Calculus Alıştırma Soruları — JAMB UTME — Sayfa 4

Evaluate the trigonometric limit:

\lim_{x \to 0} \frac{\cos(3x) - \cos(x)}{x^2}

What is the numerical value of this limit?

Cevabı ve açıklamayı göster

Cevap: -4

Cevap

The numerical value of the limit is -4.

Using either the sum-to-product identity

\cos(3x) - \cos(x) = -2 \sin(2x) \sin(x)

along with standard limits

\lim_{x \to 0} \frac{\sin(kx)}{x} = k

, or applying L'Hôpital's rule twice on the

\frac{0}{0}

form, yields the exact value

-4

.

Adım Adım Çözüm

1

Check the form of the limit by direct substitution

Substituting

x = 0

gives

\frac{\cos(0) - \cos(0)}{0^2} = \frac{1 - 1}{0} = \frac{0}{0}

, an indeterminate form.

Determines whether algebraic transformation or L'Hôpital's rule is required.

2

Transform the numerator using the sum-to-product formula

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

Converts difference of cosines into product of sines to utilize standard trigonometric limits.

3

Rewrite the fractional expression and apply limit laws

\lim_{x \to 0} \frac{-2 \sin(2x) \sin(x)}{x^2} = -2 \cdot \left(\lim_{x \to 0} \frac{\sin(2x)}{x}\right) \cdot \left(\lim_{x \to 0} \frac{\sin(x)}{x}\right)

Splits

x^2

into

x \cdot x

under each sine function.

4

Evaluate the individual standard limits

\lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin(x)}{x} = 1

Applies the known fundamental trigonometric limit rule

\lim_{u \to 0} \frac{\sin(au)}{u} = a

.

5

Calculate the final product

-2 \times 2 \times 1 = -4

Combines all factors to reach the evaluated value.

Anahtar Kavram

Trigonometric Limits and Indeterminate Forms

Alternatif Yöntem

Alternatively, apply L'Hôpital's rule twice. First derivative of numerator over denominator yields

\lim_{x \to 0} \frac{-3\sin(3x) + \sin(x)}{2x}

(still

\frac{0}{0}

). Differentiating a second time yields

\lim_{x \to 0} \frac{-9\cos(3x) + \cos(x)}{2} = \frac{-9(1) + 1}{2} = \frac{-8}{2} = -4

.

Tahmini Süre:1m 30s

Soru 62Soru →

If $y = (2x - 3)^5$ , what is the value of $\frac{dy}{dx}$ at $x = 2$ ?

5

8

10

20

Cevabı ve açıklamayı göster

Cevap: 10

Cevap

10

The derivative of

y = (2x - 3)^5

with respect to

x

requires the chain rule:

\frac{dy}{dx} = 5(2x - 3)^4 \cdot \frac{d}{dx}(2x - 3) = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4

. Substituting

x = 2

gives

10(2(2) - 3)^4 = 10(1)^4 = 10

, which makes

10

the correct value.

Adım Adım Çözüm

1

Identify the inner function

u(x)

and outer function

f(u)

Let

u = 2x - 3

, so

y = u^5

.

The chain rule states that

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

.

2

Differentiate

y

with respect to

u

and

u

with respect to

x

\frac{dy}{du} = 5u^4 = 5(2x - 3)^4

and

\frac{du}{dx} = 2

.

Apply the power rule to both functions.

3

Multiply the derivatives to find

\frac{dy}{dx}

\frac{dy}{dx} = 5(2x - 3)^4 \cdot 2 = 10(2x - 3)^4

.

Combine terms using the chain rule formula.

4

Substitute

x = 2

into the derivative

\frac{dy}{dx}\Big|_{x=2} = 10(2(2) - 3)^4 = 10(1)^4 = 10

.

Evaluate the expression numerically at the target point.

Anahtar Kavram

Chain Rule for Differentiation

Tahmini Süre:45s

Soru 63Soru →

Determine the area under the curve $y = 3x^2 - 3$ above the $x$ -axis between $x = 1$ and $x = 3$ .

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The area under the curve between

x = 1

and

x = 3

is 20 square units.

The area under the curve

y = 3x^2 - 3

from

x = 1

to

x = 3

is obtained by calculating the definite integral

\int_{1}^{3} (3x^2 - 3) \, dx = [x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20

.

Adım Adım Çözüm

1

Set up the definite integral representing the area under the curve.

