Examkin

Functions and Custom Symbol Operations

29 soru

Tüm türler29Çoktan seçmeli14Çoklu seçim7Sayısal8
Soru 1Soru →

For all real numbers xx and yy, the custom operation ▲\blacktriangle is defined by x▲y=2x2−y2+xyx \blacktriangle y = 2x^2 - y^2 + xy. The function gg is defined for all real numbers nn by g(n)=n▲3g(n) = n \blacktriangle 3. If kk is a positive constant such that g(k)=0g(k) = 0, what is the value of kk?

Cevabı ve açıklamayı göster

Cevap: 32\frac{3}{2}

Cevap

32\frac{3}{2}
Evaluating g(k)=k▲3g(k) = k \blacktriangle 3 using the rule x▲y=2x2−y2+xyx \blacktriangle y = 2x^2 - y^2 + xy gives 2k2−32+3k=2k2+3k−92k^2 - 3^2 + 3k = 2k^2 + 3k - 9. Setting this quadratic equal to zero gives (2k−3)(k+3)=0(2k - 3)(k + 3) = 0. The roots are k=32k = \frac{3}{2} and k=−3k = -3. Because kk is specified to be positive, the only valid solution is 32\frac{3}{2}.

Adım Adım Çözüm

1
Apply the definition of the custom operation ▲\blacktriangle to evaluate g(k)g(k).
g(k)=k▲3=2(k)2−(3)2+k(3)=2k2−9+3k=2k2+3k−9g(k) = k \blacktriangle 3 = 2(k)^2 - (3)^2 + k(3) = 2k^2 - 9 + 3k = 2k^2 + 3k - 9
Substitute x=kx = k and y=3y = 3 into the given expression x▲y=2x2−y2+xyx \blacktriangle y = 2x^2 - y^2 + xy.
2
Set the expression for g(k)g(k) equal to 00 and factor the quadratic equation.
2k2+3k−9=(2k−3)(k+3)=02k^2 + 3k - 9 = (2k - 3)(k + 3) = 0
We need to find values of kk for which g(k)=0g(k) = 0.
3
Solve for kk and apply the constraint that kk is a positive constant.
2k−3=0  ⟹  k=322k - 3 = 0 \implies k = \frac{3}{2} or k+3=0  ⟹  k=−3k + 3 = 0 \implies k = -3. Since k>0k > 0, k=32k = \frac{3}{2}.
Disregard the negative root k=−3k = -3 because the problem specifies that kk must be positive.

Anahtar Kavram

Combining custom symbol operations with function notation to solve quadratic equations
Tahmini Süre:1m 30s
Soru 2Soru →

For all non-zero real numbers xx, the function ff satisfies the equation f(x)+2f(1x)=3xf(x) + 2f\left(\frac{1}{x}\right) = 3x. What is the value of f(2)f(2)?

Cevabı ve açıklamayı göster

Cevap: −1-1

Cevap

−1-1
Evaluating the functional equation at x=2x = 2 gives f(2)+2f(1/2)=6f(2) + 2f(1/2) = 6. Evaluating the equation at x=1/2x = 1/2 gives f(1/2)+2f(2)=3/2f(1/2) + 2f(2) = 3/2. Substituting f(1/2)=3/2−2f(2)f(1/2) = 3/2 - 2f(2) into the first equation produces −3f(2)+3=6-3f(2) + 3 = 6, which simplifies to f(2)=−1f(2) = -1.

Adım Adım Çözüm

1
Substitute x=2x = 2 into the given functional equation f(x)+2f(1x)=3xf(x) + 2f\left(\frac{1}{x}\right) = 3x.
f(2)+2f(12)=6f(2) + 2f\left(\frac{1}{2}\right) = 6
This establishes a relation between the target value f(2)f(2) and the auxiliary value f(12)f\left(\frac{1}{2}\right).
2
Substitute x=12x = \frac{1}{2} into the original equation f(x)+2f(1x)=3xf(x) + 2f\left(\frac{1}{x}\right) = 3x.
f(12)+2f(2)=3(12)=32f\left(\frac{1}{2}\right) + 2f(2) = 3\left(\frac{1}{2}\right) = \frac{3}{2}
This provides a second independent linear equation relating f(2)f(2) and f(12)f\left(\frac{1}{2}\right).
3
Isolate f(12)f\left(\frac{1}{2}\right) in the second equation: f(12)=32−2f(2)f\left(\frac{1}{2}\right) = \frac{3}{2} - 2f(2), and substitute this expression into the first equation.
f(2)+2(32−2f(2))=6  ⟹  f(2)+3−4f(2)=6  ⟹  −3f(2)+3=6f(2) + 2\left(\frac{3}{2} - 2f(2)\right) = 6 \implies f(2) + 3 - 4f(2) = 6 \implies -3f(2) + 3 = 6
Eliminating f(12)f\left(\frac{1}{2}\right) converts the system into a single linear equation in terms of f(2)f(2).
4
Solve −3f(2)+3=6-3f(2) + 3 = 6 for f(2)f(2).
−3f(2)=3  ⟹  f(2)=−1-3f(2) = 3 \implies f(2) = -1
Subtracting 33 from both sides gives −3f(2)=3-3f(2) = 3, and dividing by −3-3 yields f(2)=−1f(2) = -1.

Anahtar Kavram

Solving Functional Equations via Variable Substitution

Alternatif Yöntem

Generalize f(x)f(x) by setting f(x)=Ax+Bxf(x) = Ax + \frac{B}{x}. Substituting into f(x)+2f(1/x)=3xf(x) + 2f(1/x) = 3x gives (Ax+B/x)+2(A/x+Bx)=(A+2B)x+(2A+B)/x=3x(Ax + B/x) + 2(A/x + Bx) = (A + 2B)x + (2A + B)/x = 3x. Equating coefficients: A+2B=3A + 2B = 3 and 2A+B=0  ⟹  B=−2A  ⟹  A−4A=3  ⟹  A=−1,B=22A + B = 0 \implies B = -2A \implies A - 4A = 3 \implies A = -1, B = 2. Thus f(x)=−x+2/xf(x) = -x + 2/x, so f(2)=−2+2/2=−1f(2) = -2 + 2/2 = -1.
Tahmini Süre:1m 30s
Soru 3Soru →

The function ff is defined for all real numbers xx by f(x)=2x2−xf(x) = 2x^2 - x. For all real numbers aa and bb, the custom operation ⋆\star is defined by a⋆b=f(a+b)−f(a−b)a \star b = f(a + b) - f(a - b). What is the value of 2⋆(−3)2 \star (-3)?

