All practice questions

Question 7141Question →

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?

Show answer & explanation

Answer: -6

Answer

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

.

Step-by-Step Solution

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.

Key Concept

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

Question 7142Question →

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

Show answer & explanation

Answer: 7

Answer

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

.

Step-by-Step Solution

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).

Key Concept

Solving polynomial equations derived from number base expansion.

Question 7143Question →

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

Show answer & explanation

Answer: 5

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

Change of base rule for logarithms:

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

Question 7144Question →

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

Show answer & explanation

Answer: 1200

Answer

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.

Step-by-Step Solution

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.

Key Concept

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

Estimated Time:1m 30s

Question 7145Question →

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}

Show answer & explanation

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

Answer

-\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}

.

Step-by-Step Solution

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

.

Key Concept

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

Question 7146Question →

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}

Show answer & explanation

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

Answer

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}

.

Step-by-Step Solution

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}

.

Key Concept

Rationalization of Binomial Denominators

Question 7147Question →

Find the value of

x

that satisfies the exponential equation

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

Show answer & explanation

Answer: 4

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

Indices and Laws of Indices

Question 7148Question →

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

Show answer & explanation

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

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

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}}

.

Question 7149Question →

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

?

Show answer & explanation

Answer: 3

Answer

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.

Step-by-Step Solution

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

.

Key Concept

Continuity of a Piecewise Function and Limit Existence

Question 7150Question →

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

7

2

5

4

Show answer & explanation

Answer: 2

Answer

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}

.

Step-by-Step Solution

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

.

Key Concept

Modular Arithmetic and Negative Remainder Reduction

Estimated Time:45s

Question 7151Question →

Find the value of

x

that satisfies the exponential equation

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

Show answer & explanation

Answer: 2

Answer

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

.

Step-by-Step Solution

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.

Key Concept

Exponential equations solvable by converting to a common base

Question 7152Question →

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

27

9

3

81

Show answer & explanation

Answer: $27$

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

Change of base formula for logarithms:

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

Estimated Time:1m 30s

Question 7153Question →

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

Show answer & explanation

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

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

Indefinite integration of polynomial and trigonometric functions

Estimated Time:1m 30s

Question 7154Question →

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

Show answer & explanation

Answer: $20$

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

Measures of Central Tendency for Ungrouped Data

Question 7155Question →

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\}

Show answer & explanation

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

Answer

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\}.

Step-by-Step Solution

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

.

Key Concept

Solving trigonometric equations using identities and factoring

Estimated Time:1m 30s

Question 7156Question →

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

Show answer & explanation

Answer: 81

Answer

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.

Step-by-Step Solution

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}

.

Key Concept

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

Question 7157Question →

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

Show answer & explanation

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

Answer

\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

.

Step-by-Step Solution

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

.

Key Concept

Indefinite Integration of Polynomial and Trigonometric Functions

Question 7158Question →

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

Show answer & explanation

Answer: 2

Answer

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

.

Step-by-Step Solution

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.

Key Concept

Factor and Remainder Theorems for Polynomials

Question 7159Question →

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}

Show answer & explanation

Answer: $3$

Answer

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

.

Step-by-Step Solution

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

.

Key Concept

Determining an unknown boundary limit of a definite integral

Question 7160Question →

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

Click a left item, then click its matching right item

Show answer & explanation

Answer

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.

Step-by-Step Solution

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.

Key Concept

Types of Structural Isomerism and Stereoisomerism

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