antiderivative of

3x^2

is

x^3

and the antiderivative of

-2\sin x

is

2\cos x

.

2

Evaluate

F(0)

using the antiderivative expression

F(0) = 0^3 + 2\cos(0) + C = 2 + C

Since

\cos(0) = 1

, the term

2\cos(0)

simplifies to

2

.

3

Solve for the integration constant

C

using

F(0) = 6

C = 4

Subtracting

2

from both sides of

2 + C = 6

yields

C = 4

.

Key Concept

Indefinite integration of polynomial and trigonometric functions with initial conditions

Question 34Question →

The total profit $P(x)$ , in thousands of Naira, obtained from producing and selling $x$ hundred units of a commodity is modeled by the function $P(x) = -x^3 + 6x^2 + 15x - 8$ , where $x \ge 0$ . What is the maximum profit achievable?

Show answer & explanation

Answer: 92

Answer

The maximum profit achievable is 92 thousand Naira.

To find the maximum profit, we find the stationary points of

P(x) = -x^3 + 6x^2 + 15x - 8

by taking the derivative

P'(x) = -3x^2 + 12x + 15

and setting it to 0. Solving

-3(x-5)(x+1) = 0

with

x \ge 0

yields

x = 5

. Testing the second derivative gives

P''(5) = -18 < 0

, confirming

x = 5

is a maximum. Substituting

x = 5

into

P(x)

yields

P(5) = 92

.

Step-by-Step Solution

1

Differentiate the profit function

P(x)

with respect to

x

P'(x) = -3x^2 + 12x + 15

Stationary points occur where the rate of change of profit (the derivative) is equal to zero.

2

Set

P'(x) = 0

and solve for

x

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

, giving critical values

x = 5

and

x = -1

Factoring the quadratic equation yields the critical values of production level.

3

Filter critical values based on physical domain constraints

x = 5

(reject

x = -1

since production

x \ge 0

)

Production quantities cannot be negative in physical real-life contexts.

4

Perform the second derivative test to confirm the nature of the stationary point

P''(x) = -6x + 12 \implies P''(5) = -6(5) + 12 = -18

Since

P''(5) < 0

, the function achieves a local maximum at

x = 5

.

5

Substitute

x = 5

back into original function

P(x)

P(5) = -(5)^3 + 6(5)^2 + 15(5) - 8 = -125 + 150 + 75 - 8 = 92

Evaluating

P(5)

gives the maximum total profit value.

Key Concept

Optimization and Maximum Values using First and Second Derivatives

Question 35Question →

What is the value of the definite integral $\int_{1}^{3} (2x + 1) \, dx$ ?

10

12

14

2

Show answer & explanation

Answer: 10

Answer

10

The antiderivative of

2x + 1

is

x^2 + x

. Evaluating at the upper boundary

x = 3

gives

3^2 + 3 = 12

, and at the lower boundary

x = 1

gives

1^2 + 1 = 2

. Subtracting the lower bound result from the upper bound result yields

12 - 2 = 10

.

Step-by-Step Solution

1

Find the indefinite integral (antiderivative) of

2x + 1

\int (2x + 1) \, dx = x^2 + x

Apply the power rule of integration:

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

.

2

Evaluate the antiderivative at the upper limit

x = 3

3^2 + 3 = 9 + 3 = 12

Substitute

x = 3

into

x^2 + x

.

3

Evaluate the antiderivative at the lower limit

x = 1

1^2 + 1 = 1 + 1 = 2

Substitute

x = 1

into

x^2 + x

.

4

Subtract the lower limit value from the upper limit value

12 - 2 = 10

According to the Fundamental Theorem of Calculus:

\int_{a}^{b} f(x)dx = F(b) - F(a)

.

Key Concept

Definite Integrals of Linear Polynomials

Question 36Question →

What is the indefinite integral $\int \left( (3x + 1)^2 - 6\sin(3x) \right) dx$ ?