Cevabı ve açıklamayı göster

Cevap: -42

Cevap

The correct answer is -42.
Evaluating a+b=2+(−3)=−1a + b = 2 + (-3) = -1 gives f(−1)=2(−1)2−(−1)=3f(-1) = 2(-1)^2 - (-1) = 3. Evaluating a−b=2−(−3)=5a - b = 2 - (-3) = 5 gives f(5)=2(5)2−5=45f(5) = 2(5)^2 - 5 = 45. Subtracting the two function values yields 3−45=−423 - 45 = -42.

Adım Adım Çözüm

1
Identify the values for variables aa and bb and evaluate the inputs for the function ff.
Here a=2a = 2 and b=−3b = -3. Therefore, a+b=2+(−3)=−1a + b = 2 + (-3) = -1 and a−b=2−(−3)=5a - b = 2 - (-3) = 5.
The definition of the custom operation requires evaluating f(a+b)f(a+b) and f(a−b)f(a-b).
2
Calculate f(−1)f(-1) using the function rule f(x)=2x2−xf(x) = 2x^2 - x.
f(−1)=2(−1)2−(−1)=2(1)+1=3f(-1) = 2(-1)^2 - (-1) = 2(1) + 1 = 3.
Substituting x=−1x = -1 into the quadratic expression.
3
Calculate f(5)f(5) using the function rule f(x)=2x2−xf(x) = 2x^2 - x.
f(5)=2(5)2−(5)=2(25)−5=45f(5) = 2(5)^2 - (5) = 2(25) - 5 = 45.
Substituting x=5x = 5 into the quadratic expression.
4
Compute the difference f(a+b)−f(a−b)f(a+b) - f(a-b).
3−45=−423 - 45 = -42.
Evaluating f(−1)−f(5)f(-1) - f(5) according to the definition of ⋆\star.

Anahtar Kavram

Evaluating nested algebraic functions and custom symbol operations by direct substitution.
Tahmini Süre:1m 30s
Soru 4Soru →
For all non-zero real numbers uu and vv, the custom operation Δ\Delta is defined by uΔv=uv−vuu \Delta v = \frac{u}{v} - \frac{v}{u} The function ff is defined for all non-zero real numbers xx by f(x)=xΔ2f(x) = x \Delta 2. Which of the following statements must be true for all non-zero real numbers aa and bb? Select all such statements.

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Cevap: aΔb=−(bΔa)a \Delta b = -(b \Delta a); f(−a)=−f(a)f(-a) = -f(a)

Cevap

The statements aΔb=−(bΔa)a \Delta b = -(b \Delta a) and f(−a)=−f(a)f(-a) = -f(a) must be true.
The custom operation is anti-commutative because swapping variables negates the difference of the fractions, making the statement aΔb=−(bΔa)a \Delta b = -(b \Delta a) true. Additionally, substituting −a-a into f(x)=x2−2xf(x) = \frac{x}{2} - \frac{2}{x} yields −f(a)-f(a), showing f(−a)=−f(a)f(-a) = -f(a) is also true.

Adım Adım Çözüm

1
Analyze the first identity aΔb=−(bΔa)a \Delta b = -(b \Delta a).
aΔb=ab−baa \Delta b = \frac{a}{b} - \frac{b}{a} and bΔa=ba−ab=−(ab−ba)=−(aΔb)b \Delta a = \frac{b}{a} - \frac{a}{b} = -\left(\frac{a}{b} - \frac{b}{a}\right) = -(a \Delta b).
Factoring −1-1 out of bΔab \Delta a proves anti-commutativity.
2
Evaluate f(x)f(x) explicitly and test f(−a)f(-a).
f(x)=x2−2xf(x) = \frac{x}{2} - \frac{2}{x}. Substituting −a-a gives f(−a)=−a2−2−a=−(a2−2a)=−f(a)f(-a) = \frac{-a}{2} - \frac{2}{-a} = -\left(\frac{a}{2} - \frac{2}{a}\right) = -f(a).
This confirms that ff is an odd function.
3
Compare f(a2)f(a^2) with (f(a))2(f(a))^2.
f(a2)=a22−2a2f(a^2) = \frac{a^2}{2} - \frac{2}{a^2}, but (f(a))2=a24−2+4a2(f(a))^2 = \frac{a^2}{4} - 2 + \frac{4}{a^2}. Testing a=1a = 1 yields f(1)=−1.5  ⟹  (f(1))2=2.25f(1) = -1.5 \implies (f(1))^2 = 2.25, whereas f(12)=−1.5≠2.25f(1^2) = -1.5 \neq 2.25.
Counterexample shows the equality does not hold in general.
4
Compare f(a+b)f(a+b) with f(a)+f(b)f(a) + f(b).
For a=1,b=1a=1, b=1, f(1+1)=f(2)=22−22=0f(1+1) = f(2) = \frac{2}{2} - \frac{2}{2} = 0, but f(1)+f(1)=−1.5+(−1.5)=−3≠0f(1) + f(1) = -1.5 + (-1.5) = -3 \neq 0.
Functions do not in general preserve additivity f(a+b)=f(a)+f(b)f(a+b) = f(a) + f(b) unless they are linear through the origin.

Anahtar Kavram

Properties of algebraic custom operations and odd functions
Soru 5Soru →
For all non-zero real numbers xx, the function ff satisfies the functional equation
f(x)+2f(−1x)=3xf(x) + 2f\left(-\frac{1}{x}\right) = 3x

Which of the following statements must be true for all x≠0x \neq 0? Select all such statements.

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Cevap: f(−1)=3f(-1) = 3; f(x)≤−22f(x) \leq -2\sqrt{2} for all x>0x > 0

Cevap

The statements stating that f(−1)=3f(-1) = 3 and that f(x)≤−22f(x) \leq -2\sqrt{2} for all x>0x > 0 are the correct choices.
Solving the functional equation yields f(x)=−x−2xf(x) = -x - \frac{2}{x}. Substituting x=−1x = -1 gives f(−1)=3f(-1) = 3, making the statement asserting f(−1)=3f(-1) = 3 correct. Furthermore, for any positive real number xx, the sum x+2x≥22x + \frac{2}{x} \geq 2\sqrt{2} by the AM-GM inequality, so f(x)=−(x+2x)≤−22f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}, making the statement asserting f(x)≤−22f(x) \leq -2\sqrt{2} for all x>0x > 0 correct.

