Tüm alıştırma soruları

13931 soru

Soru 7261Soru

The ages, in years, of five participants in a workshop are 2,4,5,7,2, 4, 5, 7, and 1212. What is the mean deviation of the ages?

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Cevap: 2.82.8

Cevap

2.82.8
The mean deviation is calculated by taking the average of the absolute differences between each data point and the mean. For the dataset 2,4,5,7,2, 4, 5, 7, and 1212, the mean is 66. The absolute deviations from 66 are 4,2,1,1,4, 2, 1, 1, and 66, which sum to 1414. Dividing 1414 by 55 gives 2.82.8.

Adım Adım Çözüm

1
Calculate the arithmetic mean (xˉ\bar{x}) of the dataset.
xˉ=2+4+5+7+125=305=6\bar{x} = \frac{2 + 4 + 5 + 7 + 12}{5} = \frac{30}{5} = 6
The mean is needed as the reference point to calculate deviations.
2
Calculate the absolute deviation xxˉ|x - \bar{x}| for each data item.
26=4,46=2,56=1,76=1,126=6|2 - 6| = 4, \quad |4 - 6| = 2, \quad |5 - 6| = 1, \quad |7 - 6| = 1, \quad |12 - 6| = 6
Mean deviation measures distance from the mean, so absolute values are taken.
3
Find the average of these absolute deviations.
\text{Mean Deviation} = \frac{4 + 2 + 1 + 1 + 6}{5} = \frac{14}{5} = 2.8
The formula for mean deviation of ungrouped data is xxˉn\frac{\sum |x - \bar{x}|}{n}.

Anahtar Kavram

Mean Deviation for Ungrouped Data
Soru 7262Soru

The frequency distribution table below shows the mass, in grams, of 2020 harvested oranges recorded during an agricultural experiment:

Mass (gg)Frequency (ff)
101910 - 1933
202920 - 2955
303930 - 3988
404940 - 4944

What is the mean mass of the harvested oranges?

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Cevap: 31.0 g31.0\text{ g}

Cevap

The mean mass of the harvested oranges is 31.0 g31.0\text{ g}.
The mean of grouped data is given by xˉ=fxf\bar{x} = \frac{\sum fx}{\sum f}. Finding the midpoints (xx) of the intervals gives 14.514.5, 24.524.5, 34.534.5, and 44.544.5. Multiplying each midpoint by its frequency yields 43.543.5, 122.5122.5, 276.0276.0, and 178.0178.0. The sum of these products is 620.0620.0, and dividing by the total frequency (2020) gives 31.0 g31.0\text{ g}.

Adım Adım Çözüm

1
Determine the class midpoint (xx) for each class interval.
Midpoints are: 10+192=14.5\frac{10+19}{2} = 14.5, 20+292=24.5\frac{20+29}{2} = 24.5, 30+392=34.5\frac{30+39}{2} = 34.5, and 40+492=44.5\frac{40+49}{2} = 44.5.
For grouped frequency data, each interval is represented by its central value (class mark or midpoint).
2
Multiply each midpoint (xx) by its corresponding frequency (ff) to calculate fxfx.
f1x1=3×14.5=43.5f_1 x_1 = 3 \times 14.5 = 43.5, f2x2=5×24.5=122.5f_2 x_2 = 5 \times 24.5 = 122.5, f3x3=8×34.5=276.0f_3 x_3 = 8 \times 34.5 = 276.0, f4x4=4×44.5=178.0f_4 x_4 = 4 \times 44.5 = 178.0.
This computes the total mass contributed by the items in each interval.
3
Calculate the total sum of fxfx and total frequency f\sum f.
fx=43.5+122.5+276.0+178.0=620.0\sum fx = 43.5 + 122.5 + 276.0 + 178.0 = 620.0 and f=3+5+8+4=20\sum f = 3 + 5 + 8 + 4 = 20.
These sums are the numerator and denominator required for the mean formula.
4
Compute the mean using xˉ=fxf\bar{x} = \frac{\sum fx}{\sum f}.
xˉ=620.020=31.0 g\bar{x} = \frac{620.0}{20} = 31.0\text{ g}.
Dividing the total weighted value by the total number of items gives the mean.

Anahtar Kavram

Calculating Mean for Grouped Frequency Data
Soru 7263Soru

A curve has the equation y=ax3+bx2+12x+1y = ax^3 + bx^2 + 12x + 1, where aa and bb are constants. If the curve has stationary points at x=1x = 1 and x=2x = 2, what is the value of aa?

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Cevap: 22

Cevap

The value of aa is 22.
To find aa, take the first derivative of the curve, yielding dydx=3ax2+2bx+12\frac{dy}{dx} = 3ax^2 + 2bx + 12. Setting this to zero gives a quadratic equation with roots x=1x = 1 and x=2x = 2. The product of roots for a quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0 is CA\frac{C}{A}. Therefore, 1×2=123a1 \times 2 = \frac{12}{3a}, which simplifies to 2=4a2 = \frac{4}{a}, giving a=2a = 2.

