Algebra

239 soru

Soru 221Soru →

In a sports academy of 120 athletes, 70 play football, 60 play basketball, and 50 play tennis. If 10 athletes play none of these three sports and 15 athletes play all three sports, how many athletes play exactly two of these sports?

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Cevap: 40

Cevap

40 athletes play exactly two of the sports.
The correct answer is 40. Subtracting the 10 athletes who play no sports from the total of 120 leaves 110 athletes playing at least one sport. Using inclusion-exclusion, the sum of pairwise intersections is S2=70+60+50+15−110=85S_2 = 70 + 60 + 50 + 15 - 110 = 85. Since S2S_2 contains the region of all three sports counted three times, subtracting 3×15=453 \times 15 = 45 gives 40 athletes who play exactly two sports.

Adım Adım Çözüm

1
Determine the cardinality of the union of all three sets
n(F∪B∪T)=120−10=110n(F \cup B \cup T) = 120 - 10 = 110
Athletes who play none of the three sports are excluded from the total universal set.
2
Apply the Principle of Inclusion-Exclusion for three sets to find the sum of 2-set intersections
S2=n(F)+n(B)+n(T)+n(F∩B∩T)−n(F∪B∪T)=70+60+50+15−110=85S_2 = n(F) + n(B) + n(T) + n(F \cap B \cap T) - n(F \cup B \cup T) = 70 + 60 + 50 + 15 - 110 = 85
The formula relates the total union, individual set cardinalities, pairwise intersections, and triple intersection.
3
Subtract three times the triple intersection from S2S_2 to isolate regions corresponding to exactly two sports
Exactly two sports = S2−3×n(F∩B∩T)=85−3(15)=40S_2 - 3 \times n(F \cap B \cap T) = 85 - 3(15) = 40
Each pairwise intersection sum S2S_2 includes the triple intersection region three times.

Anahtar Kavram

Three-set inclusion-exclusion principle and region cardinality decomposition
Tahmini Süre:1m 30s
Soru 222Soru →

The sum of the first nn terms of an arithmetic progression (A.P.) is given by Sn=2n2+3nS_n = 2n^2 + 3n. What is the 5th5^{\text{th}} term of the progression?

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Cevap: 2121

Cevap

The 5th5^{\text{th}} term of the arithmetic progression is 2121.
The nthn^{\text{th}} term of a sequence can be determined from its sum formula using Tn=Sn−Sn−1T_n = S_n - S_{n-1}. Evaluating S5=2(5)2+3(5)=65S_5 = 2(5)^2 + 3(5) = 65 and S4=2(4)2+3(4)=44S_4 = 2(4)^2 + 3(4) = 44, the difference T5=65−44=21T_5 = 65 - 44 = 21 gives the correct value of the 5th5^{\text{th}} term.

Adım Adım Çözüm

1
Calculate the sum of the first 5 terms (S5S_5)
S5=2(5)2+3(5)=2(25)+15=65S_5 = 2(5)^2 + 3(5) = 2(25) + 15 = 65
To find the sum up to the 5th5^{\text{th}} term using the given formula Sn=2n2+3nS_n = 2n^2 + 3n.
2
Calculate the sum of the first 4 terms (S4S_4)
S4=2(4)2+3(4)=2(16)+12=44S_4 = 2(4)^2 + 3(4) = 2(16) + 12 = 44
To find the cumulative total of all terms prior to the 5th5^{\text{th}} term.
3
Subtract S4S_4 from S5S_5 to isolate the 5th5^{\text{th}} term (T5T_5)
T5=S5−S4=65−44=21T_5 = S_5 - S_4 = 65 - 44 = 21
The nthn^{\text{th}} term of any sequence is given by the relation Tn=Sn−Sn−1T_n = S_n - S_{n-1}.

Anahtar Kavram

Relationship between the sum of nn terms (SnS_n) and the nthn^{\text{th}} term (TnT_n) in an Arithmetic Progression
Tahmini Süre:1m 30s
Soru 223Soru →

What is the positive value of xx that satisfies the equation 22x+1−9⋅2x+4=02^{2x+1} - 9 \cdot 2^x + 4 = 0?

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Cevap: 2

Cevap

The positive value of xx that satisfies the equation is 2.
Applying the law of indices am+n=am⋅ana^{m+n} = a^m \cdot a^n gives 22x+1=2⋅(2x)22^{2x+1} = 2 \cdot (2^x)^2. Setting y=2xy = 2^x yields the quadratic equation 2y2−9y+4=02y^2 - 9y + 4 = 0. Factoring this expression gives (2y−1)(y−4)=0(2y - 1)(y - 4) = 0, which yields roots y=12y = \frac{1}{2} and y=4y = 4. Solving 2x=122^x = \frac{1}{2} gives x=−1x = -1, and solving 2x=42^x = 4 gives x=2x = 2. The positive value is 22.

