Algebra

239 soru

Soru 101Soru

The line y=2x1y = 2x - 1 intersects the curve y=x24x+4y = x^2 - 4x + 4 at two distinct points, (x1,y1)(x_1, y_1) and (x2,y2)(x_2, y_2). Calculate the sum of the yy-coordinates of these two points of intersection, y1+y2y_1 + y_2.

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Cevap: 10

Cevap

The sum of the yy-coordinates of the points of intersection is 10.
Equating y=2x1y = 2x - 1 and y=x24x+4y = x^2 - 4x + 4 yields x26x+5=0x^2 - 6x + 5 = 0, whose solutions are x=1x = 1 and x=5x = 5. Substituting these values into y=2x1y = 2x - 1 gives y1=1y_1 = 1 and y2=9y_2 = 9. The sum y1+y2=1+9=10y_1 + y_2 = 1 + 9 = 10.

Adım Adım Çözüm

1
Equate the linear and quadratic equations to eliminate yy.
x26x+5=0x^2 - 6x + 5 = 0
Setting 2x1=x24x+42x - 1 = x^2 - 4x + 4 allows finding the xx-coordinates of the intersection points.
2
Solve the quadratic equation for xx.
x1=1x_1 = 1 and x2=5x_2 = 5
Factoring (x1)(x5)=0(x - 1)(x - 5) = 0 yields the two xx-values.
3
Determine the corresponding yy-values using y=2x1y = 2x - 1.
y1=1y_1 = 1 and y2=9y_2 = 9
Substituting x=1x = 1 yields y=1y = 1, and substituting x=5x = 5 yields y=9y = 9.
4
Calculate the sum of the yy-coordinates.
10
Adding y1+y2=1+9=10y_1 + y_2 = 1 + 9 = 10.

Anahtar Kavram

Solving simultaneous linear and quadratic equations to find coordinates of intersection
Soru 102Soru

The 4th4^{\text{th}} term of an arithmetic progression (A.P.) is 1515 and the 9th9^{\text{th}} term is 3535. Calculate the sum of the first 1010 terms of the progression.

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Cevap: 210

Cevap

The sum of the first 10 terms of the progression is 210.
Using the nthn^{\text{th}} term formula Tn=a+(n1)dT_n = a + (n-1)d, the equations a+3d=15a + 3d = 15 and a+8d=35a + 8d = 35 yield common difference d=4d = 4 and first term a=3a = 3. Substituting these into the sum formula Sn=n2[2a+(n1)d]S_n = \frac{n}{2}[2a + (n-1)d] for n=10n = 10 yields S10=5[2(3)+9(4)]=210S_{10} = 5[2(3) + 9(4)] = 210.

Adım Adım Çözüm

1
Set up equations for the given terms
a+3d=15a + 3d = 15 and a+8d=35a + 8d = 35
The nthn^{\text{th}} term of an A.P. is given by Tn=a+(n1)dT_n = a + (n-1)d.
2
Solve for the common difference dd
d=4d = 4
Subtracting (a+3d=15)(a + 3d = 15) from (a+8d=35)(a + 8d = 35) gives 5d=205d = 20, so d=4d = 4.
3
Solve for the first term aa
a=3a = 3
Substituting d=4d = 4 into a+3(4)=15a + 3(4) = 15 yields a=1512=3a = 15 - 12 = 3.
4
Calculate the sum of the first 10 terms
S10=210S_{10} = 210
Applying S10=102[2(3)+9(4)]=5(6+36)=210S_{10} = \frac{10}{2}[2(3) + 9(4)] = 5(6 + 36) = 210.

Anahtar Kavram

Finding terms and sums of an Arithmetic Progression using simultaneous linear equations
Tahmini Süre:1m 30s
Soru 103Soru

What are the values of yy that satisfy the simultaneous equations 3x+y=103x + y = 10 and x2+y=14x^2 + y = 14?