Adım Adım Çözüm

1
Find the derivative of the given curve with respect to xx.
dydx=3ax2+2bx+12\frac{dy}{dx} = 3ax^2 + 2bx + 12
Stationary points occur where the first derivative dydx=0\frac{dy}{dx} = 0.
2
Set the derivative to zero and substitute the stationary point locations x=1x = 1 and x=2x = 2.
The roots of the quadratic equation 3ax2+2bx+12=03ax^2 + 2bx + 12 = 0 are x1=1x_1 = 1 and x2=2x_2 = 2.
Since stationary points are given at x=1x=1 and x=2x=2, these values satisfy dydx=0\frac{dy}{dx} = 0.
3
Apply the product of roots formula for a quadratic equation Ax2+Bx+C=0Ax^2 + Bx + C = 0, where x1x2=CAx_1 \cdot x_2 = \frac{C}{A}.
1×2=123a    2=4a1 \times 2 = \frac{12}{3a} \implies 2 = \frac{4}{a}
Equating the product of roots 1×2=21 \times 2 = 2 to 123a\frac{12}{3a} isolates parameter aa.
4
Solve for aa.
a=2a = 2
Multiplying both sides by aa gives 2a=42a = 4, so a=2a = 2.

Anahtar Kavram

Stationary points occur where dydx=0\frac{dy}{dx} = 0. For a cubic curve, the derivative is a quadratic equation whose roots correspond to the xx-coordinates of the stationary points.
Tahmini Süre:2m 0s
Soru 7264Soru

A pie chart is used to display the distribution of 720720 candidates registered for a competitive examination across five subjects. If the sector representing Further Mathematics has a central angle of 4545^\circ, what is the total number of candidates who registered for Further Mathematics?

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Cevap: 90

Cevap

The total number of candidates who registered for Further Mathematics is 90.
To calculate the number of candidates represented by a pie chart sector, multiply the total count by the ratio of the sector's central angle to 360 degrees: (45 / 360) * 720 = 90 candidates.

Adım Adım Çözüm

1
Determine the fraction of the total population represented by the Further Mathematics sector.
45360=18\frac{45^\circ}{360^\circ} = \frac{1}{8}
A complete pie chart circle corresponds to an angle of 360 degrees.
2
Calculate the actual number of candidates by multiplying the fraction by the total student population.
18×720=90\frac{1}{8} \times 720 = 90
The number of items in a sector is directly proportional to its central sector angle relative to 360 degrees.

Anahtar Kavram

Calculating category frequencies from pie chart sector angles
Soru 7265Soru

If (x+3)(x + 3) is a factor of the polynomial P(x)=x3+4x2+mx6P(x) = x^3 + 4x^2 + mx - 6, what is the value of mm?

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Cevap: 1

Cevap

The value of mm is 1.
According to the Factor Theorem, (x+3)(x + 3) is a factor of P(x)P(x) if P(3)=0P(-3) = 0. Substituting x=3x = -3 into P(x)=x3+4x2+mx6P(x) = x^3 + 4x^2 + mx - 6 yields (3)3+4(3)2+m(3)6=0(-3)^3 + 4(-3)^2 + m(-3) - 6 = 0, which simplifies to 27+363m6=0-27 + 36 - 3m - 6 = 0, giving 33m=03 - 3m = 0 and thus m=1m = 1.

Adım Adım Çözüm

1
Apply the Factor Theorem
P(3)=0P(-3) = 0
By the Factor Theorem, for a linear divisor (xa)(x - a) to be a factor of P(x)P(x), P(a)P(a) must equal zero. Here x+3=0    x=3x + 3 = 0 \implies x = -3.
2
Substitute x=3x = -3 into P(x)=x3+4x2+mx6P(x) = x^3 + 4x^2 + mx - 6
(-3)^3 + 4(-3)^2 + m(-3) - 6 = 0
Evaluating P(3)P(-3) sets up an equation to find the unknown coefficient mm.
3
Simplify and solve for mm
-27 + 36 - 3m - 6 = 0 \implies 3 - 3m = 0 \implies m = 1
Combine the constant terms 27+366=3-27 + 36 - 6 = 3 and solve the linear equation in terms of mm.

Anahtar Kavram

Factor Theorem
Soru 7266Soru

Using differentiation from first principles, evaluate the numerical value of the derivative of the polynomial function f(x)=2x33x2+4f(x) = 2x^3 - 3x^2 + 4 at x=2x = 2.