Adım Adım Çözüm

1
Use index laws to express the equation in terms of 2x2^x
2⋅(2x)2−9⋅(2x)+4=02 \cdot (2^x)^2 - 9 \cdot (2^x) + 4 = 0
By the product law of indices, 22x+1=22x⋅21=2⋅(2x)22^{2x+1} = 2^{2x} \cdot 2^1 = 2 \cdot (2^x)^2.
2
Substitute y=2xy = 2^x to form a quadratic equation
2y2−9y+4=02y^2 - 9y + 4 = 0
Replacing 2x2^x with a single variable simplifies the exponential equation into quadratic form.
3
Solve the quadratic equation for yy
y=12y = \frac{1}{2} or y=4y = 4
Factoring 2y2−9y+4=02y^2 - 9y + 4 = 0 gives (2y−1)(y−4)=0(2y - 1)(y - 4) = 0.
4
Substitute back y=2xy = 2^x to solve for xx
x=−1x = -1 or x=2x = 2
Since 2x=12=2−12^x = \frac{1}{2} = 2^{-1}, x=−1x = -1. Since 2x=4=222^x = 4 = 2^2, x=2x = 2.
5
Select the positive value requested by the question
x=2x = 2
x=2x = 2 is positive, whereas x=−1x = -1 is negative.

Anahtar Kavram

Reducing exponential equations to quadratic form using index laws
Tahmini Süre:1m 30s
Soru 224Soru →

Determine the smallest integer value of xx that satisfies the compound inequality 3<2x−5≤113 < 2x - 5 \le 11.

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Cevap: 5

Cevap

The smallest integer value of xx that satisfies the inequality is 5.
Adding 5 across the compound inequality 3<2x−5≤113 < 2x - 5 \le 11 gives 8<2x≤168 < 2x \le 16. Dividing by 2 yields 4<x≤84 < x \le 8. The integer values satisfying this range are 5, 6, 7, and 8. Therefore, the smallest integer solution is 5.

Adım Adım Çözüm

1
Add 5 to all parts of the compound inequality
8 < 2x <= 16
Isolates the variable term in the middle segment.
2
Divide all parts of the compound inequality by 2
4 < x <= 8
Solves for x without changing inequality signs since 2 is positive.
3
Identify integer solutions within the range (4, 8]
x in {5, 6, 7, 8}
Since the inequality at 4 is strict (<), 4 is excluded, but 8 is included (<=).
4
Find the minimum integer value in the solution set
5
5 is the smallest integer strictly greater than 4.

Anahtar Kavram

Solving Compound Linear Inequalities
Soru 225Soru →

Which of the following represents the solution set of real values of xx satisfying the inequality 4(1−x)≤3(x+6)4(1 - x) \le 3(x + 6)?

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Cevap: x≥−2x \ge -2

Cevap

The set of real values satisfying the inequality is x≥−2x \ge -2.
Expanding the given inequality yields 4−4x≤3x+184 - 4x \le 3x + 18. Subtracting 3x3x and 44 from both sides gives −7x≤14-7x \le 14. When dividing both sides by −7-7, the inequality sign must reverse direction, yielding x≥−2x \ge -2.

Adım Adım Çözüm

1
Expand both sides of the inequality
4−4x≤3x+184 - 4x \le 3x + 18
Remove brackets to group like terms.
2
Rearrange terms by moving variable terms to the left side and constant terms to the right side
−4x−3x≤18−4  ⟹  −7x≤14-4x - 3x \le 18 - 4 \implies -7x \le 14
Isolate the term containing the variable xx.
3
Divide both sides by −7-7 and reverse the inequality sign
x≥−2x \ge -2
Dividing or multiplying an inequality by a negative number flips the direction of the inequality symbol.

Anahtar Kavram

Solving linear inequalities involving bracket expansion and division by negative numbers
Soru 226Soru →

How many integer values of xx satisfy the quadratic inequality 2x2−7x−4≤02x^2 - 7x - 4 \le 0?

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Cevap: 5

Cevap

There are 5 integer values of x that satisfy the inequality.
Solving the quadratic inequality yields −12≤x≤4-\frac{1}{2} \le x \le 4. The integer solutions within this interval are 0,1,2,3,0, 1, 2, 3, and 44. Counting them gives a total of 55 valid integer values.

