Tüm alıştırma soruları

If $203_n - 114_n = 45_n$ , where $n$ represents a positive integer base, what is the value of $n$ ?

6

5

7

8

Cevabı ve açıklamayı göster

Cevap: $6$

Cevap

The correct base is

6

.

Expanding the numbers in terms of powers of

n

gives

(2n^2 + 3) - (n^2 + n + 4) = 4n + 5

. Grouping like terms results in the quadratic equation

n^2 - 5n - 6 = 0

. Factoring yields

(n - 6)(n + 1) = 0

, giving

n = 6

as the only valid positive integer solution greater than

5

.

Adım Adım Çözüm

1

Convert each term from base

n

to base 10 using positional expansion.

203_n = 2n^2 + 0n + 3 = 2n^2 + 3

,

114_n = 1n^2 + 1n + 4 = n^2 + n + 4

, and

45_n = 4n + 5

.

Converting all terms to a common decimal representation allows algebraic manipulation.

2

Substitute the expanded terms back into the original equation and simplify.

(2n^2 + 3) - (n^2 + n + 4) = 4n + 5 \implies n^2 - n - 1 = 4n + 5

.

Carefully distribute the negative sign across all terms of

(n^2 + n + 4)

.

3

Rearrange the expression into standard quadratic form

ax^2 + bx + c = 0

.

n^2 - 5n - 6 = 0

.

Subtracting

4n + 5

from both sides sets the quadratic equation to zero.

4

Factor the quadratic equation to find the valid base

n

.

(n - 6)(n + 1) = 0 \implies n = 6

or

n = -1

.

Since a base must be a positive integer strictly greater than any digit present in the equation (maximum digit is

5

),

n = 6

.

Anahtar Kavram

Solving unknown base equations using polynomial expansion in positional notation.

Soru 7422Soru →

If $y = \int (12x^2 + 6\sin(2x)) \, dx$ and $y = 10$ when $x = 0$ , what is the value of the constant of integration $C$ ?

Cevabı ve açıklamayı göster

Cevap: 13

Cevap

The value of the constant of integration

C

is

13

.

Integrating

12x^2 + 6\sin(2x)

gives

y = 4x^3 - 3\cos(2x) + C

. Substituting

x = 0

yields

y(0) = 4(0)^3 - 3\cos(0) + C = -3 + C

. Setting

-3 + C = 10

and solving for

C

gives

C = 13

.

Adım Adım Çözüm

1

Integrate the function with respect to

x

y = 4x^3 - 3\cos(2x) + C

Applying the power rule

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

and trigonometric integration rule

\int k\sin(bx) \, dx = -\frac{k}{b}\cos(bx)

.

2

Apply the initial condition

x = 0

and

y = 10

10 = 4(0)^3 - 3\cos(0) + C

Substituting the boundary values to solve for the specific constant of integration.

3

Evaluate trigonometric function and solve for

C

C = 13

Since

\cos(0) = 1

, the equation simplifies to

10 = -3 + C

, leading directly to

C = 13

.

Anahtar Kavram

Indefinite Integration of Polynomial and Trigonometric Functions with Initial Conditions

Soru 7423Soru →

Using differentiation from first principles, what is the derivative of the function $f(x) = 4 - x^2$ with respect to $x$ ?

-2x

2x

-2x - 1

4 - 2x

Cevabı ve açıklamayı göster

Cevap: $-2x$

Cevap

-2x

Differentiating from first principles involves finding the limit of

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

as

h \to 0

. For

f(x) = 4 - x^2

, expanding

f(x+h)

gives

4 - x^2 - 2xh - h^2

. Subtracting

f(x)

yields

-2xh - h^2

, and dividing by

h

gives

-2x - h

. Taking the limit as

h \to 0

leaves

-2x

.

Adım Adım Çözüm

1

Express

f(x+h)

for

f(x) = 4 - x^2

f(x+h) = 4 - (x+h)^2 = 4 - (x^2 + 2xh + h^2) = 4 - x^2 - 2xh - h^2

Substitute

x+h

into the original function definition.

