Tüm alıştırma soruları

Soru 7141Soru →

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 7142Soru →

If $24_x \times 13_x = 345_x$ , where $x$ represents a positive integer base, find the value of $x$ .

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Cevap: 7

Cevap

The value of the base

x

is 7.

Expanding

24_x

,

13_x

, and

345_x

into base 10 yields

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

. Expanding the left side gives

2x^2 + 10x + 12

. Equating and simplifying gives

x^2 - 6x - 7 = 0

, which factors as

(x - 7)(x + 1) = 0

. The positive integer solution greater than 5 is

x = 7

.

Adım Adım Çözüm

1

Convert all base

x

numbers to decimal (base 10) expressions.

24_x = 2x + 4

,

13_x = x + 3

, and

345_x = 3x^2 + 4x + 5

.

Place-value expansion allows algebraic manipulation in standard base 10.

2

Multiply the expanded factors on the left-hand side.

(2x + 4)(x + 3) = 2x^2 + 10x + 12

.

Applying the distributive property of multiplication.

3

Equate the expanded left-hand side to the right-hand side and rearrange into standard quadratic form.

3x^2 + 4x + 5 - (2x^2 + 10x + 12) = 0 \implies x^2 - 6x - 7 = 0

.

Setting the quadratic expression equal to zero enables factoring.

4

Factor the quadratic equation and select the valid base.

(x - 7)(x + 1) = 0 \implies x = 7

(rejecting

x = -1

).

A number base must be a positive integer strictly greater than any individual digit in the given numbers (max digit is 5).

Anahtar Kavram

Solving polynomial equations derived from number base expansion.

Soru 7143Soru →

If $\log_2 (x - 1) + \log_4 (x - 1) + \log_{16} (x - 1) = \frac{7}{2}$ , what is the value of $x$ ?

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Cevap: 5

Cevap

The value of

x

is

5

.

Converting all terms to base 2 yields

\log_2(x-1) + \frac{1}{2}\log_2(x-1) + \frac{1}{4}\log_2(x-1) = \frac{7}{4}\log_2(x-1)

. Setting

\frac{7}{4}\log_2(x-1) = \frac{7}{2}

gives

\log_2(x-1) = 2

. Exponentiating both sides in base 2 gives

x - 1 = 2^2 = 4

, which results in

x = 5

.

Adım Adım Çözüm

1

Convert each logarithmic term to base 2 using the change of base property.

\log_4(x-1) = \frac{1}{2}\log_2(x-1) \quad \text{and} \quad \log_{16}(x-1) = \frac{1}{4}\log_2(x-1)

Bases 4 and 16 are powers of 2 (

4 = 2^2

and

16 = 2^4

), allowing transformation to a common base.

2

Substitute these equivalent base-2 terms into the original equation.

\log_2(x-1) + \frac{1}{2}\log_2(x-1) + \frac{1}{4}\log_2(x-1) = \frac{7}{2}

This consolidates the equation into a single logarithmic variable,

\log_2(x-1)

.

3

Factor out

\log_2(x-1)

and add the fractional coefficients.

\left(1 + \frac{1}{2} + \frac{1}{4}\right)\log_2(x-1) = \frac{7}{4}\log_2(x-1) = \frac{7}{2}

Summing the coefficients

1 + \frac{1}{2} + \frac{1}{4} = \frac{7}{4}

.

4

Isolate

\log_2(x-1)

and solve for

x

.

\log_2(x-1) = 2 \implies x - 1 = 2^2 = 4 \implies x = 5

Multiplying both sides by

\frac{4}{7}

yields

\log_2(x-1) = 2

, and rewriting in exponential form gives

x = 5

.

Anahtar Kavram

Change of base rule for logarithms:

\log_{b^k} a = \frac{1}{k}\log_b a

Soru 7144Soru →

Find the number of distinct arrangements of the letters of the word $\text{PARALLEL}$ such that no two letters $\text{'L'}$ are adjacent.

