Tüm alıştırma soruları

What is the value of $\log_2 81 \cdot \log_3 32$ ?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The value of the logarithmic expression is 20.

Rewriting

81

as

3^4

and

32

as

2^5

allows extraction of coefficients

4

and

5

. Utilizing the change of base property

\log_2 3 \cdot \log_3 2 = 1

reduces the expression to

4 \times 5 = 20

.

Adım Adım Çözüm

1

Rewrite arguments as prime powers

\log_2 81 = \log_2(3^4)

and

\log_3 32 = \log_3(2^5)

Simplifies terms to prime base components

2

Apply the power law of logarithms

4\log_2 3 \cdot 5\log_3 2

Brings powers out as numerical coefficients

3

Apply the change of base identity

\log_b a \cdot \log_a b = 1

4 \cdot 5 \cdot (\log_2 3 \cdot \log_3 2) = 20 \cdot 1 = 20

Reciprocal logarithm bases multiply to give 1

Anahtar Kavram

Logarithms and Change of Base

Tahmini Süre:45s

Soru 602Soru →

If $\theta$ is an acute angle such that $\tan \theta = 2$ , calculate the exact numerical value of $\frac{3\sin \theta + \cos \theta}{\sin \theta + 2\cos \theta}$ .

Cevabı ve açıklamayı göster

Cevap: 1.75

Cevap

The exact numerical value of the trigonometric expression is 1.75.

Dividing every term in the expression by

\cos \theta

converts

\sin \theta

terms into

\tan \theta

. The quotient becomes

\frac{3\tan \theta + 1}{\tan \theta + 2}

. Substituting

\tan \theta = 2

gives

\frac{3(2) + 1}{2 + 2} = \frac{7}{4} = 1.75

.

Adım Adım Çözüm

1

Divide numerator and denominator by

\cos \theta

The expression becomes

\frac{3\left(\frac{\sin \theta}{\cos \theta}\right) + 1}{\frac{\sin \theta}{\cos \theta} + 2} = \frac{3\tan \theta + 1}{\tan \theta + 2}

.

Using the trigonometric identity

\tan \theta = \frac{\sin \theta}{\cos \theta}

simplifies the calculation without needing to evaluate the hypotenuse.

2

Substitute

\tan \theta = 2

\frac{3(2) + 1}{2 + 2} = \frac{7}{4}

Replace

\tan \theta

with its given numerical value of 2.

3

Convert fraction to decimal

1.75

Decimal representation of the exact fraction

\frac{7}{4}

.

Anahtar Kavram

Basic Trigonometric Ratios and Quotient Identity

Soru 603Soru →

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

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

24

Applying the chain rule gives

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

. Evaluating this derivative at

x = 2

yields

3(2)^2(2) = 24

.

Adım Adım Çözüm

1

Identify the composite structure and state the chain rule formula.

Let

u = x^2 - 2x + 2

, so

y = u^3

. By the chain rule,

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

.

The function is an expression raised to a power, requiring the chain rule.

2

Differentiate the outer function with respect to

u

and the inner function with respect to

x

.

\frac{dy}{du} = 3(x^2 - 2x + 2)^2 and \frac{du}{dx} = 2x - 2.

Apply the power rule to both the outer expression and inner polynomial.

3

Multiply the two derivatives to express

\frac{dy}{dx}

.

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

Product of outer and inner derivatives per the chain rule.

4

Evaluate the expression at

x = 2

.

\frac{dy}{dx}\Big|_{x=2} = 3(4 - 4 + 2)^2(4 - 2) = 3(2)^2(2) = 24.

Substitute

x = 2

into the derivative expression and simplify.

Anahtar Kavram

Chain Rule of Differentiation

Soru 604Soru →

The table below shows the cumulative frequency distribution of the masses (in grams) of $100$ cocoa beans sampled from an agricultural yield:

Mass Class Interval (g)	Cumulative Frequency
$10 - 19$	$10$
$20 - 29$	$30$
$30 - 39$	$65$
$40 - 49$	$90$
$50 - 59$	$100$

Using linear interpolation from the cumulative frequency distribution, calculate the 75th percentile ( $Q_3$ ) mass of the cocoa beans in grams.

