Tüm alıştırma soruları

1526 soru

Soru 721Soru

When the polynomial P(x)=3x42x3+ax2+bx12P(x) = 3x^4 - 2x^3 + ax^2 + bx - 12 is divided by (x24)(x^2 - 4), the remainder is 5x45x - 4. What is the value of a+ba + b?

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Cevap: 3

Cevap

The value of a+ba + b is 33.
By using the Remainder Theorem for the quadratic divisor (x24)=(x2)(x+2)(x^2 - 4) = (x - 2)(x + 2), evaluating P(2)=6P(2) = 6 gives 2a+b=72a + b = -7, and evaluating P(2)=14P(-2) = -14 gives 2ab=332a - b = -33. Solving these linear equations simultaneously yields a=10a = -10 and b=13b = 13, which sums to a+b=3a + b = 3.

Adım Adım Çözüm

1
Set up the polynomial division relation.
P(x)=(x2)(x+2)Q(x)+(5x4)P(x) = (x - 2)(x + 2)Q(x) + (5x - 4)
By the Remainder Theorem and Division Algorithm, dividing by (x24)(x^2 - 4) yields a remainder of R(x)=5x4R(x) = 5x - 4.
2
Find the values of P(2)P(2) and P(2)P(-2) from the remainder.
P(2)=6P(2) = 6 and P(2)=14P(-2) = -14
Substituting the roots of the divisor x=2x = 2 and x=2x = -2 eliminates the quotient term (x24)Q(x)(x^2 - 4)Q(x).
3
Substitute x=2x = 2 into the polynomial P(x)P(x) and set equal to 66.
2a+b=72a + b = -7
3(16)2(8)+4a+2b12=20+4a+2b=6    4a+2b=143(16) - 2(8) + 4a + 2b - 12 = 20 + 4a + 2b = 6 \implies 4a + 2b = -14.
4
Substitute x=2x = -2 into the polynomial P(x)P(x) and set equal to 14-14.
2ab=332a - b = -33
3(16)2(8)+4a2b12=52+4a2b=14    4a2b=663(16) - 2(-8) + 4a - 2b - 12 = 52 + 4a - 2b = -14 \implies 4a - 2b = -66.
5
Solve the system of equations for aa and bb.
a=10a = -10, b=13b = 13, and a+b=3a + b = 3
Adding the two linear equations gives 4a=40    a=104a = -40 \implies a = -10. Substituting a=10a = -10 into 2a+b=72a + b = -7 gives b=13b = 13.

Anahtar Kavram

Polynomial Remainder Theorem for Non-Linear Divisors
Tahmini Süre:2m 30s
Soru 722Soru

An arithmetic progression (A.P.) has a first term of 77 and a common difference of 55. What is the 12th12^{\text{th}} term of the progression?

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Cevap: 62

Cevap

The 12th12^{\text{th}} term of the progression is 6262.
Using the formula for the nthn^{\text{th}} term of an arithmetic progression, Tn=a+(n1)dT_n = a + (n - 1)d, with a=7a = 7, d=5d = 5, and n=12n = 12, we calculate T12=7+(121)×5=7+55=62T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62.

Adım Adım Çözüm

1
Identify known parameters from the question
a=7a = 7, d=5d = 5, n=12n = 12
These parameters are given in the problem statement.
2
Use the general formula for the nthn^{\text{th}} term of an arithmetic progression
Tn=a+(n1)dT_n = a + (n - 1)d
This formula connects the first term, common difference, and term number to the value of the term.
3
Substitute values and evaluate
T12=7+(121)×5=7+55=62T_{12} = 7 + (12 - 1) \times 5 = 7 + 55 = 62
Performing basic arithmetic gives the final result.

Anahtar Kavram

Finding the nthn^{\text{th}} term of an Arithmetic Progression
Tahmini Süre:45s
Soru 723Soru

An isosceles triangle has a perimeter of 36 cm36\text{ cm} and a base of length 16 cm16\text{ cm}. Calculate the area of the triangle in cm2\text{cm}^2.

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Cevap: 48

Cevap

The area of the isosceles triangle is 48 cm248\text{ cm}^2.
The two equal sides of the isosceles triangle measure 36162=10 cm\frac{36 - 16}{2} = 10\text{ cm} each. An altitude dropped perpendicularly to the base bisects the 16 cm16\text{ cm} base into two 8 cm8\text{ cm} segments. By the Pythagorean theorem, the perpendicular height is h=10282=6 cmh = \sqrt{10^2 - 8^2} = 6\text{ cm}. Therefore, the area is 12×16×6=48 cm2\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2.

