Function Definitions, Evaluation, and Custom Operators

25 soru

Soru 1Soru →

For all real numbers xx and yy, the custom operation ⊙\odot is defined by x⊙y=x2−2yx \odot y = x^2 - 2y. What is the value of (3⊙2)⊙4(3 \odot 2) \odot 4?

Cevabı ve açıklamayı göster

Cevap: 17

Cevap

17
Following the explicit operator rule x⊙y=x2−2yx \odot y = x^2 - 2y, first calculate the inner expression (3⊙2)=32−2(2)=5(3 \odot 2) = 3^2 - 2(2) = 5. Next, substitute 5 into the outer expression to compute 5⊙4=52−2(4)=25−8=175 \odot 4 = 5^2 - 2(4) = 25 - 8 = 17. Thus, the value 17 is correct.

Adım Adım Çözüm

1
Evaluate the inner function expression (3⊙2)(3 \odot 2) using the rule x⊙y=x2−2yx \odot y = x^2 - 2y
3⊙2=32−2(2)=9−4=53 \odot 2 = 3^2 - 2(2) = 9 - 4 = 5
Parentheses dictate that inner custom operations must be evaluated first.
2
Substitute the result of step 1 back into the expression to evaluate 5⊙45 \odot 4
5⊙4=52−2(4)=25−8=175 \odot 4 = 5^2 - 2(4) = 25 - 8 = 17
Applying the custom operator rule with x=5x = 5 and y=4y = 4 computes the final target value.

Anahtar Kavram

Custom Operator Evaluation and Order of Operations
Tahmini Süre:1m 15s
Soru 2Soru →
For all real numbers xx except 00 and 11, the function ff satisfies the functional equation
f(x)+2f(11−x)=9x+3f(x) + 2f\left(\frac{1}{1-x}\right) = 9x + 3
What is the value of f(2)f(2)?
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Cevap: 7

Cevap

The value of f(2)f(2) is 77.
Evaluating the functional equation iteratively at the points of the cycle x=2x = 2, x=−1x = -1, and x=12x = \frac{1}{2} produces a system of three linear equations in terms of f(2)f(2), f(−1)f(-1), and f(12)f\left(\frac{1}{2}\right). Solving this linear system gives f(2)=7f(2) = 7.

Adım Adım Çözüm

1
Evaluate the functional equation at x=2x = 2
f(2)+2f(11−2)=9(2)+3  ⟹  f(2)+2f(−1)=21f(2) + 2f\left(\frac{1}{1-2}\right) = 9(2) + 3 \implies f(2) + 2f(-1) = 21
This establishes a relationship between f(2)f(2) and f(−1)f(-1).
2
Evaluate the functional equation at x=−1x = -1
f(−1)+2f(11−(−1))=9(−1)+3  ⟹  f(−1)+2f(12)=−6f(-1) + 2f\left(\frac{1}{1-(-1)}\right) = 9(-1) + 3 \implies f(-1) + 2f\left(\frac{1}{2}\right) = -6
This links f(−1)f(-1) to f(12)f\left(\frac{1}{2}\right) to build a closed system of transformations under the mapping g(x)=11−xg(x) = \frac{1}{1-x}.
3
Evaluate the functional equation at x=12x = \frac{1}{2}
f(12)+2f(11−1/2)=9(12)+3  ⟹  f(12)+2f(2)=152f\left(\frac{1}{2}\right) + 2f\left(\frac{1}{1-1/2}\right) = 9\left(\frac{1}{2}\right) + 3 \implies f\left(\frac{1}{2}\right) + 2f(2) = \frac{15}{2}
This completes the cycle since g(12)=2g\left(\frac{1}{2}\right) = 2, creating a system of three linear equations with three unknown values.
4
Solve the system of three linear equations for f(2)f(2)
Let a=f(2)a = f(2), b=f(−1)b = f(-1), and c=f(12)c = f\left(\frac{1}{2}\right). From equation (3), c=152−2ac = \frac{15}{2} - 2a. Substituting cc into equation (2) gives b+2(152−2a)=−6  ⟹  b=4a−21b + 2\left(\frac{15}{2} - 2a\right) = -6 \implies b = 4a - 21. Substituting bb into equation (1) gives a+2(4a−21)=21  ⟹  9a−42=21  ⟹  9a=63  ⟹  a=7a + 2(4a - 21) = 21 \implies 9a - 42 = 21 \implies 9a = 63 \implies a = 7.
Algebraic substitution yields the exact value of f(2)f(2).

Anahtar Kavram

Cyclic Functional Equations and System Substitution
Tahmini Süre:2m 30s
Soru 3Soru →

For all real numbers aa and bb, the custom operation ⋆\star is defined by a⋆b=(a+b)2−3ba \star b = (a + b)^2 - 3b. What is the value of 4⋆24 \star 2?

Cevabı ve açıklamayı göster

Cevap: 30

Cevap

30
Substituting a=4a = 4 and b=2b = 2 into the definition (a+b)2−3b(a + b)^2 - 3b gives (4+2)2−3(2)=62−6=36−6=30(4 + 2)^2 - 3(2) = 6^2 - 6 = 36 - 6 = 30.

Adım Adım Çözüm

1
Identify the values assigned to the custom operator variables
a=4a = 4 and b=2b = 2
The expression 4⋆24 \star 2 corresponds to evaluating a⋆ba \star b where aa is the first operand and bb is the second operand.
2
Substitute a=4a = 4 and b=2b = 2 into the operator definition (a+b)2−3b(a + b)^2 - 3b
(4+2)2−3(2)(4 + 2)^2 - 3(2)
Direct substitution into the given algebraic definition.
3
Perform the operations in parentheses and exponentiation
62−6=36−66^2 - 6 = 36 - 6
First add inside the parentheses (4+2=64 + 2 = 6), then square the sum (62=366^2 = 36), and multiply 3×2=63 \times 2 = 6.
4
Subtract to find the final value
30
Subtracting 66 from 3636 yields 3030.