Cevabı ve açıklamayı göster

Cevap: 1200

Cevap

1200

To ensure no two letters 'L' are adjacent, we use the gap method. First, arrange the 5 non-L letters (P, A, R, A, E). Because 'A' repeats twice, there are 5! / 2! = 60 distinct arrangements. These 5 letters form 6 available gaps (including the ends). Selecting 3 of these 6 gaps to insert the 3 identical 'L's can be done in C(6, 3) = 20 ways. Multiplying these gives 60 × 20 = 1200 valid arrangements.

Adım Adım Çözüm

1

Count the frequency of each letter in the word PARALLEL.

The word has 8 letters: 1 P, 2 A's, 1 R, 3 L's, and 1 E.

Recognizing repeated elements is essential for permutations with duplicates.

2

Arrange the non-restricted letters (P, A, R, A, E).

Number of arrangements = 5! / 2! = 60.

The letter 'A' is repeated twice, so we divide 5! by 2!.

3

Calculate the number of available gaps for placing the 3 'L's so that no two are adjacent.

5 arranged letters create 6 gap positions. Choosing 3 gaps gives C(6, 3) = (6 × 5 × 4) / (3 × 2 × 1) = 20 ways.

Placing at most one 'L' per gap guarantees that no two 'L's will be adjacent.

4

Multiply the number of arrangements of non-L letters by the gap choices.

Total arrangements = 60 × 20 = 1200.

By the fundamental counting principle, total arrangements equal the product of independent choices.

Anahtar Kavram

Permutations with repeated elements and non-adjacency constraints using the Gap Method

Tahmini Süre:1m 30s

Soru 7145Soru →

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 7146Soru →

What is the simplified form of the surd expression $\frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}}$ after rationalizing the denominator?

5 + 2\sqrt{6}

5 + 12\sqrt{6}

5

5\sqrt{6}

Cevabı ve açıklamayı göster

Cevap: $5 + 2\sqrt{6}$

Cevap

5 + 2\sqrt{6}

To rationalize

\frac{3\sqrt{2} + 2\sqrt{3}}{3\sqrt{2} - 2\sqrt{3}}

, we multiply both the numerator and denominator by the conjugate of the denominator,

3\sqrt{2} + 2\sqrt{3}

. Expanding the numerator yields

(3\sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 18 + 12\sqrt{6} + 12 = 30 + 12\sqrt{6}

. Expanding the denominator gives

(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6

. Dividing

30 + 12\sqrt{6}

by

6

gives

5 + 2\sqrt{6}

.

Adım Adım Çözüm

1

Identify the conjugate of the denominator

The conjugate of

3\sqrt{2} - 2\sqrt{3}

is

3\sqrt{2} + 2\sqrt{3}

.

Multiplying by the conjugate converts the binomial denominator into a rational number using the difference of squares identity

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

.

2

Multiply the numerator and denominator by the conjugate

\frac{(3\sqrt{2} + 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} - 2\sqrt{3})(3\sqrt{2} + 2\sqrt{3})}

This maintains expression equivalence while eliminating radicals from the denominator.

3

Expand the numerator and denominator independently

Numerator:

(3\sqrt{2})^2 + 2(3\sqrt{2})(2\sqrt{3}) + (2\sqrt{3})^2 = 18 + 12\sqrt{6} + 12 = 30 + 12\sqrt{6}

. Denominator:

(3\sqrt{2})^2 - (2\sqrt{3})^2 = 18 - 12 = 6

.

Apply algebraic expansion

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

and simplification of radicals.

4

Divide each term in the numerator by the denominator

\frac{30 + 12\sqrt{6}}{6} = \frac{30}{6} + \frac{12\sqrt{6}}{6} = 5 + 2\sqrt{6}.

Simplify the fraction to express the answer in standard surd form

a + b\sqrt{c}

.

Anahtar Kavram

Rationalization of Binomial Denominators

Soru 7147Soru →

Find the value of

x

that satisfies the exponential equation

\sqrt{\frac{125^{x+1}}{5^{x-1}}} = 25^{x-1}

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The value of

x

is 4.