2

Set up the difference

f(x+h) - f(x)

f(x+h) - f(x) = (4 - x^2 - 2xh - h^2) - (4 - x^2) = -2xh - h^2

Subtract

f(x)

to find the net change in

y

.

3

Divide the difference by

h

to form the difference quotient

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

Divide each term in the numerator by

h

.

4

Take the limit as

h \to 0

f'(x) = \lim_{h \to 0} (-2x - h) = -2x

Evaluate the derivative by letting

h

approach zero.

Anahtar Kavram

Differentiation from First Principles

Soru 7424Soru →

The rate of heat transfer $Q$ across a building wall varies directly as the surface area $A$ of the wall and the temperature difference $\Delta T$ between the interior and exterior, and inversely as the wall thickness $d$ . When the surface area is $4\text{ m}^2$ , the temperature difference is $15^\circ\text{C}$ , and the thickness is $0.05\text{ m}$ , the heat transfer rate is $1200\text{ W}$ . What is the heat transfer rate in watts when the surface area is $6\text{ m}^2$ , the temperature difference is $20^\circ\text{C}$ , and the thickness is $0.08\text{ m}$ ?

Cevabı ve açıklamayı göster

Cevap: 1500

Cevap

The heat transfer rate is

1500\text{ W}

.

Establishing the variation constant

k = 1

using the initial given values and substituting the new parameters yields

Q = \frac{1 \times 6 \times 20}{0.08} = 1500\text{ W}

.

Adım Adım Çözüm

1

Formulate the variation equation

Q = \frac{k A \Delta T}{d}

Direct variation means multiplying by

A

and

\Delta T

, while inverse variation means dividing by

d

.

2

Calculate the constant of variation

k

k = 1

Substituting

Q=1200

,

A=4

,

\Delta T=15

, and

d=0.05

gives

1200 = \frac{60k}{0.05} = 1200k

, so

k = 1

.

3

Compute the target heat transfer rate

Q

1500\text{ W}

Substituting

k=1

,

A=6

,

\Delta T=20

, and

d=0.08

gives

Q = \frac{1 \times 6 \times 20}{0.08} = \frac{120}{0.08} = 1500\text{ W}

.

Anahtar Kavram

Joint and Inverse Variation

Soru 7425Soru →

In a department of $70$ university lecturers, $40$ publish research in Journal $A$ , $30$ publish in Journal $B$ , and $25$ publish in Journal $C$ . It is known that $15$ publish in both Journals $A$ and $B$ , $12$ publish in both Journals $B$ and $C$ , $10$ publish in both Journals $A$ and $C$ , and $5$ publish in all three journals. How many lecturers do not publish in any of these three journals?

7

12

17

63

Cevabı ve açıklamayı göster

Cevap: $7$

Cevap

The number of lecturers who do not publish in any of the three journals is

7

.

Using the inclusion-exclusion principle for three sets,

n(A \cup B \cup C) = 40 + 30 + 25 - 15 - 12 - 10 + 5 = 63

. The number of lecturers publishing in none of the journals is the complement of this union relative to the universal set of

70

, which is

70 - 63 = 7

.

Adım Adım Çözüm

1

Apply the Principle of Inclusion-Exclusion for three sets to find the total number of lecturers who publish in at least one journal,

n(A \cup B \cup C)

.

n(A \cup B \cup C) = n(A) + n(B) + n(C) - [n(A \cap B) + n(B \cap C) + n(A \cap C)] + n(A \cap B \cap C)

Elements counted multiple times in pairwise intersections must be subtracted, and the central triple intersection must be added back.

2

Substitute the given numerical values into the formula.

n(A \cup B \cup C) = 40 + 30 + 25 - (15 + 12 + 10) + 5 = 95 - 37 + 5 = 63

To evaluate the total cardinality of the union.