Adım Adım Çözüm

1
Set up a system of functional equations by substituting x→−1xx \to -\frac{1}{x}.
Replacing xx with −1x-\frac{1}{x} in the original equation f(x)+2f(−1x)=3xf(x) + 2f\left(-\frac{1}{x}\right) = 3x gives f(−1x)+2f(x)=3(−1x)=−3xf\left(-\frac{1}{x}\right) + 2f(x) = 3\left(-\frac{1}{x}\right) = -\frac{3}{x}.
This creates a system of two linear algebraic equations with two unknowns: f(x)f(x) and f(−1x)f\left(-\frac{1}{x}\right).
2
Solve the system of equations for f(x)f(x).
Multiply the second equation by 22 to obtain 4f(x)+2f(−1x)=−6x4f(x) + 2f\left(-\frac{1}{x}\right) = -\frac{6}{x}. Subtracting the first equation f(x)+2f(−1x)=3xf(x) + 2f\left(-\frac{1}{x}\right) = 3x from this gives 3f(x)=−3x−6x3f(x) = -3x - \frac{6}{x}, so f(x)=−x−2xf(x) = -x - \frac{2}{x}.
Eliminating the composite term f(−1x)f\left(-\frac{1}{x}\right) yields an explicit expression for f(x)f(x).
3
Evaluate f(−1)f(-1).
f(−1)=−(−1)−2−1=1+2=3f(-1) = -(-1) - \frac{2}{-1} = 1 + 2 = 3.
This confirms that the statement asserting f(−1)=3f(-1) = 3 is correct.
4
Analyze the maximum value of f(x)f(x) for x>0x > 0 using the AM-GM inequality.
For x>0x > 0, x+2x≥2x⋅2x=22x + \frac{2}{x} \geq 2\sqrt{x \cdot \frac{2}{x}} = 2\sqrt{2}. Therefore, f(x)=−(x+2x)≤−22f(x) = -\left(x + \frac{2}{x}\right) \leq -2\sqrt{2}.
This confirms that the statement asserting f(x)≤−22f(x) \leq -2\sqrt{2} for all x>0x > 0 is correct.

Anahtar Kavram

Solving functional equations via variable substitution and analyzing function bounds via the AM-GM inequality.
Tahmini Süre:2m 30s
Soru 6Soru →

For the function f(x)=x2−5xf(x) = x^2 - 5x, what is the value of f(−3)f(-3)?

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Cevap: 24

Cevap

24
Substituting x=−3x = -3 into the function yields (−3)2−5(−3)=9+15=24(-3)^2 - 5(-3) = 9 + 15 = 24, which correctly evaluates both the exponent and linear terms following standard sign rules.

Adım Adım Çözüm

1
Substitute x=−3x = -3 into the function expression f(x)=x2−5xf(x) = x^2 - 5x.
f(−3)=(−3)2−5(−3)f(-3) = (-3)^2 - 5(-3)
Function notation f(−3)f(-3) requires replacing every instance of variable xx with −3-3.
2
Evaluate the exponent term (−3)2(-3)^2.
9
Squaring any negative number produces a positive result: (−3)×(−3)=9(-3) \times (-3) = 9.
3
Evaluate the product term −5(−3)-5(-3).
15
Multiplying two negative numbers yields a positive product: −5×(−3)=15-5 \times (-3) = 15.
4
Sum the evaluated components.
9 + 15 = 24
Combine the results to obtain the final value of f(−3)f(-3).

Anahtar Kavram

Evaluating functions with negative input values
Tahmini Süre:45s
Soru 7Soru →

For all non-zero real numbers xx and yy, the custom operation ⋄\diamond is defined by x⋄y=x2+y2xyx \diamond y = \frac{x^2 + y^2}{xy}. Which of the following statements must be true for all non-zero real numbers aa and bb? Select all such statements.

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Cevap: a⋄b=b⋄aa \diamond b = b \diamond a; a⋄(−a)=−2a \diamond (-a) = -2

Cevap

The correct statements are that a⋄b=b⋄aa \diamond b = b \diamond a and that a⋄(−a)=−2a \diamond (-a) = -2.
The operation is symmetric with respect to aa and bb, making a⋄b=b⋄aa \diamond b = b \diamond a true. Substituting −a-a into the operation yields 2a2−a2=−2\frac{2a^2}{-a^2} = -2, which makes a⋄(−a)=−2a \diamond (-a) = -2 true as well.

Adım Adım Çözüm

1
Evaluate the commutative statement a⋄ba \diamond b
a⋄b=a2+b2aba \diamond b = \frac{a^2 + b^2}{ab} and b⋄a=b2+a2ba=a2+b2abb \diamond a = \frac{b^2 + a^2}{ba} = \frac{a^2 + b^2}{ab}.
Addition and multiplication of real numbers are commutative, so the expressions are identical.
2
Evaluate the statement a⋄(−a)a \diamond (-a)
a⋄(−a)=a2+(−a)2a(−a)=a2+a2−a2=2a2−a2=−2a \diamond (-a) = \frac{a^2 + (-a)^2}{a(-a)} = \frac{a^2 + a^2}{-a^2} = \frac{2a^2}{-a^2} = -2.
Squaring a negative quantity (−a)2(-a)^2 yields positive a2a^2, while the denominator evaluates to −a2-a^2.
3
Evaluate the scaling statement (2a)⋄(2b)(2a) \diamond (2b)
(2a)⋄(2b)=4a2+4b24ab=a2+b2ab=a⋄b(2a) \diamond (2b) = \frac{4a^2 + 4b^2}{4ab} = \frac{a^2 + b^2}{ab} = a \diamond b.
The factor of 4 in the numerator and denominator cancels out, showing (2a)⋄(2b)=a⋄b(2a) \diamond (2b) = a \diamond b, which is not equal to 2(a⋄b)2(a \diamond b).
4
Evaluate the self-operation statement a⋄aa \diamond a
a⋄a=a2+a2a2=2a2a2=2a \diamond a = \frac{a^2 + a^2}{a^2} = \frac{2a^2}{a^2} = 2.
Summing identical squared terms in the numerator produces 2a22a^2, which divides by a2a^2 to give 2.

Anahtar Kavram

Evaluating custom binary operations by algebraic substitution and simplifying terms.
Soru 8Soru →

For all real numbers pp and qq, the custom operation ⊙\odot is defined by p⊙q=2p2−3qp \odot q = 2p^2 - 3q. What is the value of 3⊙(−4)3 \odot (-4)?

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Cevap: 30

Cevap

The value of 3⊙(−4)3 \odot (-4) is 3030.
By applying the definition of the custom operation p⊙q=2p2−3qp \odot q = 2p^2 - 3q with p=3p = 3 and q=−4q = -4, we obtain 2(3)2−3(−4)=2(9)+12=18+12=302(3)^2 - 3(-4) = 2(9) + 12 = 18 + 12 = 30.

Adım Adım Çözüm

1
Substitute p=3p = 3 and q=−4q = -4 into the expression 2p2−3q2p^2 - 3q.
2(3)2−3(−4)2(3)^2 - 3(-4)
The custom operation defines how to process the two inputs pp and qq.
2
Simplify the powers and products according to order of operations.
2(9)−(−12)=18+122(9) - (-12) = 18 + 12
Exponents must be calculated before multiplication, and multiplying two negative numbers yields a positive value.
3
Add the terms together to get the final numerical result.
3030
18+12=3018 + 12 = 30.