Anahtar Kavram

Custom Operator Evaluation
Tahmini Süre:45s
Soru 4Soru →

For all non-zero real numbers aa and bb, the custom operator Δ\Delta is defined by aΔb=a2−b2aba \Delta b = \frac{a^2 - b^2}{ab}. A function f(x)f(x) is defined for all x≠0x \neq 0 and x≠−1x \neq -1 by f(x)=xΔ(x+1)f(x) = x \Delta (x + 1). If kk is a positive constant such that f(k)=−712f(k) = -\frac{7}{12}, what is the value of (k+2)Δk(k + 2) \Delta k?

Cevabı ve açıklamayı göster

Cevap: 1615\frac{16}{15}

Cevap

The value of (k+2)Δk(k + 2) \Delta k is 1615\frac{16}{15}.
Applying the custom operator to f(x)f(x) yields f(x)=x2−(x+1)2x(x+1)=−2x−1x2+xf(x) = \frac{x^2 - (x+1)^2}{x(x+1)} = \frac{-2x - 1}{x^2 + x}. Setting f(k)=−712f(k) = -\frac{7}{12} gives 2k+1k2+k=712\frac{2k + 1}{k^2 + k} = \frac{7}{12}, which simplifies to 7k2−17k−12=07k^2 - 17k - 12 = 0. Factoring (7k+4)(k−3)=0(7k + 4)(k - 3) = 0 and selecting the positive root gives k=3k = 3. Evaluating (3+2)Δ3=5Δ3=52−325⋅3=1615(3+2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \cdot 3} = \frac{16}{15}.

Adım Adım Çözüm

1
Express f(x)f(x) using the definition of the custom operator Δ\Delta.
f(x)=xΔ(x+1)=x2−(x+1)2x(x+1)=x2−(x2+2x+1)x(x+1)=−2x−1x2+xf(x) = x \Delta (x + 1) = \frac{x^2 - (x + 1)^2}{x(x + 1)} = \frac{x^2 - (x^2 + 2x + 1)}{x(x + 1)} = \frac{-2x - 1}{x^2 + x}
Substitute a=xa = x and b=x+1b = x + 1 into aΔb=a2−b2aba \Delta b = \frac{a^2 - b^2}{ab}.
2
Set f(k)=−712f(k) = -\frac{7}{12} and solve for the positive constant kk.
−2k−1k2+k=−712  ⟹  2k+1k2+k=712  ⟹  12(2k+1)=7(k2+k)  ⟹  7k2−17k−12=0\frac{-2k - 1}{k^2 + k} = -\frac{7}{12} \implies \frac{2k + 1}{k^2 + k} = \frac{7}{12} \implies 12(2k + 1) = 7(k^2 + k) \implies 7k^2 - 17k - 12 = 0
Cross-multiply and rearrange into a standard quadratic equation.
3
Factor the quadratic equation to find the positive root kk.
(7k+4)(k−3)=0  ⟹  k=−47(7k + 4)(k - 3) = 0 \implies k = -\frac{4}{7} or k=3k = 3. Since k>0k > 0, k=3k = 3.
The question specifies that kk is a positive constant.
4
Evaluate (k+2)Δk(k + 2) \Delta k for k=3k = 3.
(3+2)Δ3=5Δ3=52−325×3=25−915=1615(3 + 2) \Delta 3 = 5 \Delta 3 = \frac{5^2 - 3^2}{5 \times 3} = \frac{25 - 9}{15} = \frac{16}{15}
Substitute a=5a = 5 and b=3b = 3 into the custom operator definition.

Anahtar Kavram

Function definition substitution, non-commutative custom operators, and quadratic solving.
Tahmini Süre:2m 30s
Soru 5Soru →

For all real numbers xx and yy, the custom operator Δ\Delta is defined by xΔy=x2−2yx \Delta y = x^2 - 2y. If kk is a real number such that (3Δk)Δ4=17(3 \Delta k) \Delta 4 = 17, what is the sum of all possible values of kk?

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Cevap: 9

Cevap

The sum of all possible values of kk is 9.
Applying the custom operation rule xΔy=x2−2yx \Delta y = x^2 - 2y sequentially, the inner evaluation gives 3Δk=9−2k3 \Delta k = 9 - 2k. Applying the operation again to (9−2k)Δ4(9 - 2k) \Delta 4 gives (9−2k)2−8=17(9 - 2k)^2 - 8 = 17, which simplifies to (9−2k)2=25(9 - 2k)^2 = 25. Taking the square root gives two valid solutions for 9−2k9 - 2k: 55 and −5-5. Solving 9−2k=59 - 2k = 5 yields k=2k = 2, and solving 9−2k=−59 - 2k = -5 yields k=7k = 7. The sum of these two values is 2+7=92 + 7 = 9.

