Tüm alıştırma soruları

Soru 661Soru →

An infinite geometric progression of positive terms has a sum to infinity of $16$ , and the sum of its first two terms is $12$ . An arithmetic progression has its first term equal to the first term of this geometric progression, and its $5^{\text{th}}$ term equal to the sum to infinity of the geometric progression. Calculate the $10^{\text{th}}$ term of the arithmetic progression.

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Cevap: 26

Cevap

The 10th term of the arithmetic progression is 26.

By solving the geometric progression system, we find the common ratio

r = \frac{1}{2}

and first term

a = 8

. Using

a = 8

as the first term of the arithmetic progression and setting its

5^{\text{th}}

term

A_5 = 16

, we determine the common difference

d = 2

. Calculating

A_{10} = 8 + 9(2)

yields

26

.

Adım Adım Çözüm

1

Formulate equations for the geometric progression using the sum to infinity and sum of the first two terms

a = 16(1 - r)

and

a(1 + r) = 12

The standard formula for the sum to infinity of a GP is

S_\infty = \frac{a}{1-r}

and the sum of the first two terms is

S_2 = a + ar = a(1+r)

.

2

Solve for the common ratio

r

and first term

a

of the geometric progression

r = 0.5

and

a = 8

Substituting

a = 16(1-r)

yields

16(1-r^2) = 12 \implies r^2 = \frac{1}{4} \implies r = \frac{1}{2}

. Then

a = 16(1 - 0.5) = 8

.

3

Determine the common difference

d

of the arithmetic progression

d = 2

The first term of the AP is

A_1 = a = 8

and the 5th term is

A_5 = S_\infty = 16

. Using

A_5 = A_1 + 4d \implies 8 + 4d = 16 \implies d = 2

.

4

Calculate the 10th term of the arithmetic progression

A_{10} = 26

Using the AP

n^{\text{th}}

term formula

A_n = A_1 + (n-1)d

:

A_{10} = 8 + 9(2) = 26

.

Anahtar Kavram

Combining geometric progression parameters (sum to infinity and sum of terms) with arithmetic progression term formulas

Soru 662Soru →

A vessel contains a mixture of two liquids, $A$ and $B$ , in the ratio $5 : 3$ . If $16\text{ litres}$ of the mixture is drawn off and replaced with an equal volume of liquid $B$ , the ratio of liquid $A$ to liquid $B$ in the vessel becomes $1 : 1$ . What was the initial total volume of the mixture in the vessel, in litres?

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Cevap: 80

Cevap

The initial total volume of the mixture in the vessel was

80\text{ litres}

.

Let the initial volume of the mixture be

V

litres. Liquid

A

initially comprises

\frac{5}{8}V

litres and liquid

B

comprises

\frac{3}{8}V

litres. When

16\text{ litres}

of mixture is removed, the volume of liquid

A

removed is

\frac{5}{8} \times 16 = 10\text{ litres}

, and the volume of liquid

B

removed is

\frac{3}{8} \times 16 = 6\text{ litres}

. After adding

16\text{ litres}

of pure liquid

B

, the new volume of liquid

A

is

\frac{5}{8}V - 10

and the new volume of liquid

B

is

\frac{3}{8}V - 6 + 16 = \frac{3}{8}V + 10

. Since the new ratio is

1 : 1

, setting

\frac{5}{8}V - 10 = \frac{3}{8}V + 10

gives

\frac{2}{8}V = 20

, which simplifies to

V = 80\text{ litres}

.

Adım Adım Çözüm

1

Define total initial volume as

V

and express the initial quantities of liquids

A

and

B

.

Liquid

A = \frac{5}{8}V

litres, Liquid

B = \frac{3}{8}V

litres.

The total ratio parts equal

5 + 3 = 8

.

2

Calculate the volume of each liquid removed in

16\text{ litres}

of mixture.

Volume of

A

removed

= 10

litres; Volume of

B

removed

= 6

litres.

The drawn-off mixture retains the original

5 : 3

ratio of liquids.

3

Formulate expressions for the quantities of

A

and

B

after adding

16\text{ litres}

of pure liquid

B

.

New amount of

A = \frac{5}{8}V - 10

; New amount of

B = \frac{3}{8}V + 10

.

16

litres of liquid

B

is added to the remaining quantity of

B

, which was

\frac{3}{8}V - 6

.

4

Equate the new quantities of

A

and

B

since the final ratio is

1 : 1

.