Anahtar Kavram

Custom Symbol Operations
Soru 9Soru →

For all real numbers xx, the function gg is defined by g(x)=5−2x2g(x) = 5 - 2x^2. What is the value of g(g(2))g(g(2))?

Cevabı ve açıklamayı göster

Cevap: -13

Cevap

-13
Evaluating the inner expression g(2)g(2) gives 5−2(2)2=5−8=−35 - 2(2)^2 = 5 - 8 = -3. Substituting −3-3 back into the function g(x)g(x) gives g(−3)=5−2(−3)2=5−2(9)=−13g(-3) = 5 - 2(-3)^2 = 5 - 2(9) = -13. Thus, the correct value is −13-13.

Adım Adım Çözüm

1
Evaluate the inner function g(2)g(2)
g(2)=5−2(2)2=5−2(4)=5−8=−3g(2) = 5 - 2(2)^2 = 5 - 2(4) = 5 - 8 = -3
Substitute x=2x = 2 into the definition of g(x)g(x) and follow standard order of operations (exponents before multiplication).
2
Substitute the result into the outer function to evaluate g(−3)g(-3)
g(−3)=5−2(−3)2=5−2(9)=5−18=−13g(-3) = 5 - 2(-3)^2 = 5 - 2(9) = 5 - 18 = -13
Squaring a negative number yields a positive result, so (−3)2=9(-3)^2 = 9, which is then multiplied by 22.

Anahtar Kavram

Nested Function Evaluation and Order of Operations
Tahmini Süre:45s
Soru 10Soru →

For all non-zero real numbers xx, the function ff satisfies the relation 2f(x)+f(1x)=3x2f(x) + f\left(\frac{1}{x}\right) = 3x. What is the value of f(2)f(2)?

Cevabı ve açıklamayı göster

Cevap: 72\frac{7}{2}

Cevap

The value of f(2)f(2) is 72\frac{7}{2}.
To solve for f(2)f(2), set up a system of equations by evaluating the given relation 2f(x)+f(1/x)=3x2f(x) + f(1/x) = 3x at x=2x = 2 and at x=1/2x = 1/2. Substituting x=2x = 2 gives 2f(2)+f(1/2)=62f(2) + f(1/2) = 6, and substituting x=1/2x = 1/2 gives 2f(1/2)+f(2)=3/22f(1/2) + f(2) = 3/2. Expressing f(1/2)f(1/2) from the first equation as 6−2f(2)6 - 2f(2) and substituting it into the second yields 2(6−2f(2))+f(2)=3/22(6 - 2f(2)) + f(2) = 3/2, which simplifies to −3f(2)=−21/2-3f(2) = -21/2, so f(2)=7/2f(2) = 7/2.

Adım Adım Çözüm

1
Substitute x=2x = 2 into the functional relation.
2f(2)+f(12)=3(2)=62f(2) + f\left(\frac{1}{2}\right) = 3(2) = 6
This yields a linear relationship involving f(2)f(2) and f(12)f\left(\frac{1}{2}\right).
2
Substitute x=12x = \frac{1}{2} into the original functional relation.
2f(12)+f(2)=3(12)=322f\left(\frac{1}{2}\right) + f(2) = 3\left(\frac{1}{2}\right) = \frac{3}{2}
This provides a second independent linear equation with the same two unknown quantities, f(2)f(2) and f(12)f\left(\frac{1}{2}\right).
3
Multiply the second equation by 22 to eliminate f(12)f\left(\frac{1}{2}\right).
4f(12)+2f(2)=34f\left(\frac{1}{2}\right) + 2f(2) = 3
Preparing to eliminate f(2)f(2) or f(12)f\left(\frac{1}{2}\right) using elimination.
4
Solve the system of equations for f(2)f(2).
From 2f(2)+f(12)=62f(2) + f\left(\frac{1}{2}\right) = 6, we get f(12)=6−2f(2)f\left(\frac{1}{2}\right) = 6 - 2f(2). Substituting this into the second equation 2(6−2f(2))+f(2)=32  ⟹  12−4f(2)+f(2)=32  ⟹  −3f(2)=32−12=−212  ⟹  f(2)=722(6 - 2f(2)) + f(2) = \frac{3}{2} \implies 12 - 4f(2) + f(2) = \frac{3}{2} \implies -3f(2) = \frac{3}{2} - 12 = -\frac{21}{2} \implies f(2) = \frac{7}{2}.
Algebraic reduction isolates f(2)f(2).

Anahtar Kavram

Functional Equations and System of Equations via Variable Substitution
Soru 11Soru →

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

Cevabı ve açıklamayı göster

Cevap: −1-1

Cevap

−1-1
Substituting a=4a = 4 and b=−3b = -3 into the given operation gives 4((−3)+2)−(−3)=4(−1)+3=−4+3=−14((-3) + 2) - (-3) = 4(-1) + 3 = -4 + 3 = -1. The value of the expression is −1-1.

Adım Adım Çözüm

1
Substitute a=4a = 4 and b=−3b = -3 into the custom operation formula a⋆b=a(b+2)−ba \star b = a(b + 2) - b.
4⋆(−3)=4((−3)+2)−(−3)4 \star (-3) = 4((-3) + 2) - (-3)
The definition specifies replacing variable aa with 44 and variable bb with −3-3.
2
Evaluate the expression inside the parentheses.
−3+2=−1-3 + 2 = -1, simplifying the expression to 4(−1)−(−3)4(-1) - (-3)
Perform operations inside grouping symbols first according to the standard order of operations.
3
Perform the multiplication and simplify the double negative.
4(−1)=−44(-1) = -4 and −(−3)=+3-(-3) = +3, yielding −4+3=−1-4 + 3 = -1
Multiplying positive and negative yields negative, and subtracting a negative number is equivalent to addition.

Anahtar Kavram

Evaluating Custom Binary Operations
Tahmini Süre:45s
Soru 12Soru →
For all real numbers xx such that x≠0x \neq 0 and x≠1x \neq 1, the function ff satisfies the equation
f(x)+f(x−1x)=1+xf(x) + f\left(\frac{x - 1}{x}\right) = 1 + x
What is the value of f(2)f(2)?
Cevabı ve açıklamayı göster

Cevap: 0.75

Cevap

The value of f(2)f(2) is 0.750.75 (or 34\frac{3}{4}).
Evaluating the given equation at x=2x = 2, x=1/2x = 1/2, and x=−1x = -1 generates a system of three linear equations: f(2)+f(1/2)=3f(2) + f(1/2) = 3, f(1/2)+f(−1)=1.5f(1/2) + f(-1) = 1.5, and f(−1)+f(2)=0f(-1) + f(2) = 0. Adding all three equations yields 2(f(2)+f(1/2)+f(−1))=4.52(f(2) + f(1/2) + f(-1)) = 4.5, which simplifies to f(2)+f(1/2)+f(−1)=2.25f(2) + f(1/2) + f(-1) = 2.25. Subtracting the equation f(1/2)+f(−1)=1.5f(1/2) + f(-1) = 1.5 from 2.252.25 gives f(2)=0.75f(2) = 0.75.