Adım Adım Çözüm

1
Evaluate the inner custom operation 3Δk3 \Delta k.
3Δk=32−2k=9−2k3 \Delta k = 3^2 - 2k = 9 - 2k
Apply the definition xΔy=x2−2yx \Delta y = x^2 - 2y with x=3x = 3 and y=ky = k.
2
Substitute the inner result into the outer expression (9−2k)Δ4=17(9 - 2k) \Delta 4 = 17.
(9−2k)2−2(4)=17  ⟹  (9−2k)2−8=17(9 - 2k)^2 - 2(4) = 17 \implies (9 - 2k)^2 - 8 = 17
Apply the definition xΔy=x2−2yx \Delta y = x^2 - 2y with x=9−2kx = 9 - 2k and y=4y = 4.
3
Isolate the squared term and solve for 9−2k9 - 2k.
(9−2k)2=25  ⟹  9−2k=5(9 - 2k)^2 = 25 \implies 9 - 2k = 5 or 9−2k=−59 - 2k = -5
Adding 8 to both sides gives 25; taking the square root requires considering both positive and negative roots.
4
Solve each linear equation for kk and calculate their sum.
Case 1: 9−2k=5  ⟹  2k=4  ⟹  k=29 - 2k = 5 \implies 2k = 4 \implies k = 2.
Case 2: 9−2k=−5  ⟹  2k=14  ⟹  k=79 - 2k = -5 \implies 2k = 14 \implies k = 7.
Sum = 2+7=92 + 7 = 9.
Solving both equations yields all possible values for kk.

Anahtar Kavram

Custom Operator Evaluation and Quadratic Equation Root Extraction
Tahmini Süre:2m 0s
Soru 6Soru →

For all real numbers pp and qq, the operation ∇\nabla is defined by p∇q=2p+3q−pqp \nabla q = 2p + 3q - pq. What is the value of 3∇43 \nabla 4?

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Cevap: 6

Cevap

The value of 3∇43 \nabla 4 is 6.
Substituting p=3p = 3 and q=4q = 4 directly into the expression 2p+3q−pq2p + 3q - pq gives 2(3)+3(4)−(3)(4)=6+12−12=62(3) + 3(4) - (3)(4) = 6 + 12 - 12 = 6.

Adım Adım Çözüm

1
Identify the assigned values for the operator inputs.
p=3p = 3 and q=4q = 4.
The expression 3∇43 \nabla 4 matches p∇qp \nabla q with pp as the first argument and qq as the second.
2
Substitute p=3p = 3 and q=4q = 4 into the custom operation formula 2p+3q−pq2p + 3q - pq.
2(3)+3(4)−(3)(4)2(3) + 3(4) - (3)(4)
Direct evaluation of the defined function rule.
3
Perform the multiplication and algebraic addition/subtraction.
6+12−12=66 + 12 - 12 = 6
Simplifying numerical operations gives the final result.

Anahtar Kavram

Custom Operator Evaluation
Soru 7Soru →

For all real numbers xx and yy, the custom operation ♢\diamondsuit is defined by x♢y=3x−2y+1x \diamondsuit y = 3x - 2y + 1. What is the value of 5♢45 \diamondsuit 4?

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Cevap: 8

Cevap

The value of 5♢45 \diamondsuit 4 is 8.
Substituting x=5x = 5 and y=4y = 4 into the operation x♢y=3x−2y+1x \diamondsuit y = 3x - 2y + 1 yields 3(5)−2(4)+1=15−8+1=83(5) - 2(4) + 1 = 15 - 8 + 1 = 8.

Adım Adım Çözüm

1
Substitute the operands into the custom operator rule
5 \diamondsuit 4 = 3(5) - 2(4) + 1
The first term 55 corresponds to xx and the second term 44 corresponds to yy in the formula.
2
Simplify the numeric expression
15 - 8 + 1 = 8
Multiply first, then perform subtraction and addition from left to right.

Anahtar Kavram

Function Definitions, Evaluation, and Custom Operators
Soru 8Soru →

For all non-zero real numbers xx, the function ff is defined by f(x)=x2−42xf(x) = \frac{x^2 - 4}{2x}, and the function gg is defined by g(x)=3x−1g(x) = 3x - 1. What is the positive value of aa for which f(g(a))=0f(g(a)) = 0?

Cevabı ve açıklamayı göster

Cevap: 1

Cevap

The positive value of aa for which f(g(a))=0f(g(a)) = 0 is 1.
To solve f(g(a))=0f(g(a)) = 0, first find the values of yy where f(y)=0f(y) = 0. The numerator of f(y)=y2−42yf(y) = \frac{y^2 - 4}{2y} equals zero when y2−4=0y^2 - 4 = 0, yielding y=2y = 2 and y=−2y = -2. Setting g(a)=3a−1g(a) = 3a - 1 equal to these roots gives 3a−1=2  ⟹  a=13a - 1 = 2 \implies a = 1 and 3a−1=−2  ⟹  a=−1/33a - 1 = -2 \implies a = -1/3. Since aa must be positive, the correct value is 1.

Adım Adım Çözüm

1
Determine the values of the argument yy that satisfy f(y)=0f(y) = 0.
y=2y = 2 or y=−2y = -2.
A fraction equals zero when its numerator is zero and its denominator is non-zero. Setting y2−4=0y^2 - 4 = 0 gives y=±2y = \pm 2.
2
Substitute g(a)=3a−1g(a) = 3a - 1 into yy to solve for aa.
Solving 3a−1=23a - 1 = 2 yields a=1a = 1; solving 3a−1=−23a - 1 = -2 yields a=−1/3a = -1/3.
Setting the expression for g(a)g(a) equal to each root of f(y)=0f(y) = 0 identifies all potential values for aa.
3
Select the value of aa matching the positivity condition.
a=1a = 1.
The question explicitly specifies finding the positive value of aa.

Anahtar Kavram

Function composition and evaluation of nested functional equations.
Soru 9Soru →

For all non-zero real numbers aa and bb, the binary operator ⋄\diamond is defined by a⋄b=a−baa \diamond b = a - \frac{b}{a}. If kk is a positive integer such that (k⋄3)⋄3=12(k \diamond 3) \diamond 3 = \frac{1}{2}, what is the value of kk?