3

Subtract

n(A \cup B \cup C)

from the universal set size

n(U)

.

n((A \cup B \cup C)') = n(U) - n(A \cup B \cup C) = 70 - 63 = 7

The number of lecturers publishing in none of the journals corresponds to the complement of the union of all three sets.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets and Set Complement

Soru 7426Soru →

The table below shows the distribution of ages (in years) of trees in a forest reserve, presented alongside their frequency densities for a histogram representation:

Age Interval (years)	Frequency Density
$0 \le x < 10$	$3.0$
$10 \le x < 25$	$2.0$
$25 \le x < 45$	$2.25$
$45 \le x < 80$	$2.0$

If the dataset is instead displayed using a pie chart, what is the sector angle representing the age interval $25 \le x < 45$ ?

90^\circ

70^\circ

60^\circ

80^\circ

Cevabı ve açıklamayı göster

Cevap: $90^\circ$

Cevap

The sector angle representing the age interval

25 \le x < 45

is

90^\circ

.

The frequency of each interval is found by multiplying its frequency density by its class width (

3.0 \times 10 = 30

,

2.0 \times 15 = 30

,

2.25 \times 20 = 45

, and

2.0 \times 35 = 75

). The total frequency is

180

. The sector angle for

25 \le x < 45

(frequency

45

) is

\frac{45}{180} \times 360^\circ = 90^\circ

.

Adım Adım Çözüm

1

Calculate the class width for each interval.

Widths are:

10 - 0 = 10

,

25 - 10 = 15

,

45 - 25 = 20

, and

80 - 45 = 35

.

For histograms with unequal class widths, frequency density is defined as frequency divided by class width.

2

Calculate the frequency (

f

) for each interval using

f = \text{Frequency Density} \times \text{Class Width}

.

Interval

0 \le x < 10

:

f_1 = 3.0 \times 10 = 30

.
Interval

10 \le x < 25

:

f_2 = 2.0 \times 15 = 30

.
Interval

25 \le x < 45

:

f_3 = 2.25 \times 20 = 45

.
Interval

45 \le x < 80

:

f_4 = 2.0 \times 35 = 75

.

The frequency of a class in a histogram corresponds to the area of its bar.

3

Compute the total frequency (

N

).

N = 30 + 30 + 45 + 75 = 180

.

The total frequency represents the entire dataset needed for pie chart sector calculations.

4

Calculate the sector angle (

\,\theta\,

) for the interval

25 \le x < 45

.

\theta = \frac{f_3}{N} \times 360^\circ = \frac{45}{180} \times 360^\circ = 90^\circ

.

The sector angle in a pie chart is proportional to the relative frequency of the class out of

360^\circ

.

Anahtar Kavram

Conversion between Histogram Frequency Density and Pie Chart Sector Angles

Tahmini Süre:2m 0s

Soru 7427Soru →

What is the gradient of the normal line to the curve $y = x^2 - 3x + 5$ at the point where $x = 1$ ?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The gradient of the normal line to the curve at

x = 1

is 1.

Differentiating

y = x^2 - 3x + 5

gives

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

. Evaluating this derivative at

x = 1

gives the tangent gradient

m_t = -1

. Because the normal line is perpendicular to the tangent, its gradient is

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

.

Adım Adım Çözüm

1

Differentiate the function with respect to x

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

The derivative of a function gives the slope of the tangent line at any given x-coordinate.

2

Evaluate the derivative at x = 1

m_t = -1

Substituting the given point's x-coordinate into the gradient function yields the slope of the tangent.

3

Calculate the negative reciprocal of the tangent slope

m_n = 1

Since the normal line is perpendicular to the tangent line, its gradient is m_n = -1 / m_t.

Anahtar Kavram

Gradient of a Normal Line

Soru 7428Soru →

An arithmetic progression (A.P.) has a first term of $5$ and a common difference of $-3$ . What is the $7^{\text{th}}$ term of the progression?

-13

-16

-19

23

Cevabı ve açıklamayı göster

Cevap: $-13$

Cevap

-13

Using the

n^{\text{th}}

term formula

T_n = a + (n - 1)d

with

a = 5

,

d = -3

, and

n = 7

, we get

T_7 = 5 + (6)(-3) = -13

.

Adım Adım Çözüm

1

Identify the given parameters of the A.P.

First term

a = 5

, common difference

d = -3

, and position

n = 7

.

These parameters are directly specified in the problem.

2

Apply the

n^{\text{th}}

term formula

T_n = a + (n - 1)d

.