Cevabı ve açıklamayı göster

Cevap: 43.5

Cevap

The 75th percentile (

Q_3

) mass of the cocoa beans is

43.5\text{ g}

.

To find the 75th percentile (

Q_3

) from cumulative frequency data, calculate the rank

\frac{75}{100} \times 100 = 75

. This falls into the

40 - 49

class interval (boundaries

39.5 - 49.5

). Applying

Q_3 = L + \left(\frac{75 - F}{f}\right)c

yields

39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 43.5\text{ g}

.

Adım Adım Çözüm

1

Calculate the percentile position rank

Rank position is

75

The 75th percentile corresponds to

75\%

of the total sample size

N = 100

, giving

\frac{75}{100} \times 100 = 75

.

2

Locate the 75th percentile class interval and its boundaries

Class interval is

40 - 49

, with lower boundary

L = 39.5

and upper boundary

49.5

Cumulative frequency before

40 - 49

is

65

, and up to

40 - 49

is

90

. Since

65 < 75 \leq 90

, the 75th item falls in this interval.

3

Identify class parameters for interpolation

L = 39.5

,

F = 65

,

f = 25

,

c = 10

Lower boundary

L = 39.5

, previous cumulative frequency

F = 65

, class frequency

f = 90 - 65 = 25

, class width

c = 49.5 - 39.5 = 10

.

4

Compute

Q_3

using the linear interpolation formula

Q_3 = 43.5\text{ g}

Q_3 = 39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 39.5 + 4 = 43.5

.

Anahtar Kavram

Linear Interpolation of Percentiles from Cumulative Frequency Data

Soru 605Soru →

Find the value of $x$ that satisfies the logarithmic equation $\log_2(x^2 - 1) - \log_2(x - 1) = 3$ .

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of

x

is

7

.

Applying the logarithm quotient law reduces

\log_2(x^2 - 1) - \log_2(x - 1)

to

\log_2\left(\frac{x^2 - 1}{x - 1}\right)

. Factoring

x^2 - 1

into

(x - 1)(x + 1)

and canceling

(x - 1)

leaves

\log_2(x + 1) = 3

. Converting this to exponential form gives

x + 1 = 2^3 = 8

, which yields

x = 7

.

Adım Adım Çözüm

1

Apply the quotient rule for logarithms to combine the left-hand side into a single logarithmic expression.

\log_2\left(\frac{x^2 - 1}{x - 1}\right) = 3

According to logarithmic rules, the difference of two logarithms of the same base is equal to the logarithm of their quotient:

\log_b A - \log_b B = \log_b\left(\frac{A}{B}\right)

.

2

Factor the numerator as a difference of two squares and simplify the rational expression.

\log_2(x + 1) = 3

Since

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

, dividing by

(x - 1)

leaves

(x + 1)

, given

x \neq 1

.

3

Rewrite the logarithmic equation in exponential form.

x + 1 = 2^3 = 8

By definition of logarithms,

\log_b y = z \iff b^z = y

.

4

Solve the resulting linear equation for

x

.

x = 7

Subtracting

1

from both sides gives

x = 7

.

Anahtar Kavram

Quotient Law of Logarithms and Logarithmic Equations

Tahmini Süre:1m 30s

Soru 606Soru →

The power consumption $P$ (in watts) of a variable-speed motor is partly constant and partly varies directly as the square of its operational speed $v$ (in revolutions per second). If $P = 250\text{ W}$ when $v = 10\text{ rev/s}$ and $P = 700\text{ W}$ when $v = 20\text{ rev/s}$ , calculate the value of $P$ (in watts) when $v = 15\text{ rev/s}$ .

Cevabı ve açıklamayı göster

Cevap: 437.5

Cevap

437.5 W

The partial variation equation is

P = k_1 + k_2 v^2

. Setting up simultaneous equations

k_1 + 100 k_2 = 250

and

k_1 + 400 k_2 = 700

yields

k_1 = 100

and

k_2 = 1.5

. Substituting

v = 15

into

P = 100 + 1.5(15^2)

gives

P = 437.5\text{ W}

.