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Cevap: 3

Cevap

The positive integer value of kk is 3.
Applying the non-commutative custom operator definition a⋄b=a−baa \diamond b = a - \frac{b}{a} to the nested expression (k⋄3)⋄3=12(k \diamond 3) \diamond 3 = \frac{1}{2} requires working from the inside out. Letting u=k⋄3=k−3ku = k \diamond 3 = k - \frac{3}{k}, the outer equation becomes u⋄3=u−3u=12u \diamond 3 = u - \frac{3}{u} = \frac{1}{2}. Multiplying through by 2u2u gives 2u2−u−6=02u^2 - u - 6 = 0, which factors as (2u+3)(u−2)=0(2u + 3)(u - 2) = 0, yielding u=2u = 2 or u=−32u = -\frac{3}{2}. Substituting u=2u = 2 back into k−3k=2k - \frac{3}{k} = 2 produces k2−2k−3=0k^2 - 2k - 3 = 0, which factors as (k−3)(k+1)=0(k - 3)(k + 1) = 0. Since kk must be a positive integer, k=3k = 3. Testing u=−32u = -\frac{3}{2} gives 2k2+3k−6=02k^2 + 3k - 6 = 0, which has no integer solutions.

Adım Adım Çözüm

1
Substitute a temporary variable for the inner custom operator expression.
Let u=k⋄3=k−3ku = k \diamond 3 = k - \frac{3}{k}.
Simplifies the nested expression (k⋄3)⋄3=12(k \diamond 3) \diamond 3 = \frac{1}{2} to a single variable operation u⋄3=12u \diamond 3 = \frac{1}{2}.
2
Apply the operator definition to u⋄3u \diamond 3 and solve the resulting quadratic equation for uu.
u−3u=12  ⟹  2u2−u−6=0  ⟹  (2u+3)(u−2)=0u - \frac{3}{u} = \frac{1}{2} \implies 2u^2 - u - 6 = 0 \implies (2u + 3)(u - 2) = 0, so u=2u = 2 or u=−32u = -\frac{3}{2}.
Clearing the denominator uu transforms the custom operator equation into a standard quadratic equation.
3
Solve for positive integer kk using the obtained values of uu.
Setting k−3k=2k - \frac{3}{k} = 2 yields k2−2k−3=0  ⟹  (k−3)(k+1)=0k^2 - 2k - 3 = 0 \implies (k - 3)(k + 1) = 0. Since kk must be a positive integer, k=3k = 3. Setting k−3k=−32k - \frac{3}{k} = -\frac{3}{2} yields 2k2+3k−6=02k^2 + 3k - 6 = 0, which has non-integer discriminant 57\sqrt{57}.
Only u=2u = 2 yields an integer solution for kk, satisfying all problem constraints.

Anahtar Kavram

Custom Operator Evaluation and Nested Functional Equations
Soru 10Soru →

For all non-zero real numbers xx, the function ff satisfies the equation 2f(x)+f(1x)=6x2f(x) + f\left(\frac{1}{x}\right) = 6x. What is the value of f(2)f(2)?

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Cevap: 7

Cevap

The value of f(2)f(2) is 7.
Substituting x=2x = 2 into the functional equation gives 2f(2)+f(12)=122f(2) + f\left(\frac{1}{2}\right) = 12. Substituting x=12x = \frac{1}{2} gives 2f(12)+f(2)=32f\left(\frac{1}{2}\right) + f(2) = 3. Multiplying the first equation by 2 and subtracting the second equation eliminates f(12)f\left(\frac{1}{2}\right), producing 3f(2)=213f(2) = 21, so f(2)=7f(2) = 7.

Adım Adım Çözüm

1
Substitute x=2x = 2 into the functional equation
2f(2)+f(12)=122f(2) + f\left(\frac{1}{2}\right) = 12
This establishes a linear relation between f(2)f(2) and f(12)f\left(\frac{1}{2}\right).
2
Substitute x=12x = \frac{1}{2} into the functional equation
2f(12)+f(2)=32f\left(\frac{1}{2}\right) + f(2) = 3
This establishes a second linear relation involving the reciprocal input.
3
Eliminate f(12)f\left(\frac{1}{2}\right) by solving the system of equations
f(2)=7f(2) = 7
Multiplying the first equation by 2 yields 4f(2)+2f(12)=244f(2) + 2f\left(\frac{1}{2}\right) = 24. Subtracting the second equation gives 3f(2)=213f(2) = 21, which simplifies to f(2)=7f(2) = 7.

Anahtar Kavram

Evaluating functional equations by reciprocal variable substitution to form and solve a linear system.
Tahmini Süre:2m 0s
Soru 11Soru →

For all real numbers xx except 00 and 11, the function f(x)f(x) satisfies the relation f(x)+2f(11−x)=9xf(x) + 2f\left(\frac{1}{1-x}\right) = 9x. What is the value of f(2)f(2)?

Cevabı ve açıklamayı göster

Cevap: 6

Cevap

The value of f(2)f(2) is 66.
Evaluating the functional equation f(x)+2f(11−x)=9xf(x) + 2f\left(\frac{1}{1-x}\right) = 9x at the values x=2x = 2, x=−1x = -1, and x=12x = \frac{1}{2} creates a system of three linear equations in terms of f(2)f(2), f(−1)f(-1), and f(12)f\left(\frac{1}{2}\right). Solving this system yields 9f(2)=549f(2) = 54, which simplifies directly to f(2)=6f(2) = 6.