Adım Adım Çözüm

1
Determine the length of the two equal sides
Each equal side is 10 cm10\text{ cm}
Subtract the base length from the total perimeter (3616=20 cm36 - 16 = 20\text{ cm}) and divide by 22.
2
Calculate the perpendicular height (altitude) to the base
Height h=6 cmh = 6\text{ cm}
The perpendicular altitude bisects the base into two 8 cm8\text{ cm} segments, creating right-angled triangles with hypotenuse 10 cm10\text{ cm}. Use Pythagoras' theorem: h=10282=6 cmh = \sqrt{10^2 - 8^2} = 6\text{ cm}.
3
Calculate the area of the triangle
Area = 48 cm248\text{ cm}^2
Multiply half the base by the perpendicular height: 12×16×6=48 cm2\frac{1}{2} \times 16 \times 6 = 48\text{ cm}^2.

Anahtar Kavram

Perimeter and Area of Isosceles Triangles using Pythagorean Theorem
Tahmini Süre:1m 30s
Soru 724Soru

Two straight paths diverge from a junction JJ at an angle of 120120^\circ. A person walks 5 km5\text{ km} along the first path to point AA, and another person walks 16 km16\text{ km} along the second path to point BB. What is the direct distance between AA and BB in kilometres?

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Cevap: 19

Cevap

The direct distance between points AA and BB is 19 km19\text{ km}.
Using the Cosine Rule c2=a2+b22abcosCc^2 = a^2 + b^2 - 2ab \cos C with sides 5 km5\text{ km} and 16 km16\text{ km} and included angle 120120^\circ yields c2=52+1622(5)(16)(0.5)=25+256+80=361c^2 = 5^2 + 16^2 - 2(5)(16)(-0.5) = 25 + 256 + 80 = 361. Taking the square root gives 19 km19\text{ km}.

Adım Adım Çözüm

1
Identify given values and setup the Cosine Rule formula
Side a=5a = 5, side b=16b = 16, and included angle θ=120\theta = 120^\circ
The scenario provides two sides and the included angle (SAS configuration), which requires the Cosine Rule to find the third side.
2
Substitute the values into c2=a2+b22abcosθc^2 = a^2 + b^2 - 2ab \cos \theta
c2=52+1622(5)(16)cos(120)c^2 = 5^2 + 16^2 - 2(5)(16) \cos(120^\circ)
Populating the formula allows evaluation of the unknown distance squared.
3
Evaluate the trigonometric term and simplify
c2=25+256160(0.5)=281+80=361c^2 = 25 + 256 - 160(-0.5) = 281 + 80 = 361
The cosine of an obtuse angle in the second quadrant (120120^\circ) is negative: cos(120)=0.5\cos(120^\circ) = -0.5.
4
Take the square root to find the distance cc
c=361=19 kmc = \sqrt{361} = 19\text{ km}
Taking the positive square root gives the physical distance between the two points.

Anahtar Kavram

Applying the Cosine Rule to find the length of an unknown side in a non-right triangle given two sides and the included angle (SAS).
Soru 725Soru

Evaluate the definite integral 03(x2+2)dx\int_{0}^{3} (x^2 + 2) \, dx.

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Cevap: 15

Cevap

The value of the definite integral is 15.
Integrating x2+2x^2 + 2 with respect to xx yields [x33+2x]03\left[\frac{x^3}{3} + 2x\right]_{0}^{3}. Substituting the upper bound x=3x = 3 gives 273+6=15\frac{27}{3} + 6 = 15. Substituting the lower bound x=0x = 0 gives 00. The net value is 150=1515 - 0 = 15.

Adım Adım Çözüm

1
Find the antiderivative of x2+2x^2 + 2
x33+2x\frac{x^3}{3} + 2x
Apply the power rule of integration to each term.
2
Substitute the upper limit x=3x = 3
15
\frac{3^3}{3} + 2(3) = 9 + 6 = 15
3
Substitute the lower limit x=0x = 0
0
033+2(0)=0\frac{0^3}{3} + 2(0) = 0
4
Compute the difference between upper and lower limit values
15
15 - 0 = 15

Anahtar Kavram

Definite Integration of Polynomial Functions
Tahmini Süre:45s
Soru 726Soru

A bag contains 55 red balls, 33 blue balls, and nn green balls. The theoretical probability of drawing a blue ball at random from the bag is 320\frac{3}{20}. In a probability experiment, a ball is drawn at random from the bag and replaced 500500 times. If a green ball is observed 340340 times, calculate the absolute difference between the experimental probability and the theoretical probability of drawing a green ball.

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Cevap: 0.08

Cevap

The absolute difference between the experimental probability and the theoretical probability of drawing a green ball is 0.080.08 (or 225\frac{2}{25}).
The theoretical probability of drawing a blue ball establishes that the bag contains 2020 total balls, meaning there are 1212 green balls. The theoretical probability of picking a green ball is therefore 1220=0.60\frac{12}{20} = 0.60. The experimental probability is 340500=0.68\frac{340}{500} = 0.68. Taking the positive difference gives 0.680.60=0.08|0.68 - 0.60| = 0.08.