Adım Adım Çözüm

1
Factor the quadratic expression.
(2x+1)(x−4)≤0(2x + 1)(x - 4) \le 0
Factoring allows determination of the critical boundary points.
2
Find the critical values (roots of the equation).
x=−12x = -\frac{1}{2} and x=4x = 4
The roots divide the number line into test intervals.
3
Determine the solution set of the inequality.
−12≤x≤4-\frac{1}{2} \le x \le 4
Since the quadratic coefficient is positive, the quadratic expression is non-positive between its roots.
4
List and count the integers within the range.
The integers are 0,1,2,3,40, 1, 2, 3, 4, making a total of 55 integers.
Counting only whole numbers in the closed interval [−0.5,4][ -0.5, 4 ].

Anahtar Kavram

Quadratic Inequalities and Integer Solution Counting
Soru 227Soru →

Given that (x+2)(x + 2) is a factor of the polynomial P(x)=2x3−x2+ax+bP(x) = 2x^3 - x^2 + ax + b, and that dividing P(x)P(x) by (2x−1)(2x - 1) leaves a remainder of 1515, what is the value of a+ba + b?

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Cevap: 1414

Cevap

The value of a+ba + b is 1414.
Applying the Factor Theorem with x=−2x = -2 gives the equation −2a+b=20-2a + b = 20. Applying the Remainder Theorem with x=12x = \frac{1}{2} gives the equation a+2b=30a + 2b = 30. Solving these simultaneous equations yields a=−2a = -2 and b=16b = 16. Adding aa and bb gives 1414.

Adım Adım Çözüm

1
Apply the Factor Theorem for (x+2)(x + 2)
−2a+b=20-2a + b = 20
Since (x+2)(x + 2) is a factor of P(x)P(x), P(−2)=0P(-2) = 0. Substituting x=−2x = -2 yields 2(−2)3−(−2)2+a(−2)+b=02(-2)^3 - (-2)^2 + a(-2) + b = 0, which simplifies to −16−4−2a+b=0-16 - 4 - 2a + b = 0 or −2a+b=20-2a + b = 20.
2
Apply the Remainder Theorem for (2x−1)(2x - 1)
a+2b=30a + 2b = 30
Dividing P(x)P(x) by (2x−1)(2x - 1) leaves a remainder of 1515, so P(12)=15P\left(\frac{1}{2}\right) = 15. Substituting x=12x = \frac{1}{2} yields 2(18)−14+a2+b=152\left(\frac{1}{8}\right) - \frac{1}{4} + \frac{a}{2} + b = 15, which simplifies to a2+b=15\frac{a}{2} + b = 15 or a+2b=30a + 2b = 30.
3
Solve the system of linear equations for aa and bb
a=−2a = -2 and b=16b = 16
From step 1, b=2a+20b = 2a + 20. Substituting this into step 2 gives a+2(2a+20)=30  ⟹  5a+40=30  ⟹  5a=−10  ⟹  a=−2a + 2(2a + 20) = 30 \implies 5a + 40 = 30 \implies 5a = -10 \implies a = -2. Substituting a=−2a = -2 into b=2a+20b = 2a + 20 gives b=16b = 16.
4
Calculate a+ba + b
1414
Summing the calculated values gives a+b=−2+16=14a + b = -2 + 16 = 14.

Anahtar Kavram

Factor Theorem and Remainder Theorem for Polynomials
Soru 228Soru →

A binary operation ⋆\star defined on the set of real numbers R\mathbb{R} is given by x⋆y=x+yx−y+1x \star y = \frac{x + y}{x - y + 1} for x−y≠−1x - y \neq -1. If 5⋆p=35 \star p = 3, what is the value of pp?

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Cevap: 3.25

Cevap

The value of pp is 3.253.25 (or 134\frac{13}{4}).
Applying the operation rule x⋆y=x+yx−y+1x \star y = \frac{x + y}{x - y + 1} with x=5x = 5 and y=py = p gives 5+p6−p=3\frac{5 + p}{6 - p} = 3. Cross-multiplying gives 5+p=18−3p5 + p = 18 - 3p, which simplifies to 4p=134p = 13 or p=3.25p = 3.25.

Adım Adım Çözüm

1
Substitute x=5x = 5 and y=py = p into the given binary operation definition
5+p5−p+1=3\frac{5 + p}{5 - p + 1} = 3
This sets up the equation for the given condition 5⋆p=35 \star p = 3.
2
Simplify the denominator in the algebraic fraction
5+p6−p=3\frac{5 + p}{6 - p} = 3
Combining the constants 5+1=65 + 1 = 6 simplifies the denominator.
3
Multiply both sides by (6−p)(6 - p) and expand
5+p=18−3p5 + p = 18 - 3p
Eliminating the denominator allows linear terms in pp to be collected.
4
Rearrange terms to solve for pp
4p=13  ⟹  p=3.254p = 13 \implies p = 3.25
Adding 3p3p to both sides and subtracting 55 gives 4p=134p = 13.