Adım Adım Çözüm

1
Evaluate the functional equation at x=2x = 2
f(2)+2f(11−2)=9(2)  ⟹  f(2)+2f(−1)=18f(2) + 2f\left(\frac{1}{1-2}\right) = 9(2) \implies f(2) + 2f(-1) = 18
Applying the input x=2x = 2 creates an equation connecting f(2)f(2) and f(−1)f(-1).
2
Evaluate the functional equation at x=−1x = -1
f(−1)+2f(11−(−1))=9(−1)  ⟹  f(−1)+2f(12)=−9f(-1) + 2f\left(\frac{1}{1-(-1)}\right) = 9(-1) \implies f(-1) + 2f\left(\frac{1}{2}\right) = -9
Evaluating at the new input x=−1x = -1 generates a second equation connecting f(−1)f(-1) and f(12)f\left(\frac{1}{2}\right).
3
Evaluate the functional equation at x=12x = \frac{1}{2}
f(12)+2f(11−1/2)=9(12)  ⟹  f(12)+2f(2)=4.5f\left(\frac{1}{2}\right) + 2f\left(\frac{1}{1-1/2}\right) = 9\left(\frac{1}{2}\right) \implies f\left(\frac{1}{2}\right) + 2f(2) = 4.5
Evaluating at x=12x = \frac{1}{2} completes the cyclic chain by linking back to f(2)f(2).
4
Solve the system of three linear equations for f(2)f(2)
From equation 3, f(12)=4.5−2f(2)f\left(\frac{1}{2}\right) = 4.5 - 2f(2). Substituting into equation 2 yields f(−1)=4f(2)−18f(-1) = 4f(2) - 18. Substituting into equation 1 gives f(2)+2(4f(2)−18)=18  ⟹  9f(2)=54  ⟹  f(2)=6f(2) + 2(4f(2) - 18) = 18 \implies 9f(2) = 54 \implies f(2) = 6.
Eliminating f(12)f\left(\frac{1}{2}\right) and f(−1)f(-1) isolated f(2)f(2) to find its exact numerical value.

Anahtar Kavram

Cyclic Functional Equations and Substitution Systems
Soru 12Soru →

For all real numbers aa and bb such that a+b≠0a + b \neq 0, the custom operator ⋆\star is defined by a⋆b=a−ba+ba \star b = \frac{a - b}{a + b}. What is the value of (3⋆1)⋆2(3 \star 1) \star 2?

Cevabı ve açıklamayı göster

Cevap: −35-\frac{3}{5}

Cevap

−35-\frac{3}{5}
Evaluating the expression step-by-step according to parentheses yields 3⋆1=3−13+1=123 \star 1 = \frac{3-1}{3+1} = \frac{1}{2}. Substituting 12\frac{1}{2} as the first input and 22 as the second input gives 1/2−21/2+2=−3/25/2=−35\frac{1/2 - 2}{1/2 + 2} = \frac{-3/2}{5/2} = -\frac{3}{5}.

Adım Adım Çözüm

1
Evaluate the inner custom operation inside parentheses: 3⋆13 \star 1.
3⋆1=3−13+1=24=123 \star 1 = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}.
Follow the order of operations by resolving the grouped expression first using the definition a⋆b=a−ba+ba \star b = \frac{a - b}{a + b} with a=3a = 3 and b=1b = 1.
2
Substitute the result 12\frac{1}{2} back into the main expression to compute (12)⋆2(\frac{1}{2}) \star 2.
(12)⋆2=12−212+2(\frac{1}{2}) \star 2 = \frac{\frac{1}{2} - 2}{\frac{1}{2} + 2}.
Apply the definition of the custom operator again, where the left operand is 12\frac{1}{2} and the right operand is 22.
3
Simplify the complex fraction.
\frac{\frac{1}{2} - \frac{4}{2}}{\frac{1}{2} + \frac{4}{2}} = \frac{-\frac{3}{2}}{\frac{5}{2}} = -\frac{3}{5}.
Combine the fractions in the numerator and denominator, then divide.

Anahtar Kavram

Custom Operators and Order of Operations
Tahmini Süre:1m 30s
Soru 13Soru →

For any real numbers uu and vv, the binary operator ⋆\star is defined by u⋆v=2u2−3vu \star v = 2u^2 - 3v. A function ff is defined by f(x)=x⋆4f(x) = x \star 4, and a function gg is defined by g(x)=3x+1g(x) = 3x + 1. If mm is a real number such that f(g(m))=38f(g(m)) = 38, what is the product of all possible values of mm?

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Cevap: −83-\frac{8}{3}

Cevap

The product of all possible values of mm is −83-\frac{8}{3}.
First, evaluate f(x)=x⋆4=2x2−3(4)=2x2−12f(x) = x \star 4 = 2x^2 - 3(4) = 2x^2 - 12. Next, substitute g(m)=3m+1g(m) = 3m + 1 into f(x)f(x) to get f(g(m))=2(3m+1)2−12f(g(m)) = 2(3m + 1)^2 - 12. Setting this expression equal to 3838 gives 2(3m+1)2−12=382(3m + 1)^2 - 12 = 38, which simplifies to (3m+1)2=25(3m + 1)^2 = 25. Taking the square root gives two possible linear equations: 3m+1=53m + 1 = 5 (which gives m=43m = \frac{4}{3}) and 3m+1=−53m + 1 = -5 (which gives m=−2m = -2). Multiplying these two solutions yields (43)×(−2)=−83\left(\frac{4}{3}\right) \times (-2) = -\frac{8}{3}. Thus, the option equal to −83-\frac{8}{3} is correct.