Converting all terms to base 5 yields

5^{x+2}

on the left-hand side and

5^{2x-2}

on the right-hand side. Setting the exponents equal gives

x + 2 = 2x - 2

, which solves to

x = 4

.

Adım Adım Çözüm

1

Express all terms with a common base of 5

125 = 5^3

and

25 = 5^2

Converting terms to prime base 5 allows the application of standard laws of indices.

2

Simplify the fraction inside the square root

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

Subtract the denominator exponent from the numerator exponent when dividing like bases.

3

Apply the square root as a fractional exponent

\sqrt{5^{2x+4}} = (5^{2x+4})^{1/2} = 5^{x+2}

Taking the square root of a power is equivalent to multiplying the exponent by 1/2.

4

Equate the simplified exponents of both sides

x + 2 = 2x - 2

With identical bases of 5 on both sides, the exponents must be equal.

5

Solve the linear equation for x

x = 4

Rearranging terms gives

2x - x = 2 + 2

, which yields

x = 4

.

Anahtar Kavram

Indices and Laws of Indices

Soru 7148Soru →

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 7149Soru →

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 7150Soru →

What is the value of $(17 - 42) \pmod{9}$ expressed in standard non-negative remainder form?

7

2

5

4

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

(17 - 42) \pmod{9}

in standard non-negative remainder form is

2

.

Evaluating

17 - 42

gives

-25

. Reducing

-25

modulo

9

requires finding a non-negative remainder

r

such that

-25 = 9q + r

with

0 \le r < 9

. Choosing

q = -3

yields

-25 = 9(-3) + 2

, making the correct non-negative remainder

2

. Alternatively,

17 \equiv 8 \pmod{9}

and

42 \equiv 6 \pmod{9}

, so

8 - 6 = 2 \pmod{9}

.

Adım Adım Çözüm

1

Perform the subtraction within the parentheses.

17 - 42 = -25

Evaluate the arithmetic expression inside the modulo operation first.

2

Express

-25

in terms of the modulus

9

using the division algorithm

a = qn + r

, where

0 \le r < n

.

-25 = 9 \times (-3) + 2

The remainder

r

must be non-negative (

0 \le r < 9

).

3

Identify the canonical non-negative remainder.

r = 2

Alternatively,

-25 \equiv -7 \pmod{9}

, and adding the modulus gives

-7 + 9 = 2

.

Anahtar Kavram

Modular Arithmetic and Negative Remainder Reduction

Tahmini Süre:45s

Soru 7151Soru →

Find the value of

x

that satisfies the exponential equation

\frac{100^{x+1}}{1000^{x-1}} = 10^{2x-1}

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

x

that satisfies the equation is 2.

Rewriting the terms in base 10 gives

\frac{10^{2x+2}}{10^{3x-3}} = 10^{2x-1}

. Using the division rule of indices yields

10^{-x+5} = 10^{2x-1}

. Equating exponents gives

-x+5 = 2x-1

, which simplifies to

3x = 6

, so

x = 2

.

Adım Adım Çözüm

1

Convert each power to base 10

100^{x+1} = (10^2)^{x+1} = 10^{2x+2}

and

1000^{x-1} = (10^3)^{x-1} = 10^{3x-3}

Converting all non-prime composite bases to powers of a common fundamental base allows exponent comparison.

2

Apply the quotient rule of indices to the left side

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

According to the index quotient law

\frac{a^m}{a^n} = a^{m-n}

, subtract the denominator's exponent from the numerator's exponent.

3

Equate exponents of equal bases

-x + 5 = 2x - 1

If

a^f(x) = a^g(x)

for

a > 0, a \neq 1

, then

f(x) = g(x)

.

4

Solve the linear equation for

x

3x = 6 \implies x = 2

Isolate the variable

x

to find its value.

Anahtar Kavram

Exponential equations solvable by converting to a common base

Soru 7152Soru →

If $\log_3 x + \log_9 x + \log_{27} x = \frac{11}{2}$ , what is the value of $x$ ?

