Calculus Practice Questions — JAMB UTME — Page 3

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

\frac{5}{27}

\frac{13}{9}

\frac{25}{81}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{5}{27}$

Answer

The value of

\frac{dy}{dx}

at

x = 1

is

\frac{5}{27}

.

Applying the quotient rule

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

with

u = 3x - 1

and

v = (x + 2)^2

yields

\frac{dy}{dx} = \frac{8 - 3x}{(x + 2)^3}

. Substituting

x = 1

results in

\frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{27}

.

Step-by-Step Solution

1

Identify the numerator

u(x)

and denominator

v(x)

for the quotient rule.

Let

u = 3x - 1

and

v = (x + 2)^2

.

The given function

y = \frac{u}{v}

requires the quotient rule

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

.

2

Differentiate

u(x)

and

v(x)

with respect to

x

.

u' = 3

and, using the chain rule,

v' = 2(x + 2)(1) = 2(x + 2)

.

The derivative of the inner term

(x + 2)

is

1

, giving

v' = 2(x + 2)

.

3

Substitute

u, v, u', v'

into the quotient rule formula and simplify.

\frac{dy}{dx} = \frac{3(x + 2)^2 - (3x - 1) \cdot 2(x + 2)}{(x + 2)^4} = \frac{(x + 2)[3(x + 2) - 2(3x - 1)]}{(x + 2)^4} = \frac{3x + 6 - 6x + 2}{(x + 2)^3} = \frac{8 - 3x}{(x + 2)^3}

.

Factoring out

(x + 2)

simplifies the algebraic expression.

4

Evaluate the derivative at

x = 1

.

\frac{dy}{dx}\Big|_{x=1} = \frac{8 - 3(1)}{(1 + 2)^3} = \frac{5}{3^3} = \frac{5}{27}

.

Substituting

x = 1

gives the final numerical derivative value.

Key Concept

Quotient and Chain Rules of Differentiation

Question 42Question →

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

Show answer & explanation

Answer: 1

Answer

The numerical value of

\frac{dy}{dx}

at

x = 1

is 1.

Applying the quotient rule

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

alongside the chain rule for the denominator yields

\frac{dy}{dx} = \frac{2(x^2 + 1)^2 - (2x - 3) \cdot 4x(x^2 + 1)}{(x^2 + 1)^4}

. Substituting

x = 1

evaluates to

\frac{2(4) - (-1)(8)}{16} = \frac{16}{16} = 1

.

Step-by-Step Solution

1

Identify the numerator and denominator functions

Let

u(x) = 2x - 3

and

v(x) = (x^2 + 1)^2

.

The given function is structured as a quotient

y = \frac{u}{v}

, requiring the quotient rule.

2

Find the derivatives

u'(x)

and

v'(x)

u'(x) = 2

and

v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)

.

Differentiating

u(x)

follows standard polynomial rules;

v(x)

requires the chain rule.

3

Evaluate

u(1), u'(1), v(1),

and

v'(1)

at

x = 1

u(1) = -1

,

u'(1) = 2

,

v(1) = 4

, and

v'(1) = 8

.

Evaluating components before substitution simplifies the arithmetic.

4

Apply the quotient rule formula

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

at

x = 1

\frac{dy}{dx} = \frac{(2)(4) - (-1)(8)}{4^2} = \frac{8 + 8}{16} = 1.

Substitute the calculated component values into the quotient rule formula.

Key Concept

Quotient Rule and Chain Rule of Differentiation

Estimated Time:1m 30s

Question 43Question →

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

-\frac{5}{4}

\frac{1}{4}

\frac{13}{4}

\frac{1}{3}

Show answer & explanation

Answer: $-\frac{5}{4}$

Answer

-\frac{5}{4}

Applying the quotient rule

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

with

u = x^2 + 2x

(

u' = 2x + 2

) and

v = (3x - 1)^2

(

v' = 6(3x - 1)

) yields

\frac{dy}{dx} = \frac{16 - 36}{16} = -\frac{5}{4}

at

x = 1

.

Step-by-Step Solution

1

Identify the numerator and denominator functions for the quotient rule

y = \frac{u}{v}

.

u = x^2 + 2x

and

v = (3x - 1)^2

.

The function is structured as a quotient of two algebraic expressions.

2

Differentiate

u

and

v

with respect to

x

.

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

and

\frac{dv}{dx} = 2(3x - 1) \cdot 3 = 6(3x - 1)

.

Use the power rule for

u

and the chain rule for

v

.

3

Apply the quotient rule formula

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

.

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

.

The quotient rule formula combines the expressions and their derivatives.