Adım Adım Çözüm

1
Find the total number of balls and the value of nn
n=12n = 12, Total balls = 2020
The theoretical probability of picking a blue ball is 38+n=320\frac{3}{8+n} = \frac{3}{20}, giving 8+n=208 + n = 20.
2
Find the theoretical probability of drawing a green ball
P(Green)theo=0.60P(\text{Green})_{\text{theo}} = 0.60
There are 1212 green balls out of 2020 total balls, so 1220=0.60\frac{12}{20} = 0.60.
3
Find the experimental probability of drawing a green ball
P(Green)exp=0.68P(\text{Green})_{\text{exp}} = 0.68
In 500500 trials, a green ball was observed 340340 times, so 340500=0.68\frac{340}{500} = 0.68.
4
Calculate the absolute difference
0.680.60=0.08|0.68 - 0.60| = 0.08
Subtract the theoretical probability from the experimental probability and take the absolute value.

Anahtar Kavram

Comparison of theoretical probability based on outcome sample space and experimental probability based on observed trial relative frequencies.
Soru 727Soru

If y=extanx+ln(2x+1)y = e^{-x} \tan x + \ln(2x + 1), find the value of dydx\frac{dy}{dx} at x=0x = 0.

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Cevap: 3

Cevap

The value of dydx\frac{dy}{dx} at x=0x = 0 is 3.
Applying the product rule to extanxe^{-x}\tan x gives extanx+exsec2x-e^{-x}\tan x + e^{-x}\sec^2 x, and applying the chain rule to ln(2x+1)\ln(2x+1) gives 22x+1\frac{2}{2x+1}. Evaluating dydx=extanx+exsec2x+22x+1\frac{dy}{dx} = -e^{-x}\tan x + e^{-x}\sec^2 x + \frac{2}{2x+1} at x=0x = 0 yields e0(0)+e0(1)+21=3-e^0(0) + e^0(1) + \frac{2}{1} = 3.

Adım Adım Çözüm

1
Differentiate u(x)=extanxu(x) = e^{-x} \tan x using the product rule.
\frac{du}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x
The derivative of exe^{-x} is ex-e^{-x} and the derivative of tanx\tan x is \sec^2 x.
2
Differentiate v(x)=ln(2x+1)v(x) = \ln(2x + 1) using the chain rule.
dvdx=22x+1\frac{dv}{dx} = \frac{2}{2x + 1}
The derivative of ln(g(x))\ln(g(x)) is g(x)g(x)\frac{g'(x)}{g(x)}, where g(x)=2x+1g(x) = 2x + 1 and g(x)=2g'(x) = 2.
3
Sum the derivatives to find the complete expression for dydx\frac{dy}{dx}.
\frac{dy}{dx} = -e^{-x} \tan x + e^{-x} \sec^2 x + \frac{2}{2x + 1}
The derivative of a sum is equal to the sum of the derivatives.
4
Evaluate the derivative at x=0x = 0.
\frac{dy}{dx}\Big|_{x=0} = -e^0(0) + e^0(1)^2 + \frac{2}{1} = 3
Since tan(0)=0\tan(0) = 0, e0=1e^0 = 1, and sec(0)=1\sec(0) = 1, substituting x=0x = 0 simplifies the derivative to 0+1+2=30 + 1 + 2 = 3.

Anahtar Kavram

Differentiation of Transcendental Functions (Product and Chain Rules)
Soru 728Soru

Find the value of xx, in degrees, for 0x900^\circ \le x \le 90^\circ that satisfies the trigonometric equation sin2x=cos(x+30)\sin 2x = \cos(x + 30^\circ).

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Cevap: 20

Cevap

The value of xx in the interval 0x900^\circ \le x \le 90^\circ satisfying the equation is 2020^\circ.
Using the co-function identity cosα=sin(90α)\cos \alpha = \sin(90^\circ - \alpha), we convert the right-hand side to sin(90(x+30))=sin(60x)\sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x). Equating the arguments gives 2x=60x2x = 60^\circ - x, which simplifies to 3x=603x = 60^\circ, yielding x=20x = 20^\circ.

Adım Adım Çözüm

1
Apply the co-function trigonometric identity
cos(x+30)=sin(90(x+30))=sin(60x)\cos(x + 30^\circ) = \sin(90^\circ - (x + 30^\circ)) = \sin(60^\circ - x)
Converting cosine to sine allows direct comparison of sine functions on both sides of the equation.
2
Set up the equation equating the angle expressions
2x=60x2x = 60^\circ - x
Since sin(2x)=sin(60x)\sin(2x) = \sin(60^\circ - x) and x[0,90]x \in [0^\circ, 90^\circ], equating the principal angle arguments gives the primary solution.
3
Solve the linear equation for xx
3x=60    x=203x = 60^\circ \implies x = 20^\circ
Adding xx to both sides gives 3x=603x = 60^\circ, and dividing by 3 yields x=20x = 20^\circ.