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Cevap: 12

Cevap

The numerical value of the derivative at x=2x = 2 is 12.
Applying the first principles limit formula limh0f(x+h)f(x)h\lim_{h \to 0} \frac{f(x+h) - f(x)}{h} to f(x)=2x33x2+4f(x) = 2x^3 - 3x^2 + 4 yields f(x)=6x26xf'(x) = 6x^2 - 6x. Evaluating this derivative function at x=2x = 2 yields 6(4)12=126(4) - 12 = 12.

Adım Adım Çözüm

1
Set up the difference quotient using the definition of differentiation from first principles
f(x)=limh0[2(x+h)33(x+h)2+4][2x33x2+4]hf'(x) = \lim_{h \to 0} \frac{[2(x+h)^3 - 3(x+h)^2 + 4] - [2x^3 - 3x^2 + 4]}{h}
Differentiation from first principles evaluates the limit of the rate of change as the increment hh approaches zero.
2
Expand (x+h)3(x+h)^3 and (x+h)2(x+h)^2 and subtract f(x)f(x)
f(x+h)f(x)=6x2h+6xh2+2h36xh3h2f(x+h) - f(x) = 6x^2h + 6xh^2 + 2h^3 - 6xh - 3h^2
Expanding the terms allows cancellation of all terms not containing hh.
3
Divide by hh and evaluate the limit as h0h \to 0
f(x)=limh0(6x2+6xh+2h26x3h)=6x26xf'(x) = \lim_{h \to 0} (6x^2 + 6xh + 2h^2 - 6x - 3h) = 6x^2 - 6x
Dividing by hh eliminates the indeterminate form 00\frac{0}{0}, allowing direct substitution of h=0h = 0.
4
Substitute x=2x = 2 into the derivative function f(x)f'(x)
f(2)=6(2)26(2)=2412=12f'(2) = 6(2)^2 - 6(2) = 24 - 12 = 12
Evaluating at the given point gives the slope of the tangent line at x=2x = 2.

Anahtar Kavram

Differentiation from first principles using limit of difference quotient
Soru 7267Soru

In a convex polygon of nn sides, three of its interior angles are right angles, and each of the remaining interior angles is equal to 150150^\circ. What is the total number of diagonals in this polygon?

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Cevap: 9

Cevap

9 diagonals
The total sum of interior angles of an nn-sided polygon is (n2)×180(n - 2) \times 180^\circ. Setting this equal to 3(90)+(n3)(150)3(90^\circ) + (n - 3)(150^\circ) simplifies to 180n360=150n180180n - 360 = 150n - 180, which solves to n=6n = 6. Substituting n=6n = 6 into the diagonal formula D=n(n3)2D = \frac{n(n - 3)}{2} yields 6×32=9\frac{6 \times 3}{2} = 9 diagonals.

Adım Adım Çözüm

1
Formulate the interior angle sum equation for the nn-sided convex polygon.
The total interior angle sum is (n2)×180(n - 2) \times 180^\circ. Summing the individual angles gives 3×90+(n3)×1503 \times 90^\circ + (n - 3) \times 150^\circ. Thus: (n2)×180=270+(n3)×150(n - 2) \times 180^\circ = 270^\circ + (n - 3) \times 150^\circ.
The sum of all interior angles of any nn-sided polygon is given by (n2)×180(n - 2) \times 180^\circ.
2
Solve the equation for the number of sides nn.
180n360=270+150n450    180n360=150n180    30n=180    n=6180n - 360 = 270 + 150n - 450 \implies 180n - 360 = 150n - 180 \implies 30n = 180 \implies n = 6.
Expanding terms and isolating nn reveals that the polygon has 6 sides (a hexagon).
3
Calculate the total number of diagonals using the formula D=n(n3)2D = \frac{n(n - 3)}{2}.
D=6(63)2=6×32=9D = \frac{6(6 - 3)}{2} = \frac{6 \times 3}{2} = 9.
Each of the nn vertices can connect to n3n - 3 non-adjacent vertices, and dividing by 2 avoids double-counting each diagonal line segment.

Anahtar Kavram

Polygon interior angle sum theorem and diagonal calculation
Soru 7268Soru

What is the derivative of the function f(x)=sin(2x)(1+x2)2f(x) = \frac{\sin(2x)}{(1 + x^2)^2} with respect to xx?

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Cevap: 2[(1+x2)cos(2x)2xsin(2x)](1+x2)3\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}

Cevap

2[(1+x2)cos(2x)2xsin(2x)](1+x2)3\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}
Applying the quotient rule uvuvv2\frac{u'v - uv'}{v^2} with u=sin(2x)u = \sin(2x) and v=(1+x2)2v = (1 + x^2)^2 yields u=2cos(2x)u' = 2\cos(2x) and v=4x(1+x2)v' = 4x(1 + x^2). Substituting these into the formula gives 2cos(2x)(1+x2)24xsin(2x)(1+x2)(1+x2)4\frac{2\cos(2x)(1+x^2)^2 - 4x\sin(2x)(1+x^2)}{(1+x^2)^4}. Factoring out (1+x2)(1+x^2) from numerator and denominator reduces the expression to 2[(1+x2)cos(2x)2xsin(2x)](1+x2)3\frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}.