T_7 = 5 + (7 - 1)(-3) = 5 + 6(-3)

.

The common difference is added

(n-1)

times to the first term.

3

Perform the multiplication and addition.

T_7 = 5 - 18 = -13

.

Simplifying the arithmetic gives the exact value of the

7^{\text{th}}

term.

Anahtar Kavram

Calculating the

n^{\text{th}}

term of an Arithmetic Progression

Soru 7429Soru →

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

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The numerical value of the derivative at

x = 1

is 2.

To find

\frac{dy}{dx}

at

x = 1

for

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

, we use the quotient rule

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

. Setting

u = (x + 1)^3

gives

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

, and setting

v = x^2 + 1

gives

\frac{dv}{dx} = 2x

. At

x = 1

,

u = 8

,

\frac{du}{dx} = 12

,

v = 2

, and

\frac{dv}{dx} = 2

. Substituting into the quotient formula gives

\frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2

.

Adım Adım Çözüm

1

Identify the components of the quotient rule

Let

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

and

v(x) = x^2 + 1

.

The given function is a quotient of two functions of

x

.

2

Differentiate the numerator using the chain rule and the denominator using standard rules

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

and

v'(x) = 2x

.

The chain rule states that

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

.

3

Evaluate all function components at

x = 1

u(1) = 8

,

u'(1) = 12

,

v(1) = 2

, and

v'(1) = 2

.

Substituting

x = 1

simplifies the calculation before applying the full quotient expression.

4

Apply the quotient rule formula to calculate the final derivative value

\frac{dy}{dx}\Big|_{x=1} = \frac{v(1)u'(1) - u(1)v'(1)}{[v(1)]^2} = \frac{(2)(12) - (8)(2)}{2^2} = \frac{24 - 16}{4} = 2

.

Substituting the numerical values yields the final result.

Anahtar Kavram

Combining the Quotient Rule and Chain Rule for differentiation

Soru 7430Soru →

A point $P(x, y)$ moves such that its perpendicular distance from the straight line $L_1: 4x - 3y + 5 = 0$ is equal to its perpendicular distance from the straight line $L_2: 3x + 4y - 10 = 0$ . Which of the following equations represents one of the straight lines constituting the locus of $P$ ?

x - 7y + 15 = 0

7x - y - 5 = 0

x + 7y + 15 = 0

7x + y + 5 = 0

Cevabı ve açıklamayı göster

Cevap: $x - 7y + 15 = 0$

Cevap

The equation

x - 7y + 15 = 0

represents one of the lines constituting the locus.

The locus of a point equidistant from two intersecting straight lines is the pair of angle bisectors between those lines. Setting the perpendicular distance formulas equal yields

4x - 3y + 5 = \pm(3x + 4y - 10)

. Solving the positive branch yields

x - 7y + 15 = 0

, which correctly represents one of the component lines of the locus.

Adım Adım Çözüm

1

Write the perpendicular distance formulas from point

P(x, y)

to both given lines.

d_1 = \frac{|4x - 3y + 5|}{\sqrt{4^2 + (-3)^2}} = \frac{|4x - 3y + 5|}{5}

and

d_2 = \frac{|3x + 4y - 10|}{\sqrt{3^2 + 4^2}} = \frac{|3x + 4y - 10|}{5}

.

The locus of points equidistant from two intersecting lines consists of the angle bisectors of the angles between the lines.

2

Set the two perpendicular distances equal to each other.

\frac{|4x - 3y + 5|}{5} = \frac{|3x + 4y - 10|}{5} \implies |4x - 3y + 5| = |3x + 4y - 10|.

Since the point is equidistant from both lines,

d_1 = d_2

.

3

Remove absolute values by considering both positive and negative cases.

4x - 3y + 5 = \pm(3x + 4y - 10)

.

Absolute value equality

|A| = |B|

implies

A = B

or

A = -B

.

4

Evaluate Case 1 (positive sign) to find the first line equation.

4x - 3y + 5 = 3x + 4y - 10 \implies (4x - 3x) + (-3y - 4y) + (5 + 10) = 0 \implies x - 7y + 15 = 0

.