Anahtar Kavram

Solving linear equations derived from non-commutative binary operations
Soru 229Soru →

Given the matrices A=(2103)A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} and B=(142−1)B = \begin{pmatrix} 1 & 4 \\ 2 & -1 \end{pmatrix}, what is the product matrix ABAB?

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Cevap: (476−3)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}

Cevap

The matrix product ABAB is (476−3)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}.
To find ABAB, each element cijc_{ij} is formed by taking the dot product of row ii of matrix AA and column jj of matrix BB. Performing these steps gives c11=2(1)+1(2)=4c_{11} = 2(1) + 1(2) = 4, c12=2(4)+1(−1)=7c_{12} = 2(4) + 1(-1) = 7, c21=0(1)+3(2)=6c_{21} = 0(1) + 3(2) = 6, and c22=0(4)+3(−1)=−3c_{22} = 0(4) + 3(-1) = -3, yielding (476−3)\begin{pmatrix} 4 & 7 \\ 6 & -3 \end{pmatrix}.

Adım Adım Çözüm

1
Multiply the first row of AA by the first column of BB to find element (1,1)(1,1).
2(1)+1(2)=2+2=42(1) + 1(2) = 2 + 2 = 4
Matrix multiplication rule requires taking the dot product of rows from the first matrix and columns from the second matrix.
2
Multiply the first row of AA by the second column of BB to find element (1,2)(1,2).
2(4)+1(−1)=8−1=72(4) + 1(-1) = 8 - 1 = 7
Evaluates the element in row 1, column 2 of the resulting matrix.
3
Multiply the second row of AA by the first column of BB to find element (2,1)(2,1).
0(1)+3(2)=0+6=60(1) + 3(2) = 0 + 6 = 6
Evaluates the element in row 2, column 1 of the resulting matrix.
4
Multiply the second row of AA by the second column of BB to find element (2,2)(2,2).
0(4)+3(−1)=0−3=−30(4) + 3(-1) = 0 - 3 = -3
Evaluates the element in row 2, column 2 of the resulting matrix.

Anahtar Kavram

Matrix Multiplication
Tahmini Süre:1m 0s
Soru 230Soru →

A binary operation Δ\Delta defined on the set of real numbers R\mathbb{R} is given by aΔb=a+b−ab4a \Delta b = a + b - \frac{ab}{4}. What is the inverse of 22 under this operation?

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Cevap: −4-4

Cevap

The inverse of 22 under the binary operation is −4-4.
To find the inverse of 22, we first determine the identity element ee using aΔe=aa \Delta e = a, which yields e=0e = 0. Setting 2Δx=02 \Delta x = 0 gives 2+x−2x4=02 + x - \frac{2x}{4} = 0, which simplifies to 2+x2=02 + \frac{x}{2} = 0, resulting in x=−4x = -4.

Adım Adım Çözüm

1
Determine the identity element ee of the operation.
e=0e = 0
By definition, aΔe=aa \Delta e = a. Thus, a+e−ae4=a  ⟹  e(1−a4)=0a + e - \frac{ae}{4} = a \implies e\left(1 - \frac{a}{4}\right) = 0, which gives e=0e = 0 for all real aa.
2
Set up the inverse equation for the element 22.
2+x−2x4=02 + x - \frac{2x}{4} = 0
Let xx be the inverse of 22. By definition, 2Δx=e2 \Delta x = e, where e=0e = 0.
3
Simplify and solve for xx.
x=−4x = -4
Simplify 2+x−x2=0  ⟹  2+x2=0  ⟹  x2=−2  ⟹  x=−42 + x - \frac{x}{2} = 0 \implies 2 + \frac{x}{2} = 0 \implies \frac{x}{2} = -2 \implies x = -4.

Anahtar Kavram

Inverse Element in Binary Operations
Soru 231Soru →

An arithmetic progression (A.P.) and a geometric progression (G.P.) both have a first term of 22. The common difference of the A.P. is 44. If the 5th5^{\text{th}} term of the A.P. is equal to the 3rd3^{\text{rd}} term of the G.P., and the common ratio of the G.P. is positive, what is the 4th4^{\text{th}} term of the G.P.?