A = \int_{1}^{3} (3x^2 - 3) \, dx

The area bounded by a non-negative curve

y = f(x)

, the

x

-axis, and vertical lines

x = a

and

x = b

is given by

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

.

2

Integrate the polynomial function term by term.

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

Applying the power rule of integration:

\int 3x^2 dx = x^3

and

\int 3 dx = 3x

.

3

Apply the Fundamental Theorem of Calculus by substituting the limits of integration.

[x^3 - 3x]_{1}^{3} = (3^3 - 3(3)) - (1^3 - 3(1)) = 18 - (-2) = 20

Evaluating

F(b) - F(a)

gives

(27 - 9) - (1 - 3) = 18 - (-2) = 20

.

Anahtar Kavram

Definite Integrals and Area Under Curves

Soru 64Soru →

A cubic curve defined by $y = ax^3 + bx^2 + cx + d$ has a local maximum at $(-1, 10)$ and a point of inflexion at $(1, 2)$ . What is the value of $y$ at the local minimum of the curve?

Cevabı ve açıklamayı göster

Cevap: -6

Cevap

The local minimum value of

y

on the curve is

-6

.

By setting up a system of equations using the conditions for the local maximum at

(-1, 10)

and the point of inflexion at

(1, 2)

, the cubic curve is uniquely determined as

y = 0.5x^3 - 1.5x^2 - 4.5x + 7.5

. Setting the derivative

\frac{dy}{dx} = 1.5(x-3)(x+1) = 0

gives

x = 3

as the

x

-coordinate of the local minimum. Evaluating

y(3)

gives

-6

.

Adım Adım Çözüm

1

Differentiate the general cubic equation to obtain expressions for the first and second derivatives.

\frac{dy}{dx} = 3ax^2 + 2bx + c

and

\frac{d^2y}{dx^2} = 6ax + 2b

.

Stationary points require

\frac{dy}{dx} = 0

and points of inflexion require

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

.

2

Apply the point of inflexion conditions at

(1, 2)

.

6a(1) + 2b = 0 \implies b = -3a

, and

a(1)^3 + b(1)^2 + c(1) + d = 2 \implies -2a + c + d = 2

.

At a point of inflexion, the second derivative is zero, and the point lies on the curve.

3

Apply the stationary point and coordinate conditions at the local maximum

(-1, 10)

.

3a(-1)^2 + 2b(-1) + c = 0 \implies 9a + c = 0 \implies c = -9a

, and

-a + b - c + d = 10 \implies 5a + d = 10

.

At a local maximum, the first derivative is zero, and the point lies on the curve.

4

Solve the system of linear equations for coefficients

a, b, c, d

.

a = 0.5

,

b = -1.5

,

c = -4.5

,

d = 7.5

.

Combining

-11a + d = 2

and

5a + d = 10

yields

16a = 8 \implies a = 0.5

.

5

Find the

x

-coordinate of the local minimum by solving

\frac{dy}{dx} = 0

.

1.5x^2 - 3x - 4.5 = 0 \implies 1.5(x - 3)(x + 1) = 0 \implies x = 3

(since

x = -1

is the local maximum).

Evaluating

\frac{d^2y}{dx^2}(3) = 6(0.5)(3) + 2(-1.5) = 6 > 0

confirms a local minimum at

x = 3

.

6

Calculate the value of

y

at

x = 3

.

y = 0.5(3)^3 - 1.5(3)^2 - 4.5(3) + 7.5 = -6

.

Substituting

x = 3

into the curve equation gives the value of

y

at the local minimum.

Anahtar Kavram

Determining polynomial coefficients from stationary and inflexion point conditions to find extreme values.

Soru 65Soru →

Using differentiation from first principles, what is the derivative of the function $f(x) = \frac{3}{x}$ for $x \neq 0$ ?

-\frac{3}{x^2}

\frac{3}{x^2}

-\frac{3}{x}

-\frac{3}{x^3}

Cevabı ve açıklamayı göster

Cevap: $-\frac{3}{x^2}$

Cevap

-\frac{3}{x^2}

Using the first principles formula

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

, substituting

f(x) = \frac{3}{x}

gives

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

.

Adım Adım Çözüm

1

Set up the first principles difference quotient

f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = \lim_{h \to 0} \frac{\frac{3}{x+h} - \frac{3}{x}}{h}

Definition of differentiation from first principles.