4

Substitute

x = 1

into the derivative expression and simplify.

\frac{dy}{dx} = \frac{(2)^2 (4) - (3) \cdot 6(2)}{(2)^4} = \frac{16 - 36}{16} = -\frac{20}{16} = -\frac{5}{4}

.

Evaluating at

x = 1

yields the numerical derivative value.

Key Concept

Quotient Rule and Chain Rule of Differentiation

Question 44Question →

What is the indefinite integral $\int (3x^2 - 5\sin(5x) + 2) \, dx$ ?

x^3 + \cos(5x) + 2x + C

x^3 - \cos(5x) + 2x + C

x^3 + 25\cos(5x) + 2x + C

x^3 + \cos(5x) + 2x

Show answer & explanation

Answer: $x^3 + \cos(5x) + 2x + C$

Answer

x^3 + \cos(5x) + 2x + C

Integrating term-by-term yields

\int 3x^2 dx = x^3

,

\int -5\sin(5x) dx = \cos(5x)

, and

\int 2 dx = 2x

. Adding the arbitrary constant

C

produces

x^3 + \cos(5x) + 2x + 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 = 3 \cdot \frac{x^3}{3} = x^3

Applying the power rule for integration.

2

Integrate the trigonometric term

-5\sin(5x)

using

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

\int -5\sin(5x) dx = -5 \left(-\frac{1}{5}\cos(5x)\right) = \cos(5x)

Integration of the sine function reverses differentiation with a positive sign change for negative sine.

3

Integrate the constant term

2

and append the constant of integration

C

\int 2 dx = 2x

, giving total antiderivative

x^3 + \cos(5x) + 2x + C

Indefinite integrals always require an arbitrary constant of integration

C

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 45Question →

What is the x-coordinate of the maximum stationary point of the curve $y = \sin x + \cos x$ in the interval $0 \le x \le \pi$ ?

\frac{\pi}{4}

\frac{3\pi}{4}

\frac{\pi}{2}

\pi

Show answer & explanation

Answer: $\frac{\pi}{4}$

Answer

The x-coordinate of the maximum stationary point is

\frac{\pi}{4}

.

To find stationary points, we differentiate

y = \sin x + \cos x

to get

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

. Setting

\frac{dy}{dx} = 0

gives

\sin x = \cos x

, or

\tan x = 1

. Within

0 \le x \le \pi

, the solution is

x = \frac{\pi}{4}

. Evaluating the second derivative

\frac{d^2y}{dx^2} = -\sin x - \cos x

at

x = \frac{\pi}{4}

yields

-\sqrt{2} < 0

, confirming a local maximum.

Step-by-Step Solution

1

Differentiate the function

y = \sin x + \cos x

with respect to

x

.

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

Stationary points occur where the first derivative

\frac{dy}{dx} = 0

.

2

Set the first derivative to zero and solve for

x

in the interval

0 \le x \le \pi

.

\cos x - \sin x = 0 \implies \sin x = \cos x \implies \tan x = 1 \implies x = \frac{\pi}{4}

Dividing both sides by

\cos x

gives

\tan x = 1

, which has the solution

x = \frac{\pi}{4}

in the given domain.

3

Verify the nature of the stationary point using the second derivative test.

\frac{d^2y}{dx^2} = -\sin x - \cos x \implies \left.\frac{d^2y}{dx^2}\right|_{x=\frac{\pi}{4}} = -\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2} = -\sqrt{2} < 0

A negative second derivative confirms that

x = \frac{\pi}{4}

is a local maximum point.

Key Concept

Finding stationary points and determining their nature using derivatives of trigonometric functions.

Question 46Question →

Find the positive constant $p$ such that the value of the definite integral $\int_{0}^{p} (3x^2 + 2x) \, dx$ is equal to $12$ .

Show answer & explanation

Answer: 2

Answer

The positive constant

p

is equal to

2

.

Evaluating the definite integral

\int_{0}^{p} (3x^2 + 2x) \, dx

gives

p^3 + p^2

. Setting this expression equal to

12

forms the cubic equation

p^3 + p^2 - 12 = 0

. Solving for the positive real solution yields

p = 2

, since

2^3 + 2^2 = 8 + 4 = 12

.

Step-by-Step Solution

1

Find the antiderivative of the integrand

\int (3x^2 + 2x) \, dx = x^3 + x^2 + C

Applying the power rule for integration

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

term by term gives

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

.

2

Evaluate the antiderivative at the limits

0

and

p

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

Using the Fundamental Theorem of Calculus: substitute the upper limit

p

and subtract the value at the lower limit

0

.