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Cevap: 2-2 or 1313

Cevap

The values of yy that satisfy the simultaneous equations are 2-2 or 1313.
Rearranging the linear equation gives y=103xy = 10 - 3x. Substituting this into x2+y=14x^2 + y = 14 yields x2+103x=14x^2 + 10 - 3x = 14, which simplifies to x23x4=0x^2 - 3x - 4 = 0. Factoring gives (x4)(x+1)=0(x - 4)(x + 1) = 0, so x=4x = 4 or x=1x = -1. Substituting x=4x = 4 into y=103xy = 10 - 3x gives y=2y = -2, and substituting x=1x = -1 gives y=13y = 13. Thus, the required values of yy are 2-2 or 1313.

Adım Adım Çözüm

1
Express yy in terms of xx from the linear equation
y=103xy = 10 - 3x
Isolation of variable yy allows straightforward substitution into the quadratic equation.
2
Substitute y=103xy = 10 - 3x into the quadratic equation x2+y=14x^2 + y = 14
x2+(103x)=14    x23x4=0x^2 + (10 - 3x) = 14 \implies x^2 - 3x - 4 = 0
Creates a quadratic equation in terms of a single variable xx.
3
Solve the quadratic equation x23x4=0x^2 - 3x - 4 = 0 for xx
(x4)(x+1)=0    x=4 or x=1(x - 4)(x + 1) = 0 \implies x = 4 \text{ or } x = -1
Factoring determines the xx-coordinates of the solution points.
4
Substitute each xx-value back into y=103xy = 10 - 3x to find the corresponding yy-values
For x=4x = 4: y=103(4)=2y = 10 - 3(4) = -2; for x=1x = -1: y=103(1)=13y = 10 - 3(-1) = 13
Evaluates the exact values of yy requested by the question.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations
Soru 104Soru

When the polynomial P(x)=x32x2+ax+8P(x) = x^3 - 2x^2 + ax + 8 is divided by (x3)(x - 3), the remainder is 1414. What is the value of the constant aa?

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Cevap: -1

Cevap

The value of the constant aa is 1-1.
According to the Remainder Theorem, dividing P(x)P(x) by (x3)(x - 3) means the remainder is P(3)P(3). Evaluating P(3)=332(3)2+3a+8=2718+3a+8=17+3aP(3) = 3^3 - 2(3)^2 + 3a + 8 = 27 - 18 + 3a + 8 = 17 + 3a. Setting this equal to the remainder 1414 gives 17+3a=1417 + 3a = 14, which simplifies to 3a=33a = -3 and yields a=1a = -1.

Adım Adım Çözüm

1
Apply the Remainder Theorem
The remainder when P(x)P(x) is divided by (x3)(x - 3) is equal to P(3)P(3).
The Remainder Theorem states that dividing a polynomial P(x)P(x) by (xc)(x - c) yields a remainder equal to P(c)P(c).
2
Substitute x=3x = 3 into the polynomial and set equal to the given remainder
332(3)2+a(3)+8=143^3 - 2(3)^2 + a(3) + 8 = 14
Setting the value of P(3)P(3) equal to 1414 allows us to form a linear equation for the unknown constant aa.
3
Simplify numerical terms in the equation
2718+3a+8=14    17+3a=1427 - 18 + 3a + 8 = 14 \implies 17 + 3a = 14
Evaluate exponents and multiplication to isolate the term containing aa.
4
Solve for aa
3a=3    a=13a = -3 \implies a = -1
Subtract 1717 from both sides and divide by 33.

Anahtar Kavram

Polynomial Remainder Theorem
Soru 105Soru

If 6+262\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}} is expressed in the form a+b3a + b\sqrt{3}, where aa and bb are integers, what is the value of a+ba + b?

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Cevap: 3

Cevap

The value of a+ba + b is 3.
Multiplying both numerator and denominator by the conjugate (6+2)(\sqrt{6} + \sqrt{2}) yields 8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}. Comparing this to a+b3a + b\sqrt{3} gives a=2a = 2 and b=1b = 1, so a+b=3a + b = 3.