27

9

3

81

Cevabı ve açıklamayı göster

Cevap: $27$

Cevap

The value of

x

is

27

.

Applying the change of base identity yields

\log_9 x = \frac{1}{2}\log_3 x

and

\log_{27} x = \frac{1}{3}\log_3 x

. Combining like terms gives

\frac{11}{6}\log_3 x = \frac{11}{2}

, which reduces to

\log_3 x = 3

. Expressing in exponential form yields

x = 3^3 = 27

.

Adım Adım Çözüm

1

Express all logarithms in base 3 using the change of base rule

\log_{a^k} x = \frac{1}{k} \log_a x

.

\log_9 x = \frac{1}{2} \log_3 x \quad \text{and} \quad \log_{27} x = \frac{1}{3} \log_3 x

Converting all terms to a common base allows them to be combined algebraically.

2

Substitute these expressions back into the original equation.

\log_3 x + \frac{1}{2} \log_3 x + \frac{1}{3} \log_3 x = \frac{11}{2}

This creates a single linear equation in terms of

\log_3 x

.

3

Combine the coefficients of

\log_3 x

.

\left(1 + \frac{1}{2} + \frac{1}{3}\right) \log_3 x = \frac{6 + 3 + 2}{6} \log_3 x = \frac{11}{6} \log_3 x = \frac{11}{2}

Adding the fractions gives a single coefficient.

4

Solve for

\log_3 x

and evaluate

x

.

\log_3 x = \frac{11}{2} \times \frac{6}{11} = 3 \implies x = 3^3 = 27

Converting from logarithmic to exponential form gives the value of

x

.

Anahtar Kavram

Change of base formula for logarithms:

\log_{b^k} x = \frac{1}{k} \log_b x

Tahmini Süre:1m 30s

Soru 7153Soru →

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 7154Soru →

A set of six numbers arranged in ascending order is given as $3, 8, x, 14, y, 25$ . If the median of the data set is $11$ and its arithmetic mean is $13$ , what is the value of $y$ ?

17

20

22

16

Cevabı ve açıklamayı göster

Cevap: $20$

Cevap

The value of

y

is

20

.

The median of an even number of values arranged in order is the average of the two central terms, which gives

\frac{x + 14}{2} = 11

, solving to

x = 8

. Since the mean of the six numbers is

13

, their total sum is

6 \times 13 = 78

. Adding all terms gives

3 + 8 + 8 + 14 + y + 25 = 58 + y = 78

, which yields

y = 20

.

Adım Adım Çözüm

1

Find the value of

x

using the median of the six ordered numbers.

\frac{x + 14}{2} = 11 \implies x + 14 = 22 \implies x = 8

For an even number of observations (

n = 6

), the median is the average of the 3rd term (

x

) and 4th term (

14

).

2

Calculate the required total sum of all six numbers using the given mean.

\text{Total Sum} = 6 \times 13 = 78

The mean of

n

numbers is equal to the sum of the numbers divided by

n

.

3

Sum all six terms and solve for

y

.

3 + 8 + 8 + 14 + y + 25 = 78 \implies 58 + y = 78 \implies y = 20

Substitute

x = 8

into the dataset and set the sum of all elements equal to

78

.

Anahtar Kavram

Measures of Central Tendency for Ungrouped Data

Soru 7155Soru →

Which set contains all the solutions to the trigonometric equation $\sin 2x = \cos x$ for $0^\circ \le x \le 180^\circ$ ?

\{30^\circ, 90^\circ, 150^\circ\}

\{30^\circ, 150^\circ\}

\{30^\circ, 90^\circ\}

\{60^\circ, 90^\circ, 120^\circ\}

Cevabı ve açıklamayı göster

Cevap: $\{30^\circ, 90^\circ, 150^\circ\}$

Cevap

The correct set of solutions is \{30^\circ, 90^\circ, 150^\circ\}.