Adım Adım Çözüm

1
Determine the cyclic sequence of inputs starting at x=2x = 2
Evaluating the inner transformation g(x)=x−1xg(x) = \frac{x-1}{x} yields g(2)=12g(2) = \frac{1}{2}, g(12)=−1g\left(\frac{1}{2}\right) = -1, and g(−1)=2g(-1) = 2. The input sequence cycles every 3 steps.
Identifying that g(g(g(x)))=xg(g(g(x))) = x allows constructing a closed system of linear equations.
2
Set up a system of three linear equations using the cycle values
For x=2x = 2: f(2)+f(12)=3f(2) + f\left(\frac{1}{2}\right) = 3.
For x=12x = \frac{1}{2}: f(12)+f(−1)=32f\left(\frac{1}{2}\right) + f(-1) = \frac{3}{2}.
For x=−1x = -1: f(−1)+f(2)=0f(-1) + f(2) = 0.
Substituting each element of the cycle into the given relation produces three linear equations in three unknown function values.
3
Sum the equations and isolate f(2)f(2)
Summing all three equations gives 2[f(2)+f(12)+f(−1)]=922\left[f(2) + f\left(\frac{1}{2}\right) + f(-1)\right] = \frac{9}{2}, so f(2)+f(12)+f(−1)=94f(2) + f\left(\frac{1}{2}\right) + f(-1) = \frac{9}{4}. Subtracting f(12)+f(−1)=32f\left(\frac{1}{2}\right) + f(-1) = \frac{3}{2} yields f(2)=94−64=34=0.75f(2) = \frac{9}{4} - \frac{6}{4} = \frac{3}{4} = 0.75.
Symmetric summation provides the total sum of the three function values, allowing direct elimination of the other two variables.

Anahtar Kavram

Solving cyclic functional equations by forming a system of linear equations
Soru 13Soru →

For all real numbers aa and bb, the custom operation ♢\diamondsuit is defined by a♢b=a2−b2+2aba \diamondsuit b = a^2 - b^2 + 2ab. The functions ff and gg are defined by f(x)=x♢2f(x) = x \diamondsuit 2 and g(x)=2♢xg(x) = 2 \diamondsuit x. If kk is a positive real number such that f(k)=g(k)f(k) = g(k), what is the value of f(g(−1))f(g(-1))?

Cevabı ve açıklamayı göster

Cevap: −7-7

Cevap

−7-7
Applying the custom symbol rule gives f(x)=x2+4x−4f(x) = x^2 + 4x - 4 and g(x)=−x2+4x+4g(x) = -x^2 + 4x + 4. Equating f(k)=g(k)f(k) = g(k) yields 2k2=82k^2 = 8, so the positive solution is k=2k = 2. Evaluating the inner function gives g(−1)=−(−1)2+4(−1)+4=−1g(-1) = -(-1)^2 + 4(-1) + 4 = -1. Substituting this value into ff gives f(−1)=(−1)2+4(−1)−4=−7f(-1) = (-1)^2 + 4(-1) - 4 = -7.

Adım Adım Çözüm

1
Express f(x)f(x) and g(x)g(x) using the definition of the custom operation ♢\diamondsuit.
f(x)=x♢2=x2−22+2(x)(2)=x2+4x−4f(x) = x \diamondsuit 2 = x^2 - 2^2 + 2(x)(2) = x^2 + 4x - 4 and g(x)=2♢x=22−x2+2(2)(x)=−x2+4x+4g(x) = 2 \diamondsuit x = 2^2 - x^2 + 2(2)(x) = -x^2 + 4x + 4.
Applying a♢b=a2−b2+2aba \diamondsuit b = a^2 - b^2 + 2ab with (a,b)=(x,2)(a, b) = (x, 2) and (a,b)=(2,x)(a, b) = (2, x) separately.
2
Set f(k)=g(k)f(k) = g(k) to solve for the positive constant kk.
k2+4k−4=−k2+4k+4  ⟹  2k2=8  ⟹  k2=4  ⟹  k=2k^2 + 4k - 4 = -k^2 + 4k + 4 \implies 2k^2 = 8 \implies k^2 = 4 \implies k = 2 (since k>0k > 0).
Equating the two algebraic function expressions and solving the resulting quadratic equation.
3
Evaluate the inner function expression g(−1)g(-1).
g(−1)=−(−1)2+4(−1)+4=−1−4+4=−1g(-1) = -(-1)^2 + 4(-1) + 4 = -1 - 4 + 4 = -1.
Substituting x=−1x = -1 into the formula for g(x)g(x).
4
Evaluate the outer function f(g(−1))=f(−1)f(g(-1)) = f(-1).
f(−1)=(−1)2+4(−1)−4=1−4−4=−7f(-1) = (-1)^2 + 4(-1) - 4 = 1 - 4 - 4 = -7.
Substituting the result from Step 3 into the formula for f(x)f(x).

Anahtar Kavram

Evaluating algebraic custom operations, solving functional equalities, and applying nested function compositions.
Soru 14Soru →

For all real numbers x≠0x \neq 0 and x≠1x \neq 1, the function ff is defined by f(x)=x−1xf(x) = \frac{x - 1}{x}. The composite function notation fn(x)f^n(x) is defined by f1(x)=f(x)f^1(x) = f(x) and fn(x)=f(fn−1(x))f^n(x) = f\left(f^{n-1}(x)\right) for all integers n≥2n \ge 2. What is the value of f2026(5)f^{2026}(5)?

Cevabı ve açıklamayı göster

Cevap: 45\frac{4}{5}

Cevap

45\frac{4}{5}
Evaluating successive compositions of f(x)=x−1xf(x) = \frac{x-1}{x} starting at x=5x = 5 yields f1(5)=45f^1(5) = \frac{4}{5}, f2(5)=−14f^2(5) = -\frac{1}{4}, and f3(5)=5f^3(5) = 5. This produces a repeating sequence of period 3. Since 2026≡1(mod3)2026 \equiv 1 \pmod 3, f2026(5)f^{2026}(5) equals f1(5)=45f^1(5) = \frac{4}{5}.