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Cevap: 54

Cevap

54
First, evaluate the 5th term of the A.P. using T5=2+(5−1)4=18T_5 = 2 + (5-1)4 = 18. Next, set the 3rd term of the G.P. equal to 18: 2r2=182 r^2 = 18, giving r2=9r^2 = 9 and r=3r = 3. Finally, find the 4th term of the G.P. using T4=2×33=54T_4 = 2 \times 3^3 = 54.

Adım Adım Çözüm

1
Calculate the 5th term of the A.P.
T5(A.P.)=a+(5−1)d=2+4(4)=18T_5^{(A.P.)} = a + (5-1)d = 2 + 4(4) = 18
The formula for the nthn^{\text{th}} term of an A.P. is Tn=a+(n−1)dT_n = a + (n-1)d.
2
Find the common ratio rr of the G.P. by equating the 3rd term of the G.P. to 18
T3(G.P.)=ar3−1=2r2=18  ⟹  r2=9  ⟹  r=3T_3^{(G.P.)} = a r^{3-1} = 2 r^2 = 18 \implies r^2 = 9 \implies r = 3 (since r>0r > 0)
The formula for the nthn^{\text{th}} term of a G.P. is Tn=arn−1T_n = a r^{n-1}.
3
Compute the 4th term of the G.P.
T4(G.P.)=ar4−1=2×33=2×27=54T_4^{(G.P.)} = a r^{4-1} = 2 \times 3^3 = 2 \times 27 = 54
Substitute a=2a = 2, r=3r = 3, and n=4n = 4 into Tn=arn−1T_n = a r^{n-1}.

Anahtar Kavram

Connecting terms of Arithmetic and Geometric Progressions using their nthn^{\text{th}} term formulas
Soru 232Soru →

If the determinant of the matrix A=(k123−10241)A = \begin{pmatrix} k & 1 & 2 \\ 3 & -1 & 0 \\ 2 & 4 & 1 \end{pmatrix} is equal to 1717, what is the value of kk?

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Cevap: 8

Cevap

The value of kk is 88.
Expanding the matrix determinant along the first row gives det⁡(A)=k(−1)−1(3)+2(14)=−k+25\det(A) = k(-1) - 1(3) + 2(14) = -k + 25. Setting −k+25=17-k + 25 = 17 leads to −k=−8-k = -8, so k=8k = 8.

Adım Adım Çözüm

1
Perform cofactor expansion along the first row of matrix AA.
\det(A) = k((-1)(1) - (0)(4)) - 1((3)(1) - (0)(2)) + 2((3)(4) - (-1)(2))
Expanding along the first row uses the formula \det(A) = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13}.
2
Evaluate the products and simplify the algebraic expression for the determinant.
\det(A) = -k - 3 + 28 = -k + 25
Simplifying each sub-determinant term yields a linear expression in kk.
3
Equate the expression to 1717 and solve for kk.
-k + 25 = 17 \implies k = 8
Subtracting 25 from both sides gives −k=−8-k = -8, which simplifies to k=8k = 8.

Anahtar Kavram

Determinant of a 3x3 Matrix
Soru 233Soru →

In an Arithmetic Progression (A.P.), the sum of the first 1010 terms is 120120 and the sum of the next 1010 terms is 320320. What is the common difference of the progression?

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Cevap: 22

Cevap

The common difference of the progression is 22.
The sum of an A.P. is Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d]. For the first 10 terms, S10=5(2a+9d)=120S_{10} = 5(2a + 9d) = 120, simplifying to 2a+9d=242a + 9d = 24. The sum of the first 20 terms is 120+320=440120 + 320 = 440, so S20=10(2a+19d)=440S_{20} = 10(2a + 19d) = 440, simplifying to 2a+19d=442a + 19d = 44. Subtracting the two equations gives 10d=2010d = 20, which yields the common difference d=2d = 2.

Adım Adım Çözüm

1
Express the sum of the first 10 terms using the formula Sn=n2[2a+(n−1)d]S_n = \frac{n}{2}[2a + (n-1)d].
S10=102[2a+9d]=5(2a+9d)=120  ⟹  2a+9d=24S_{10} = \frac{10}{2}[2a + 9d] = 5(2a + 9d) = 120 \implies 2a + 9d = 24.
Relating the given sum of the first 10 terms to the first term aa and common difference dd generates the first linear equation.
2
Determine the sum of the first 20 terms (S20S_{20}) and set up the second linear equation.
S20=S10+sum of next 10 terms=120+320=440S_{20} = S_{10} + \text{sum of next 10 terms} = 120 + 320 = 440. Thus, S20=202[2a+19d]=10(2a+19d)=440  ⟹  2a+19d=44S_{20} = \frac{20}{2}[2a + 19d] = 10(2a + 19d) = 440 \implies 2a + 19d = 44.
The sum of the next 10 terms added to the sum of the first 10 terms gives the total sum of the first 20 terms.
3
Solve the system of simultaneous linear equations for dd.
(2a+19d)−(2a+9d)=44−24  ⟹  10d=20  ⟹  d=2(2a + 19d) - (2a + 9d) = 44 - 24 \implies 10d = 20 \implies d = 2.
Subtracting equation (1) from equation (2) eliminates aa to solve directly for the common difference dd.