Adım Adım Çözüm

1
Identify the numerator u(x)u(x) and denominator v(x)v(x) for the quotient rule.
u(x)=sin(2x)u(x) = \sin(2x) and v(x)=(1+x2)2v(x) = (1 + x^2)^2
The quotient rule states that ddx(uv)=uvuvv2\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}.
2
Find the derivatives u(x)u'(x) and v(x)v'(x) using the chain rule.
u(x)=2cos(2x)u'(x) = 2\cos(2x) and v(x)=2(1+x2)(2x)=4x(1+x2)v'(x) = 2(1 + x^2) \cdot (2x) = 4x(1 + x^2)
By the chain rule, ddxsin(2x)=cos(2x)2\frac{d}{dx}\sin(2x) = \cos(2x) \cdot 2 and ddx(1+x2)2=2(1+x2)2x\frac{d}{dx}(1 + x^2)^2 = 2(1 + x^2) \cdot 2x.
3
Substitute u,u,v,vu, u', v, v' into the quotient rule formula.
f(x)=2cos(2x)(1+x2)2sin(2x)4x(1+x2)((1+x2)2)2f'(x) = \frac{2\cos(2x)(1 + x^2)^2 - \sin(2x) \cdot 4x(1 + x^2)}{((1 + x^2)^2)^2}
Applying uvuvv2\frac{u'v - uv'}{v^2} directly.
4
Factor out (1+x2)(1 + x^2) from the numerator and simplify the denominator.
f(x)=(1+x2)[2cos(2x)(1+x2)4xsin(2x)](1+x2)4=2(1+x2)cos(2x)4xsin(2x)(1+x2)3f'(x) = \frac{(1 + x^2)[2\cos(2x)(1 + x^2) - 4x\sin(2x)]}{(1 + x^2)^4} = \frac{2(1 + x^2)\cos(2x) - 4x\sin(2x)}{(1 + x^2)^3}
Canceling the common factor (1+x2)(1 + x^2) from numerator and denominator.
5
Factor out the constant 22 from the numerator.
f(x)=2[(1+x2)cos(2x)2xsin(2x)](1+x2)3f'(x) = \frac{2[(1 + x^2)\cos(2x) - 2x\sin(2x)]}{(1 + x^2)^3}
Putting the derivative into simplified factored form.

Anahtar Kavram

Combining Quotient Rule and Chain Rule for Composite Functions
Soru 7269Soru

What is the gradient of the tangent line to the curve y=4x27x+2y = 4x^2 - 7x + 2 at the point where x=3x = 3?

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Cevap: 17

Cevap

The gradient of the tangent line to the curve at x=3x = 3 is 17.
Differentiating y=4x27x+2y = 4x^2 - 7x + 2 gives dydx=8x7\frac{dy}{dx} = 8x - 7. Substituting x=3x = 3 yields 8(3)7=178(3) - 7 = 17.

Adım Adım Çözüm

1
Find the derivative of the curve's equation
dydx=8x7\frac{dy}{dx} = 8x - 7
The gradient of the curve at any point is given by its first derivative with respect to x.
2
Evaluate the derivative at x = 3
m = 8(3) - 7 = 17
Substituting the given x-coordinate into the derivative gives the specific slope of the tangent line at that point.

Anahtar Kavram

The gradient of the tangent to a curve y=f(x)y = f(x) at x=ax = a is the value of the first derivative f(a)f'(a).
Soru 7270Soru

For what set of real values of kk does the quadratic inequality (k+2)x22(k1)x+(k+5)>0(k + 2)x^2 - 2(k - 1)x + (k + 5) > 0 hold for all real values of xx?

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Cevap: k>1k > -1

Cevap

The condition holds for k>1k > -1.
A quadratic function is strictly positive for all real values of xx if and only if its parabola opens upwards (a>0a > 0) and it has no real x-intercepts (Δ<0\Delta < 0). Here a=k+2>0a = k + 2 > 0 gives k>2k > -2. The discriminant Δ=[2(k1)]24(k+2)(k+5)=36(k+1)\Delta = [-2(k-1)]^2 - 4(k+2)(k+5) = -36(k+1). Setting 36(k+1)<0-36(k+1) < 0 requires dividing by 36-36 and reversing the inequality sign, giving k+1>0    k>1k + 1 > 0 \implies k > -1. The intersection of k>2k > -2 and k>1k > -1 is k>1k > -1.