Anahtar Kavram

Co-function identities and simple trigonometric equations
Soru 729Soru

Two independent weather forecasting stations, AA and BB, operate in a region. The probability that station AA makes an accurate forecast on any given day is 0.800.80, and the probability that station BB makes an accurate forecast is 0.750.75. What is the probability that at least one of the two stations makes an accurate forecast on a given day?

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Cevap: 0.95

Cevap

The probability that at least one of the two stations makes an accurate forecast is 0.950.95.
The probability of at least one event occurring is given by P(AB)=P(A)+P(B)P(AB)P(A \cup B) = P(A) + P(B) - P(A \cap B). Because events AA and BB are independent, P(AB)=P(A)×P(B)=0.80×0.75=0.60P(A \cap B) = P(A) \times P(B) = 0.80 \times 0.75 = 0.60. Substituting into the addition rule yields 0.80+0.750.60=0.950.80 + 0.75 - 0.60 = 0.95. Alternatively, using the complement rule: 1P(A)P(B)=1(10.80)(10.75)=1(0.20×0.25)=10.05=0.951 - P(A')P(B') = 1 - (1 - 0.80)(1 - 0.75) = 1 - (0.20 \times 0.25) = 1 - 0.05 = 0.95.

Adım Adım Çözüm

1
Calculate the probability of both events occurring simultaneously using the multiplication law for independent events.
P(AB)=P(A)×P(B)=0.80×0.75=0.60P(A \cap B) = P(A) \times P(B) = 0.80 \times 0.75 = 0.60
Since the two forecasting stations operate independently, the joint probability is the product of their individual probabilities.
2
Apply the general addition law of probability to calculate the probability of at least one station making an accurate forecast.
P(AB)=P(A)+P(B)P(AB)=0.80+0.750.60=0.95P(A \cup B) = P(A) + P(B) - P(A \cap B) = 0.80 + 0.75 - 0.60 = 0.95
The probability of compound event 'at least one' corresponds to the union of the two events.

Anahtar Kavram

Addition and Multiplication Laws of Probability for Independent Events
Soru 730Soru

What is the simplified numerical value of 123+1+1231\frac{\sqrt{12}}{\sqrt{3} + 1} + \frac{\sqrt{12}}{\sqrt{3} - 1}?

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Cevap: 6

Cevap

The simplified numerical value of the given expression is 6.
Combining the fractions using their common conjugate denominator (3+1)(31)=2(\sqrt{3}+1)(\sqrt{3}-1) = 2 leads to a numerator of 23[(31)+(3+1)]=23(23)=122\sqrt{3}[(\sqrt{3}-1)+(\sqrt{3}+1)] = 2\sqrt{3}(2\sqrt{3}) = 12. Dividing 12 by 2 yields the final answer of 6.

Adım Adım Çözüm

1
Simplify the surd in the numerator
\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3}
Simplifying surds into basic form makes subsequent calculations simpler.
2
Combine the fractions by finding a common denominator
23(31)+23(3+1)(3+1)(31)\frac{2\sqrt{3}(\sqrt{3}-1) + 2\sqrt{3}(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}
Multiplying denominators forms a conjugate pair, which rationalizes the combined denominator.
3
Simplify the numerator and the denominator
\text{Numerator} = 2\sqrt{3}(\sqrt{3}-1 + \sqrt{3}+1) = 2\sqrt{3}(2\sqrt{3}) = 12; \quad \text{Denominator} = (\sqrt{3})^2 - 1^2 = 3 - 1 = 2
Expanding the numerator combines like surd terms, and using the difference of squares simplifies the denominator to a rational integer.
4
Perform the final division
122=6\frac{12}{2} = 6
Dividing the simplified numerator by the rationalized denominator gives the final integer value.

Anahtar Kavram

Rationalization of binomial denominators using conjugate surds
Soru 731Soru

Find the total area of the region bounded by the curve y=3x26xy = 3x^2 - 6x, the xx-axis, and the vertical lines x=0x = 0 and x=3x = 3.

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Cevap: 8

Cevap

The total area bounded by the curve and the x-axis between x = 0 and x = 3 is 8 square units.
To find the total geometric area bounded by a curve and the x-axis, we must split the integral at any x-intercepts within the domain. For y=3x26xy = 3x^2 - 6x, the x-intercepts are x=0x = 0 and x=2x = 2. Between x=0x = 0 and x=2x = 2, the curve lies below the x-axis, giving an area magnitude of 44. Between x=2x = 2 and x=3x = 3, the curve lies above the x-axis, giving an area magnitude of 44. Summing these positive magnitudes gives a total area of 88.