Adım Adım Çözüm

1
Rationalise the denominator by multiplying the numerator and denominator by the conjugate of the denominator, (6+2)(\sqrt{6} + \sqrt{2}).
(6+2)(6+2)(62)(6+2)\frac{(\sqrt{6} + \sqrt{2})(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}
Multiplying by the conjugate eliminates radicals from the denominator using the difference of squares.
2
Expand the numerator (6+2)2(\sqrt{6} + \sqrt{2})^2 using (x+y)2=x2+2xy+y2(x+y)^2 = x^2 + 2xy + y^2.
(6)2+212+(2)2=6+2(23)+2=8+43(\sqrt{6})^2 + 2\sqrt{12} + (\sqrt{2})^2 = 6 + 2(2\sqrt{3}) + 2 = 8 + 4\sqrt{3}
Simplifying 12=4×3=23\sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} converts compound surds to standard form.
3
Expand the denominator using the difference of two squares (xy)(x+y)=x2y2(x-y)(x+y) = x^2 - y^2.
(6)2(2)2=62=4(\sqrt{6})^2 - (\sqrt{2})^2 = 6 - 2 = 4
Squaring each square root leaves rational integers in the denominator.
4
Divide the expanded numerator by the denominator to find aa and bb.
8+434=2+3\frac{8 + 4\sqrt{3}}{4} = 2 + \sqrt{3}
Dividing each term by 4 gives a=2a = 2 and b=1b = 1.
5
Compute the sum a+ba + b.
2+1=32 + 1 = 3
Adding the coefficients aa and bb yields the required value.

Anahtar Kavram

Rationalisation of Binomial Denominators
Soru 106Soru

Let the universal set U={xZ+:x60}\mathcal{U} = \{x \in \mathbb{Z}^+ : x \le 60\}. Subsets AA, BB, and CC of U\mathcal{U} are defined as:

A={xU:x is a multiple of 3}A = \{x \in \mathcal{U} : x \text{ is a multiple of } 3\}
B={xU:x is a multiple of 4}B = \{x \in \mathcal{U} : x \text{ is a multiple of } 4\}
C={xU:x is a factor of 60}C = \{x \in \mathcal{U} : x \text{ is a factor of } 60\}

What is the cardinality of the set (AΔB)C(A \Delta B) \setminus C', where AΔBA \Delta B represents the symmetric difference of AA and BB, and CC' is the complement of CC in U\mathcal{U}?

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Cevap: 6

Cevap

6
Using the set identity XY=XYX \setminus Y' = X \cap Y, the expression (AΔB)C(A \Delta B) \setminus C' simplifies directly to (AΔB)C(A \Delta B) \cap C. The set CC consists of all factors of 60: {1,2,3,4,5,6,10,12,15,20,30,60}\{1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\}. The symmetric difference AΔBA \Delta B consists of numbers that are divisible by 3 or 4, but not both (excluding multiples of 12). Checking the 12 factors of 60 against this rule, the numbers 3, 6, 15, and 30 are multiples of 3 but not 4, while 4 and 20 are multiples of 4 but not 3. The numbers 12 and 60 are multiples of both 3 and 4, so they are excluded. Thus, the resulting set is {3,4,6,15,20,30}\{3, 4, 6, 15, 20, 30\}, which contains exactly 6 elements.

Adım Adım Çözüm

1
Simplify the set expression using complement and relative difference identities.
(AΔB)C=(AΔB)(C)=(AΔB)C(A \Delta B) \setminus C' = (A \Delta B) \cap (C')' = (A \Delta B) \cap C
By definition of set difference, XY=XYX \setminus Y = X \cap Y'. Replacing YY with CC' gives X(C)=XCX \cap (C')' = X \cap C.
2
List all elements in set CC (factors of 60).
C={1,2,3,4,5,6,10,12,15,20,30,60}C = \{1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60\}
60 prime factorizes as 22×31×512^2 \times 3^1 \times 5^1, giving (2+1)(1+1)(1+1)=12(2+1)(1+1)(1+1) = 12 positive factors.
3
Recall the definition of symmetric difference AΔBA \Delta B.
AΔB=(AB)(AB)A \Delta B = (A \cup B) \setminus (A \cap B)
An element belongs to AΔBA \Delta B if it is a multiple of 3 or a multiple of 4, but NOT a multiple of both 3 and 4 (i.e. not a multiple of 12).
4
Evaluate which elements of CC belong to AΔBA \Delta B.
The elements in CC that are multiples of 3 only (not 12) are {3,6,15,30}\{3, 6, 15, 30\}. The elements in CC that are multiples of 4 only (not 12) are {4,20}\{4, 20\}. Elements {12,60}\{12, 60\} are multiples of 12, so they are excluded.
Testing each factor of 60 against the criteria for symmetric difference.
5
Form the set (AΔB)C(A \Delta B) \cap C and find its cardinality.
(AΔB)C={3,4,6,15,20,30}(A \Delta B) \cap C = \{3, 4, 6, 15, 20, 30\}, so n((AΔB)C)=6n((A \Delta B) \cap C) = 6.
Counting the elements in the intersection.