Grouping like terms yields the linear equation for the first angle bisector.

5

Evaluate Case 2 (negative sign) to find the second line equation.

4x - 3y + 5 = -(3x + 4y - 10) \implies 4x - 3y + 5 = -3x - 4y + 10 \implies 7x + y - 5 = 0

.

Grouping like terms yields the linear equation for the second angle bisector.

Anahtar Kavram

Locus equidistant from two intersecting lines (Angle Bisectors)

Tahmini Süre:2m 0s

Soru 7431Soru →

What is the exact numerical value of the trigonometric expression $\frac{6\sin^2 60^\circ - 2\cos^2 45^\circ}{\tan^2 30^\circ + \sec^2 45^\circ}$ ?

Cevabı ve açıklamayı göster

Cevap: 1.5

Cevap

1.5

Substituting the exact special angle values gives a numerator of

6\left(\frac{3}{4}\right) - 2\left(\frac{1}{2}\right) = \frac{7}{2}

and a denominator of

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

. Dividing

\frac{7}{2}

by

\frac{7}{3}

yields

\frac{3}{2} = 1.5

.

Adım Adım Çözüm

1

Substitute the exact values for the trigonometric ratios of the special angles.

\sin 60^\circ = \frac{\sqrt{3}}{2}

,

\cos 45^\circ = \frac{1}{\sqrt{2}}

,

\tan 30^\circ = \frac{1}{\sqrt{3}}

, and

\sec 45^\circ = \sqrt{2}

.

Exact surd forms for special angles

30^\circ

,

45^\circ

, and

60^\circ

must be used.

2

Evaluate and simplify the numerator expression

6\sin^2 60^\circ - 2\cos^2 45^\circ

.

6\left(\frac{3}{4}\right) - 2\left(\frac{1}{2}\right) = \frac{9}{2} - 1 = \frac{7}{2}

.

Square each trigonometric ratio first, multiply by the coefficients, and then subtract.

3

Evaluate and simplify the denominator expression

\tan^2 30^\circ + \sec^2 45^\circ

.

\left(\frac{1}{\sqrt{3}}\right)^2 + (\sqrt{2})^2 = \frac{1}{3} + 2 = \frac{7}{3}

.

Square each trigonometric ratio and simplify the sum into a single improper fraction.

4

Divide the numerator result by the denominator result.

\frac{7/2}{7/3} = \frac{7}{2} \times \frac{3}{7} = \frac{3}{2} = 1.5

.

Dividing by a fraction is equivalent to multiplying by its reciprocal.

Anahtar Kavram

Evaluation of Trigonometric Expressions using Special Angles

Soru 7432Soru →

A curve is defined by the equation $y = 4x + \frac{9}{x}$ for $x > 0$ . What is the $y$ -value at the minimum stationary point of the curve?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The

y

-value at the minimum stationary point is 12.

To find the minimum value of

y = 4x + \frac{9}{x}

for

x > 0

, set the first derivative

\frac{dy}{dx} = 4 - \frac{9}{x^2}

equal to

0

, yielding

x = 1.5

. The second derivative

\frac{d^2y}{dx^2} = \frac{18}{x^3}

is positive at

x = 1.5

, confirming a minimum stationary point. Evaluating the original equation at

x = 1.5

gives

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

.

Adım Adım Çözüm

1

Differentiate the given function

y = 4x + 9x^{-1}

with respect to

x

.

\frac{dy}{dx} = 4 - 9x^{-2} = 4 - \frac{9}{x^2}

Stationary points occur where the gradient (first derivative) equals zero.

2

Set the first derivative to zero and solve for

x

given the domain constraint

x > 0

.

4 - \frac{9}{x^2} = 0 \implies 4x^2 = 9 \implies x^2 = \frac{9}{4} \implies x = \frac{3}{2} = 1.5

Solving

\frac{dy}{dx} = 0

yields the

x

-coordinate of the turning point.

3

Evaluate the second derivative to confirm the turning point is a local minimum.