Adım Adım Çözüm

1
Find the roots of the curve y=3x26xy = 3x^2 - 6x within the given interval [0,3][0, 3].
Setting 3x26x=03x^2 - 6x = 0 yields 3x(x2)=03x(x - 2) = 0, giving x=0x = 0 and x=2x = 2.
Roots inside the integration boundaries indicate where the curve crosses the x-axis, changing the sign of yy.
2
Determine the position of the curve relative to the x-axis on each sub-interval.
On [0,2][0, 2], y0y \le 0 (below the x-axis). On [2,3][2, 3], y0y \ge 0 (above the x-axis).
Geometric area must be non-negative, so regions below the x-axis require integrating y-y or taking the absolute value of the integral.
3
Evaluate the area A1A_1 for the region below the x-axis from x=0x = 0 to x=2x = 2.
A1=02(6x3x2)dx=[3x2x3]02=(3(4)8)0=4A_1 = \int_{0}^{2} (6x - 3x^2) \, dx = \left[ 3x^2 - x^3 \right]_{0}^{2} = (3(4) - 8) - 0 = 4.
Integrating y=6x3x2-y = 6x - 3x^2 yields the positive magnitude of the area below the x-axis.
4
Evaluate the area A2A_2 for the region above the x-axis from x=2x = 2 to x=3x = 3.
A2=23(3x26x)dx=[x33x2]23=(333(32))(233(22))=0(4)=4A_2 = \int_{2}^{3} (3x^2 - 6x) \, dx = \left[ x^3 - 3x^2 \right]_{2}^{3} = (3^3 - 3(3^2)) - (2^3 - 3(2^2)) = 0 - (-4) = 4.
Direct integration of yy on [2,3][2, 3] gives the area above the x-axis.
5
Combine the areas of both sub-regions.
Total Area =A1+A2=4+4=8= A_1 + A_2 = 4 + 4 = 8.
The total geometric area is the sum of the magnitudes of the areas of all separate bounded regions.

Anahtar Kavram

Calculating area under curves crossing the x-axis by splitting definite integrals at real roots
Soru 732Soru

An insulated neutral conductor gains 5.0×10135.0 \times 10^{13} electrons during a electrostatic charging process. Given that the magnitude of the elementary charge is e=1.6×1019 Ce = 1.6 \times 10^{-19}\text{ C}, what is the magnitude of the net charge acquired by the conductor in microcoulombs (μC\mu\text{C})?

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Cevap: 8

Cevap

The magnitude of the net electric charge acquired by the conductor is 8.0 μC8.0\ \mu\text{C}.
According to the principle of charge quantization, the total magnitude of charge QQ acquired by gaining nn electrons is given by Q=neQ = n e. Substituting n=5.0×1013n = 5.0 \times 10^{13} and e=1.6×1019 Ce = 1.6 \times 10^{-19}\text{ C} gives Q=8.0×106 CQ = 8.0 \times 10^{-6}\text{ C}. Expressed in microcoulombs, 8.0×106 C=8.0 μC8.0 \times 10^{-6}\text{ C} = 8.0\ \mu\text{C}.

Adım Adım Çözüm

1
Apply the principle of quantization of electric charge formula
Formula Q=neQ = n e established
Electric charge is quantized and exists in integer multiples of the elementary charge.
2
Multiply the number of electrons by the elementary charge value
Q=8.0×106 CQ = 8.0 \times 10^{-6}\text{ C}
Calculates total electrostatic charge in base SI units.
3
Convert the value from Coulombs to microcoulombs
8.0 μC8.0\ \mu\text{C}
The unit 1 μC1\ \mu\text{C} equals 106 C10^{-6}\text{ C}.

Anahtar Kavram

Quantization of Electric Charge
Soru 733Soru

Given that the matrix A=(12k0k3211)A = \begin{pmatrix} 1 & 2 & k \\ 0 & k & 3 \\ 2 & -1 & 1 \end{pmatrix} is singular, what is the positive value of kk?

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Cevap: 3

Cevap

The positive value of kk is 33.
A square matrix is singular when its determinant is zero. Expanding det(A)\det(A) along the first column yields 1(k+3)+2(6k2)=2k2+k+151 \cdot (k + 3) + 2 \cdot (6 - k^2) = -2k^2 + k + 15. Equating this to zero gives the quadratic equation 2k2k15=02k^2 - k - 15 = 0, which factors as (2k+5)(k3)=0(2k + 5)(k - 3) = 0. The roots are k=2.5k = -2.5 and k=3k = 3. The positive value is 33.