3x^3 + 3x^2 + x + 2\cos(3x) + C

3x^3 + 3x^2 + x - 2\cos(3x) + C

3x^3 + 3x^2 + x + 2\cos(3x)

3x^3 + 3x^2 + x + 6\cos(3x) + C

Show answer & explanation

Answer: $3x^3 + 3x^2 + x + 2\cos(3x) + C$

Answer

3x^3 + 3x^2 + x + 2\cos(3x) + C

Expanding

(3x+1)^2

yields

9x^2 + 6x + 1

. Integrating

9x^2 + 6x + 1

gives

3x^3 + 3x^2 + x

. Integrating

-6\sin(3x)

gives

+2\cos(3x)

. Combining these results with the mandatory constant of integration

+ C

gives

3x^3 + 3x^2 + x + 2\cos(3x) + C

.

Step-by-Step Solution

1

Expand the squared polynomial binomial term

(3x + 1)^2 = 9x^2 + 6x + 1

Expanding the expression allows for simple term-by-term integration using the power rule.

2

Integrate the polynomial component term by term

\int (9x^2 + 6x + 1) dx = 3x^3 + 3x^2 + x

Apply the power rule

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

to each polynomial term.

3

Integrate the trigonometric component

\int -6\sin(3x) dx = -6 \left(-\frac{1}{3}\cos(3x)\right) = +2\cos(3x)

The integral of

\sin(kx)

is

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

. Multiplying by

-6

yields

+2\cos(3x)

.

4

Combine terms and add the constant of integration

3x^3 + 3x^2 + x + 2\cos(3x) + C

Indefinite integrals must include an arbitrary constant

+ C

to account for all antiderivatives.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 37Question →

The area of the region bounded by the parabola $y = kx - x^2$ (where $k > 0$ ) and the $x$ -axis between its $x$ -intercepts at $x = 0$ and $x = k$ is equal to $36$ square units. What is the value of the positive constant $k$ ?

Show answer & explanation

Answer: 6

Answer

The value of the positive constant

k

is 6.

The area bounded by

y = kx - x^2

and the x-axis from

x = 0

to

x = k

is obtained by integrating

kx - x^2

, which yields

\frac{k^3}{6}

. Setting

\frac{k^3}{6} = 36

gives

k^3 = 216

, whose cube root is

k = 6

.

Step-by-Step Solution

1

Set up the definite integral representing the area bounded by the curve and the x-axis between the intercepts

x = 0

and

x = k

.

\int_{0}^{k} (kx - x^2) \, dx = 36

The area under a curve

y = f(x)

above the x-axis from

x = a

to

x = b

is given by

\int_{a}^{b} f(x) \, dx

.

2

Find the antiderivative and evaluate it at the limits

x = k

and

x = 0

.

\left[ \frac{kx^2}{2} - \frac{x^3}{3} \right]_{0}^{k} = \left(\frac{k(k)^2}{2} - \frac{k^3}{3}\right) - 0 = \frac{k^3}{6}

Applying the integration power rule

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

and simplifying

\frac{k^3}{2} - \frac{k^3}{3} = \frac{k^3}{6}

.

3

Set the evaluated expression equal to 36 and solve for

k

.

\frac{k^3}{6} = 36 \implies k^3 = 216 \implies k = 6

Multiplying both sides by 6 yields

k^3 = 216

, and taking the cube root gives

k = 6

.

Key Concept

Definite Integral and Area Under Curve

Question 38Question →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{x^2 + x - 6}{x - 2}, & x \neq 2 \\ 2k - 1, & x = 2 \end{cases}

If

f(x)

is continuous at

x = 2

, what is the value of the constant

k

?

3

2

5

-1

Show answer & explanation

Answer: 3

Answer

The value of the constant

k

is

3

.

By definition, a function

f(x)

is continuous at

x = c

if

\lim_{x \to c} f(x) = f(c)

. Factoring the numerator gives

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

. Canceling the common factor

(x - 2)

for

x \neq 2

, the limit as

x \to 2

is

2 + 3 = 5

. Equating

f(2) = 2k - 1

to 5 yields

2k - 1 = 5

, which solves to

k = 3

.