2

Combine the fractions in the numerator using a common denominator

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

Algebraic fraction subtraction.

3

Divide by

h

and cancel the common factor

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

Simplifying the difference quotient before taking the limit.

4

Evaluate the limit as

h \to 0

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

Direct substitution of

h=0

after canceling

h

.

Anahtar Kavram

Differentiation of a reciprocal function using the first principles limit definition.

Soru 66Soru →

The line $y = 5x + c$ is tangent to the curve $y = x^3 - 7x + 15$ at a point $P$ located in the first quadrant. What is the equation of the normal line to the curve at point $P$ ?

x + 5y - 47 = 0

5x - y - 1 = 0

x + 5y - 43 = 0

5x + y - 19 = 0

Cevabı ve açıklamayı göster

Cevap: $x + 5y - 47 = 0$

Cevap

The equation of the normal line to the curve at point P is

x + 5y - 47 = 0

.

Differentiating the curve yields

\frac{dy}{dx} = 3x^2 - 7

. Equating this derivative to the tangent slope of 5 gives

x = 2

for the first quadrant. Evaluating the curve equation at

x = 2

gives

y = 9

, locating point

P(2, 9)

. The normal gradient is the negative reciprocal of the tangent slope, giving

m_n = -\frac{1}{5}

. Applying the point-slope formula with

(2, 9)

yields

y - 9 = -\frac{1}{5}(x - 2)

, which simplifies to

x + 5y - 47 = 0

.

Adım Adım Çözüm

1

Find the x-coordinate of the point of tangency P

x = 2

The gradient of the curve

\frac{dy}{dx} = 3x^2 - 7

must equal the gradient of the given line

m = 5

. Solving

3x^2 - 7 = 5

gives

3x^2 = 12 \implies x^2 = 4 \implies x = 2

(since

P

is in the first quadrant).

2

Calculate the y-coordinate of point P

y = 9

Substitute

x = 2

into the curve equation

y = 2^3 - 7(2) + 15 = 8 - 14 + 15 = 9

. Thus,

P = (2, 9)

.

3

Determine the gradient of the normal line

m_n = -\frac{1}{5}

The normal line is perpendicular to the tangent line. Since the tangent gradient

m_t = 5

, the normal gradient is

m_n = -\frac{1}{m_t} = -\frac{1}{5}

.

4

Formulate the equation of the normal line

x + 5y - 47 = 0

Using the point-slope equation

y - y_1 = m_n(x - x_1)

with point

(2, 9)

and gradient

-\frac{1}{5}

:

y - 9 = -\frac{1}{5}(x - 2) \implies 5(y - 9) = -(x - 2) \implies 5y - 45 = -x + 2 \implies x + 5y - 47 = 0

.

Anahtar Kavram

The gradient of the normal to a curve at a given point is the negative reciprocal of the derivative (tangent gradient) at that point:

m_{normal} = -\frac{1}{\frac{dy}{dx}}

.

Soru 67Soru →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{x^2 + kx - 10}{x - 2}, & x \neq 2 \\ 7, & x = 2 \end{cases}

If

f(x)

is continuous at

x = 2

, what is the numerical value of the constant

k

?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The numerical value of the constant

k

is 3.

By definition of continuity,

f(x)

is continuous at

x = 2

if

\lim_{x \to 2} f(x) = f(2) = 7

. As

x \to 2

, the denominator

x - 2

approaches

0

. For the quotient to have a finite limit, the numerator

x^2 + kx - 10

must also evaluate to

0

at

x = 2

, yielding

2^2 + 2k - 10 = 0

. Solving this gives

2k = 6

, so

k = 3

. Substituting

k = 3

gives

\lim_{x \to 2} \frac{(x-2)(x+5)}{x-2} = 7

, confirming that

k = 3

is correct.

Adım Adım Çözüm

1

Apply the definition of continuity at

x = 2

\lim_{x \to 2} f(x) = f(2) = 7

A function

f(x)

is continuous at

x = a

if and only if

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

.

2

Set the numerator to zero at the point of discontinuity

x = 2

2^2 + k(2) - 10 = 0

Because the denominator

(x - 2) \to 0

as

x \to 2

, the limit can only exist if the numerator also approaches

0

, forming an indeterminate form

\frac{0}{0}

that can be simplified.