Adım Adım Çözüm

1
Calculate the determinant of matrix A
\det(A) = -2k^2 + k + 15
Expanding along the first column simplifies the calculation because of the zero entry.
2
Set the determinant to zero for singularity
2k^2 - k - 15 = 0
A matrix is singular if and only if its determinant equals zero.
3
Solve the quadratic equation for k
k = 3 or k = -2.5
Factoring 2k2k15=(2k+5)(k3)=02k^2 - k - 15 = (2k + 5)(k - 3) = 0 yields two roots.
4
Choose the positive solution
k = 3
The question explicitly requires the positive value of kk.

Anahtar Kavram

Singular matrix definition and 3x3 determinant evaluation
Soru 734Soru

If aa and bb are real numbers satisfying the simultaneous equations a+b=10a + b = 10 and a2b2=40a^2 - b^2 = 40, what is the value of aa?

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Cevap: 7

Cevap

The value of aa is 7.
Using the difference of squares identity, a2b2=(a+b)(ab)a^2 - b^2 = (a+b)(a-b). Substituting a+b=10a+b = 10 into a2b2=40a^2 - b^2 = 40 gives 10(ab)=4010(a-b) = 40, which simplifies to ab=4a-b = 4. Adding the two linear equations a+b=10a+b = 10 and ab=4a-b = 4 eliminates bb, giving 2a=142a = 14, hence a=7a = 7.

Adım Adım Çözüm

1
Factorize the quadratic expression a2b2a^2 - b^2.
(a+b)(ab)=40(a + b)(a - b) = 40
Apply the difference of two squares identity.
2
Substitute a+b=10a + b = 10 into the factorized equation.
10(ab)=40    ab=410(a - b) = 40 \implies a - b = 4
Simplifying yields a second linear equation.
3
Solve the system of linear equations a+b=10a + b = 10 and ab=4a - b = 4 for aa.
(a+b)+(ab)=10+4    2a=14    a=7(a + b) + (a - b) = 10 + 4 \implies 2a = 14 \implies a = 7
Adding the two equations eliminates bb directly.

Anahtar Kavram

Simultaneous linear and quadratic equations involving difference of squares
Soru 735Soru

The parallel lines L1:3x4y+25=0L_1: 3x - 4y + 25 = 0 and L2:3x4y=0L_2: 3x - 4y = 0 are intersected by a straight line L3L_3 with gradient m>1m > 1. If the length of the line segment of L3L_3 intercepted between L1L_1 and L2L_2 is 555\sqrt{5} units, find the value of mm.

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Cevap: 2

Cevap

The value of the gradient mm is 2.
The perpendicular distance between the parallel lines L1:3x4y+25=0L_1: 3x - 4y + 25 = 0 and L2:3x4y=0L_2: 3x - 4y = 0 is d=2532+(4)2=5d = \frac{25}{\sqrt{3^2 + (-4)^2}} = 5 units. The acute angle ϕ\phi between L3L_3 and the parallel lines satisfies sin(ϕ)=555=15\sin(\phi) = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}, which gives tan(ϕ)=12\tan(\phi) = \frac{1}{2}. The gradient of L1L_1 and L2L_2 is m1=34m_1 = \frac{3}{4}. Using the tangent formula for the angle between two lines, tan(ϕ)=mm11+mm1    12=4m34+3m\tan(\phi) = \left|\frac{m - m_1}{1 + m m_1}\right| \implies \frac{1}{2} = \left|\frac{4m - 3}{4 + 3m}\right|, which yields m=2m = 2 or m=211m = \frac{2}{11}. Under the constraint m>1m > 1, the unique value of mm is 22.

Adım Adım Çözüm

1
Calculate the perpendicular distance dd between the parallel lines L1L_1 and L2L_2
d=25032+(4)2=255=5d = \frac{|25 - 0|}{\sqrt{3^2 + (-4)^2}} = \frac{25}{5} = 5 units
The perpendicular distance between two parallel lines Ax+By+C1=0Ax + By + C_1 = 0 and Ax+By+C2=0Ax + By + C_2 = 0 is given by C1C2A2+B2\frac{|C_1 - C_2|}{\sqrt{A^2 + B^2}}.
2
Determine the trigonometric relationship between the perpendicular distance, intercepted segment, and intersection angle ϕ\phi
sin(ϕ)=perpendicular distanceintercepted segment=555=15\sin(\phi) = \frac{\text{perpendicular distance}}{\text{intercepted segment}} = \frac{5}{5\sqrt{5}} = \frac{1}{\sqrt{5}}
The perpendicular distance forms the opposite side of a right triangle whose hypotenuse is the intercepted segment.
3
Calculate tan(ϕ)\tan(\phi) using right-triangle trigonometry
Since sin(ϕ)=15\sin(\phi) = \frac{1}{\sqrt{5}}, cos(ϕ)=1sin2(ϕ)=25\cos(\phi) = \sqrt{1 - \sin^2(\phi)} = \frac{2}{\sqrt{5}}, so tan(ϕ)=12\tan(\phi) = \frac{1}{2}
The angle between lines formula requires tan(ϕ)\tan(\phi).
4
Apply the angle between two lines formula and solve for mm
tan(ϕ)=mm11+mm1    12=4m34+3m\tan(\phi) = \left| \frac{m - m_1}{1 + m m_1} \right| \implies \frac{1}{2} = \left| \frac{4m - 3}{4 + 3m} \right| where m1=34m_1 = \frac{3}{4}. Case 1: 4m34+3m=12    8m6=4+3m    5m=10    m=2\frac{4m - 3}{4 + 3m} = \frac{1}{2} \implies 8m - 6 = 4 + 3m \implies 5m = 10 \implies m = 2. Case 2: 4m34+3m=12    8m6=43m    11m=2    m=211\frac{4m - 3}{4 + 3m} = -\frac{1}{2} \implies 8m - 6 = -4 - 3m \implies 11m = 2 \implies m = \frac{2}{11}.
Evaluating the absolute value produces two potential solutions.
5
Apply the domain constraint m>1m > 1
m=2m = 2
The problem restricts m>1m > 1, which excludes m=211m = \frac{2}{11}.