Adım Adım Çözüm

1
Evaluate the function f(x)f(x) using the definition of the custom operator ⋆\star.
f(x)=x⋆4=2x2−3(4)=2x2−12f(x) = x \star 4 = 2x^2 - 3(4) = 2x^2 - 12
Substitute u=xu = x and v=4v = 4 into the formula u⋆v=2u2−3vu \star v = 2u^2 - 3v.
2
Express the nested function f(g(m))f(g(m)) in terms of mm.
f(g(m))=2(g(m))2−12=2(3m+1)2−12f(g(m)) = 2(g(m))^2 - 12 = 2(3m + 1)^2 - 12
Substitute g(m)=3m+1g(m) = 3m + 1 into f(x)f(x).
3
Set f(g(m))f(g(m)) equal to 3838 and solve for (3m+1)2(3m + 1)^2.
2(3m+1)2−12=38  ⟹  2(3m+1)2=50  ⟹  (3m+1)2=252(3m + 1)^2 - 12 = 38 \implies 2(3m + 1)^2 = 50 \implies (3m + 1)^2 = 25
Isolate the squared binomial term using basic algebraic manipulation.
4
Take the square root of both sides to find all possible values of mm.
3m+1=5  ⟹  m=433m + 1 = 5 \implies m = \frac{4}{3} or 3m+1=−5  ⟹  m=−23m + 1 = -5 \implies m = -2
A positive real number has both positive and negative square roots.
5
Calculate the product of the two solutions for mm.
(43)×(−2)=−83\left(\frac{4}{3}\right) \times (-2) = -\frac{8}{3}
Multiply the two roots together as requested by the question stem.

Anahtar Kavram

Evaluating custom binary operators and nested composite functions, solving quadratic equations, and finding products of roots.
Soru 14Soru →

For all non-zero real numbers xx and yy, the custom operation ⋆\star is defined by x⋆y=xy−yxx \star y = \frac{x}{y} - \frac{y}{x}. If pp and qq are positive numbers such that p+q=5p + q = 5 and p⋆q=56p \star q = \frac{5}{6}, what is the value of (p+1)⋆q(p + 1) \star q?

Cevabı ve açıklamayı göster

Cevap: 32\frac{3}{2}

Cevap

32\frac{3}{2}
By simplifying p⋆q=p2−q2pq=(p−q)(p+q)pqp \star q = \frac{p^2 - q^2}{pq} = \frac{(p - q)(p + q)}{pq}, we substitute p+q=5p + q = 5 to get 5(p−q)pq=56\frac{5(p - q)}{pq} = \frac{5}{6}, which reduces to 6(p−q)=pq6(p - q) = pq. Substituting q=5−pq = 5 - p yields the quadratic equation p2+7p−30=0p^2 + 7p - 30 = 0. Since p>0p > 0, we find p=3p = 3 and q=2q = 2. Evaluating (3+1)⋆2=4⋆2=42−24=32(3 + 1) \star 2 = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}.

Adım Adım Çözüm

1
Express the custom operation using algebraic fractions and factor the numerator
p⋆q=pq−qp=p2−q2pq=(p−q)(p+q)pqp \star q = \frac{p}{q} - \frac{q}{p} = \frac{p^2 - q^2}{pq} = \frac{(p - q)(p + q)}{pq}
Finding a common denominator converts the expression into a form where known identities can be substituted.
2
Substitute p+q=5p + q = 5 and p⋆q=56p \star q = \frac{5}{6} into the simplified expression
5(p−q)pq=56  ⟹  p−qpq=16  ⟹  6(p−q)=pq\frac{5(p - q)}{pq} = \frac{5}{6} \implies \frac{p - q}{pq} = \frac{1}{6} \implies 6(p - q) = pq
Simplifying by dividing both sides by 5 establishes a direct relation between the difference and product of p and q.
3
Substitute q=5−pq = 5 - p to form and solve a quadratic equation for pp
6(2p−5)=p(5−p)  ⟹  12p−30=5p−p2  ⟹  p2+7p−30=0  ⟹  (p+10)(p−3)=06(2p - 5) = p(5 - p) \implies 12p - 30 = 5p - p^2 \implies p^2 + 7p - 30 = 0 \implies (p + 10)(p - 3) = 0
Since p and q are positive numbers, p must be positive, giving p = 3 and q = 2.
4
Evaluate (p+1)⋆q(p + 1) \star q with p=3p = 3 and q=2q = 2
(3+1)⋆2=4⋆2=42−24=2−12=32(3 + 1) \star 2 = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = 2 - \frac{1}{2} = \frac{3}{2}
Applying the custom operator definition to inputs 4 and 2 yields the final value.

Anahtar Kavram

Custom Operator Evaluation and Quadratic System Solving
Soru 15Soru →

For any real number xx, the function hh is defined by h(x)=3x2−5x+4h(x) = 3x^2 - 5x + 4. What is the value of h(3)h(3)?

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Cevap: 16

Cevap

16
To find h(3)h(3), substitute x=3x = 3 into the function h(x)=3x2−5x+4h(x) = 3x^2 - 5x + 4. First, compute 32=93^2 = 9. Multiplying by 3 yields 27. Subtracting 5(3)=155(3) = 15 gives 12, and adding 4 results in the final value of 16.

Adım Adım Çözüm

1
Substitute the value x=3x = 3 into the definition of h(x)h(x).
h(3)=3(3)2−5(3)+4h(3) = 3(3)^2 - 5(3) + 4
To evaluate a function at a specific value, replace all instances of the input variable with that value.
2
Evaluate the exponent and carry out the multiplications.
h(3)=3(9)−15+4=27−15+4h(3) = 3(9) - 15 + 4 = 27 - 15 + 4
Follow standard order of operations: compute powers before performing multiplication.
3
Perform the remaining addition and subtraction.
h(3)=16h(3) = 16
Complete the calculation from left to right.