3

Set up the polynomial equation using the given integral value

p^3 + p^2 = 12 \implies p^3 + p^2 - 12 = 0

The problem states that the definite integral equals 12.

4

Solve for the positive real value of

p

p = 2

Testing

p=2

:

2^3 + 2^2 = 8 + 4 = 12

. Factoring gives

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

, where the quadratic factor has a negative discriminant

3^2 - 4(1)(6) = -15 < 0

, making

p=2

the only real solution.

Key Concept

Definite integration with an unknown limit of integration

Question 47Question →

What is the value of the definite integral $\int_{0}^{\frac{\pi}{3}} \cos(2x) \, dx$ ?

\frac{\sqrt{3}}{4}

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

-\frac{\sqrt{3}}{4}

\frac{1}{4}

Show answer & explanation

Answer: $\frac{\sqrt{3}}{4}$

Answer

\frac{\sqrt{3}}{4}

The antiderivative of

\cos(2x)

is

\frac{1}{2}\sin(2x)

. Substituting the limits gives

\frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0) = \frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}

, which is correct.

Step-by-Step Solution

1

Find the antiderivative of

\cos(2x)

\int \cos(2x) \, dx = \frac{1}{2}\sin(2x) + C

Using standard trigonometric integration rules,

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

.

2

Substitute the upper limit

x = \frac{\pi}{3}

and lower limit

x = 0

\left[\frac{1}{2}\sin(2x)\right]_{0}^{\frac{\pi}{3}} = \frac{1}{2}\sin\left(\frac{2\pi}{3}\right) - \frac{1}{2}\sin(0)

Apply the Fundamental Theorem of Calculus:

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

.

3

Evaluate the trigonometric values and simplify

\frac{1}{2}\left(\frac{\sqrt{3}}{2}\right) - 0 = \frac{\sqrt{3}}{4}

\sin\left(\frac{2\pi}{3}\right) = \sin(120^\circ) = \frac{\sqrt{3}}{2}

and

\sin(0) = 0

.

Key Concept

Definite Integration of Trigonometric Functions

Question 48Question →

What is the value of the definite integral $\int_{1}^{2} (4x^3 - 3x^2 + 2) \, dx$ ?

8

10

12

14

Show answer & explanation

Answer: 10

Answer

10

Integrating each term yields

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

. Substituting the upper boundary gives

F(2) = 12

and the lower boundary gives

F(1) = 2

. Calculating

F(2) - F(1)

gives

12 - 2 = 10

, which is the exact value of the definite integral.

Step-by-Step Solution

1

Find the indefinite antiderivative of the integrand

4x^3 - 3x^2 + 2

.

F(x) = \int (4x^3 - 3x^2 + 2) \, dx = x^4 - x^3 + 2x

Apply the power rule of integration

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

to each term.

2

Evaluate the antiderivative at the upper limit

x = 2

.

F(2) = (2)^4 - (2)^3 + 2(2) = 16 - 8 + 4 = 12

Substitute

x = 2

into

F(x)

.

3

Evaluate the antiderivative at the lower limit

x = 1

.

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

Substitute

x = 1

into

F(x)

.

4

Subtract the lower limit value from the upper limit value.

F(2) - F(1) = 12 - 2 = 10

Apply the Fundamental Theorem of Calculus:

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

.

Key Concept

Definite Integration of Polynomial Functions

Estimated Time:1m 30s

Question 49Question →

Find the value of the definite integral $\int_{1}^{2} \left(x + \frac{1}{x^2}\right) dx$ .

Show answer & explanation

Answer: 2

Answer

The value of the definite integral is 2.

Integrating

x + x^{-2}

gives

\frac{x^2}{2} - \frac{1}{x}

. Evaluating from

x=1

to

x=2

yields

\left(2 - \frac{1}{2}\right) - \left(\frac{1}{2} - 1\right) = \frac{3}{2} - \left(-\frac{1}{2}\right) = 2

.

Step-by-Step Solution

1

Find the antiderivative of

f(x) = x + x^{-2}

F(x) = \frac{x^2}{2} - \frac{1}{x} + C

Apply the power rule of integration

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

for

n \neq -1

.

2

Evaluate

F(x)

at the upper boundary

x = 2

F(2) = \frac{2^2}{2} - \frac{1}{2} = 2 - \frac{1}{2} = \frac{3}{2}

Substitute

x = 2

into the antiderivative.

3

Evaluate

F(x)

at the lower boundary

x = 1

F(1) = \frac{1^2}{2} - \frac{1}{1} = \frac{1}{2} - 1 = -\frac{1}{2}

Substitute

x = 1

into the antiderivative.