Anahtar Kavram

Distance between parallel lines and angle of intersection between straight lines
Soru 736Soru

In a survey of 180180 cloud computing engineers regarding their proficiency in three major platforms—AWS (AA), Azure (BB), and Google Cloud (CC)—it was found that 9595 are proficient in AWS, 8080 in Azure, and 7575 in Google Cloud. Furthermore, 4040 are proficient in both AWS and Azure, 3535 in both Azure and Google Cloud, 3030 in both AWS and Google Cloud, and 1515 are not proficient in any of the three platforms. How many engineers are proficient in exactly one of these platforms?

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Cevap: 100

Cevap

100 engineers are proficient in exactly one platform.
Using the 3-set inclusion-exclusion formula, the number of engineers proficient in all three platforms is solved as 20. Subtracting the relevant overlap regions from each set gives 45 for AWS only, 25 for Azure only, and 30 for Google Cloud only. Adding these single-set values yields a total of 100 engineers proficient in exactly one platform.

Adım Adım Çözüm

1
Determine the cardinality of the union of all three sets.
n(ABC)=18015=165n(A \cup B \cup C) = 180 - 15 = 165.
Subtracting the engineers who are not proficient in any of the three platforms from the universal set.
2
Apply the Principle of Inclusion-Exclusion to calculate the triple intersection n(ABC)n(A \cap B \cap C).
n(ABC)=20n(A \cap B \cap C) = 20.
Substituting known values gives 165=250105+n(ABC)165 = 250 - 105 + n(A \cap B \cap C), which simplifies to n(ABC)=20n(A \cap B \cap C) = 20.
3
Calculate the counts for regions representing exactly two platforms.
AWS & Azure only = 20, Azure & GCP only = 15, AWS & GCP only = 10.
Subtracting the triple intersection count (20) from each pairwise intersection.
4
Calculate the single-set exclusive regions.
AWS only = 45, Azure only = 25, GCP only = 30.
Subtracting all multi-platform overlap regions from each total set size.
5
Sum the single-set exclusive regions.
Total = 45+25+30=10045 + 25 + 30 = 100.
Combining the counts of engineers proficient in AWS only, Azure only, and Google Cloud only.

Anahtar Kavram

Principle of Inclusion-Exclusion for Three Sets and Venn Diagram Region Partitioning
Soru 737Soru

In triangle PQRPQR, the side length p=12 cmp = 12\text{ cm}, side length q=18 cmq = 18\text{ cm}, and sinP=0.4\sin P = 0.4. What is the exact value of sinQ\sin Q?

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Cevap: 0.6

Cevap

The value of sinQ\sin Q is 0.6.
Applying the Sine Rule psinP=qsinQ\frac{p}{\sin P} = \frac{q}{\sin Q} with p=12 cmp = 12\text{ cm}, q=18 cmq = 18\text{ cm}, and sinP=0.4\sin P = 0.4 gives 120.4=18sinQ\frac{12}{0.4} = \frac{18}{\sin Q}. Evaluating 120.4=30\frac{12}{0.4} = 30 leads to 30=18sinQ30 = \frac{18}{\sin Q}, which yields sinQ=1830=0.6\sin Q = \frac{18}{30} = 0.6.