Anahtar Kavram

Function Evaluation
Soru 16Soru →

For all real numbers xx and yy, the custom operation Δ\Delta is defined by xΔy=(x+1)(y−2)x \Delta y = (x + 1)(y - 2). If the function gg is defined by g(n)=n2−3g(n) = n^2 - 3 for all real numbers nn, what is the value of g(3)Δg(2)g(3) \Delta g(2)?

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Cevap: -7

Cevap

The value of g(3)Δg(2)g(3) \Delta g(2) is −7-7.
Evaluating g(3)g(3) gives 32−3=63^2 - 3 = 6, and evaluating g(2)g(2) gives 22−3=12^2 - 3 = 1. Substituting these values as x=6x = 6 and y=1y = 1 into the custom operator definition xΔy=(x+1)(y−2)x \Delta y = (x + 1)(y - 2) yields (6+1)(1−2)=7×(−1)=−7(6 + 1)(1 - 2) = 7 \times (-1) = -7.

Adım Adım Çözüm

1
Evaluate the inner function g(n)g(n) at n=3n = 3
g(3)=32−3=9−3=6g(3) = 3^2 - 3 = 9 - 3 = 6
Before evaluating the custom operator, the inputs g(3)g(3) and g(2)g(2) must be explicitly calculated.
2
Evaluate the inner function g(n)g(n) at n=2n = 2
g(2)=22−3=4−3=1g(2) = 2^2 - 3 = 4 - 3 = 1
Determine the second operand for the custom operation.
3
Apply the custom operation Δ\Delta to the results from Step 1 and Step 2
6Δ1=(6+1)(1−2)=7×(−1)=−76 \Delta 1 = (6 + 1)(1 - 2) = 7 \times (-1) = -7
Substitute x=6x = 6 and y=1y = 1 into the formula xΔy=(x+1)(y−2)x \Delta y = (x + 1)(y - 2).

Anahtar Kavram

Evaluating functions and applying non-commutative custom operators
Soru 17Soru →

For all non-zero real numbers aa and bb, the custom binary operator ⋆\star is defined by a⋆b=a2−b2aba \star b = \frac{a^2 - b^2}{ab}. A function ff is defined for all positive numbers xx by f(x)=(x+1)⋆xf(x) = (x + 1) \star x. What is the value of f(3)⋆f(2)f(3) \star f(2)?

Cevabı ve açıklamayı göster

Cevap: −5170-\frac{51}{70}

Cevap

−5170-\frac{51}{70}
First, evaluate the inner function values: f(3)=4⋆3=42−324⋅3=712f(3) = 4 \star 3 = \frac{4^2 - 3^2}{4 \cdot 3} = \frac{7}{12} and f(2)=3⋆2=32−223⋅2=56=1012f(2) = 3 \star 2 = \frac{3^2 - 2^2}{3 \cdot 2} = \frac{5}{6} = \frac{10}{12}. Next, evaluate the outer operation (712)⋆(1012)=(7/12)2−(10/12)2(7/12)(10/12)=49−10070=−5170\left(\frac{7}{12}\right) \star \left(\frac{10}{12}\right) = \frac{(7/12)^2 - (10/12)^2}{(7/12)(10/12)} = \frac{49 - 100}{70} = -\frac{51}{70}. Hence, the correct response is −5170-\frac{51}{70}.

Adım Adım Çözüm

1
Evaluate f(3)f(3) using the definition of f(x)f(x) and the custom operator ⋆\star.
f(3)=(3+1)⋆3=4⋆3=42−324×3=16−912=712f(3) = (3 + 1) \star 3 = 4 \star 3 = \frac{4^2 - 3^2}{4 \times 3} = \frac{16 - 9}{12} = \frac{7}{12}
By definition, f(x)=(x+1)⋆xf(x) = (x+1) \star x. Setting x=3x=3 yields 4⋆34 \star 3.
2
Evaluate f(2)f(2) using the definition of f(x)f(x) and the custom operator ⋆\star.
f(2)=(2+1)⋆2=3⋆2=32−223×2=9−46=56f(2) = (2 + 1) \star 2 = 3 \star 2 = \frac{3^2 - 2^2}{3 \times 2} = \frac{9 - 4}{6} = \frac{5}{6}
Setting x=2x=2 in f(x)f(x) yields 3⋆23 \star 2.
3
Evaluate f(3)⋆f(2)=(712)⋆(56)f(3) \star f(2) = \left(\frac{7}{12}\right) \star \left(\frac{5}{6}\right).
(712)⋆(1012)=(712)2−(1012)2(712)(1012)=49−10014470144=−5170\left(\frac{7}{12}\right) \star \left(\frac{10}{12}\right) = \frac{\left(\frac{7}{12}\right)^2 - \left(\frac{10}{12}\right)^2}{\left(\frac{7}{12}\right)\left(\frac{10}{12}\right)} = \frac{\frac{49 - 100}{144}}{\frac{70}{144}} = -\frac{51}{70}
Apply the rule u⋆v=u2−v2uvu \star v = \frac{u^2 - v^2}{uv} with u=712u = \frac{7}{12} and v=1012v = \frac{10}{12}.

Anahtar Kavram

Custom non-commutative binary operators and function evaluation
Tahmini Süre:2m 0s
Soru 18Soru →

For all real numbers aa and bb, the custom binary operation ⋆\star is defined by a⋆b=a2−2b2+aba \star b = a^2 - 2b^2 + ab. The function gg is defined by g(x)=x⋆3g(x) = x \star 3. What is the positive value of xx for which g(x)=10g(x) = 10?