Adım Adım Çözüm

1
Identify the conditions required for a quadratic expression ax2+bx+cax^2 + bx + c to be strictly positive for all real xx.
The coefficient of x2x^2 must be positive (a>0a > 0) and the discriminant must be strictly negative (Δ<0\Delta < 0).
If a<0a < 0, the parabola opens downwards and yields negative values. If Δ0\Delta \ge 0, the quadratic has real roots or a double root, taking non-positive values.
2
Apply the first condition a>0a > 0 to the coefficient of x2x^2.
k+2>0    k>2k + 2 > 0 \implies k > -2.
Ensures the parabola opens upwards.
3
Calculate the discriminant Δ=b24ac\Delta = b^2 - 4ac and set Δ<0\Delta < 0.
Δ=[2(k1)]24(k+2)(k+5)=4(k22k+1)4(k2+7k+10)=4(9k9)=36(k+1)\Delta = [-2(k - 1)]^2 - 4(k + 2)(k + 5) = 4(k^2 - 2k + 1) - 4(k^2 + 7k + 10) = 4(-9k - 9) = -36(k + 1). Setting 36(k+1)<0-36(k + 1) < 0.
Ensures the quadratic equation has no real roots and does not cross or touch the x-axis.
4
Solve the linear inequality 36(k+1)<0-36(k + 1) < 0 and combine with the first condition.
Dividing by 36-36 reverses the inequality: k+1>0    k>1k + 1 > 0 \implies k > -1. Taking the intersection of k>2k > -2 and k>1k > -1 yields k>1k > -1.
Dividing an inequality by a negative number flips the inequality symbol.

Anahtar Kavram

Conditions for positive definiteness of quadratic expressions
Soru 7271Soru

What is the value of log281log332\log_2 81 \cdot \log_3 32?

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Cevap: 20

Cevap

The value of the logarithmic expression is 20.
Rewriting 8181 as 343^4 and 3232 as 252^5 allows extraction of coefficients 44 and 55. Utilizing the change of base property log23log32=1\log_2 3 \cdot \log_3 2 = 1 reduces the expression to 4×5=204 \times 5 = 20.

Adım Adım Çözüm

1
Rewrite arguments as prime powers
log281=log2(34)\log_2 81 = \log_2(3^4) and log332=log3(25)\log_3 32 = \log_3(2^5)
Simplifies terms to prime base components
2
Apply the power law of logarithms
4log235log324\log_2 3 \cdot 5\log_3 2
Brings powers out as numerical coefficients
3
Apply the change of base identity logbalogab=1\log_b a \cdot \log_a b = 1
45(log23log32)=201=204 \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 7272Soru

Let the universal set be U={xZ:1x36}\mathcal{U} = \{x \in \mathbb{Z} : 1 \le x \le 36\}. Subsets AA, BB, and CC of U\mathcal{U} are defined as follows:
A={xU:x is a perfect square}A = \{x \in \mathcal{U} : x \text{ is a perfect square}\},
B={xU:x is a multiple of 3}B = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}, and
C={xU:x is an even number}C = \{x \in \mathcal{U} : x \text{ is an even number}\}.
What is the cardinality of the set (AB)C(A \cup B)' \cap C?

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Cevap: 10

Cevap

The cardinality of the set (AB)C(A \cup B)' \cap C is 10.
The set (AB)C(A \cup B)' \cap C represents elements in CC that are neither in AA nor in BB. Set CC contains 18 even integers up to 36. Among these 18 numbers, three are perfect squares (4, 16, 36) and six are multiples of 3 (6, 12, 18, 24, 30, 36). Since 36 satisfies both conditions, the total number of even numbers in ABA \cup B is 3+61=83 + 6 - 1 = 8. Subtracting these 8 elements from 18 leaves 10 elements.

Adım Adım Çözüm

1
Identify the elements of set C (even numbers in the universal set)
C={2,4,6,8,10,12,14,16,18,20,22,24,26,28,30,32,34,36}C = \{2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34, 36\}, so n(C)=18n(C) = 18.
The complement set operation is restricted to elements within set C.
2
Find the elements of set C that belong to set A (even perfect squares)
CA={4,16,36}C \cap A = \{4, 16, 36\}, so n(CA)=3n(C \cap A) = 3.
Identify which elements in C must be excluded due to being perfect squares.
3
Find the elements of set C that belong to set B (even multiples of 3, i.e., multiples of 6)
CB={6,12,18,24,30,36}C \cap B = \{6, 12, 18, 24, 30, 36\}, so n(CB)=6n(C \cap B) = 6.
Identify which elements in C must be excluded due to being multiples of 3.
4
Find the overlap between the excluded sets within C
CAB={36}C \cap A \cap B = \{36\}, so n(CAB)=1n(C \cap A \cap B) = 1.
Avoid double-counting 36 when subtracting excluded elements.
5
Calculate n(C(AB))n(C \cap (A \cup B)) using inclusion-exclusion and find the remaining complement elements in C
n(C(AB))=3+61=8n(C \cap (A \cup B)) = 3 + 6 - 1 = 8. Therefore, n((AB)C)=188=10n((A \cup B)' \cap C) = 18 - 8 = 10.
Subtracting the 8 excluded even numbers from the total 18 even numbers yields the final cardinality.