Adım Adım Çözüm

1

Express the partial variation mathematically.

P = k_1 + k_2 v^2

, where

k_1

and

k_2

are constants.

The total power consumption is the sum of a fixed baseline constant

k_1

and a variable component proportional to

v^2

.

2

Form simultaneous linear equations using the provided data points.

k_1 + 100 k_2 = 250

and

k_1 + 400 k_2 = 700

.

Substituting

v = 10

gives

10^2 = 100

, and substituting

v = 20

gives

20^2 = 400

.

3

Solve for the constants

k_1

and

k_2

.

k_2 = 1.5

and

k_1 = 100

.

Subtracting the two equations eliminates

k_1

, yielding

300 k_2 = 450 \implies k_2 = 1.5

. Substituting

k_2 = 1.5

back into

k_1 + 100 k_2 = 250

gives

k_1 = 100

.

4

Calculate the value of

P

at

v = 15\text{ rev/s}

.

P = 437.5\text{ W}

.

Substitute

v = 15

,

k_1 = 100

, and

k_2 = 1.5

into the governing formula

P = 100 + 1.5(15^2) = 100 + 337.5 = 437.5

.

Anahtar Kavram

Partial Variation and Simultaneous Linear Equations

Tahmini Süre:1m 30s

Soru 607Soru →

A point $P(x, y)$ moves in a Cartesian plane such that the sum of the squares of its distances from two fixed points $A(0, 0)$ and $B(8, 0)$ is equal to $82$ , defining a locus $L_1$ . A second locus $L_2$ is the set of all points equidistant from the parallel lines $y = -1$ and $y = 7$ . Given that $L_1$ and $L_2$ intersect at two distinct points $M$ and $N$ , what is the length of the line segment $MN$ ?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The length of the line segment

MN

is 8 units.

The locus

L_1

simplifies to the circle

(x - 4)^2 + y^2 = 25

with center

(4,0)

and radius

5

. The locus

L_2

is the horizontal line

y = 3

. Substituting

y = 3

into the circle equation yields

(x - 4)^2 = 16

, giving intersection points at

(0,3)

and

(8,3)

. The distance between these points is

8

units.

Adım Adım Çözüm

1

Formulate the algebraic expression for locus

L_1

The distance squared

PA^2 = x^2 + y^2

and

PB^2 = (x - 8)^2 + y^2

. Setting

PA^2 + PB^2 = 82

gives

x^2 + y^2 + (x - 8)^2 + y^2 = 82

.

By the distance formula, the sum of squared distances from

P(x, y)

to

A(0,0)

and

B(8,0)

equals 82.

2

Simplify and complete the square for locus

L_1

Expanding yields

2x^2 - 16x + 64 + 2y^2 = 82 \implies 2x^2 - 16x + 2y^2 = 18 \implies x^2 - 8x + y^2 = 9

. Completing the square gives

(x - 4)^2 + y^2 = 25

.

This shows

L_1

is a circle with center

(4, 0)

and radius

R = 5

.

3

Find the equation for locus

L_2

The locus of points equidistant from two parallel lines

y = -1

and

y = 7

is the line

y = \frac{-1 + 7}{2} = 3

.

The locus equidistant from two parallel lines is a parallel line running midway between them.

4

Find the intersection points

M

and

N

Substitute

y = 3

into

(x - 4)^2 + y^2 = 25

:

(x - 4)^2 + 3^2 = 25 \implies (x - 4)^2 = 16 \implies x - 4 = \pm 4

, so

x = 8

or

x = 0

. Thus,

M = (0, 3)

and

N = (8, 3)

.

Solving the circle and line equations simultaneously gives the

x

-coordinates of the intersection points.

5

Compute the distance between

M

and

N

Since both points share the ordinate

y = 3

, distance

MN = 8 - 0 = 8

.

The length of a horizontal segment is the difference between its endpoints'

x

-coordinates.