Adım Adım Çözüm

1
Evaluate the first iteration f1(5)f^1(5).
f1(5)=f(5)=5−15=45f^1(5) = f(5) = \frac{5 - 1}{5} = \frac{4}{5}.
Substitute x=5x = 5 into the definition of f(x)f(x).
2
Evaluate the second iteration f2(5)=f(f1(5))f^2(5) = f(f^1(5)).
f2(5)=f(45)=45−145=−1545=−14f^2(5) = f\left(\frac{4}{5}\right) = \frac{\frac{4}{5} - 1}{\frac{4}{5}} = \frac{-\frac{1}{5}}{\frac{4}{5}} = -\frac{1}{4}.
Substitute x=45x = \frac{4}{5} into the function definition.
3
Evaluate the third iteration f3(5)=f(f2(5))f^3(5) = f(f^2(5)).
f3(5)=f(−14)=−14−1−14=−54−14=5f^3(5) = f\left(-\frac{1}{4}\right) = \frac{-\frac{1}{4} - 1}{-\frac{1}{4}} = \frac{-\frac{5}{4}}{-\frac{1}{4}} = 5.
Substitute x=−14x = -\frac{1}{4} into the function definition.
4
Identify the periodic pattern and evaluate f2026(5)f^{2026}(5).
Since f3(5)=5f^3(5) = 5, the function values repeat in a cycle of length 3: (45,−14,5)(\frac{4}{5}, -\frac{1}{4}, 5). Dividing 2026 by 3 yields 2026=3×675+12026 = 3 \times 675 + 1, giving a remainder of 1. Therefore, f2026(5)=f1(5)=45f^{2026}(5) = f^1(5) = \frac{4}{5}.
The remainder determines the equivalent position in the 3-element repeating sequence.

Anahtar Kavram

Nested Function Composition and Periodicity
Soru 15Soru →

For all real numbers xx, the function ff is defined by f(x)=x2−4x+7f(x) = x^2 - 4x + 7 and the function gg is defined by g(x)=2x+1g(x) = 2x + 1. If f(g(k))=19f(g(k)) = 19 and k<0k < 0, what is the value of kk?

Cevabı ve açıklamayı göster

Cevap: -1.5

Cevap

The value of kk is −1.5-1.5.
To solve for kk, substitute g(k)=2k+1g(k) = 2k + 1 into f(x)f(x) to get f(g(k))=(2k+1)2−4(2k+1)+7f(g(k)) = (2k + 1)^2 - 4(2k + 1) + 7. Expanding and combining like terms yields 4k2−4k+44k^2 - 4k + 4. Setting this equal to 1919 gives 4k2−4k−15=04k^2 - 4k - 15 = 0. Factoring the quadratic expression yields (2k−5)(2k+3)=0(2k - 5)(2k + 3) = 0, which gives k=52=2.5k = \frac{5}{2} = 2.5 or k=−32=−1.5k = -\frac{3}{2} = -1.5. Since the problem specifies that k<0k < 0, the value of kk must be −1.5-1.5.

Adım Adım Çözüm

1
Express the nested composite function f(g(k))f(g(k)) in terms of kk
f(g(k))=(2k+1)2−4(2k+1)+7f(g(k)) = (2k + 1)^2 - 4(2k + 1) + 7
Substituting g(k)=2k+1g(k) = 2k + 1 for every occurrence of xx in f(x)f(x) sets up the composite function evaluation.
2
Expand terms and simplify the algebraic expression
f(g(k))=4k2−4k+4f(g(k)) = 4k^2 - 4k + 4
Expanding (2k+1)2=4k2+4k+1(2k + 1)^2 = 4k^2 + 4k + 1 and distributing −4(2k+1)=−8k−4-4(2k + 1) = -8k - 4, then combining like terms yields 4k2−4k+44k^2 - 4k + 4.
3
Set the composite function expression equal to 19 and form a standard quadratic equation
4k2−4k−15=04k^2 - 4k - 15 = 0
Subtracting 1919 from both sides produces a standard quadratic equation ax2+bx+c=0ax^2 + bx + c = 0.
4
Solve the quadratic equation and apply the condition k<0k < 0
k=−1.5k = -1.5
Factoring (2k−5)(2k+3)=0(2k - 5)(2k + 3) = 0 gives k=2.5k = 2.5 or k=−1.5k = -1.5. The constraint k<0k < 0 selects k=−1.5k = -1.5.

Anahtar Kavram

Composite function evaluation f(g(x))f(g(x)) combined with quadratic equation solving under domain constraints.
Soru 16Soru →

For all real numbers xx and yy, the custom operation ⊙\odot is defined by x⊙y=x2y−y2xx \odot y = x^2 y - y^2 x. The function ff is defined by f(t)=t⊙3f(t) = t \odot 3. If tt is a positive real number such that f(f(t))=0f(f(t)) = 0 and f(t)≠0f(t) \neq 0, what is the value of tt?

Cevabı ve açıklamayı göster

Cevap: 3+132\frac{3 + \sqrt{13}}{2}

Cevap

The correct value of tt is 3+132\frac{3 + \sqrt{13}}{2}.
Applying the custom binary operator gives f(t)=3t2−9tf(t) = 3t^2 - 9t. Substituting u=f(t)u = f(t) into f(u)=0f(u) = 0 yields 3u(u−3)=03u(u - 3) = 0, so u=0u = 0 or u=3u = 3. Because f(t)≠0f(t) \neq 0, it must be that f(t)=3f(t) = 3. Setting 3t2−9t=33t^2 - 9t = 3 leads to t2−3t−1=0t^2 - 3t - 1 = 0. Applying the quadratic formula yields the positive value 3+132\frac{3 + \sqrt{13}}{2}.

Adım Adım Çözüm

1
Express f(t)f(t) using the custom symbol definition.
f(t)=t⊙3=t2(3)−(3)2t=3t2−9tf(t) = t \odot 3 = t^2(3) - (3)^2 t = 3t^2 - 9t.
Apply the rule x⊙y=x2y−y2xx \odot y = x^2 y - y^2 x with x=tx = t and y=3y = 3.
2
Analyze the nested function condition f(f(t))=0f(f(t)) = 0.
Let u=f(t)u = f(t). Then f(u)=3u2−9u=3u(u−3)=0f(u) = 3u^2 - 9u = 3u(u - 3) = 0, which yields u=0u = 0 or u=3u = 3.
Evaluate the outer function ff at the argument u=f(t)u = f(t).
3
Apply the problem constraints to determine the exact value of f(t)f(t).
Since f(t)≠0f(t) \neq 0, u=f(t)=3u = f(t) = 3. Thus, 3t2−9t=33t^2 - 9t = 3.
Eliminate f(t)=0f(t) = 0 based on the explicit condition given in the problem.
4
Solve the quadratic equation for t>0t > 0.
Dividing 3t2−9t−3=03t^2 - 9t - 3 = 0 by 3 gives t2−3t−1=0t^2 - 3t - 1 = 0. Using the quadratic formula, t=−(−3)±(−3)2−4(1)(−1)2(1)=3±132t = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)} = \frac{3 \pm \sqrt{13}}{2}. Since t>0t > 0, t=3+132t = \frac{3 + \sqrt{13}}{2}.
Find the positive real root of the simplified quadratic equation.