4

Calculate

F(2) - F(1)

\frac{3}{2} - \left(-\frac{1}{2}\right) = \frac{3}{2} + \frac{1}{2} = 2

Apply the Fundamental Theorem of Calculus:

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

.

Key Concept

Definite Integration using the Power Rule

Question 50Question →

What is the $y$ -intercept of the normal line to the curve $y = \frac{2x - 1}{x + 1}$ at the point where the curve crosses the $x$ -axis?

\frac{3}{8}

-\frac{2}{3}

\frac{9}{8}

\frac{2}{3}

Show answer & explanation

Answer: $\frac{3}{8}$

Answer

The

y

-intercept of the normal line is

\frac{3}{8}

.

To find the

y

-intercept of the normal line, set

y = 0

to find the point of contact on the

x

-axis, which gives

\left(\frac{1}{2}, 0\right)

. Differentiating

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

gives

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

. At

x = \frac{1}{2}

, the tangent slope is

\frac{4}{3}

, making the normal slope

-\frac{3}{4}

. The equation of the normal line is

y = -\frac{3}{4}x + \frac{3}{8}

, so its

y

-intercept is

\frac{3}{8}

.

Step-by-Step Solution

1

Find the point of intersection of the curve with the

x

-axis

Set

y = 0 \implies \frac{2x - 1}{x + 1} = 0 \implies 2x - 1 = 0 \implies x = \frac{1}{2}

. The point is

\left(\frac{1}{2}, 0\right)

.

The normal line is drawn at the point where the curve crosses the

x

-axis.

2

Differentiate

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

using the quotient rule

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

The derivative provides the gradient function of the tangent line to the curve.

3

Evaluate the gradient of the tangent and normal lines at

x = \frac{1}{2}

Tangent gradient

m_t = \frac{3}{\left(\frac{1}{2} + 1\right)^2} = \frac{3}{\frac{9}{4}} = \frac{4}{3}

. Normal gradient

m_n = -\frac{1}{m_t} = -\frac{3}{4}

.

The normal line is perpendicular to the tangent line at the point of contact.

4

Determine the equation of the normal line and evaluate its

y

-intercept

Using

y - 0 = -\frac{3}{4}\left(x - \frac{1}{2}\right) \implies y = -\frac{3}{4}x + \frac{3}{8}

. Setting

x = 0

gives

y = \frac{3}{8}

.

The

y

-intercept is the value of

y

when

x = 0

on the line.

Key Concept

Tangents and Normals to Curves

Question 51Question →

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 52Question →

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 53Question →

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 54Question →

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 55Question →

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 56Question →

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 57Question →

Evaluate the algebraic limit:

\lim_{x \to 4} \frac{\sqrt{x} - 2}{x^2 - 16}

What is the exact value of this limit?

\frac{1}{32}

\frac{1}{16}

\frac{1}{8}

0

Show answer & explanation

Answer: $\frac{1}{32}$

Answer

The exact value of the limit is

\frac{1}{32}

.

The limit presents an indeterminate form

\frac{0}{0}

at

x = 4

. Factoring

x^2 - 16

into

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

allows the factor

(\sqrt{x} - 2)

to be cancelled from both the numerator and denominator. Substituting

x = 4

into the simplified expression

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

yields

\frac{1}{(2+2)(4+4)} = \frac{1}{32}

.

Step-by-Step Solution

1

Identify the form of the limit by direct substitution.

Substituting

x = 4

into

\frac{\sqrt{4} - 2}{4^2 - 16}

gives

\frac{0}{0}

, which is an indeterminate form.

Direct substitution yields

\frac{0}{0}

, requiring algebraic simplification.

2

Factor the denominator

x^2 - 16

.

x^2 - 16 = (x - 4)(x + 4)

Use the difference of two squares identity

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

.

3

Factor

(x - 4)

as a difference of squares involving square roots.

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

This exposes the vanishing factor

(\sqrt{x} - 2)

in the denominator.

4

Cancel the common factor

(\sqrt{x} - 2)

and evaluate the limit.

\lim_{x \to 4} \frac{\sqrt{x} - 2}{(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)} = \lim_{x \to 4} \frac{1}{(\sqrt{x} + 2)(x + 4)} = \frac{1}{(\sqrt{4} + 2)(4 + 4)} = \frac{1}{4 \times 8} = \frac{1}{32}

Cancelling the factor removing the

\frac{0}{0}

condition allows direct evaluation.

Key Concept

Resolution of indeterminate limits of the form 0/0 using algebraic factorization and conjugate radical identities.