Using the identity

\sin 2x = 2\sin x \cos x

, the equation becomes

2\sin x \cos x - \cos x = 0

. Factoring out

\cos x

gives

\cos x(2\sin x - 1) = 0

. Setting each factor to zero yields

\cos x = 0

(giving

x = 90^\circ

) and

\sin x = \frac{1}{2}

(giving

x = 30^\circ

and

x = 150^\circ

). Thus, the complete set of solutions in the given interval is \{30^\circ, 90^\circ, 150^\circ\}.

Adım Adım Çözüm

1

Apply the double-angle identity for sine.

Substitute

\sin 2x = 2\sin x \cos x

into the equation to get

2\sin x \cos x = \cos x

.

This expresses the equation in terms of single angle

x

.

2

Rearrange and factor the equation.

2\sin x \cos x - \cos x = 0 \implies \cos x(2\sin x - 1) = 0

.

Factoring prevents losing valid roots that occur when a variable factor equals zero.

3

Set each factor to zero and solve for

x

in the interval

0^\circ \le x \le 180^\circ

.

First factor:

\cos x = 0 \implies x = 90^\circ

.
Second factor:

2\sin x - 1 = 0 \implies \sin x = \frac{1}{2} \implies x = 30^\circ

or

x = 180^\circ - 30^\circ = 150^\circ

.

Finding all principal and secondary angles within the specified domain.

4

Combine all valid solutions into a set.

x \in \{30^\circ, 90^\circ, 150^\circ\}

.

All three values satisfy the original equation and lie within

0^\circ \le x \le 180^\circ

.

Anahtar Kavram

Solving trigonometric equations using identities and factoring

Tahmini Süre:1m 30s

Soru 7156Soru →

Find the positive integer value of $x$ that satisfies the logarithmic equation $x^{\log_3 x} = 81x^3$ .

Cevabı ve açıklamayı göster

Cevap: 81

Cevap

The positive integer value of

x

is 81.

Taking the logarithm base 3 on both sides transforms

x^{\log_3 x} = 81x^3

into

(\log_3 x)^2 = 4 + 3\log_3 x

. Substituting

u = \log_3 x

yields the quadratic equation

u^2 - 3u - 4 = 0

, which factors as

(u-4)(u+1) = 0

. The roots are

u = 4

(

x = 3^4 = 81

) and

u = -1

(

x = 3^{-1} = 1/3

). Since

x

must be a positive integer, the correct value is 81.

Adım Adım Çözüm

1

Take logarithm base 3 of both sides of the equation

x^{\log_3 x} = 81x^3

\log_3(x^{\log_3 x}) = \log_3(81x^3)

Taking the logarithm with base 3 allows us to simplify the exponent containing

\log_3 x

.

2

Apply logarithmic identities

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

and

\log_b(mn) = \log_b m + \log_b n

(\log_3 x)^2 = \log_3 81 + 3\log_3 x = 4 + 3\log_3 x

Expanding the products and powers reduces the equation into a single logarithmic variable

\log_3 x

.

3

Substitute

u = \log_3 x

to create a quadratic equation

u^2 - 3u - 4 = 0

Setting

u = \log_3 x

converts the equation into standard quadratic form.

4

Factor the quadratic expression

(u - 4)(u + 1) = 0 \implies u = 4 \text{ or } u = -1

Factoring allows us to find all possible real values for

u

.

5

Convert back to

x

using

x = 3^u

and select the positive integer root

x = 3^4 = 81

or

x = 3^{-1} = \frac{1}{3}

. The positive integer solution is

x = 81

.

The question specifically requests the positive integer solution, eliminating

x = \frac{1}{3}

.

Anahtar Kavram

Solving equations with variable exponents by taking logarithms and reducing to a quadratic form.

Soru 7157Soru →

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 7158Soru →

When the polynomial $P(x) = 2x^3 + 3x^2 - px + q$ is divided by $(x - 1)$ , the remainder is $3$ . Given that $(x + 2)$ is a factor of $P(x)$ , what is the value of $p + q$ ?

2

4

6

-2

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

p + q

is

2

.