Anahtar Kavram

Sum of an Arithmetic Progression and Simultaneous Equations
Soru 234Soru →

If (x,y)(x, y) is a pair of positive integers satisfying the simultaneous equations x+2y=7x + 2y = 7 and x2+y2=13x^2 + y^2 = 13, what is the value of 2x−y2x - y?

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Cevap: 4

Cevap

The value of 2x−y2x - y is 4.
Substituting x=7−2yx = 7 - 2y into x2+y2=13x^2 + y^2 = 13 yields the quadratic 5y2−28y+36=05y^2 - 28y + 36 = 0, which factors as (5y−18)(y−2)=0(5y - 18)(y - 2) = 0. Since xx and yy must both be positive integers, the valid solution pair is (x,y)=(3,2)(x, y) = (3, 2). Evaluating 2x−y2x - y gives 2(3)−2=42(3) - 2 = 4.

Adım Adım Çözüm

1
Express xx in terms of yy using the linear equation x+2y=7x + 2y = 7.
x=7−2yx = 7 - 2y
Isolating xx allows substitution into the quadratic equation.
2
Substitute x=7−2yx = 7 - 2y into x2+y2=13x^2 + y^2 = 13 and expand.
(7−2y)2+y2=13  ⟹  49−28y+4y2+y2=13(7 - 2y)^2 + y^2 = 13 \implies 49 - 28y + 4y^2 + y^2 = 13
This reduces the system to a single quadratic equation in terms of yy.
3
Rearrange and solve the resulting quadratic equation 5y2−28y+36=05y^2 - 28y + 36 = 0.
(5y−18)(y−2)=0  ⟹  y=2(5y - 18)(y - 2) = 0 \implies y = 2 or y=3.6y = 3.6
Factoring provides the possible values for yy.
4
Identify the integer solution pair and compute the target expression 2x−y2x - y.
Since yy must be a positive integer, y=2y = 2 and x=3x = 3. Therefore, 2x−y=2(3)−2=42x - y = 2(3) - 2 = 4.
The non-integer solution is rejected based on the positive integer constraint.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations
Tahmini Süre:1m 30s
Soru 235Soru →

A binary operation ∗\ast is defined on the set of real numbers R\mathbb{R} by a∗b=a+b−4a \ast b = a + b - 4. If x−1x^{-1} represents the inverse of an element xx under the operation ∗\ast, what is the value of xx such that x∗3=10−1x \ast 3 = 10^{-1}?

Cevabı ve açıklamayı göster

Cevap: −1-1

Cevap

The value of xx is −1-1.
First, the identity element ee is determined using a∗e=aa \ast e = a, which gives a+e−4=a  ⟹  e=4a + e - 4 = a \implies e = 4. Next, the inverse 10−110^{-1} is calculated from 10∗10−1=410 \ast 10^{-1} = 4, yielding 10+10−1−4=4  ⟹  10−1=−210 + 10^{-1} - 4 = 4 \implies 10^{-1} = -2. Finally, setting x∗3=−2x \ast 3 = -2 gives x+3−4=−2  ⟹  x−1=−2  ⟹  x=−1x + 3 - 4 = -2 \implies x - 1 = -2 \implies x = -1.

Adım Adım Çözüm

1
Find the identity element ee under the operation ∗\ast.
e=4e = 4
By definition of an identity element, a∗e=a  ⟹  a+e−4=a  ⟹  e=4a \ast e = a \implies a + e - 4 = a \implies e = 4.
2
Determine the inverse of 1010, denoted 10−110^{-1}, under ∗\ast.
10−1=−210^{-1} = -2
By definition of an inverse element, 10∗10−1=e  ⟹  10+10−1−4=4  ⟹  6+10−1=4  ⟹  10−1=−210 \ast 10^{-1} = e \implies 10 + 10^{-1} - 4 = 4 \implies 6 + 10^{-1} = 4 \implies 10^{-1} = -2.
3
Substitute 10−1=−210^{-1} = -2 into the given equation x∗3=10−1x \ast 3 = 10^{-1} and solve for xx.
x=−1x = -1
Using the definition of the operation, x∗3=x+3−4=x−1x \ast 3 = x + 3 - 4 = x - 1. Setting x−1=−2x - 1 = -2 yields x=−1x = -1.