Anahtar Kavram

Nested Function Evaluation and Custom Symbol Operations
Tahmini Süre:2m 30s
Soru 17Soru →
For all real numbers xx and yy, the custom operation ♢\diamondsuit is defined by x♢y=x2−y2−2x+2yx \diamondsuit y = x^2 - y^2 - 2x + 2y The function ff is defined for all real numbers tt by f(t)=t+1f(t) = t + 1. Which of the following statements must be true for all real numbers aa and bb? Select all such statements.

Geçerli olan tümünü seçin

Cevabı ve açıklamayı göster

Cevap: If a≠ba \neq b, then a♢b=0a \diamondsuit b = 0 if and only if a+b=2a + b = 2.; f(a)♢f(b)=a2−b2f(a) \diamondsuit f(b) = a^2 - b^2

Cevap

The statements 'If a≠ba \neq b, then a♢b=0a \diamondsuit b = 0 if and only if a+b=2a + b = 2' and 'f(a)♢f(b)=a2−b2f(a) \diamondsuit f(b) = a^2 - b^2' must be true.
The statement claiming that if a≠ba \neq b, a♢b=0a \diamondsuit b = 0 if and only if a+b=2a + b = 2 is correct because factoring a♢ba \diamondsuit b gives (a−b)(a+b−2)=0(a - b)(a + b - 2) = 0, which requires a+b=2a + b = 2 when a−b≠0a - b \neq 0. The statement claiming f(a)♢f(b)=a2−b2f(a) \diamondsuit f(b) = a^2 - b^2 is also correct because substituting f(a)=a+1f(a) = a + 1 and f(b)=b+1f(b) = b + 1 into the operation yields [(a+1)−(b+1)][(a+1)+(b+1)−2]=(a−b)(a+b)=a2−b2[(a + 1) - (b + 1)][(a + 1) + (b + 1) - 2] = (a - b)(a + b) = a^2 - b^2.

Adım Adım Çözüm

1
Factor the custom operation definition
x♢y=(x2−y2)−2(x−y)=(x−y)(x+y)−2(x−y)=(x−y)(x+y−2)x \diamondsuit y = (x^2 - y^2) - 2(x - y) = (x - y)(x + y) - 2(x - y) = (x - y)(x + y - 2)
Factoring groups terms into products that simplify statement testing.
2
Analyze the condition for a♢b=0a \diamondsuit b = 0 when a≠ba \neq b
(a−b)(a+b−2)=0  ⟹  a+b−2=0  ⟹  a+b=2(a - b)(a + b - 2) = 0 \implies a + b - 2 = 0 \implies a + b = 2
Since a≠ba \neq b, a−b≠0a - b \neq 0, allowing division by (a−b)(a - b) to isolate a+b=2a + b = 2.
3
Evaluate f(a)♢f(b)f(a) \diamondsuit f(b) using f(t)=t+1f(t) = t + 1
f(a)=a+1,f(b)=b+1  ⟹  f(a)♢f(b)=[(a+1)−(b+1)][(a+1)+(b+1)−2]=(a−b)(a+b)=a2−b2f(a) = a + 1, f(b) = b + 1 \implies f(a) \diamondsuit f(b) = [(a + 1) - (b + 1)][(a + 1) + (b + 1) - 2] = (a - b)(a + b) = a^2 - b^2
Direct substitution into the factored form simplifies the evaluation.
4
Evaluate remaining statements to identify counterexamples
f(a+b)=a+b+1≠a+b+2=f(a)+f(b)f(a + b) = a + b + 1 \neq a + b + 2 = f(a) + f(b); a♢(−a)=−4a≠0a \diamondsuit (-a) = -4a \neq 0; (a+b)♢0=a2+2ab+b2−2a−2b≠a2+b2−2a−2b(a + b) \diamondsuit 0 = a^2 + 2ab + b^2 - 2a - 2b \neq a^2 + b^2 - 2a - 2b
Confirming false statements ensures accurate multi-selection.

Anahtar Kavram

Algebraic manipulation of custom binary operations and function evaluation
Soru 18Soru →
For all real numbers xx, the function ff is defined by f(x)=x2−2xf(x) = x^2 - 2x. The custom binary operation ⊗\otimes is defined for all real numbers aa and bb by a⊗b=f(a+b)−f(a−b)a \otimes b = f(a + b) - f(a - b) If kk is a constant such that (k⊗3)⊗2=88(k \otimes 3) \otimes 2 = 88, what is the value of kk?
Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The value of kk is 2.
Expanding the definition a⊗b=f(a+b)−f(a−b)a \otimes b = f(a+b) - f(a-b) using f(x)=x2−2xf(x) = x^2 - 2x gives [(a+b)2−2(a+b)]−[(a−b)2−2(a−b)]=(a2+2ab+b2−2a−2b)−(a2−2ab+b2−2a+2b)=4ab−4b=4b(a−1)[(a+b)^2 - 2(a+b)] - [(a-b)^2 - 2(a-b)] = (a^2 + 2ab + b^2 - 2a - 2b) - (a^2 - 2ab + b^2 - 2a + 2b) = 4ab - 4b = 4b(a-1). Evaluating k⊗3k \otimes 3 yields 12k−1212k - 12. Substituting this into (12k−12)⊗2(12k - 12) \otimes 2 yields 4(2)(12k−12−1)=96k−1044(2)(12k - 12 - 1) = 96k - 104. Setting 96k−104=8896k - 104 = 88 gives 96k=19296k = 192, so k=2k = 2.

Adım Adım Çözüm

1
Express the custom operation a⊗ba \otimes b in simplified algebraic terms.
a⊗b=f(a+b)−f(a−b)=[(a+b)2−2(a+b)]−[(a−b)2−2(a−b)]=4ab−4b=4b(a−1)a \otimes b = f(a+b) - f(a-b) = [(a+b)^2 - 2(a+b)] - [(a-b)^2 - 2(a-b)] = 4ab - 4b = 4b(a - 1).
Expanding and canceling common terms simplifies the binary operation definition.
2
Evaluate the inner operation k⊗3k \otimes 3.
k⊗3=4(3)(k−1)=12k−12k \otimes 3 = 4(3)(k - 1) = 12k - 12.
Substitute a=ka = k and b=3b = 3 into the simplified operation formula 4b(a−1)4b(a - 1).
3
Evaluate the outer operation (12k−12)⊗2(12k - 12) \otimes 2.
(12k−12)⊗2=4(2)[(12k−12)−1]=8(12k−13)=96k−104(12k - 12) \otimes 2 = 4(2)[(12k - 12) - 1] = 8(12k - 13) = 96k - 104.
Substitute a=12k−12a = 12k - 12 and b=2b = 2 into 4b(a−1)4b(a - 1).
4
Set the resulting expression equal to 88 and solve for kk.
96k−104=88  ⟹  96k=192  ⟹  k=296k - 104 = 88 \implies 96k = 192 \implies k = 2.
Linear algebraic equation solving yields the value of kk.