Anahtar Kavram

Set Complement and Relative Difference with Inclusion-Exclusion
Tahmini Süre:2m 0s
Soru 7273Soru

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

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Cevap: 1.75

Cevap

The exact numerical value of the trigonometric expression is 1.75.
Dividing every term in the expression by cosθ\cos \theta converts sinθ\sin \theta terms into tanθ\tan \theta. The quotient becomes 3tanθ+1tanθ+2\frac{3\tan \theta + 1}{\tan \theta + 2}. Substituting tanθ=2\tan \theta = 2 gives 3(2)+12+2=74=1.75\frac{3(2) + 1}{2 + 2} = \frac{7}{4} = 1.75.

Adım Adım Çözüm

1
Divide numerator and denominator by cosθ\cos \theta
The expression becomes 3(sinθcosθ)+1sinθcosθ+2=3tanθ+1tanθ+2\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 tanθ=sinθcosθ\tan \theta = \frac{\sin \theta}{\cos \theta} simplifies the calculation without needing to evaluate the hypotenuse.
2
Substitute tanθ=2\tan \theta = 2
3(2)+12+2=74\frac{3(2) + 1}{2 + 2} = \frac{7}{4}
Replace tanθ\tan \theta with its given numerical value of 2.
3
Convert fraction to decimal
1.75
Decimal representation of the exact fraction 74\frac{7}{4}.

Anahtar Kavram

Basic Trigonometric Ratios and Quotient Identity
Soru 7274Soru

If y=(x22x+2)3y = (x^2 - 2x + 2)^3, what is the numerical value of dydx\frac{dy}{dx} evaluated at x=2x = 2?

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Cevap: 24

Cevap

24
Applying the chain rule gives dydx=3(x22x+2)2(2x2)\frac{dy}{dx} = 3(x^2 - 2x + 2)^2(2x - 2). Evaluating this derivative at x=2x = 2 yields 3(2)2(2)=243(2)^2(2) = 24.

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1
Identify the composite structure and state the chain rule formula.
Let u=x22x+2u = x^2 - 2x + 2, so y=u3y = u^3. By the chain rule, dydx=dydududx\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 uu and the inner function with respect to xx.
\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 dydx\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=2x = 2.
\frac{dy}{dx}\Big|_{x=2} = 3(4 - 4 + 2)^2(4 - 2) = 3(2)^2(2) = 24.
Substitute x=2x = 2 into the derivative expression and simplify.

Anahtar Kavram

Chain Rule of Differentiation
Soru 7275Soru

If 73211=a2+b11\frac{7}{3\sqrt{2} - \sqrt{11}} = a\sqrt{2} + b\sqrt{11}, where aa and bb are rational numbers, what is the value of a+ba + b?

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Cevap: 4

Cevap

4
Multiplying both the top and bottom by the conjugate (32+11)(3\sqrt{2} + \sqrt{11}) transforms the denominator into (32)2(11)2=1811=7(3\sqrt{2})^2 - (\sqrt{11})^2 = 18 - 11 = 7. Dividing the numerator 7(32+11)7(3\sqrt{2} + \sqrt{11}) by 77 simplifies to 32+113\sqrt{2} + \sqrt{11}. Matching coefficients yields a=3a = 3 and b=1b = 1, giving a+b=4a + b = 4.

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1
Multiply the numerator and denominator by the conjugate of the denominator, (32+11)(3\sqrt{2} + \sqrt{11}).
\frac{7(3\sqrt{2} + \sqrt{11})}{(3\sqrt{2} - \sqrt{11})(3\sqrt{2} + \sqrt{11})}
Rationalising the denominator eliminates radicals from the bottom of the fraction.
2
Expand the denominator using the difference of squares formula (xy)(x+y)=x2y2(x - y)(x + y) = x^2 - y^2.
(3\sqrt{2})^2 - (\sqrt{11})^2 = (9 \times 2) - 11 = 18 - 11 = 7
Squaring each term simplifies the denominator into an integer.
3
Simplify the overall rational fraction by cancelling common factors.
\frac{7(3\sqrt{2} + \sqrt{11})}{7} = 3\sqrt{2} + \sqrt{11}
The factor of 7 in the numerator and denominator cancels out.
4
Equate 32+1113\sqrt{2} + 1\sqrt{11} with a2+b11a\sqrt{2} + b\sqrt{11} to determine the values of aa and bb, then sum them.
a = 3, b = 1 \implies a + b = 3 + 1 = 4
Comparing coefficients of corresponding surd terms gives the required values.