\frac{d^2y}{dx^2} = \frac{18}{x^3}. \text{ At } x = 1.5, \frac{d^2y}{dx^2} = \frac{18}{3.375} = 5.333 > 0

A positive second derivative indicates that the stationary point is a local minimum.

4

Substitute

x = 1.5

back into the original equation

y = 4x + \frac{9}{x}

to find the corresponding

y

-value.

y = 4(1.5) + \frac{9}{1.5} = 6 + 6 = 12

The question asks for the

y

-value of the curve at the minimum stationary point.

Anahtar Kavram

Stationary Points, Maxima, and Minima

Tahmini Süre:2m 0s

Soru 7433Soru →

The table below presents the cumulative frequency distribution of examination marks for $60$ candidates:

Mark Boundary	Cumulative Frequency
$< 20.5$	$5$
$< 40.5$	$18$
$< 60.5$	$42$
$< 80.5$	$54$
$< 100.5$	$60$

How many candidates scored between $40.5$ and $80.5$ marks?

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

36 candidates

To find the number of candidates with scores between

40.5

and

80.5

, subtract the cumulative frequency of scores below

40.5

(

18

) from the cumulative frequency of scores below

80.5

(

54

). This gives

54 - 18 = 36

.

Adım Adım Çözüm

1

Find the cumulative frequency up to the upper boundary of

80.5

Cumulative frequency (

F_{\text{upper}}

) =

54

This represents the total number of candidates scoring below

80.5

marks.

2

Find the cumulative frequency up to the lower boundary of

40.5

Cumulative frequency (

F_{\text{lower}}

) =

18

This represents the total number of candidates scoring below

40.5

marks.

3

Calculate the number of candidates within the interval

(40.5, 80.5)

by finding the difference

54 - 18 = 36

Subtracting the cumulative frequency at

40.5

from that at

80.5

isolates the count of candidates within this specific mark range.

Anahtar Kavram

Finding class frequency from cumulative frequency boundaries

Soru 7434Soru →

In a secondary school of $800$ students, $45\%$ of the students are girls. If $\frac{1}{6}$ of the girls and $\frac{1}{4}$ of the boys wear eyeglasses, what is the total number of students in the school who wear eyeglasses?

Cevabı ve açıklamayı göster

Cevap: 170

Cevap

170 students

Calculating

45\%

of

800

gives

360

girls, leaving

800 - 360 = 440

boys. Taking

\frac{1}{6}

of

360

gives

60

girls with eyeglasses, and taking

\frac{1}{4}

of

440

gives

110

boys with eyeglasses. Adding these two quantities yields

60 + 110 = 170

students.

Adım Adım Çözüm

1

Find the total number of girls

360 girls

45% of the total 800 students are girls

2

Find the total number of boys

440 boys

Subtracting the number of girls (360) from total students (800) gives the number of boys

3

Find the number of girls wearing eyeglasses

60 girls

One-sixth of the 360 girls wear eyeglasses

4

Find the number of boys wearing eyeglasses

110 boys

One-fourth of the 440 boys wear eyeglasses

5

Sum the girls and boys wearing eyeglasses

170 students

Adding 60 girls and 110 boys gives the total students wearing eyeglasses

Anahtar Kavram

Fractions and Percentages of Quantities

Soru 7435Soru →

If $y = x^2 \sqrt{3x + 1}$ , what is the value of $\frac{dy}{dx}$ evaluated at $x = 1$ ?

\frac{19}{4}

\frac{17}{4}

\frac{13}{4}

\frac{3}{2}

Cevabı ve açıklamayı göster

Cevap: $\frac{19}{4}$

Cevap

The numerical value of

\frac{dy}{dx}

at

x = 1

is

\frac{19}{4}

.

Applying the product rule to

y = x^2 \sqrt{3x + 1}

gives

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

. By the chain rule,

\frac{d}{dx}(\sqrt{3x + 1}) = \frac{3}{2\sqrt{3x + 1}}

. Substituting

x = 1

into

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

gives

\frac{3}{4} + 4 = \frac{19}{4}

. Thus, the option equal to

\frac{19}{4}

is correct.

Adım Adım Çözüm

1

Identify the main differentiation rule required and split the function into two components.