3

Solve for the unknown parameter

k

4 + 2k - 10 = 0 \implies 2k - 6 = 0 \implies k = 3

Linear algebraic equation solving.

4

Verify that the simplified limit equals

f(2)

\lim_{x \to 2} \frac{x^2 + 3x - 10}{x - 2} = \lim_{x \to 2} \frac{(x - 2)(x + 5)}{x - 2} = \lim_{x \to 2} (x + 5) = 7

Canceling the common factor

(x - 2)

yields

7

, which matches

f(2) = 7

.

Anahtar Kavram

Continuity of a Piecewise Function and Limit Existence

Soru 68Soru →

What is the indefinite integral $\int (8x^3 + 12\cos(4x) - 5) \, dx$ ?

2x^4 + 3\sin(4x) - 5x + C

2x^4 + 3\sin(4x) - 5x

2x^4 - 3\sin(4x) - 5x + C

2x^4 + 48\sin(4x) - 5x + C

Cevabı ve açıklamayı göster

Cevap: $2x^4 + 3\sin(4x) - 5x + C$

Cevap

2x^4 + 3\sin(4x) - 5x + C

Integrating term by term gives

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

,

\int 12\cos(4x) \, dx = 3\sin(4x)

, and

\int -5 \, dx = -5x

, along with the constant of integration

C

, yielding

2x^4 + 3\sin(4x) - 5x + C

.

Adım Adım Çözüm

1

Integrate the polynomial term

8x^3

\frac{8x^{3+1}}{3+1} = \frac{8x^4}{4} = 2x^4

Apply the power rule for integration:

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

.

2

Integrate the trigonometric term

12\cos(4x)

12 \cdot \frac{\sin(4x)}{4} = 3\sin(4x)

Apply the standard trigonometric integral rule:

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

.

3

Integrate the constant term

-5

-5x

The integral of a constant

k

with respect to

x

is

kx

.

4

Combine all integrated terms and add the arbitrary constant of integration

2x^4 + 3\sin(4x) - 5x + C

An indefinite integral represents a family of functions and requires the addition of

+ C

.

Anahtar Kavram

Indefinite integration of polynomial and trigonometric functions

Tahmini Süre:1m 30s

Soru 69Soru →

What is the indefinite integral $\int \left( (2x - 3)^2 + 12\sin(4x) - 6\cos(3x) \right) dx$ ?

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

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

\frac{4}{3}x^3 - 6x^2 + 9x + 3\cos(4x) - 2\sin(3x) + C

\frac{4}{3}x^3 - 6x^2 + 9x - 48\cos(4x) + 18\sin(3x) + C

Cevabı ve açıklamayı göster

Cevap: $\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C$

Cevap

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

Expanding

(2x-3)^2

yields

4x^2 - 12x + 9

. Integrating term by term:

\int 4x^2 dx = \frac{4}{3}x^3

,

\int -12x dx = -6x^2

,

\int 9 dx = 9x

,

\int 12\sin(4x) dx = -3\cos(4x)

, and

\int -6\cos(3x) dx = -2\sin(3x)

. Summing these results and adding the constant of integration

C

gives

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

.

Adım Adım Çözüm

1

Expand the squared binomial inside the integrand.

(2x - 3)^2 = 4x^2 - 12x + 9

Expanding algebraic terms simplifies the integrand into basic power terms ready for integration.

2

Integrate the polynomial terms using the power rule

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

.

\int (4x^2 - 12x + 9) dx = \frac{4}{3}x^3 - 6x^2 + 9x

Applying the power rule to each term individually gives

\frac{4x^3}{3} - \frac{12x^2}{2} + 9x

.

3

Integrate the trigonometric terms using standard integration rules

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

and

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

.

\int 12\sin(4x) dx = -\frac{12}{4}\cos(4x) = -3\cos(4x) and \int -6\cos(3x) dx = -\frac{6}{3}\sin(3x) = -2\sin(3x)

Integrating sine gives negative cosine divided by the inner linear coefficient, and integrating cosine gives sine divided by the inner linear coefficient.

4

Combine all integrated terms and append the constant of integration

C

.

\frac{4}{3}x^3 - 6x^2 + 9x - 3\cos(4x) - 2\sin(3x) + C

Indefinite integrals require the addition of an arbitrary constant of integration

C

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 70Soru →

Given that $\int_{0}^{p} (6x - 4) \, dx = 15$ and $p > 0$ , what is the value of $p$ ?