Anahtar Kavram

Intersection of loci involving circles and parallel lines

Soru 608Soru →

A point $P(x, y)$ moves such that it is equidistant from two parallel lines $3x - 4y + 11 = 0$ and $3x - 4y - 1 = 0$ , defining locus $L_1$ . A second locus $L_2$ consists of all points that are at a constant distance of $5$ units from the fixed point $(1, 7)$ . Calculate the distance between the two points of intersection of locus $L_1$ and locus $L_2$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The distance between the two points of intersection of locus L1 and locus L2 is 6 units.

Locus L1 is the line midway between the two given parallel lines, having the equation 3x - 4y + 5 = 0. Locus L2 is a circle centered at (1, 7) with radius 5. The perpendicular distance from the center (1, 7) to line L1 is 4 units. Applying Pythagoras' theorem to the right triangle formed by the radius, perpendicular distance, and half-chord gives a half-chord length of sqrt(5^2 - 4^2) = 3. Therefore, the distance between the two intersection points (the full chord length) is 2 * 3 = 6 units.

Adım Adım Çözüm

1

Determine the equation of locus L1

3x - 4y + 5 = 0

The locus of points equidistant from two parallel lines ax + by + c1 = 0 and ax + by + c2 = 0 is a line parallel to both and midway between them, given by ax + by + (c1 + c2)/2 = 0.

2

Identify the geometric representation of locus L2

A circle centered at (1, 7) with radius r = 5

The locus of points at a constant distance r from a fixed point (x0, y0) is a circle.

3

Calculate the perpendicular distance d from the circle's center (1, 7) to line L1

d = 4 units

Using the perpendicular distance formula d = |ax0 + by0 + c| / sqrt(a^2 + b^2), we get d = |3(1) - 4(7) + 5| / sqrt(3^2 + (-4)^2) = |-20| / 5 = 4.

4

Compute the total distance between the two intersection points

6 units

The line intersects the circle along a chord. Using Pythagoras' theorem, half the chord length is sqrt(r^2 - d^2) = sqrt(5^2 - 4^2) = 3. Multiplying by 2 gives the full distance of 6 units.

Anahtar Kavram

Intersection of Loci (Parallel Line Bisector and Circle)

Tahmini Süre:2m 30s

Soru 609Soru →

An environmental research station recorded the daily particulate matter concentration (in $\mu\text{g/m}^3$ ) near an urban center over a period of $50$ days. The observations are summarized in the table below:

Particulate Matter ( $\mu\text{g/m}^3$ )	Number of Days ( $f$ )
$20 - 29$	$6$
$30 - 39$	$10$
$40 - 49$	$15$
$50 - 59$	$11$
$60 - 69$	$8$

Find the estimated mean particulate matter concentration, in $\mu\text{g/m}^3$ , for the 50-day period.

Cevabı ve açıklamayı göster

Cevap: 45.5

Cevap

The estimated mean particulate matter concentration is

45.5\text{ }\mu\text{g/m}^3

.

To calculate the mean of grouped data, determine the midpoint (

x

) of each class interval:

24.5

,

34.5

,

44.5

,

54.5

, and

64.5

. Multiply each midpoint by its frequency (

f

) to get the products

147

,

345

,

667.5

,

599.5

, and

516

. The sum of these products (

\sum fx

) is

2275

. Dividing

\sum fx

by the total frequency (

\sum f = 50

) yields the estimated mean of

45.5\text{ }\mu\text{g/m}^3

.

Adım Adım Çözüm

1

Calculate the class midpoints (

x

) for each class interval

Midpoints are

24.5

,

34.5

,

44.5

,

54.5

, and

64.5

.

Grouped data uses the midpoint of each interval to represent all values falling within that interval.

2

Compute the product of frequency and midpoint (

fx

) for each class

6 \times 24.5 = 147

,

10 \times 34.5 = 345

,

15 \times 44.5 = 667.5

,

11 \times 54.5 = 599.5

,

8 \times 64.5 = 516

.

This determines the estimated sum of data values within each group.

3

Sum all products

\sum fx

and total frequency

\sum f

\sum fx = 147 + 345 + 667.5 + 599.5 + 516 = 2275

and

\sum f = 50

.

These totals are required for the mean formula.