Anahtar Kavram

Symmetric Difference and Set Complement Identities
Tahmini Süre:2m 30s
Soru 107Soru

The time, tt hours, required to complete a road maintenance project varies inversely as the number of workers, ww, assigned to the project. If 8 workers can finish the project in 15 hours, calculate the time, in hours, required for 12 workers to finish the same project.

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Cevap: 10

Cevap

10 hours
Because the time tt varies inversely as the number of workers ww, the total worker-hours required for the project is constant: k=8×15=120k = 8 \times 15 = 120 worker-hours. Dividing this total work by 12 workers gives 12012=10\frac{120}{12} = 10 hours.

Adım Adım Çözüm

1
Set up the inverse variation equation
t=kwt = \frac{k}{w}, where kk is the constant of variation.
Inverse variation implies that as the number of workers increases, the time required decreases proportionally.
2
Determine the value of the constant of variation kk
k=t×w=15×8=120k = t \times w = 15 \times 8 = 120.
Substitute the known pair of values (w=8,t=15w = 8, t = 15) into the equation.
3
Compute the new value of tt for 12 workers
t=12012=10t = \frac{120}{12} = 10 hours.
Substitute w=12w = 12 and k=120k = 120 into t=kwt = \frac{k}{w}.

Anahtar Kavram

Inverse Variation
Soru 108Soru

The total energy loss EE in Joules per minute in a magnetic core circuit is partly constant and partly varies directly as the square of the frequency ff in Hz of the alternating current. Given that E=120 JE = 120\text{ J} when f=10 Hzf = 10\text{ Hz}, and E=360 JE = 360\text{ J} when f=20 Hzf = 20\text{ Hz}, what is the value of EE in Joules when f=15 Hzf = 15\text{ Hz}?

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Cevap: 220

Cevap

220
The relation describes a partial variation model E=c+kf2E = c + k f^2. Substituting the two given conditions (f=10,E=120f=10, E=120 and f=20,E=360f=20, E=360) yields the system of equations c+100k=120c + 100k = 120 and c+400k=360c + 400k = 360. Solving this system gives k=0.8k = 0.8 and c=40c = 40. Evaluating E=40+0.8(15)2E = 40 + 0.8(15)^2 results in 220 J220\text{ J}.

Adım Adım Çözüm

1
Write the general equation for partial variation involving a constant term and a term proportional to f2f^2
E=c+kf2E = c + k f^2
Partial variation consists of a sum of a constant component and a variable component.
2
Substitute the known conditions into the variation equation to create a system of linear equations
Equation 1: c+100k=120c + 100k = 120; Equation 2: c+400k=360c + 400k = 360
Two pairs of values are provided to determine the two unknown constants cc and kk.
3
Solve the system of simultaneous equations for kk and cc
k=0.8k = 0.8 and c=40c = 40
Subtracting Equation 1 from Equation 2 eliminates cc, allowing direct calculation of kk, after which cc is found by substitution.
4
Calculate the required value of EE when f=15f = 15
E=40+0.8(152)=40+0.8(225)=220E = 40 + 0.8(15^2) = 40 + 0.8(225) = 220
Applying the discovered constants to the target frequency yields the final energy loss value.

Anahtar Kavram

Partial Variation and Simultaneous Equations
Tahmini Süre:2m 0s
Soru 109Soru

If yy varies directly as xx, and y=20y = 20 when x=4x = 4, what is the value of yy when x=10x = 10?