3

\frac{5}{3}

5

\frac{19}{6}

Cevabı ve açıklamayı göster

Cevap: $3$

Cevap

The value of

p

is

3

.

Integrating

6x - 4

yields

3x^2 - 4x

. Substituting the limits from

0

to

p

gives

3p^2 - 4p

. Equating this to

15

produces the quadratic equation

3p^2 - 4p - 15 = 0

, which factors as

(3p + 5)(p - 3) = 0

. Since

p > 0

, the only valid solution is

3

.

Adım Adım Çözüm

1

Find the indefinite integral of the integrand

f(x) = 6x - 4

.

\int (6x - 4) \, dx = 3x^2 - 4x

Apply the power rule of integration:

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

.

2

Evaluate the definite integral from lower limit

0

to upper limit

p

.

[3x^2 - 4x]_{0}^{p} = (3p^2 - 4p) - (0) = 3p^2 - 4p

Substitute the upper and lower limits into the antiderivative.

3

Set the evaluated definite integral equal to the given value of

15

and form a quadratic equation.

3p^2 - 4p - 15 = 0

Equate the definite integral value to 15.

4

Factor the quadratic equation to solve for

p

.

(3p + 5)(p - 3) = 0 \implies p = -\frac{5}{3}

or

p = 3

Find two numbers that multiply to

-45

and add up to

-4

, which are

-9

and

5

.

5

Select the valid positive value of

p

as specified in the question (

p > 0

).

p = 3

Disregard the negative root since

p > 0

.

Anahtar Kavram

Determining an unknown boundary limit of a definite integral

Soru 71Soru →

A function

f(x)

is defined by

f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & x \neq 3 \\ k + 2, & x = 3 \end{cases}

If

f(x)

is continuous at

x = 3

, what is the value of the constant

k

?

3

5

7

-2

Cevabı ve açıklamayı göster

Cevap: $5$

Cevap

The constant value is

k = 5

.

For the function to be continuous at

x = 3

, the limit as

x \to 3

must equal the value of the function at

x = 3

, which is

f(3) = k + 2

. Factoring the numerator gives

2x^2 - 5x - 3 = (2x + 1)(x - 3)

. Canceling the common factor

(x - 3)

leaves

\lim_{x \to 3}(2x + 1) = 7

. Setting

k + 2 = 7

yields

k = 5

.

Adım Adım Çözüm

1

Evaluate the limit of

f(x)

as

x

approaches

3

\lim_{x \to 3} \frac{2x^2 - 5x - 3}{x - 3} = \lim_{x \to 3} \frac{(2x + 1)(x - 3)}{x - 3} = \lim_{x \to 3} (2x + 1) = 2(3) + 1 = 7

Direct substitution yields the indeterminate form

\frac{0}{0}

, so the numerator must be factored to cancel the common term

(x - 3)

.

2

Apply the definition of continuity at

x = 3

f(3) = \lim_{x \to 3} f(x) \implies k + 2 = 7

For a function to be continuous at a point

x = c

, the function value

f(c)

must equal the limit of

f(x)

as

x \to c

.

3

Solve for the constant

k

k = 7 - 2 = 5

Subtract

2

from both sides of the equation.

Anahtar Kavram

Continuity of a Piecewise Function at a Point

Soru 72Soru →

What is the $x$ -intercept of the normal line to the curve $y = x^2 - 4x + 5$ at the point where $x = 3$ ?

2

7

-1

-7

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The

x

-intercept of the normal line is 7.

Differentiating

y = x^2 - 4x + 5

gives

\frac{dy}{dx} = 2x - 4

. Substituting

x = 3

yields a tangent gradient of

2

, so the normal gradient is

-\frac{1}{2}

. The curve passes through

(3, 2)

at

x = 3

. The equation of the normal is

y - 2 = -\frac{1}{2}(x - 3)

, which simplifies to

x + 2y - 7 = 0

. Setting

y = 0

gives

x = 7

.

Adım Adım Çözüm

1

Find the

y

-coordinate corresponding to

x = 3

y = 3^2 - 4(3) + 5 = 9 - 12 + 5 = 2

. The point on the curve is

(3, 2)

.

The normal line passes through the specific point of tangency on the curve.