4

Calculate the mean concentration using

\bar{x} = \frac{\sum fx}{\sum f}

\bar{x} = \frac{2275}{50} = 45.5

.

Dividing the total estimated sum by the total number of days gives the estimated mean.

Anahtar Kavram

Calculation of Mean for Grouped Frequency Data

Soru 610Soru →

Given that

x

and

y

are real numbers satisfying the simultaneous exponential equations

3^x \times 9^y = 81

and

8^x \times 4^y = 256

find the value of

x^2 + y^2

.

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of

x^2 + y^2

is 5.

Converting all terms to their prime bases (

3

for the first equation and

2

for the second equation) produces the simultaneous linear equations

x + 2y = 4

and

3x + 2y = 8

. Solving these gives

x = 2

and

y = 1

. Substituting these values into

x^2 + y^2

gives

2^2 + 1^2 = 5

.

Adım Adım Çözüm

1

Convert all terms in the first equation to powers of base 3.

x + 2y = 4

Since

9 = 3^2

and

81 = 3^4

, applying the product law of indices

a^m \times a^n = a^{m+n}

gives

3^{x+2y} = 3^4

. Equating the exponents gives

x + 2y = 4

.

2

Convert all terms in the second equation to powers of base 2.

3x + 2y = 8

Since

8 = 2^3

,

4 = 2^2

, and

256 = 2^8

, applying the laws of indices yields

2^{3x} \times 2^{2y} = 2^8 \implies 2^{3x+2y} = 2^8

. Equating exponents gives

3x + 2y = 8

.

3

Solve the system of simultaneous linear equations for

x

and

y

.

x = 2

and

y = 1

Subtracting

x + 2y = 4

from

3x + 2y = 8

yields

2x = 4 \implies x = 2

. Substituting

x = 2

into

x + 2y = 4

gives

2 + 2y = 4 \implies y = 1

.

4

Evaluate the target expression

x^2 + y^2

.

5

Substitute

x = 2

and

y = 1

into

x^2 + y^2

to obtain

2^2 + 1^2 = 4 + 1 = 5

.

Anahtar Kavram

Converting exponential terms to common prime bases to reduce exponential equations into linear equations.

Soru 611Soru →

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

Cevabı ve açıklamayı göster

Cevap: 24

Cevap

The numerical value of the derivative

\frac{dy}{dx}

at

x = 1

is 24.

Evaluating the derivative requires combined application of the quotient rule and chain rule. Differentiating

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

via the chain rule yields

u'(x) = 12x(2x^2 + 1)^2

. At

x = 1

,

u(1) = 27

,

u'(1) = 108

,

v(1) = 3

, and

v'(1) = 4

. Substituting these into the quotient rule

\frac{u'v - uv'}{v^2}

produces

\frac{324 - 108}{9} = 24

.

Adım Adım Çözüm

1

Decompose the function for the Quotient Rule

Let

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

and

v(x) = 4x - 1

.

The given expression is a quotient of two differentiable functions.

2

Differentiate the numerator using the Chain Rule

u'(x) = 3(2x^2 + 1)^2 \cdot 4x = 12x(2x^2 + 1)^2

The chain rule states that

\frac{d}{dx}[g(x)]^n = n[g(x)]^{n-1} \cdot g'(x)

.

3

Differentiate the denominator and evaluate all components at

x = 1

v'(1) = 4

,

v(1) = 3

,

u(1) = 27

, and

u'(1) = 108

Evaluating each term individually reduces algebraic complexity prior to applying the quotient rule.

4

Apply the Quotient Rule formula and evaluate

\frac{dy}{dx}\Big|_{x=1} = \frac{(108)(3) - (27)(4)}{3^2} = \frac{324 - 108}{9} = 24

The quotient rule formula is

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

.

Anahtar Kavram

Combination of Quotient Rule and Chain Rule

Tahmini Süre:1m 30s

Soru 612Soru →

Using differentiation from first principles, evaluate the value of the derivative of the function $f(x) = 2x^2 + 3x - 1$ at the point where $x = 1$ .

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

The value of the derivative at

x = 1

is 7.