Anahtar Kavram

Rationalisation of binomial surd denominators using conjugates
Soru 7276Soru

The table below shows the cumulative frequency distribution of the masses (in grams) of 100100 cocoa beans sampled from an agricultural yield:

Mass Class Interval (g)Cumulative Frequency
101910 - 191010
202920 - 293030
303930 - 396565
404940 - 499090
505950 - 59100100

Using linear interpolation from the cumulative frequency distribution, calculate the 75th percentile (Q3Q_3) mass of the cocoa beans in grams.

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Cevap: 43.5

Cevap

The 75th percentile (Q3Q_3) mass of the cocoa beans is 43.5 g43.5\text{ g}.
To find the 75th percentile (Q3Q_3) from cumulative frequency data, calculate the rank 75100×100=75\frac{75}{100} \times 100 = 75. This falls into the 404940 - 49 class interval (boundaries 39.549.539.5 - 49.5). Applying Q3=L+(75Ff)cQ_3 = L + \left(\frac{75 - F}{f}\right)c yields 39.5+(756525)×10=43.5 g39.5 + \left(\frac{75 - 65}{25}\right) \times 10 = 43.5\text{ g}.

Adım Adım Çözüm

1
Calculate the percentile position rank
Rank position is 7575
The 75th percentile corresponds to 75%75\% of the total sample size N=100N = 100, giving 75100×100=75\frac{75}{100} \times 100 = 75.
2
Locate the 75th percentile class interval and its boundaries
Class interval is 404940 - 49, with lower boundary L=39.5L = 39.5 and upper boundary 49.549.5
Cumulative frequency before 404940 - 49 is 6565, and up to 404940 - 49 is 9090. Since 65<759065 < 75 \leq 90, the 75th item falls in this interval.
3
Identify class parameters for interpolation
L=39.5L = 39.5, F=65F = 65, f=25f = 25, c=10c = 10
Lower boundary L=39.5L = 39.5, previous cumulative frequency F=65F = 65, class frequency f=9065=25f = 90 - 65 = 25, class width c=49.539.5=10c = 49.5 - 39.5 = 10.
4
Compute Q3Q_3 using the linear interpolation formula
Q3=43.5 gQ_3 = 43.5\text{ g}
Q3=39.5+(756525)×10=39.5+4=43.5Q_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 7277Soru

Find the value of xx that satisfies the logarithmic equation log2(x21)log2(x1)=3\log_2(x^2 - 1) - \log_2(x - 1) = 3.

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Cevap: 7

Cevap

The value of xx is 77.
Applying the logarithm quotient law reduces log2(x21)log2(x1)\log_2(x^2 - 1) - \log_2(x - 1) to log2(x21x1)\log_2\left(\frac{x^2 - 1}{x - 1}\right). Factoring x21x^2 - 1 into (x1)(x+1)(x - 1)(x + 1) and canceling (x1)(x - 1) leaves log2(x+1)=3\log_2(x + 1) = 3. Converting this to exponential form gives x+1=23=8x + 1 = 2^3 = 8, which yields x=7x = 7.

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1
Apply the quotient rule for logarithms to combine the left-hand side into a single logarithmic expression.
log2(x21x1)=3\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: logbAlogbB=logb(AB)\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.
log2(x+1)=3\log_2(x + 1) = 3
Since x21=(x1)(x+1)x^2 - 1 = (x - 1)(x + 1), dividing by (x1)(x - 1) leaves (x+1)(x + 1), given x1x \neq 1.
3
Rewrite the logarithmic equation in exponential form.
x+1=23=8x + 1 = 2^3 = 8
By definition of logarithms, logby=z    bz=y\log_b y = z \iff b^z = y.
4
Solve the resulting linear equation for xx.
x=7x = 7
Subtracting 11 from both sides gives x=7x = 7.

Anahtar Kavram

Quotient Law of Logarithms and Logarithmic Equations
Tahmini Süre:1m 30s
Soru 7278Soru

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

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Cevap: 437.5

Cevap

437.5 W
The partial variation equation is P=k1+k2v2P = k_1 + k_2 v^2. Setting up simultaneous equations k1+100k2=250k_1 + 100 k_2 = 250 and k1+400k2=700k_1 + 400 k_2 = 700 yields k1=100k_1 = 100 and k2=1.5k_2 = 1.5. Substituting v=15v = 15 into P=100+1.5(152)P = 100 + 1.5(15^2) gives P=437.5 WP = 437.5\text{ W}.