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Cevap: 50

Cevap

The value of yy when x=10x = 10 is 50.
Direct variation means y=kxy = kx. Substituting y=20y = 20 and x=4x = 4 yields k=5k = 5. Substituting k=5k = 5 and x=10x = 10 gives y=50y = 50.

Adım Adım Çözüm

1
Set up the equation for direct variation
y=kxy = kx, where kk is the constant of variation.
Direct variation implies that yy is directly proportional to xx.
2
Substitute the given initial values (y=20y = 20, x=4x = 4) to find kk
20=k(4)    k=204=520 = k(4) \implies k = \frac{20}{4} = 5.
Determining the constant of variation allows us to write the specific relationship equation.
3
Calculate yy when x=10x = 10 using the constant k=5k = 5
y=5×10=50y = 5 \times 10 = 50.
Substituting x=10x = 10 into y=5xy = 5x yields the required value.

Anahtar Kavram

Direct Variation (y=kxy = kx)
Tahmini Süre:45s
Soru 110Soru

What is the set of real values of xx that satisfies the quadratic inequality 2x25x302x^2 - 5x - 3 \le 0?

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Cevap: 12x3-\frac{1}{2} \le x \le 3

Cevap

12x3-\frac{1}{2} \le x \le 3
Factoring 2x25x32x^2 - 5x - 3 yields (2x+1)(x3)0(2x + 1)(x - 3) \le 0. Setting the factors to zero gives roots x=1/2x = -1/2 and x=3x = 3. Because the coefficient of x2x^2 is positive, the quadratic curve opens upwards and is less than or equal to zero in the closed interval between the roots, resulting in 12x3-\frac{1}{2} \le x \le 3.

Adım Adım Çözüm

1
Factor the quadratic expression 2x25x32x^2 - 5x - 3
(2x+1)(x3)0(2x + 1)(x - 3) \le 0
Finding the factors helps identify the critical points (roots) of the inequality.
2
Determine the critical points by setting each factor equal to zero
x=12x = -\frac{1}{2} and x=3x = 3
The critical points divide the real number line into test intervals.
3
Determine the region where (2x+1)(x3)0(2x + 1)(x - 3) \le 0
12x3-\frac{1}{2} \le x \le 3
Since the quadratic coefficient is positive (2>02 > 0), the parabola opens upwards, so the function values are less than or equal to zero between the two roots.

Anahtar Kavram

Solving Quadratic Inequalities by Factorisation
Soru 111Soru

Find the smallest positive integer kk for which the inequality (k2)x2+8x+k+4>0(k - 2)x^2 + 8x + k + 4 > 0 holds for all real values of xx.

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Cevap: 5

Cevap

5
For the quadratic expression (k2)x2+8x+k+4(k - 2)x^2 + 8x + k + 4 to be positive for all real values of xx, two conditions must be satisfied simultaneously: the leading coefficient must be positive (k2>0    k>2k - 2 > 0 \implies k > 2) and the discriminant must be strictly negative (Δ<0\Delta < 0). Calculating the discriminant gives Δ=824(k2)(k+4)=968k4k2\Delta = 8^2 - 4(k - 2)(k + 4) = 96 - 8k - 4k^2. Setting 968k4k2<096 - 8k - 4k^2 < 0 and dividing by 4-4 (reversing the inequality) yields k2+2k24>0k^2 + 2k - 24 > 0, which factors as (k+6)(k4)>0(k + 6)(k - 4) > 0. This gives k<6k < -6 or k>4k > 4. Intersecting with k>2k > 2 results in k>4k > 4. The smallest integer greater than 4 is 5.

Adım Adım Çözüm

1
Determine the conditions for positivity for all real numbers
k2>0k - 2 > 0 and Δ<0\Delta < 0
A quadratic Ax2+Bx+CAx^2 + Bx + C remains above the x-axis for all real xx if and only if its parabola opens upwards (A>0A > 0) and has no real roots (Δ<0\Delta < 0).
2
Set up and solve the discriminant inequality
k2+2k24>0    (k+6)(k4)>0k^2 + 2k - 24 > 0 \implies (k + 6)(k - 4) > 0
Expanding 824(k2)(k+4)<08^2 - 4(k - 2)(k + 4) < 0 gives 644(k2+2k8)<064 - 4(k^2 + 2k - 8) < 0, which simplifies to k2+2k24>0k^2 + 2k - 24 > 0 after dividing by 4-4 and reversing the inequality sign.
3
Intersect solution sets and find the smallest integer
k=5k = 5
The intersection of k>2k > 2 and (k<6 or k>4)(k < -6 \text{ or } k > 4) gives k>4k > 4. The smallest integer strictly greater than 4 is 5.