Let

u = x^2

and

v = (3x + 1)^{1/2}

, so

y = u \cdot v

.

The function is a product of

x^2

and a composite function

\sqrt{3x + 1}

.

2

Differentiate

u

with respect to

x

, and use the chain rule to differentiate

v

with respect to

x

.

\frac{du}{dx} = 2x

and

\frac{dv}{dx} = \frac{1}{2}(3x + 1)^{-1/2} \cdot 3 = \frac{3}{2\sqrt{3x + 1}}

.

The chain rule requires multiplying the derivative of the outer square root function by the derivative of the inner linear function

3x + 1

.

3

Apply the product rule formula

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

.

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

.

The derivative of a product

uv

is

u v' + v u'

.

4

Substitute

x = 1

into the derivative expression.

\frac{dy}{dx}\Big|_{x=1} = (1)^2 \left( \frac{3}{2\sqrt{3(1) + 1}} \right) + \sqrt{3(1) + 1} \cdot 2(1) = 1 \cdot \left( \frac{3}{4} \right) + 2 \cdot 2 = \frac{3}{4} + 4 = \frac{19}{4}

.

Evaluating at

x = 1

requires simplifying

\sqrt{4} = 2

and combining fractions.

Anahtar Kavram

Combining Product Rule and Chain Rule for Differentiation

Soru 7436Soru →

A panel of $5$ members is to be selected from $10$ eligible candidates. If $2$ specific candidates refuse to serve on the panel together, in how many different ways can the panel be formed?

Cevabı ve açıklamayı göster

Cevap: 196

Cevap

The panel can be formed in 196 different ways.

To find the number of valid panels, use complementary counting. First, compute the total number of ways to pick any 5 candidates from 10 without restrictions:

\binom{10}{5} = 252

. Next, find the number of invalid panels that contain both restricted candidates; since 2 candidates are already placed, pick the remaining 3 members from the remaining 8 candidates:

\binom{8}{3} = 56

. Subtracting these invalid panels from the total gives

252 - 56 = 196

.

Adım Adım Çözüm

1

Calculate the total possible combinations without any restrictions.

\binom{10}{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252

The combinations formula

\binom{n}{r} = \frac{n!}{r!(n-r)!}

applies since the order of selection does not matter.

2

Determine the number of invalid combinations where both specific candidates are included.

\binom{8}{3} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56

If both specific candidates are already placed on the panel, 3 more members must be chosen from the remaining 8 candidates.

3

Apply complementary counting to subtract invalid selections from total selections.

252 - 56 = 196

Subtracting the restricted combinations from the total combinations gives the number of valid panel configurations.

Anahtar Kavram

Combinations with mutual exclusion (Complementary Counting)

Soru 7437Soru →

Find the area of the region bounded by the curve $y = 3x^2$ , the $x$ -axis, and the vertical lines $x = 1$ and $x = 3$ .

Cevabı ve açıklamayı göster

Cevap: 26

Cevap

The area of the bounded region is 26 square units.

The area under

y = 3x^2

from

x = 1

to

x = 3

is calculated using the definite integral

\int_{1}^{3} 3x^2 \, dx = [x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26

square units.

Adım Adım Çözüm

1

Set up the definite integral representing the bounded area.

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

The area under a non-negative curve

y = f(x)

from

x = a

to

x = b

above the

x

-axis is given by the definite integral

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

.

2

Determine the antiderivative of

3x^2

.

\int 3x^2 \, dx = 3 \cdot \frac{x^3}{3} = x^3

Applying the power rule of integration

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

gives

x^3

.

3

Evaluate the definite integral using the fundamental theorem of calculus.

[x^3]_{1}^{3} = 3^3 - 1^3 = 27 - 1 = 26

Substitute the upper limit

x = 3

and subtract the value of the function evaluated at the lower limit

x = 1

.

Anahtar Kavram

Area under a curve using definite integration

Soru 7438Soru →

If $2^{2x + 1} - 5(2^x) + 2 = 0$ , what are the possible values of $x$ ?