2

Calculate the gradient of the tangent and normal lines at

x = 3

\frac{dy}{dx} = 2x - 4

. At

x = 3

,

m_t = 2(3) - 4 = 2

. Therefore,

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

.

The normal line is perpendicular to the tangent line.

3

Determine the equation of the normal line

y - 2 = -\frac{1}{2}(x - 3) \implies 2(y - 2) = -(x - 3) \implies x + 2y - 7 = 0

.

Use the point-slope form

y - y_1 = m(x - x_1)

to write the linear equation.

4

Find the

x

-intercept of the normal line

Set

y = 0

:

x + 2(0) - 7 = 0 \implies x = 7

.

The

x

-intercept occurs where

y = 0

.

Anahtar Kavram

Tangents and Normals to Curves

Tahmini Süre:1m 30s

Soru 73Soru →

What is the result of evaluating the indefinite integral $\int \left( (2x - 1)^2 - 6\cos(2x) \right) \, dx$ ?

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

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

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

\frac{4}{3}x^3 - 2x^2 + x - 12\sin(2x) + C

Cevabı ve açıklamayı göster

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

Cevap

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

Expanding the squared term

(2x - 1)^2

yields

4x^2 - 4x + 1

. Integrating term by term using the power rule gives

\frac{4}{3}x^3 - 2x^2 + x

. The integral of

-6\cos(2x)

is

-\frac{6}{2}\sin(2x) = -3\sin(2x)

. Adding the arbitrary constant

C

results in

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

.

Adım Adım Çözüm

1

Expand the squared binomial term inside the integral

(2x - 1)^2 = 4x^2 - 4x + 1

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

2

Integrate each term of the expanded polynomial

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

Apply the power rule for integration: \int x^n \, dx = \frac{x^{n+1}}{n+1}.

3

Integrate the trigonometric term

\int -6\cos(2x) \, dx = -6 \cdot \frac{1}{2}\sin(2x) = -3\sin(2x)

The standard integral of \cos(kx) is \frac{1}{k}\sin(kx).

4

Combine all integrated terms and append the constant of integration

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

Indefinite integration requires an arbitrary constant of integration C.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions

Soru 74Soru →

Using differentiation from first principles, what is the derivative $\frac{\mathrm{d}y}{\mathrm{d}x}$ of the function $f(x) = 3x^2 - 2x$ ?

6x - 2

6x + 2

3x - 2

6x^2 - 2

Cevabı ve açıklamayı göster

Cevap: $6x - 2$

Cevap

The derivative of

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

with respect to

x

is

6x - 2

.

Applying the first principles definition

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

to

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

yields

\frac{3x^2 + 6xh + 3h^2 - 2x - 2h - (3x^2 - 2x)}{h} = \frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2

. As

h \to 0

, this simplifies directly to

6x - 2

.

Adım Adım Çözüm

1

Set up the definition of the derivative from first principles.

\frac{\mathrm{d}y}{\mathrm{d}x} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}

The definition of the derivative is the limit of the difference quotient as

h

approaches zero.

2

Substitute

f(x+h)

and

f(x)

into the difference quotient.

f(x+h) = 3(x+h)^2 - 2(x+h) = 3(x^2 + 2xh + h^2) - 2x - 2h = 3x^2 + 6xh + 3h^2 - 2x - 2h

Expand

(x+h)^2

completely using algebraic identity.

3

Subtract

f(x)

from

f(x+h)

to find the numerator.

f(x+h) - f(x) = (3x^2 + 6xh + 3h^2 - 2x - 2h) - (3x^2 - 2x) = 6xh + 3h^2 - 2h

Cancel out identical terms

3x^2

and

-2x

.

4

Divide by

h

and evaluate the limit as

h \to 0

.

\frac{6xh + 3h^2 - 2h}{h} = 6x + 3h - 2; \quad \lim_{h \to 0} (6x + 3h - 2) = 6x - 2

Factor out

h

to cancel the denominator, then set

h = 0

.

Anahtar Kavram

Differentiation from First Principles

Tahmini Süre:1m 30s

Soru 75Soru →

Evaluate the definite integral $\int_{1}^{3} (4x^3 - 6x) \, dx$ .

Cevabı ve açıklamayı göster

Cevap: 56

Cevap

The value of the definite integral is 56.