Using first principles, the derivative is defined as

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

. Evaluating at

x = 1

, we find

f(1) = 4

and

f(1+h) = 4 + 7h + 2h^2

. Subtracting

f(1)

leaves

7h + 2h^2

, and dividing by

h

yields

7 + 2h

. Taking the limit as

h \to 0

yields the final result of 7.

Adım Adım Çözüm

1

Calculate the value of the function at

x = 1

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

This establishes the base value needed for the difference quotient.

2

Expand and simplify

f(1+h)

f(1+h) = 2(1+h)^2 + 3(1+h) - 1 = 4 + 7h + 2h^2

This gives the value of the function at the incremented point

x + h

.

3

Form and simplify the difference quotient

\frac{f(1+h) - f(1)}{h}

\frac{4 + 7h + 2h^2 - 4}{h} = 7 + 2h

Dividing by

h

eliminates the indeterminate form before taking the limit.

4

Evaluate the limit as

h \to 0

f'(1) = \lim_{h \to 0} (7 + 2h) = 7

Taking

h = 0

in the simplified quotient gives the exact rate of change at

x = 1

.

Anahtar Kavram

Differentiation from First Principles

Tahmini Süre:1m 0s

Soru 613Soru →

How many positive integer values of $x$ satisfy the quadratic inequality $x^2 - 4x - 5 < 0$ ?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

There are 4 positive integer values of

x

that satisfy the inequality.

Factoring

x^2 - 4x - 5 < 0

yields

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

, giving the real solution interval

-1 < x < 5

. Filtering for positive integers (

x \in \mathbb{Z}^+

) yields the set

\{1, 2, 3, 4\}

, which contains exactly 4 values.

Adım Adım Çözüm

1

Factor the quadratic inequality

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

Factoring helps find the boundary roots of the quadratic equation.

2

Determine the solution set interval

-1 < x < 5

For a quadratic inequality of the form

(x - a)(x - b) < 0

with

a < b

, the solution interval is

a < x < b

.

3

List the positive integer solutions in the range

-1 < x < 5

x \in \{1, 2, 3, 4\}

Positive integers are whole numbers strictly greater than 0.

4

Count the number of positive integers

4

Counting the elements in the set

\{1, 2, 3, 4\}

gives a total of 4.

Anahtar Kavram

Solving quadratic inequalities and finding valid integer solutions within a target domain.

Tahmini Süre:45s

Soru 614Soru →

If $y = (3x - 1)^4$ , what is the numerical value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

Cevabı ve açıklamayı göster

Cevap: 96

Cevap

96

Applying the chain rule

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

to

y = (3x - 1)^4

gives

\frac{dy}{dx} = 4(3x - 1)^3 \cdot 3 = 12(3x - 1)^3

. Evaluating this expression at

x = 1

yields

12(3(1) - 1)^3 = 12(2)^3 = 96

.

Adım Adım Çözüm

1

Differentiate

y = (3x - 1)^4

with respect to

x

using the chain rule.

\frac{dy}{dx} = 12(3x - 1)^3

Differentiate the outer power function to get

4(3x - 1)^3

and multiply by the derivative of the inner linear function

3x - 1

, which is 3.

2

Substitute

x = 1

into the derivative function.

96

Evaluating

12(3(1) - 1)^3

gives

12(2)^3 = 12 \times 8 = 96

.

Anahtar Kavram

Chain Rule of Differentiation

Soru 615Soru →

If $27^{x - 1} = 9^{x + 1}$ , determine the value of $x$ .

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The value of

x

is 5.

Rewriting 27 as

3^3

and 9 as

3^2

transforms the given equation into

3^{3(x - 1)} = 3^{2(x + 1)}

. Equating exponents gives

3x - 3 = 2x + 2

, which simplifies directly to

x = 5

.

Adım Adım Çözüm

1

Express numbers in terms of a common base

(3^3)^{x - 1} = (3^2)^{x + 1}

Both 27 and 9 are powers of 3, allowing reduction to a single base.

2

Apply power of a power index law

3^{3x - 3} = 3^{2x + 2}

Multiply the base power by the expression in the exponent:

3 \times (x - 1) = 3x - 3

and

2 \times (x + 1) = 2x + 2

.