Step-by-Step Solution

1

Evaluate the limit of

f(x)

as

x

approaches

2

.

\lim_{x \to 2} \frac{x^2 + x - 6}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 3)}{x - 2} = \lim_{x \to 2} (x + 3) = 5

Direct substitution gives the indeterminate form

\frac{0}{0}

, so factor the numerator to simplify.

2

Apply the definition of continuity at a point.

f(2) = \lim_{x \to 2} f(x) \implies 2k - 1 = 5

For

f(x)

to be continuous at

x = 2

, the value of the function at

x = 2

must equal its limit as

x \to 2

.

3

Solve the linear equation for

k

.

2k = 6 \implies k = 3

Add 1 to both sides and divide by 2.

Key Concept

Continuity of a Piecewise Function at a Point

Alternative Method

Alternatively, use L'Hôpital's rule to evaluate the limit:

\lim_{x \to 2} \frac{\frac{d}{dx}(x^2+x-6)}{\frac{d}{dx}(x-2)} = \lim_{x \to 2} \frac{2x+1}{1} = 5

. Then set

2k - 1 = 5

to find

k = 3

.

Estimated Time:1m 30s

Question 39Question →

Evaluate the limit $\lim_{x \to 0} \frac{\sqrt{x + 4} - 2}{x}$ . What is the numerical value of this limit?

Show answer & explanation

Answer: 0.25

Answer

The numerical value of the limit is 0.25 (or

\frac{1}{4}

).

When direct substitution into

\frac{\sqrt{x + 4} - 2}{x}

yields the indeterminate form

\frac{0}{0}

, rationalizing the numerator by multiplying by its conjugate

\sqrt{x + 4} + 2

simplifies the expression to

\frac{1}{\sqrt{x + 4} + 2}

. Taking the limit as

x \to 0

gives

\frac{1}{4} = 0.25

.

Step-by-Step Solution

1

Check direct substitution

Indeterminate form

\frac{0}{0}

Directly evaluating at

x = 0

yields zero in both numerator and denominator.

2

Multiply by the conjugate of the numerator

\frac{(\sqrt{x + 4} - 2)(\sqrt{x + 4} + 2)}{x(\sqrt{x + 4} + 2)} = \frac{x}{x(\sqrt{x + 4} + 2)}

Rationalizing the radical in the numerator allows cancellation of the term causing the zero denominator.

3

Cancel common factors and evaluate limit

\frac{1}{\sqrt{0 + 4} + 2} = 0.25

Canceling

x

removes the zero factor, permitting direct evaluation.

Key Concept

Limits of indeterminate algebraic expressions using surd rationalization

Question 40Question →

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

Show answer & explanation

Answer: 34

Answer

The value of

\frac{dy}{dx}

at

x = 2

is

34

.

Applying the product rule

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

alongside the chain rule gives

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

. Evaluating at

x = 2

yields

4(1) + 30(1) = 34

.

Step-by-Step Solution

1

Identify component functions for the product rule

Let

u(x) = x^2 + 1

and

v(x) = (2x - 3)^3

.

The function

y

is a product of two differentiable functions.

2

Differentiate each component function

u'(x) = 2x

and

v'(x) = 3(2x - 3)^2 \cdot 2 = 6(2x - 3)^2

.

The power rule gives

u'(x)

and the chain rule gives

v'(x)

by multiplying by the derivative of the inner function

(2x - 3)

.

3

Apply the product rule formula

\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)

\frac{dy}{dx} = 2x(2x - 3)^3 + 6(x^2 + 1)(2x - 3)^2

.

To find the general derivative of a product of functions.

4

Evaluate the derivative at

x = 2

\frac{dy}{dx}\Big|_{x=2} = 2(2)(2(2) - 3)^3 + 6(2^2 + 1)(2(2) - 3)^2 = 4(1) + 30(1) = 34

.

To calculate the specific numerical value of the derivative at

x = 2

.

Key Concept

Product and Chain Rules of Differentiation

Estimated Time:1m 30s

Calculus

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