\frac{5}{8}V - 10 = \frac{3}{8}V + 10 \implies \frac{2}{8}V = 20 \implies V = 80

.

A final ratio of

1 : 1

implies equal quantities of both liquids.

Anahtar Kavram

Ratio modification through mixture removal and replacement

Tahmini Süre:2m 0s

Soru 663Soru →

What is the value of $\frac{\log_4 81}{\log_4 3}$ ?

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Cevap: 4

Cevap

4

Applying the change of base property

\frac{\log_c b}{\log_c a} = \log_a b

, the ratio

\frac{\log_4 81}{\log_4 3}

reduces directly to

\log_3 81

. Since

3^4 = 81

, the value is 4.

Adım Adım Çözüm

1

Apply the change of base formula to rewrite the ratio

\frac{\log_4 81}{\log_4 3} = \log_3 81

By the change of base identity,

\frac{\log_c b}{\log_c a} = \log_a b

.

2

Evaluate the logarithm

4

Since

3^4 = 81

,

\log_3 81 = 4

.

Anahtar Kavram

Change of Base Formula for Logarithms

Soru 664Soru →

A vessel initially contains $80\text{ litres}$ of water. After standing in the sun, $12\text{ litres}$ of water evaporates. What percentage of the original volume of water evaporated?

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Cevap: 15

Cevap

The percentage of the original volume of water that evaporated is

15\%

.

To find the percentage of evaporated water, divide the evaporated amount (

12\text{ litres}

) by the original total volume (

80\text{ litres}

) and multiply by

100\%

. This yields

\frac{12}{80} \times 100\% = 15\%

.

Adım Adım Çözüm

1

Determine the fractional portion of the evaporated water relative to the initial total volume.

The fraction is

\frac{12}{80} = \frac{3}{20}

.

Percentage loss must be evaluated relative to the original starting amount.

2

Convert the resulting fraction into a percentage.

\frac{3}{20} \times 100\% = 15\%.

Multiplying a dimensionless ratio by 100 yields its percentage equivalent.

Anahtar Kavram

Calculating percentage change or loss relative to an initial amount

Tahmini Süre:45s

Soru 665Soru →

If $123_x = 38_{10}$ , what is the value of the base $x$ ?

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Cevap: 5

Cevap

The base

x

is 5.

Expanding

123_x

in terms of powers of

x

yields

1 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3

. Setting this equal to the decimal value 38 produces the quadratic equation

x^2 + 2x + 3 = 38

, which simplifies to

x^2 + 2x - 35 = 0

. Factoring gives

(x + 7)(x - 5) = 0

, yielding solutions

x = -7

and

x = 5

. Since a number base must be a positive integer, the correct value for

x

is 5.

Adım Adım Çözüm

1

Expand the base

x

number into decimal form using place values

123_x = 1 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3

Each digit position in base

x

corresponds to a power of

x

, starting from

x^0

on the right.

2

Set up and rearrange the quadratic equation

x^2 + 2x + 3 = 38 \implies x^2 + 2x - 35 = 0

Subtracting 38 from both sides converts the equation into standard quadratic form

ax^2 + bx + c = 0

.

3

Solve the quadratic equation for

x

(x + 7)(x - 5) = 0 \implies x = -7 \text{ or } x = 5

Factoring gives the roots of the quadratic equation.

4

Select the valid positive base

x = 5

A base must be a positive integer greater than the largest digit appearing in the number (which is 3).

Anahtar Kavram

Place value expansion and base conversion to base 10

Soru 666Soru →

The mean of five numbers $3, 5, 7, x,$ and $y$ is $6$ . Given that the variance of the numbers is $8$ and $x < y$ , calculate the value of $y$ .

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Cevap: 11

Cevap

11

The total sum of the five numbers is

5 \times 6 = 30

, giving

x + y = 15

. The sum of squared deviations from the mean

6

is

5 \times 8 = 40

. The known numbers

3, 5, 7

contribute

(-3)^2 + (-1)^2 + 1^2 = 11

to this sum, leaving

(x-6)^2 + (y-6)^2 = 29

. Substituting

y = 15 - x

yields

2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0

. Factoring gives roots

4

and

11

. Since

x < y

, we find

y = 11

.

Adım Adım Çözüm

1

Use the definition of the arithmetic mean to write a linear relationship between

x

and

y

.

x + y = 15

, or

y = 15 - x

.