3

Equate exponents and solve for

x

3x - 3 = 2x + 2 \implies x = 5

Equal bases imply that the index powers must be equal.

Anahtar Kavram

Equating exponential expressions using a common base

Soru 616Soru →

Find the $x$ -intercept of the normal line to the curve $y = x^3 - 3x^2 + 4x - 1$ at the point where $x = 1$ .

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The

x

-intercept of the normal line is

2

.

At

x = 1

, the point on the curve

y = x^3 - 3x^2 + 4x - 1

is

(1, 1)

. The derivative is

\frac{dy}{dx} = 3x^2 - 6x + 4

, which equals

1

at

x = 1

. The normal gradient is therefore

-1

. The normal line equation is

y - 1 = -1(x - 1)

, which simplifies to

y = -x + 2

. Setting

y = 0

gives

x = 2

.

Adım Adım Çözüm

1

Find the y-coordinate at x = 1

y = 1, so the point on the curve is (1, 1)

The point of contact is needed to construct the equation of the normal line.

2

Differentiate the function with respect to x

dy/dx = 3x^2 - 6x + 4

The derivative gives the expression for the gradient of the tangent to the curve.

3

Find the gradient of the tangent and normal at x = 1

Gradient of tangent m_t = 1; gradient of normal m_n = -1

The normal line is perpendicular to the tangent line, so m_n = -1 / m_t.

4

Determine the equation of the normal line

y - 1 = -1(x - 1) => y = -x + 2

Applying the straight-line equation y - y_1 = m(x - x_1).

5

Find the x-intercept of the normal line

x = 2

Setting y = 0 in the normal line equation yields the x-intercept.

Anahtar Kavram

Equation and axis intercepts of a normal line to a curve

Soru 617Soru →

A solid right circular cylinder has a height of $10\text{ cm}$ and a total surface area of $192\pi\text{ cm}^2$ . What is the volume of the cylinder in $\text{cm}^3$ expressed in terms of $\pi$ ?

Cevabı ve açıklamayı göster

Cevap: 360

Cevap

360

The total surface area of a closed cylinder is

A = 2\pi r^2 + 2\pi r h

. Given

A = 192\pi

and

h = 10\text{ cm}

, we have

2\pi r^2 + 20\pi r = 192\pi

. Dividing by

2\pi

yields

r^2 + 10r - 96 = 0

. Factoring the quadratic gives

(r + 16)(r - 6) = 0

, so

r = 6\text{ cm}

(since radius must be positive). Substituting

r = 6\text{ cm}

and

h = 10\text{ cm}

into the volume formula

V = \pi r^2 h

yields

V = \pi (6^2)(10) = 360\pi\text{ cm}^3

. The numerical coefficient of

\pi

is

360

.

Adım Adım Çözüm

1

Set up the equation for the total surface area of the cylinder

2\pi r^2 + 2\pi r(10) = 192\pi

The total surface area of a solid cylinder comprises the areas of the top and bottom circular bases (

2\pi r^2

) plus the curved surface area (

2\pi r h

).

2

Simplify and solve the resulting quadratic equation for radius

r

r^2 + 10r - 96 = 0 \implies (r + 16)(r - 6) = 0 \implies r = 6\text{ cm}

Dividing the equation by

2\pi

yields

r^2 + 10r - 96 = 0

. Since radius must be positive,

r = 6\text{ cm}

.

3

Calculate the volume of the cylinder using the radius and height

V = \pi r^2 h = \pi (6)^2 (10) = 360\pi\text{ cm}^3

The volume of a cylinder is calculated using the formula

V = \pi r^2 h

.

Anahtar Kavram

Surface Area and Volume of 3D Solids

Soru 618Soru →

A piece of wire of length $88\text{ cm}$ is bent to form the perimeter of a sector of a circle of radius $21\text{ cm}$ . What is the area of the sector in $\text{cm}^2$ ?

Cevabı ve açıklamayı göster

Cevap: 483

Cevap

The area of the sector is

483\text{ cm}^2

.