Anahtar Kavram

Identity and Inverse Elements in Binary Operations
Soru 236Soru →

A 2×22 \times 2 matrix is given by A=(x+234x−2)A = \begin{pmatrix} x + 2 & 3 \\ 4 & x - 2 \end{pmatrix}. If AA is a singular matrix, what is the positive value of xx?

Cevabı ve açıklamayı göster

Cevap: 4

Cevap

The positive value of xx is 44.
A matrix is singular when its determinant is zero. Expanding det⁡(A)=(x+2)(x−2)−12=0\det(A) = (x+2)(x-2) - 12 = 0 yields x2−16=0x^2 - 16 = 0. Solving x2=16x^2 = 16 gives x=±4x = \pm 4. The positive value is 44.

Adım Adım Çözüm

1
Apply the condition for a matrix to be singular
det⁡(A)=0\det(A) = 0
By definition, a matrix is singular if and only if its determinant is zero.
2
Formulate the determinant equation for matrix AA
(x+2)(x−2)−(3)(4)=0(x + 2)(x - 2) - (3)(4) = 0
The determinant of a 2×22 \times 2 matrix (abcd)\begin{pmatrix} a & b \\ c & d \end{pmatrix} is calculated as ad−bcad - bc.
3
Expand and simplify the algebraic equation
x^2 - 4 - 12 = 0 \implies x^2 - 16 = 0
Expanding (x+2)(x−2)(x+2)(x-2) gives x2−4x^2 - 4, and subtracting 1212 yields x2−16x^2 - 16.
4
Solve for xx and select the positive root
x^2 = 16 \implies x = 4 \text{ or } x = -4 \implies x = 4
Taking the square root gives two roots, and the problem explicitly specifies the positive value.

Anahtar Kavram

Singular Matrices and Determinants
Tahmini Süre:1m 15s
Soru 237Soru →

A binary operation ∘\circ defined on the set of real numbers R\mathbb{R} is given by a∘b=3a+2b−1a \circ b = 3a + 2b - 1. What is the value of xx for which (2∘x)∘3=26(2 \circ x) \circ 3 = 26?

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

The value of xx for which (2∘x)∘3=26(2 \circ x) \circ 3 = 26 is 11.
Applying the binary operation rule a∘b=3a+2b−1a \circ b = 3a + 2b - 1 to the inner expression gives 2∘x=2x+52 \circ x = 2x + 5. Then applying the rule to (2x+5)∘3(2x + 5) \circ 3 gives 3(2x+5)+2(3)−1=6x+203(2x + 5) + 2(3) - 1 = 6x + 20. Equating 6x+20=266x + 20 = 26 yields 6x=66x = 6, giving x=1x = 1.

Adım Adım Çözüm

1
Evaluate the inner binary operation expression 2∘x2 \circ x using the rule a∘b=3a+2b−1a \circ b = 3a + 2b - 1.
2∘x=3(2)+2(x)−1=6+2x−1=2x+52 \circ x = 3(2) + 2(x) - 1 = 6 + 2x - 1 = 2x + 5
The definition of the binary operation replaces aa with 22 and bb with xx.
2
Substitute the result (2x+5)(2x + 5) into the outer binary operation expression (2∘x)∘3(2 \circ x) \circ 3.
(2x+5)∘3=3(2x+5)+2(3)−1=6x+15+6−1=6x+20(2x + 5) \circ 3 = 3(2x + 5) + 2(3) - 1 = 6x + 15 + 6 - 1 = 6x + 20
Apply the rule a∘b=3a+2b−1a \circ b = 3a + 2b - 1 where a=2x+5a = 2x + 5 and b=3b = 3.
3
Equate the simplified expression to 2626 and solve for xx.
6x+20=26  ⟹  6x=6  ⟹  x=16x + 20 = 26 \implies 6x = 6 \implies x = 1
Setting the calculated expression equal to the given value allows isolation of xx.

Anahtar Kavram

Evaluation of non-commutative composite binary operations
Soru 238Soru →

What is the sum of the yy-values of the solution pairs that satisfy the simultaneous equations y−x=2y - x = 2 and y=x2−3x+5y = x^2 - 3x + 5?