Anahtar Kavram

Custom Binary Operations and Nested Function Evaluation
Soru 19Soru →

For any real number xx, the custom unary operation △\triangle is defined by △(x)=2x−x2\triangle(x) = 2x - x^2. For all real numbers uu and vv, the custom binary operation ⊙\odot is defined by u⊙v=△(u)+△(v)+uvu \odot v = \triangle(u) + \triangle(v) + uv. The function ff is defined for all real numbers xx by f(x)=x⊙(3−x)f(x) = x \odot (3 - x). What is the maximum value of f(x)f(x)?

Cevabı ve açıklamayı göster

Cevap: 3.75

Cevap

The maximum value of f(x)f(x) is 3.753.75 (or 154\frac{15}{4}).
Expanding f(x)=x⊙(3−x)f(x) = x \odot (3 - x) yields △(x)+△(3−x)+x(3−x)\triangle(x) + \triangle(3 - x) + x(3 - x). Substituting △(t)=2t−t2\triangle(t) = 2t - t^2 gives (2x−x2)+(6−2x−(9−6x+x2))+(3x−x2)=−3x2+9x−3(2x - x^2) + (6 - 2x - (9 - 6x + x^2)) + (3x - x^2) = -3x^2 + 9x - 3. The vertex of this quadratic function occurs at x=96=1.5x = \frac{9}{6} = 1.5, where f(1.5)=−3(2.25)+13.5−3=3.75f(1.5) = -3(2.25) + 13.5 - 3 = 3.75.

Adım Adım Çözüm

1
Evaluate the custom unary operation △\triangle for each argument
△(x)=2x−x2\triangle(x) = 2x - x^2 and △(3−x)=2(3−x)−(3−x)2=−x2+4x−3\triangle(3-x) = 2(3-x) - (3-x)^2 = -x^2 + 4x - 3
Substitute xx and 3−x3-x into the definition △(t)=2t−t2\triangle(t) = 2t - t^2 and expand carefully.
2
Compute the product term uvuv
x(3 - x) = 3x - x^2
The binary definition u⊙vu \odot v includes an additive product term uvuv.
3
Sum all components to construct the explicit quadratic expression for f(x)f(x)
f(x) = (2x - x^2) + (-x^2 + 4x - 3) + (3x - x^2) = -3x^2 + 9x - 3
Combine like terms for x2x^2, xx, and the constant.
4
Determine the vertex of the downward-opening parabola f(x)=−3x2+9x−3f(x) = -3x^2 + 9x - 3
x = \frac{3}{2} = 1.5 ,yielding, yielding f(1.5) = 3.75$
Since the coefficient of x2x^2 is negative (−3<0-3 < 0), the maximum occurs at x=−b2a=96=1.5x = -\frac{b}{2a} = \frac{9}{6} = 1.5.

Anahtar Kavram

Custom operations combined with quadratic function optimization
Tahmini Süre:2m 0s
Soru 20Soru →

For all real numbers xx and yy, the custom binary operation ⊙\odot is defined by x⊙y=x+y−xyx \odot y = x + y - xy. Which of the following statements must be true for all real numbers aa, bb, and cc? Select all such statements.

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Cevap: a⊙b=b⊙aa \odot b = b \odot a; a⊙1=1a \odot 1 = 1; (a⊙b)⊙c=a⊙(b⊙c)(a \odot b) \odot c = a \odot (b \odot c)

Cevap

The statements a⊙b=b⊙aa \odot b = b \odot a, a⊙1=1a \odot 1 = 1, and (a⊙b)⊙c=a⊙(b⊙c)(a \odot b) \odot c = a \odot (b \odot c) must be true for all real numbers aa, bb, and cc.
The custom operation is commutative (a⊙b=b⊙aa \odot b = b \odot a), maps any real number operated with 11 to 11 (a⊙1=1a \odot 1 = 1), and is associative ((a⊙b)⊙c=a⊙(b⊙c)(a \odot b) \odot c = a \odot (b \odot c)). These three properties hold universally for all real numbers.

Adım Adım Çözüm

1
Evaluate a⊙ba \odot b and b⊙ab \odot a to test commutativity.
a⊙b=a+b−aba \odot b = a + b - ab and b⊙a=b+a−bab \odot a = b + a - ba. Since addition and multiplication of real numbers are commutative, a+b−ab=b+a−baa + b - ab = b + a - ba.
Verify if order of operands alters the result.
2
Evaluate a⊙1a \odot 1 using the operation definition.
a⊙1=a+1−a(1)=a+1−a=1a \odot 1 = a + 1 - a(1) = a + 1 - a = 1.
Test the behavior of operating with the constant 11.
3
Evaluate a⊙0a \odot 0 to verify the zero property statement.
a⊙0=a+0−a(0)=aa \odot 0 = a + 0 - a(0) = a. This equals aa, not 00 for general values of aa.
Check if operating with 00 results in 00.
4
Evaluate both sides of (a⊙b)⊙c=a⊙(b⊙c)(a \odot b) \odot c = a \odot (b \odot c) to test associativity.
Left side: (a⊙b)⊙c=(a+b−ab)⊙c=(a+b−ab)+c−(a+b−ab)c=a+b+c−ab−ac−bc+abc(a \odot b) \odot c = (a + b - ab) \odot c = (a + b - ab) + c - (a + b - ab)c = a + b + c - ab - ac - bc + abc. Right side: a⊙(b⊙c)=a⊙(b+c−bc)=a+(b+c−bc)−a(b+c−bc)=a+b+c−bc−ab−ac+abca \odot (b \odot c) = a \odot (b + c - bc) = a + (b + c - bc) - a(b + c - bc) = a + b + c - bc - ab - ac + abc. Both sides are identical.
Verify whether grouping alters the result.
5
Evaluate a⊙aa \odot a.
a⊙a=a+a−a(a)=2a−a2a \odot a = a + a - a(a) = 2a - a^2. This is not equal to a2a^2 except when 2a−a2=a2  ⟹  2a2−2a=0  ⟹  a=02a - a^2 = a^2 \implies 2a^2 - 2a = 0 \implies a=0 or a=1a=1.
Determine if the self-operation yields a2a^2 for all real numbers.

Anahtar Kavram

Evaluating algebraic properties (commutativity, identity, associativity) of custom binary operations.
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