The total sum of 5 numbers with a mean of 6 is

5 \times 6 = 30

. Subtracting the known numbers

3 + 5 + 7 = 15

leaves

x + y = 15

.

2

Apply the variance formula for population data.

\sum (x_i - 6)^2 = 40

.

Variance is the mean of squared deviations from the mean:

\frac{\sum (x_i - \bar{x})^2}{5} = 8 \implies \sum (x_i - 6)^2 = 40

.

3

Compute the sum of squared deviations for the known elements and simplify the variance equation.

(x - 6)^2 + (y - 6)^2 = 29

.

The squared deviations for

3, 5, 7

are

(-3)^2 = 9

,

(-1)^2 = 1

, and

1^2 = 1

. Subtracting

9 + 1 + 1 = 11

from

40

gives

29

.

4

Substitute

y = 15 - x

into the simplified equation and solve the resulting quadratic equation.

x = 4

or

x = 11

.

Substituting

y = 15 - x

gives

(x - 6)^2 + (9 - x)^2 = 29 \implies 2x^2 - 30x + 88 = 0 \implies x^2 - 15x + 44 = 0 \implies (x - 4)(x - 11) = 0

.

5

Select the correct pair

(x, y)

using the condition

x < y

.

x = 4

and

y = 11

.

Since

x < y

,

x

must be the smaller value (

4

) and

y

must be the larger value (

11

).

Anahtar Kavram

Calculation of variance and mean for ungrouped data containing unknown elements

Soru 667Soru →

The operating lifespans (in hours) of a sample of $60$ newly manufactured micro-components tested under laboratory conditions are presented in the frequency table below:

Lifespan (hours)	Frequency ( $f$ )
$10 - 19$	$5$
$20 - 29$	$8$
$30 - 39$	$12$
$40 - 49$	$20$
$50 - 59$	$15$

Using linear interpolation on class boundaries, calculate the median lifespan of the components in hours.

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Cevap: 42

Cevap

The median lifespan of the micro-components is 42 hours.

To calculate the median of grouped data, first determine cumulative frequencies: 5, 13, 25, 45, 60. Total frequency is

N = 60

, placing the median at

N/2 = 30

. The median class is

40 - 49

with lower class boundary

L = 39.5

, class frequency

f = 20

, cumulative frequency prior to median class

c.f. = 25

, and class width

c = 10

. Substituting into

\text{Median} = L + \left(\frac{N/2 - c.f.}{f}\right) \times c

gives

39.5 + \left(\frac{30 - 25}{20}\right) \times 10 = 39.5 + 2.5 = 42

.

Adım Adım Çözüm

1

Calculate the total frequency

N

and the median rank

N/2

.

N = 5 + 8 + 12 + 20 + 15 = 60

, so

N/2 = 30

.

The median corresponds to the value at position 30 in the ordered cumulative frequency distribution.

2

Determine cumulative frequencies to locate the median class.

Cumulative frequencies are 5, 13, 25, 45, 60. The 30th value falls in the interval

40 - 49

.

The cumulative frequency first reaches or exceeds 30 at the

40 - 49

class.

3

Identify boundary parameters for the median class.

Lower boundary

L = 39.5

, class width

c = 10

, class frequency

f = 20

, and cumulative frequency of preceding class

c.f. = 25

.

Continuous data analysis requires using lower class boundaries rather than class limits.

4

Apply the grouped median linear interpolation formula.

\text{Median} = 39.5 + \left(\frac{30 - 25}{20}\right) \times 10 = 39.5 + 2.5 = 42

.

Linear interpolation within the median class yields the precise median value.

Anahtar Kavram

Grouped Data Median via Class Boundary Interpolation

Tahmini Süre:2m 0s

Soru 668Soru →

A rectangular lawn was measured to have a length of $15.0\text{ m}$ and a width of $8.0\text{ m}$ . If the actual length of the lawn is $16.0\text{ m}$ and the measured width is exact, what is the percentage error in the calculated area of the lawn?

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Cevap: 6.25

Cevap

The percentage error in the calculated area is

6.25\%

.

The actual area of the lawn is

16.0\text{ m} \times 8.0\text{ m} = 128\text{ m}^2

, while the measured area is

15.0\text{ m} \times 8.0\text{ m} = 120\text{ m}^2

. The error is

128 - 120 = 8\text{ m}^2

. Expressed as a percentage of the actual area,

\frac{8}{128} \times 100\% = 6.25\%

.