Anahtar Kavram

Condition for Positive Definite Quadratic Inequalities
Soru 112Soru

If the polynomial P(x)=x4+ax37x2+bx+12P(x) = x^4 + ax^3 - 7x^2 + bx + 12 is completely divisible by x22x3x^2 - 2x - 3, what is the value of aba - b?

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Cevap: 10-10

Cevap

The value of aba - b is 10-10.
Factoring x22x3x^2 - 2x - 3 gives (x3)(x+1)(x - 3)(x + 1). By the Factor Theorem, P(3)=0P(3) = 0 and P(1)=0P(-1) = 0. Substituting these into P(x)P(x) produces the linear system 9a+b=109a + b = -10 and a+b=6a + b = 6. Solving this system yields a=2a = -2 and b=8b = 8. Subtracting gives ab=28=10a - b = -2 - 8 = -10.

Adım Adım Çözüm

1
Factor the quadratic divisor to find the roots.
x22x3=(x3)(x+1)x^2 - 2x - 3 = (x - 3)(x + 1), so the roots are x=3x = 3 and x=1x = -1.
By the Factor Theorem, if a polynomial is divisible by a quadratic expression, P(x)P(x) must evaluate to zero at each root of the divisor.
2
Set up equations by evaluating P(3)=0P(3) = 0 and P(1)=0P(-1) = 0.
For x=3x = 3: 34+a(3)37(3)2+b(3)+12=0    81+27a63+3b+12=0    9a+b=103^4 + a(3)^3 - 7(3)^2 + b(3) + 12 = 0 \implies 81 + 27a - 63 + 3b + 12 = 0 \implies 9a + b = -10.
For x=1x = -1: (1)4+a(1)37(1)2+b(1)+12=0    1a7b+12=0    a+b=6(-1)^4 + a(-1)^3 - 7(-1)^2 + b(-1) + 12 = 0 \implies 1 - a - 7 - b + 12 = 0 \implies a + b = 6.
Evaluating the polynomial at each root yields a system of two linear equations in variables aa and bb.
3
Solve the simultaneous equations for aa and bb.
Subtracting (a+b=6)(a + b = 6) from (9a+b=10)(9a + b = -10) gives 8a=16    a=28a = -16 \implies a = -2.
Substituting a=2a = -2 into a+b=6a + b = 6 gives 2+b=6    b=8-2 + b = 6 \implies b = 8.
Elimination isolates aa, allowing both aa and bb to be uniquely determined.
4
Calculate aba - b.
ab=28=10a - b = -2 - 8 = -10.
This computes the required expression value.

Anahtar Kavram

Factor Theorem for quadratic divisors
Tahmini Süre:2m 0s
Soru 113Soru

Let the universal set U={xZ:10x99}\mathcal{U} = \{x \in \mathbb{Z} : 10 \le x \le 99\}. Consider two subsets PP and QQ of U\mathcal{U}, where P={10a+bU:a>b}P = \{10a + b \in \mathcal{U} : a > b\} and Q={10a+bU:ab is odd}Q = \{10a + b \in \mathcal{U} : a \cdot b \text{ is odd}\}, with aa and bb representing the tens and units digits of xx respectively. What is the cardinality of the set PQP \setminus Q?

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Cevap: 35; 35 elements

Cevap

35
To evaluate n(PQ)n(P \setminus Q), we calculate n(P)n(PQ)n(P) - n(P \cap Q). Set PP consists of all two-digit numbers where the tens digit exceeds the units digit, giving 1+2++9=451 + 2 + \dots + 9 = 45 numbers. Set QQ consists of numbers with two odd digits. The intersection PQP \cap Q contains pairs of distinct odd digits (a,b)(a,b) where a>ba > b, which corresponds to (52)=10\binom{5}{2} = 10 pairs. Thus, n(PQ)=4510=35n(P \setminus Q) = 45 - 10 = 35.