Adım Adım Çözüm

1
Express the partial variation mathematically.
P=k1+k2v2P = k_1 + k_2 v^2, where k1k_1 and k2k_2 are constants.
The total power consumption is the sum of a fixed baseline constant k1k_1 and a variable component proportional to v2v^2.
2
Form simultaneous linear equations using the provided data points.
k1+100k2=250k_1 + 100 k_2 = 250 and k1+400k2=700k_1 + 400 k_2 = 700.
Substituting v=10v = 10 gives 102=10010^2 = 100, and substituting v=20v = 20 gives 202=40020^2 = 400.
3
Solve for the constants k1k_1 and k2k_2.
k2=1.5k_2 = 1.5 and k1=100k_1 = 100.
Subtracting the two equations eliminates k1k_1, yielding 300k2=450    k2=1.5300 k_2 = 450 \implies k_2 = 1.5. Substituting k2=1.5k_2 = 1.5 back into k1+100k2=250k_1 + 100 k_2 = 250 gives k1=100k_1 = 100.
4
Calculate the value of PP at v=15 rev/sv = 15\text{ rev/s}.
P=437.5 WP = 437.5\text{ W}.
Substitute v=15v = 15, k1=100k_1 = 100, and k2=1.5k_2 = 1.5 into the governing formula P=100+1.5(152)=100+337.5=437.5P = 100 + 1.5(15^2) = 100 + 337.5 = 437.5.

Anahtar Kavram

Partial Variation and Simultaneous Linear Equations
Tahmini Süre:1m 30s
Soru 7279Soru

A point P(x,y)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)A(0, 0) and B(8,0)B(8, 0) is equal to 8282, defining a locus L1L_1. A second locus L2L_2 is the set of all points equidistant from the parallel lines y=1y = -1 and y=7y = 7. Given that L1L_1 and L2L_2 intersect at two distinct points MM and NN, what is the length of the line segment MNMN?

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Cevap: 8

Cevap

The length of the line segment MNMN is 8 units.
The locus L1L_1 simplifies to the circle (x4)2+y2=25(x - 4)^2 + y^2 = 25 with center (4,0)(4,0) and radius 55. The locus L2L_2 is the horizontal line y=3y = 3. Substituting y=3y = 3 into the circle equation yields (x4)2=16(x - 4)^2 = 16, giving intersection points at (0,3)(0,3) and (8,3)(8,3). The distance between these points is 88 units.

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1
Formulate the algebraic expression for locus L1L_1
The distance squared PA2=x2+y2PA^2 = x^2 + y^2 and PB2=(x8)2+y2PB^2 = (x - 8)^2 + y^2. Setting PA2+PB2=82PA^2 + PB^2 = 82 gives x2+y2+(x8)2+y2=82x^2 + y^2 + (x - 8)^2 + y^2 = 82.
By the distance formula, the sum of squared distances from P(x,y)P(x, y) to A(0,0)A(0,0) and B(8,0)B(8,0) equals 82.
2
Simplify and complete the square for locus L1L_1
Expanding yields 2x216x+64+2y2=82    2x216x+2y2=18    x28x+y2=92x^2 - 16x + 64 + 2y^2 = 82 \implies 2x^2 - 16x + 2y^2 = 18 \implies x^2 - 8x + y^2 = 9. Completing the square gives (x4)2+y2=25(x - 4)^2 + y^2 = 25.
This shows L1L_1 is a circle with center (4,0)(4, 0) and radius R=5R = 5.
3
Find the equation for locus L2L_2
The locus of points equidistant from two parallel lines y=1y = -1 and y=7y = 7 is the line y=1+72=3y = \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 MM and NN
Substitute y=3y = 3 into (x4)2+y2=25(x - 4)^2 + y^2 = 25: (x4)2+32=25    (x4)2=16    x4=±4(x - 4)^2 + 3^2 = 25 \implies (x - 4)^2 = 16 \implies x - 4 = \pm 4, so x=8x = 8 or x=0x = 0. Thus, M=(0,3)M = (0, 3) and N=(8,3)N = (8, 3).
Solving the circle and line equations simultaneously gives the xx-coordinates of the intersection points.
5
Compute the distance between MM and NN
Since both points share the ordinate y=3y = 3, distance MN=80=8MN = 8 - 0 = 8.
The length of a horizontal segment is the difference between its endpoints' xx-coordinates.

Anahtar Kavram

Intersection of loci involving circles and parallel lines
Soru 7280Soru

A point P(x,y)P(x, y) moves such that it is equidistant from two parallel lines 3x4y+11=03x - 4y + 11 = 0 and 3x4y1=03x - 4y - 1 = 0, defining locus L1L_1. A second locus L2L_2 consists of all points that are at a constant distance of 55 units from the fixed point (1,7)(1, 7). Calculate the distance between the two points of intersection of locus L1L_1 and locus L2L_2.

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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
ÖncekiSayfa 364 / 697Sonraki
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