Cevabı ve açıklamayı göster

Cevap: 88

Cevap

The sum of the yy-values is 8.
Rearranging the linear equation gives y=x+2y = x + 2. Equating this to the quadratic expression gives x+2=x2−3x+5x + 2 = x^2 - 3x + 5, which simplifies to x2−4x+3=0x^2 - 4x + 3 = 0. Solving for xx yields x=1x = 1 and x=3x = 3. Substituting these into y=x+2y = x + 2 gives y=3y = 3 and y=5y = 5. The sum of these yy-values is 3+5=83 + 5 = 8.

Adım Adım Çözüm

1
Express yy in terms of xx from the linear equation.
y=x+2y = x + 2
Isolating yy facilitates substitution into the quadratic equation.
2
Substitute y=x+2y = x + 2 into the quadratic equation and rearrange into standard form.
x+2=x2−3x+5  ⟹  x2−4x+3=0x + 2 = x^2 - 3x + 5 \implies x^2 - 4x + 3 = 0
Setting the two expressions for yy equal to each other forms a single quadratic equation in xx.
3
Solve the quadratic equation for xx.
(x−1)(x−3)=0  ⟹  x1=1,x2=3(x - 1)(x - 3) = 0 \implies x_1 = 1, x_2 = 3
Factoring gives the xx-coordinates of the intersection points.
4
Calculate the corresponding yy-values using y=x+2y = x + 2.
y1=1+2=3y_1 = 1 + 2 = 3 and y2=3+2=5y_2 = 3 + 2 = 5
Each xx-value must be substituted into the linear equation to find its paired yy-value.
5
Find the sum of the yy-values.
y1+y2=3+5=8y_1 + y_2 = 3 + 5 = 8
The question asks for the sum of the yy-coordinates.

Anahtar Kavram

Simultaneous Linear and Quadratic Equations

Alternatif Yöntem

Using Vieta's formulas: once the quadratic equation for xx is found as x2−4x+3=0x^2 - 4x + 3 = 0, we know x1+x2=4x_1 + x_2 = 4. Since y1=x1+2y_1 = x_1 + 2 and y2=x2+2y_2 = x_2 + 2, the sum of the yy-values is y1+y2=(x1+2)+(x2+2)=(x1+x2)+4=4+4=8y_1 + y_2 = (x_1 + 2) + (x_2 + 2) = (x_1 + x_2) + 4 = 4 + 4 = 8.
Tahmini Süre:1m 30s
Soru 239Soru →

A binary operation ⋆\star defined on the set of real numbers R∖{12}\mathbb{R} \setminus \left\{\frac{1}{2}\right\} is given by a⋆b=a+b−2aba \star b = a + b - 2ab. If x−1x^{-1} represents the inverse of an element xx under ⋆\star, what is the value of xx for which x−1=3x^{-1} = 3?

Cevabı ve açıklamayı göster

Cevap: 35\frac{3}{5}

Cevap

The value of xx is 35\frac{3}{5}.
First find the identity element ee using a⋆e=aa \star e = a, which gives a+e−2ae=a  ⟹  e(1−2a)=0  ⟹  e=0a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0. Next, use the inverse definition x⋆x−1=e  ⟹  x+x−1−2xx−1=0x \star x^{-1} = e \implies x + x^{-1} - 2x x^{-1} = 0. Substituting x−1=3x^{-1} = 3 yields x+3−2x(3)=0  ⟹  −5x+3=0  ⟹  x=35x + 3 - 2x(3) = 0 \implies -5x + 3 = 0 \implies x = \frac{3}{5}.

Adım Adım Çözüm

1
Find the identity element ee under the operation ⋆\star.
e=0e = 0
By definition, a⋆e=a  ⟹  a+e−2ae=a  ⟹  e(1−2a)=0  ⟹  e=0a \star e = a \implies a + e - 2ae = a \implies e(1 - 2a) = 0 \implies e = 0 for all a≠12a \neq \frac{1}{2}.
2
Set up the inverse equation for element xx.
x⋆x−1=0  ⟹  x+x−1−2xx−1=0x \star x^{-1} = 0 \implies x + x^{-1} - 2x x^{-1} = 0
An element operating on its inverse yields the identity element e=0e = 0.
3
Substitute x−1=3x^{-1} = 3 into the inverse equation and solve for xx.
x+3−6x=0  ⟹  −5x+3=0  ⟹  5x=3  ⟹  x=35x + 3 - 6x = 0 \implies -5x + 3 = 0 \implies 5x = 3 \implies x = \frac{3}{5}
Substitute the given inverse value and perform simple linear algebraic rearrangement.

Anahtar Kavram

Identity and Inverse Elements in Binary Operations
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Algebra Alıştırma Soruları — JAMB UTME — Sayfa 12 | Examkin