Adım Adım Çözüm

1
Calculate the cardinality of set PP, denoted as n(P)n(P).
n(P)=45n(P) = 45
For a two-digit integer 10a+b10a + b, a{1,2,,9}a \in \{1, 2, \dots, 9\} and b{0,1,,9}b \in \{0, 1, \dots, 9\}. The condition a>ba > b yields 00 valid values of bb for a=1a=1 (1 choice: 10), 2 choices for a=2a=2 (20, 21), up to 9 choices for a=9a=9 (90 through 98). The total is a=19a=9×102=45\sum_{a=1}^9 a = \frac{9 \times 10}{2} = 45.
2
Calculate the cardinality of the intersection PQP \cap Q, denoted as n(PQ)n(P \cap Q).
n(PQ)=10n(P \cap Q) = 10
Set QQ contains numbers where the product aba \cdot b is odd, which requires both aa and bb to be odd digits chosen from {1,3,5,7,9}\{1, 3, 5, 7, 9\}. For PQP \cap Q, we need a>ba > b with distinct odd digits. The number of such pairs (a,b)(a, b) is given by choosing 2 distinct odd digits out of 5: (52)=10\binom{5}{2} = 10.
3
Calculate the cardinality of the set difference PQP \setminus Q.
n(PQ)=35n(P \setminus Q) = 35
Using the identity n(PQ)=n(P)n(PQ)n(P \setminus Q) = n(P) - n(P \cap Q), we get 4510=3545 - 10 = 35.

Anahtar Kavram

Cardinality of set difference and set-builder operations
Tahmini Süre:2m 30s
Soru 114Soru

Simplify the expression 33133+1\frac{\sqrt{3}}{\sqrt{3} - 1} - \frac{\sqrt{3}}{\sqrt{3} + 1}.

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Cevap: 3\sqrt{3}

Cevap

3\sqrt{3}
Combining the fractions using the common denominator (31)(3+1)=2(\sqrt{3}-1)(\sqrt{3}+1) = 2 gives a numerator of 3(3+1)3(31)=3+33+3=23\sqrt{3}(\sqrt{3}+1) - \sqrt{3}(\sqrt{3}-1) = 3 + \sqrt{3} - 3 + \sqrt{3} = 2\sqrt{3}. Dividing 232\sqrt{3} by 22 yields the simplified answer 3\sqrt{3}.

Adım Adım Çözüm

1
Find a common denominator for the two fractions
The common denominator is (31)(3+1)=(3)2(1)2=31=2(\sqrt{3} - 1)(\sqrt{3} + 1) = (\sqrt{3})^2 - (1)^2 = 3 - 1 = 2
The denominators are conjugate surds, so their product simplifies to a rational number using the difference of two squares.
2
Express the numerator over the common denominator
Numerator =3(3+1)3(31)= \sqrt{3}(\sqrt{3} + 1) - \sqrt{3}(\sqrt{3} - 1)
Multiply each numerator by the missing factor of the common denominator.
3
Expand and simplify the numerator
Numerator =(3+3)(33)=3+33+3=23= (3 + \sqrt{3}) - (3 - \sqrt{3}) = 3 + \sqrt{3} - 3 + \sqrt{3} = 2\sqrt{3}
Distribute 3\sqrt{3} and handle the subtraction sign carefully.
4
Divide the simplified numerator by the common denominator
232=3\frac{2\sqrt{3}}{2} = \sqrt{3}
Cancel the common factor of 2.

Anahtar Kavram

Rationalisation of denominators and algebraic manipulation of surd fractions
Tahmini Süre:1m 15s
Soru 115Soru

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 116Soru

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 117Soru

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 118Soru

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.

Adım Adım Çözüm

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 119Soru

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.

Adım Adım Çözüm

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 120Soru

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
ÖncekiSayfa 6 / 12Sonraki
Algebra Alıştırma Soruları — JAMB UTME — Sayfa 6 | Examkin