Adım Adım Çözüm

1
State the Sine Rule equation for the given triangle sides and angles.
psinP=qsinQ\frac{p}{\sin P} = \frac{q}{\sin Q}
The Sine Rule relates the lengths of the sides of a triangle to the sines of its opposite angles.
2
Substitute the known numerical values into the Sine Rule equation.
120.4=18sinQ\frac{12}{0.4} = \frac{18}{\sin Q}
Substituting p=12p = 12, q=18q = 18, and sinP=0.4\sin P = 0.4 sets up an equation with a single unknown.
3
Simplify the left side of the equation and solve for sinQ\sin Q.
sinQ=1830=0.6\sin Q = \frac{18}{30} = 0.6
Dividing 12 by 0.4 yields 30, so rearranging gives sinQ=1830=0.6\sin Q = \frac{18}{30} = 0.6.

Anahtar Kavram

Using the Sine Rule to calculate an unknown sine ratio
Soru 738Soru

When the polynomial P(x)=x33x2+kx+12P(x) = x^3 - 3x^2 + kx + 12 is divided by (x2)(x - 2), the remainder is 66. What is the value of the constant kk?

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Cevap: -1

Cevap

The value of the constant kk is 1-1.
By the Remainder Theorem, the remainder when P(x)P(x) is divided by (x2)(x - 2) is P(2)P(2). Substituting x=2x = 2 into P(x)=x33x2+kx+12P(x) = x^3 - 3x^2 + kx + 12 gives P(2)=812+2k+12=8+2kP(2) = 8 - 12 + 2k + 12 = 8 + 2k. Setting 8+2k=68 + 2k = 6 yields 2k=22k = -2, so k=1k = -1.

Adım Adım Çözüm

1
Apply the Remainder Theorem
P(2)=6P(2) = 6
Dividing P(x)P(x) by (x2)(x - 2) leaves a remainder equal to evaluating P(x)P(x) at x=2x = 2.
2
Substitute x=2x = 2 into P(x)P(x) and set equal to 66
(2)33(2)2+2k+12=6(2)^3 - 3(2)^2 + 2k + 12 = 6
Set the evaluated polynomial equal to the given remainder.
3
Simplify the arithmetic terms
8+2k=68 + 2k = 6
Calculate powers and products: 812+12=88 - 12 + 12 = 8.
4
Solve the linear equation for kk
k=1k = -1
Subtract 8 from both sides to get 2k=22k = -2, then divide by 2.

Anahtar Kavram

Polynomial Remainder Theorem
Soru 739Soru

If 2+323=x+y3\frac{2 + \sqrt{3}}{2 - \sqrt{3}} = x + y\sqrt{3}, where xx and yy are integers, what is the value of x+yx + y?

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Cevap: 11

Cevap

The value of x+yx + y is 11.
Multiplying the given fraction by 2+32+3\frac{2 + \sqrt{3}}{2 + \sqrt{3}} rationalises the denominator to 1 and simplifies the numerator to 7+437 + 4\sqrt{3}. Comparing coefficients yields x=7x = 7 and y=4y = 4, making x+y=11x + y = 11.

Adım Adım Çözüm

1
Multiply numerator and denominator by the conjugate of the denominator
\frac{(2 + \sqrt{3})(2 + \sqrt{3})}{(2 - \sqrt{3})(2 + \sqrt{3})}
Rationalising the denominator eliminates the surd from the denominator using the identity (a-b)(a+b) = a^2 - b^2.
2
Expand both the numerator and the denominator
\frac{4 + 4\sqrt{3} + 3}{4 - 3} = \frac{7 + 4\sqrt{3}}{1} = 7 + 4\sqrt{3}
Simplifying algebraic surd multiplication gives integer and surd terms.
3
Compare terms with x + y\sqrt{3} and solve for x and y
x = 7, y = 4 \implies x + y = 11
Matching rational components and coefficients of \sqrt{3} yields x and y.

Anahtar Kavram

Rationalisation of Binomial Denominators containing Surds
Tahmini Süre:1m 30s
Soru 740Soru

Given the 2×22 \times 2 matrix M=(7243)M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}, what is the value of det(M)\det(M)?

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Cevap: 29

Cevap

The determinant of matrix MM is 29.
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is calculated as adbcad - bc. For the matrix M=(7243)M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}, we compute (7)(3)(2)(4)=21(8)=29(7)(3) - (-2)(4) = 21 - (-8) = 29.

Adım Adım Çözüm

1
Identify the entries a,b,c,da, b, c, d from the matrix M=(7243)M = \begin{pmatrix} 7 & -2 \\ 4 & 3 \end{pmatrix}.
a=7a = 7, b=2b = -2, c=4c = 4, and d=3d = 3.
These entries correspond to the standard 2×22 \times 2 matrix representation (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix}.
2
Compute the determinant using the formula det(M)=adbc\det(M) = ad - bc.
det(M)=(7)(3)(2)(4)=21+8=29\det(M) = (7)(3) - (-2)(4) = 21 + 8 = 29.
Multiplying the main diagonal entries and subtracting the product of the off-diagonal entries yields the determinant.

Anahtar Kavram

Determinant of a 2x2 Matrix
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