-1

or

1

\frac{1}{2}

or

2

-1

or

2

-2

or

1

Cevabı ve açıklamayı göster

Cevap: $-1$ or $1$

Cevap

-1

or

1

Using the addition law of indices,

2^{2x+1}

can be rewritten as

2^1 \cdot 2^{2x} = 2 \cdot (2^x)^2

. Substituting

y = 2^x

produces the quadratic equation

2y^2 - 5y + 2 = 0

, which factorizes into

(2y - 1)(y - 2) = 0

. This gives

y = \frac{1}{2}

or

y = 2

. Converting back to exponential equations gives

2^x = 2^{-1} \implies x = -1

and

2^x = 2^1 \implies x = 1

. Thus, the solutions for

x

are

-1

or

1

.

Adım Adım Çözüm

1

Apply index laws to express the equation in terms of

2^x

2^{2x+1} = 2^{2x} \cdot 2^1 = 2 \cdot (2^x)^2

, so the equation becomes

2 \cdot (2^x)^2 - 5(2^x) + 2 = 0

Splitting the exponent using

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

reveals a quadratic structure in

2^x

.

2

Substitute

y = 2^x

and solve the resulting quadratic equation

2y^2 - 5y + 2 = 0 \implies (2y - 1)(y - 2) = 0 \implies y = \frac{1}{2} \text{ or } y = 2

Factoring the quadratic expression yields the values for the substitution variable

y

.

3

Equate

2^x

to each solution of

y

to solve for

x

For

y = \frac{1}{2}

:

2^x = 2^{-1} \implies x = -1

. For

y = 2

:

2^x = 2^1 \implies x = 1

.

Using the negative index rule

a^{-n} = \frac{1}{a^n}

allows matching exponents when bases are identical.

Anahtar Kavram

Quadratic Equations Reducible to Index Form

Soru 7439Soru →

Using differentiation from first principles, what is the value of the derivative of the function $f(x) = x^2 + 2x$ at the point where $x = 3$ ?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The derivative of

f(x) = x^2 + 2x

evaluated at

x = 3

is 8.

Applying first principles to

f(x) = x^2 + 2x

yields

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

. Evaluating this derivative at

x = 3

gives

2(3) + 2 = 8

.

Adım Adım Çözüm

1

Apply the definition of differentiation from first principles

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

First principles uses the limit of the difference quotient to compute the instantaneous rate of change.

2

Substitute

f(x) = x^2 + 2x

into the difference quotient

\frac{[(x+h)^2 + 2(x+h)] - (x^2 + 2x)}{h} = \frac{2xh + h^2 + 2h}{h}

Expanding terms allows cancellation of non-

h

terms in the numerator.

3

Simplify the fraction by dividing by

h

2x + h + 2

Dividing out

h

removes the indeterminate form

\frac{0}{0}

.

4

Evaluate the limit as

h \to 0

and substitute

x = 3

f'(3) = 2(3) + 2 = 8

Setting

h = 0

yields the derivative function

f'(x) = 2x + 2

, which evaluates to 8 at

x = 3

.

Anahtar Kavram

Differentiation from First Principles

Soru 7440Soru →

If $(x - 1)$ is a factor of the polynomial $P(x) = x^3 + 2x^2 - 5x + k$ , what is the value of $k$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of

k

is

2

.

According to the Factor Theorem, a linear polynomial

(x - a)

is a factor of

P(x)

if and only if

P(a) = 0

. For the factor

(x - 1)

, setting

x = 1

gives

P(1) = (1)^3 + 2(1)^2 - 5(1) + k = 0

. Simplifying yields

1 + 2 - 5 + k = 0

, which simplifies further to

-2 + k = 0

, giving

k = 2

.

Adım Adım Çözüm

1

Apply the Factor Theorem

P(1) = 0

Since

(x - 1)

is a factor, setting

x = 1

makes the polynomial equal to zero.

2

Substitute

x = 1

into

P(x)

1^3 + 2(1)^2 - 5(1) + k = 0

Evaluate the polynomial expression at

x = 1

.

3

Simplify and solve for

k

k = 2

Combine constants

1 + 2 - 5 = -2

and solve

-2 + k = 0

.

Anahtar Kavram

Factor Theorem

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