The total length of the wire forms the perimeter of the sector, which consists of two radii and the arc length (

P = 2r + l

). With

P = 88\text{ cm}

and

r = 21\text{ cm}

, the arc length is

l = 88 - 2(21) = 46\text{ cm}

. Using the sector area formula

A = \frac{1}{2} r l

, the area is

\frac{1}{2} \times 21 \times 46 = 483\text{ cm}^2

.

Adım Adım Çözüm

1

Determine the arc length of the sector from the given perimeter

l = 46\text{ cm}

The total perimeter of a sector includes two straight radii and the curved arc length:

P = 2r + l

. Subtracting twice the radius (

2 \times 21 = 42\text{ cm}

) from the total wire length of

88\text{ cm}

gives an arc length of

46\text{ cm}

.

2

Calculate the area of the sector

A = 483\text{ cm}^2

The area of a sector can be computed using the formula

A = \frac{1}{2} r l

. Substituting

r = 21\text{ cm}

and

l = 46\text{ cm}

yields

A = \frac{1}{2} \times 21 \times 46 = 483\text{ cm}^2

.

Anahtar Kavram

Perimeter and Area of a Sector of a Circle

Soru 619Soru →

Evaluate the limit:

\lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x + 2} - 2}

What is the numerical value of this limit?

Cevabı ve açıklamayı göster

Cevap: 48

Cevap

The numerical value of the limit is 48.

Evaluating the limit of

\frac{x^3 - 8}{\sqrt{x + 2} - 2}

as

x \to 2

gives an indeterminate form

\frac{0}{0}

. Factorizing the numerator gives

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

, and rationalizing the denominator by multiplying numerator and denominator by

(\sqrt{x + 2} + 2)

converts the denominator to

x - 2

. Canceling

(x - 2)

leaves

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

. Evaluating at

x = 2

gives

(4 + 4 + 4)(\sqrt{4} + 2) = 12 \times 4 = 48

.

Adım Adım Çözüm

1

Identify the limit form via direct substitution

Substituting

x = 2

yields

\frac{0}{0}

.

Direct evaluation results in an indeterminate form, requiring algebraic manipulation to eliminate the zero factor.

2

Factorize the numerator using the difference of cubes formula

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Exposing the factor

(x - 2)

is essential to resolving the zero denominator.

3

Rationalize the denominator using its algebraic conjugate

Multiply top and bottom by

(\sqrt{x + 2} + 2)

to get denominator

(x + 2) - 4 = x - 2

.

Applying

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

eliminates the square root from the denominator.

4

Cancel the common factor and compute the final value

\lim_{x \to 2} (x^2 + 2x + 4)(\sqrt{x + 2} + 2) = (12)(4) = 48.

With

(x - 2)

cancelled for

x \neq 2

, direct substitution now yields a defined real number.

Anahtar Kavram

Limits of Indeterminate Forms using Difference of Cubes and Surd Rationalization

Tahmini Süre:2m 30s

Soru 620Soru →

If $y = 3e^{2x} + \ln(\cos x)$ , calculate the value of $\frac{dy}{dx}$ at $x = 0$ .

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The value of the derivative of

y = 3e^{2x} + \ln(\cos x)

at

x = 0

is 6.

Differentiating each transcendental term individually using the chain rule yields

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

. Substituting

x = 0

gives

6e^0 - \tan 0 = 6 - 0 = 6

.

Adım Adım Çözüm

1

Differentiate the exponential term

3e^{2x}

6e^{2x}

Applying the derivative rule for exponential functions

\frac{d}{dx}[a e^{kx}] = a k e^{kx}

.

2

Differentiate the logarithmic term

\ln(\cos x)

-\tan x

Applying the chain rule

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

where

u = \cos x

gives

\frac{-\sin x}{\cos x} = -\tan x

.

3

Combine terms and evaluate at

x = 0

6

Substituting

x = 0

into

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

yields

6e^0 - \tan 0 = 6(1) - 0 = 6

.

Anahtar Kavram

Differentiation of Trigonometric, Exponential, and Logarithmic Functions

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram

Sık Yapılan Hatalar

Cevap

Adım Adım Çözüm

Anahtar Kavram