Adım Adım Çözüm

1

Calculate the measured area of the lawn using the measured length and width

\text{Measured Area} = 15.0\text{ m} \times 8.0\text{ m} = 120\text{ m}^2

Area of a rectangle is length multiplied by width.

2

Calculate the true (actual) area of the lawn using the actual length and exact width

\text{Actual Area} = 16.0\text{ m} \times 8.0\text{ m} = 128\text{ m}^2

The actual dimensions determine the true surface area.

3

Determine the magnitude of the error in area

\text{Error} = |128\text{ m}^2 - 120\text{ m}^2| = 8\text{ m}^2

Error is defined as the absolute difference between actual and measured values.

4

Compute the percentage error relative to the actual area

\text{Percentage Error} = \frac{8}{128} \times 100\% = 6.25\%

Percentage error is always calculated as

\frac{\text{Error}}{\text{Actual Value}} \times 100\%

.

Anahtar Kavram

Percentage Error Calculation in Compound Quantities

Tahmini Süre:1m 30s

Soru 669Soru →

The frequency distribution table below shows the recorded speeds (in $\text{km/h}$ ) of a sample of commercial buses passing through a highway toll checkpoint:

Speed ( $\text{km/h}$ )	Frequency
$40 - 49$	$6$
$50 - 59$	$10$
$60 - 69$	$f$
$70 - 79$	$12$
$80 - 89$	$8$

If the mean speed of the buses is $66.5\text{ km/h}$ , find the value of the missing frequency $f$ .

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Cevap: 6

Cevap

The value of the missing frequency

f

is

6

.

Each class interval's midpoint is calculated by averaging its lower and upper limits. The total frequency is

\sum f = 36 + f

and the total sum of products is

\sum fx = 2382 + 64.5f

. Applying the grouped mean formula

\bar{x} = \frac{\sum fx}{\sum f} = 66.5

gives

2394 + 66.5f = 2382 + 64.5f

, which yields

2f = 12

, so

f = 6

.

Adım Adım Çözüm

1

Find the class midpoints (

x

) for each grouped interval.

Midpoints are

44.5, 54.5, 64.5, 74.5,

and

84.5

.

Midpoints represent the central values of each interval for grouped mean calculations.

2

Calculate expressions for

\sum f

and

\sum fx

.

\sum f = 36 + f

and

\sum fx = 2382 + 64.5f$.

Summing the frequencies and the products of midpoints and frequencies gives the components required for the mean equation.

3

Substitute known values into the mean formula and solve for

f

.

66.5(36 + f) = 2382 + 64.5f \implies 2f = 12 \implies f = 6$.

Equating the formula expression to the given mean of

66.5\text{ km/h}

allows solving for the unknown frequency.

Anahtar Kavram

Grouped Mean with Missing Frequency

Soru 670Soru →

If $\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} - \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} = k\sqrt{6}$ , find the value of the integer $k$ .

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Cevap: 4

Cevap

The value of the integer

k

is 4.

Combining the fractions over the common denominator

(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2}) = 3-2 = 1

yields a numerator of

(3+2+2\sqrt{6}) - (3+2-2\sqrt{6}) = 4\sqrt{6}

. Thus,

k\sqrt{6} = 4\sqrt{6}

, which gives

k = 4

.

Adım Adım Çözüm

1

Combine the fractions using their common denominator

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

Subtracting algebraic fractions requires finding the least common denominator, which is the product of the conjugate pair.

2

Expand the terms in the numerator

(\sqrt{3} + \sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6} \text{ and } (\sqrt{3} - \sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}

Use the perfect square expansion formula

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

.

3

Subtract the expanded terms in the numerator and simplify the denominator

\text{Numerator: } (5 + 2\sqrt{6}) - (5 - 2\sqrt{6}) = 4\sqrt{6}, \text{ Denominator: } (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1

Apply the difference of two squares identity

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

to the denominator and carefully distribute the negative sign across terms in the numerator.

4

Equate the simplified expression to

k\sqrt{6}

and solve for

k

k = 4

Comparing

\frac{4\sqrt{6}}{1} = 4\sqrt{6}

with

k\sqrt{6}

yields

k = 4

.

Anahtar Kavram

Rationalisation of surd denominators using conjugate pairs and difference of squares

Soru 671Soru →

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

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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 672Soru →

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 $0.08\text{ m}$ ?

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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 673Soru →

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

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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 674Soru →

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

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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 675Soru →

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 676Soru →

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 677Soru →

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 678Soru →

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 679Soru →

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 680Soru →

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

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