To evaluate the definite integral

\int_{1}^{3} (4x^3 - 6x) \, dx

, integrate the function term-by-term to obtain

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

. Evaluating

F(x)

at the upper bound

x = 3

yields

54

, and evaluating at the lower bound

x = 1

yields

-2

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

54 - (-2) = 56

.

Adım Adım Çözüm

1

Find the indefinite integral of the function

\int (4x^3 - 6x) \, dx = x^4 - 3x^2

Apply the power rule of integration \int x^n \, dx = \frac{x^{n+1}}{n+1}

2

Substitute the upper limit x = 3 into the antiderivative F(x)

F(3) = 3^4 - 3(3)^2 = 81 - 27 = 54

Evaluate F(b) for b = 3

3

Substitute the lower limit x = 1 into the antiderivative F(x)

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

Evaluate F(a) for a = 1

4

Subtract F(a) from F(b)

54 - (-2) = 56

Apply the Fundamental Theorem of Calculus: \int_{a}^{b} f(x) \, dx = F(b) - F(a)

Anahtar Kavram

Definite Integration using the Fundamental Theorem of Calculus

Soru 76Soru →

Find the $y$ -intercept of the tangent line to the curve $y = x^3 - 6x + 8$ at the point where $x = 1$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The

y

-intercept of the tangent line is 6.

Differentiating

y = x^3 - 6x + 8

yields

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

. At

x = 1

, the point on the curve is

(1, 3)

and the gradient of the tangent is

m = -3

. Substituting into the point-slope form

y - 3 = -3(x - 1)

gives

y = -3x + 6

. Setting

x = 0

identifies the

y

-intercept as 6.

Adım Adım Çözüm

1

Find the point of tangency on the curve

For

x = 1

,

y = (1)^3 - 6(1) + 8 = 3

. The point is

(1, 3)

.

The tangent line touches the curve at the point corresponding to

x = 1

.

2

Find the gradient function of the curve

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

The derivative of the curve equation gives the gradient of the tangent at any point.

3

Calculate the gradient of the tangent line at

x = 1

m = 3(1)^2 - 6 = -3

Substitute

x = 1

into the derivative.

4

Formulate the equation of the tangent line

y - 3 = -3(x - 1) \implies y = -3x + 6

Apply the point-slope line equation

y - y_1 = m(x - x_1)

using

(1, 3)

and

m = -3

.

5

Determine the

y

-intercept

Setting

x = 0

yields

y = 6

.

The

y

-intercept is the value of

y

when

x = 0

.

Anahtar Kavram

Tangents to Curves and Axis Intercepts

Soru 77Soru →

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

-3\sin(3x) + e^x

3\sin(3x) + e^x

-\sin(3x) + e^x

\sin(3x) + e^x

Cevabı ve açıklamayı göster

Cevap: $-3\sin(3x) + e^x$

Cevap

-3\sin(3x) + e^x

Differentiating

y = \cos(3x) + e^x

term-by-term yields

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

. By the chain rule,

\frac{d}{dx}\cos(3x) = -3\sin(3x)

, while the derivative of

e^x

is simply

e^x

.

Adım Adım Çözüm

1

Differentiate the trigonometric term

\cos(3x)

using the chain rule

\frac{d}{dx}(\cos(3x)) = -\sin(3x) \cdot \frac{d}{dx}(3x) = -3\sin(3x)

The derivative of

\cos(u)

is

-\sin(u) \cdot u'

2

Differentiate the exponential term

e^x

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

The exponential function

e^x

is its own derivative

3

Combine the results using the sum rule

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

The derivative of a sum of functions is the sum of their individual derivatives

Anahtar Kavram

Differentiation of trigonometric and exponential functions using the chain rule

Soru 78Soru →

A piecewise function

f(x)

is defined by

f(x) = \begin{cases} \frac{2x^2 - 5x - 3}{x - 3}, & \text{for } x \neq 3 \ a^2 - 2, & \text{for } x = 3 \end{cases}

If

f(x)

is continuous at

x = 3

and

a > 0

, what is the numerical value of

a

?

Cevabı ve açıklamayı göster

Cevap: 3

Cevap

The numerical value of a is 3.

For

f(x)

to be continuous at

x = 3

, the defined value

f(3) = a^2 - 2

must equal

\lim_{x \to 3} f(x)

. Factoring the numerator gives

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

for

x \neq 3

. Taking the limit as

x \to 3

yields

2(3) + 1 = 7

. Setting

a^2 - 2 = 7

leads to

a^2 = 9

, which gives

a = 3

under the constraint

a > 0

.