According to the Remainder Theorem,

P(1) = 3

gives

2(1)^3 + 3(1)^2 - p(1) + q = 3

, which simplifies to

p - q = 2

. According to the Factor Theorem,

(x + 2)

being a factor means

P(-2) = 0

, giving

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

, which simplifies to

2p + q = 4

. Solving these equations together gives

p = 2

and

q = 0

. Summing them yields

p + q = 2

.

Adım Adım Çözüm

1

Apply the Remainder Theorem for divisor

(x - 1)

P(1) = 2(1)^3 + 3(1)^2 - p(1) + q = 3 \implies 5 - p + q = 3 \implies p - q = 2

By the Remainder Theorem, dividing

P(x)

by

(x - a)

leaves a remainder equal to

P(a)

.

2

Apply the Factor Theorem for factor

(x + 2)

P(-2) = 2(-2)^3 + 3(-2)^2 - p(-2) + q = 0 \implies -16 + 12 + 2p + q = 0 \implies 2p + q = 4

By the Factor Theorem, if

(x - a)

is a factor of

P(x)

, then

P(a) = 0

. Here

a = -2

.

3

Solve the simultaneous linear equations for

p

and

q

Adding

p - q = 2

and

2p + q = 4

yields

3p = 6 \implies p = 2

. Substituting

p = 2

into

p - q = 2

gives

q = 0

.

Eliminating

q

allows direct calculation of

p

, followed by back-substitution for

q

.

4

Calculate the target expression

p + q

p + q = 2 + 0 = 2

Combine the values of

p

and

q

to obtain the required sum.

Anahtar Kavram

Factor and Remainder Theorems for Polynomials

Soru 7159Soru →

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 7160Soru →

Match each pair of organic compounds on the left with its corresponding type of isomerism on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Hexan-2-one and Hexan-3-one exhibit Positional isomerism; Pentane and 2,2-Dimethylpropane exhibit Chain isomerism; Ethoxyethane and Butan-1-ol exhibit Functional group isomerism; (+)-Lactic acid and (-)-Lactic acid exhibit Optical isomerism.

Hexan-2-one and Hexan-3-one differ only in the locant of the carbonyl group along an unchanged six-carbon backbone (positional isomerism). Pentane and 2,2-dimethylpropane differ in the branching of their carbon skeletons (chain isomerism). Ethoxyethane and Butan-1-ol share the formula C4H10O but contain different functional groups (functional group isomerism). (+)-Lactic acid and (-)-Lactic acid are optical enantiomers due to an asymmetric chiral carbon center.

Adım Adım Çözüm

1

Examine Hexan-2-one and Hexan-3-one

Both share the molecular formula

C_6H_{12}O

and contain the carbonyl (

C=O

) functional group. In Hexan-2-one, the carbonyl carbon is at C-2, whereas in Hexan-3-one, it is at C-3.

Molecules with identical functional groups located at different positions on the carbon chain are positional isomers.

2

Examine Pentane and 2,2-Dimethylpropane

Both share the formula

C_5H_{12}

. Pentane is a straight 5-carbon chain (

CH_3-CH_2-CH_2-CH_2-CH_3

), whereas 2,2-Dimethylpropane consists of a 3-carbon chain with two methyl branches,

C(CH_3)_4

.

Molecules with the same molecular formula but different carbon chain structures are chain isomers.

3

Examine Ethoxyethane and Butan-1-ol

Both share the molecular formula

C_4H_{10}O

. Ethoxyethane (

C_2H_5-O-C_2H_5

) is an ether, while Butan-1-ol (

C_4H_9OH

) is a primary alkanol.

Molecules possessing the same molecular formula but belonging to different homologous series with distinct functional groups are functional group isomers.

4

Examine (+)-Lactic acid and (-)-Lactic acid

Lactic acid (

2

-hydroxypropanoic acid) features a central carbon atom bonded to four distinct groups:

-H

,

-OH

,

-CH_3

, and

-COOH

. This chiral center generates two non-superimposable mirror-image forms.

Stereoisomers that rotate plane-polarized light in opposite directions due to molecular chirality are optical isomers.

Anahtar Kavram

Types of Structural Isomerism and Stereoisomerism

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