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Cevap: 4

Cevap

The positive value of xx for which g(x)=10g(x) = 10 is 44.
Evaluating g(x)=x⋆3g(x) = x \star 3 gives x2−2(3)2+x(3)=x2+3x−18x^2 - 2(3)^2 + x(3) = x^2 + 3x - 18. Setting this expression equal to 1010 produces the quadratic equation x2+3x−28=0x^2 + 3x - 28 = 0. Factoring yields (x+7)(x−4)=0(x + 7)(x - 4) = 0, giving solutions x=−7x = -7 and x=4x = 4. Since xx must be positive, the correct value is 44.

Adım Adım Çözüm

1
Apply the definition of the custom binary operator ⋆\star to find g(x)g(x).
g(x)=x⋆3=x2−2(3)2+x(3)=x2−18+3x=x2+3x−18g(x) = x \star 3 = x^2 - 2(3)^2 + x(3) = x^2 - 18 + 3x = x^2 + 3x - 18
Substitute a=xa = x and b=3b = 3 into the given formula a⋆b=a2−2b2+aba \star b = a^2 - 2b^2 + ab.
2
Set the function expression equal to 1010 and rearrange into standard quadratic form.
x2+3x−18=10  ⟹  x2+3x−28=0x^2 + 3x - 18 = 10 \implies x^2 + 3x - 28 = 0
Subtract 1010 from both sides to form a quadratic equation equal to zero.
3
Factor the quadratic equation to find the values of xx.
(x+7)(x−4)=0  ⟹  x=−7 or x=4(x + 7)(x - 4) = 0 \implies x = -7 \text{ or } x = 4
Find two numbers that multiply to −28-28 and add to 33, which are 77 and −4-4.
4
Select the positive value of xx as specified in the stem.
x=4x = 4
The question asks specifically for the positive value of xx.

Anahtar Kavram

Evaluating non-commutative custom operators and solving polynomial equations resulting from function evaluations.
Tahmini Süre:2m 0s
Soru 19Soru →

For all real numbers aa and bb, the custom binary operator ⋆\star is defined by a⋆b=2a−ba \star b = 2a - b. The function ff is defined for all real numbers by f(x)=x2−1f(x) = x^2 - 1. If (x⋆3)⋆f(3)=6(x \star 3) \star f(3) = 6, what is the value of xx?

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Cevap: 5

Cevap

5
First, evaluate the function f(3)=32−1=8f(3) = 3^2 - 1 = 8. Next, evaluate the first custom operation x⋆3=2x−3x \star 3 = 2x - 3. Substituting these into the overall expression yields (2x−3)⋆8=2(2x−3)−8=4x−14(2x - 3) \star 8 = 2(2x - 3) - 8 = 4x - 14. Setting this equal to 6 gives 4x−14=64x - 14 = 6, which simplifies to 4x=204x = 20, so x=5x = 5.

Adım Adım Çözüm

1
Evaluate the inner function f(3)f(3)
f(3)=32−1=9−1=8f(3) = 3^2 - 1 = 9 - 1 = 8
Substitute x=3x = 3 into the given definition f(x)=x2−1f(x) = x^2 - 1.
2
Evaluate the inner custom operation x⋆3x \star 3
x⋆3=2x−3x \star 3 = 2x - 3
Apply the custom operator rule a⋆b=2a−ba \star b = 2a - b with a=xa = x and b=3b = 3.
3
Evaluate the outer custom operation (x⋆3)⋆f(3)(x \star 3) \star f(3)
(2x−3)⋆8=2(2x−3)−8=4x−6−8=4x−14(2x - 3) \star 8 = 2(2x - 3) - 8 = 4x - 6 - 8 = 4x - 14
Substitute the results from Step 1 and Step 2 into the custom operator definition.
4
Solve the algebraic equation for xx
4x−14=6  ⟹  4x=20  ⟹  x=54x - 14 = 6 \implies 4x = 20 \implies x = 5
Set the simplified expression equal to 6 as given in the problem stem and isolate xx.

Anahtar Kavram

Custom Operator Evaluation and Nested Functions
Tahmini Süre:1m 30s
Soru 20Soru →
The function ff is defined for all real numbers by
f(x)={3x−1if x≤4x2−kif x>4f(x) = \begin{cases} 3x - 1 & \text{if } x \le 4 \\ x^2 - k & \text{if } x > 4 \end{cases}
where kk is a constant. If f(f(3))=14f(f(3)) = 14, what is the value of kk?
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Cevap: 50

Cevap

50
To solve for kk, evaluate the composite function f(f(3))f(f(3)) step-by-step. First, find f(3)f(3). Since 3≤43 \le 4, we use the first branch of the definition: f(3)=3(3)−1=8f(3) = 3(3) - 1 = 8. Next, evaluate f(8)f(8). Since 8>48 > 4, we use the second branch of the definition: f(8)=82−k=64−kf(8) = 8^2 - k = 64 - k. Given that f(f(3))=14f(f(3)) = 14, we set 64−k=1464 - k = 14, which yields k=50k = 50.

Adım Adım Çözüm

1
Evaluate the inner function value f(3)
f(3) = 8
Because 3 is less than or equal to 4, the definition f(x) = 3x - 1 applies, yielding f(3) = 3(3) - 1 = 8.
2
Express f(f(3)) in terms of k
f(8) = 64 - k
Since the inner value is 8 and 8 > 4, the piecewise condition specifies using f(x) = x^2 - k, giving f(8) = 8^2 - k = 64 - k.
3
Solve the linear equation for k
k = 50
Setting the calculated expression equal to the given value 14 gives 64 - k = 14, which simplifies to k = 50.

Anahtar Kavram

Piecewise Function Evaluation and Parameter Resolution
Tahmini Süre:1m 30s
Sayfa 1 / 2Sonraki
Function Definitions, Evaluation, and Custom Operators Alıştırma Soruları — GMAT | Examkin