Adım Adım Çözüm

1

Evaluate the limit of

f(x)

as

x \to 3

Factor the numerator

2x^2 - 5x - 3 = (2x + 1)(x - 3)

. For

x \neq 3

,

f(x) = 2x + 1

. Thus,

\lim_{x \to 3} f(x) = 2(3) + 1 = 7

.

Direct substitution gives an indeterminate form

\frac{0}{0}

, so canceling the common factor

(x - 3)

allows direct evaluation of the limit.

2

Apply the definition of continuity at

x = 3

f(3) = a^2 - 2 = 7

.

For a function to be continuous at a point

c

, the function value

f(c)

must equal the limit

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

.

3

Solve for the parameter

a

a^2 = 9 \implies a = 3

(since

a > 0

).

Solving

a^2 = 9

gives solutions

3

and

-3

. The condition

a > 0

specifies the positive root.

Anahtar Kavram

Continuity of a Piecewise Function at a Point

Soru 79Soru →

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

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 1

is

11

.

Applying the product rule

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

along with the chain rule for the expression

(2x - 1)^4

gives

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

. Evaluating this at

x = 1

gives

8(1)^3(1)^3 + 3(1)^2(1)^4 = 8 + 3 = 11

.

Adım Adım Çözüm

1

Set up the product rule for

y = u \cdot v

, where

u = x^3

and

v = (2x - 1)^4

.

u = x^3

and

v = (2x - 1)^4

The function is expressed as the product of two algebraic terms.

2

Find the derivative of each function component.

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

and

\frac{dv}{dx} = 8(2x - 1)^3

The power rule gives

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

, and applying the chain rule to

(2x - 1)^4

yields

4(2x - 1)^3 \cdot 2 = 8(2x - 1)^3

.

3

Substitute components into the product rule formula

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

.

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

Combining

u\frac{dv}{dx}

and

v\frac{du}{dx}

provides the full expression for the derivative.

4

Evaluate the derivative at

x = 1

.

\frac{dy}{dx}\Big|_{x=1} = 8(1)^3(2(1) - 1)^3 + 3(1)^2(2(1) - 1)^4 = 8 + 3 = 11

Substituting

x = 1

simplifies the terms to

8(1) + 3(1) = 11

.

Anahtar Kavram

Rules of Differentiation (Product and Chain Rules)

Soru 80Soru →

If $y = \ln(\sec 3x + \tan 3x)$ , what is $\frac{dy}{dx}$ ?

\sec 3x

3\sec 3x

3\tan 3x

\frac{3}{\sec 3x + \tan 3x}

Cevabı ve açıklamayı göster

Cevap: $3\sec 3x$

Cevap

3\sec 3x

Using the chain rule for

y = \ln u

where

u = \sec 3x + \tan 3x

, we find

u' = 3\sec 3x\tan 3x + 3\sec^2 3x = 3\sec 3x(\tan 3x + \sec 3x)

. Dividing

u'

by

u

cancels out

(\sec 3x + \tan 3x)

, leaving

3\sec 3x

.

Adım Adım Çözüm

1

Apply the chain rule for logarithmic functions

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

.

Set

u = \sec 3x + \tan 3x

, so

\frac{dy}{dx} = \frac{1}{\sec 3x + \tan 3x} \cdot \frac{d}{dx}(\sec 3x + \tan 3x)

.

The function is a composite function of the form

y = \ln(u(x))

.

2

Differentiate the inner function

u = \sec 3x + \tan 3x

using the chain rule.

\frac{du}{dx} = 3\sec 3x \tan 3x + 3\sec^2 3x.

The derivative of

\sec(ax)

is

a\sec(ax)\tan(ax)

and the derivative of

\tan(ax)

is

a\sec^2(ax)

.

3

Factor out common terms in the numerator and simplify the expression.

\frac{dy}{dx} = \frac{3\sec 3x(\tan 3x + \sec 3x)}{\sec 3x + \tan 3x} = 3\sec 3x.

The term

(\tan 3x + \sec 3x)

in the numerator cancels with the denominator

(\sec 3x + \tan 3x)

.

Anahtar Kavram

Differentiation of Logarithmic and Trigonometric Functions via the Chain Rule

Tahmini Süre:1m 30s

Calculus

Cevap

Adım Adım Çözüm

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