Estimated Time:2m 0s

Question 58Question →

A curve has the equation $y = \frac{1}{3}x^3 - 2x^2 + 3x + 1$ . What is the positive $x$ -coordinate of the point on the curve where the tangent line is parallel to the line $y = 8x - 5$ ?

Show answer & explanation

Answer: 5

Answer

The positive

x

-coordinate of the point where the tangent is parallel to the line is

5

.

The gradient of the line

y = 8x - 5

is

8

. Differentiating

y = \frac{1}{3}x^3 - 2x^2 + 3x + 1

gives

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

. Setting

\frac{dy}{dx} = 8

leads to

x^2 - 4x - 5 = 0

, which factors as

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

. The solutions are

x = 5

and

x = -1

. Selecting the positive value gives

x = 5

.

Step-by-Step Solution

1

Find the gradient (slope) of the given straight line.

The line equation is in slope-intercept form

y = mx + c

, where the slope

m = 8

.

Parallel lines have equal slopes, so the gradient of the tangent to the curve must equal 8.

2

Differentiate the curve equation to find the gradient function

\frac{dy}{dx}

.

\frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{3}x^3 - 2x^2 + 3x + 1\right) = x^2 - 4x + 3

.

The first derivative of a curve represents the gradient of the tangent at any point

x

.

3

Equate the derivative to the slope of the line and solve for

x

.

x^2 - 4x + 3 = 8 \implies x^2 - 4x - 5 = 0 \implies (x - 5)(x + 1) = 0

. The roots are

x = 5

and

x = -1

.

Solving the quadratic equation gives all

x

-values where the tangent line has a slope of 8.

4

Select the positive

x

-coordinate as requested.

x = 5

.

The question specifically asks for the positive value among the solutions.

Key Concept

Finding points on a curve where the tangent is parallel to a given line

Question 59Question →

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

Show answer & explanation

Answer: 32

Answer

The numerical value of

\frac{dy}{dx}

evaluated at

x = 1

is

32

.

Using the product rule together with the chain rule for composite functions, we find

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

and

v'(x) = 4x(x^2 + 1)

. Evaluating at

x = 1

gives

u(1)=1

,

u'(1)=6

,

v(1)=4

, and

v'(1)=8

. Calculating

\frac{dy}{dx} = u'(1)v(1) + u(1)v'(1) = 6(4) + 1(8) = 32

.

Step-by-Step Solution

1

Set up the product rule for

y = u(x)v(x)

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

and

v(x) = (x^2 + 1)^2

The given function is a product of two composite expressions.

2

Differentiate

u(x)

using the chain rule

u'(x) = 3(2x - 1)^2 \cdot 2 = 6(2x - 1)^2

Differentiating the outer power function and multiplying by the derivative of the inner function

2x - 1

.

3

Differentiate

v(x)

using the chain rule

v'(x) = 2(x^2 + 1) \cdot 2x = 4x(x^2 + 1)

Differentiating the outer power function and multiplying by the derivative of the inner function

x^2 + 1

.

4

Apply the product rule formula

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

and evaluate at

x = 1

u(1) = 1

,

u'(1) = 6

,

v(1) = 4

,

v'(1) = 8

, giving

\frac{dy}{dx}\Big|_{x=1} = (6)(4) + (1)(8) = 32

Substituting

x = 1

into each individual term simplifies the arithmetic before combining.

Key Concept

Combined application of the Product Rule and Chain Rule

Question 60Question →

Evaluate the definite integral $\int_{0}^{2} (3x^2 + 4) \, dx$ .

Show answer & explanation

Answer: 16

Answer

The value of the definite integral is

16

.

Integrating

3x^2 + 4

with respect to

x

gives the antiderivative

F(x) = x^3 + 4x

. Evaluating this antiderivative at the upper limit

x = 2

yields

2^3 + 4(2) = 16

, and at the lower limit

x = 0

yields

0^3 + 4(0) = 0

. Subtracting the lower boundary value from the upper boundary value gives

16 - 0 = 16

.

Step-by-Step Solution

1

Integrate the polynomial term by term

\int (3x^2 + 4) dx = x^3 + 4x

Apply the power rule of integration \int x^n dx = \frac{x^{n+1}}{n+1} and \int k dx = kx.

2

Apply the fundamental theorem of calculus with limits 0 and 2

[x^3 + 4x]_0^2 = (2^3 + 4(2)) - (0^3 + 4(0)) = 16 - 0 = 16

Evaluate F(b) - F(a) where F(x) is the antiderivative.

Key Concept

Definite Integral Evaluation using the Fundamental Theorem of Calculus

Calculus

Answer

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