Functions and Custom Symbol Operations

29 soru

Soru 21Soru →

For all positive real numbers xx and yy, the custom operation ⋆\star is defined by x⋆y=x2+y2xyx \star y = \frac{x^2 + y^2}{xy}. The function gg is defined by g(t)=4(t⋆3)−7g(t) = 4(t \star 3) - 7. If g(k)=3g(k) = 3, where k>3k > 3, what is the value of kk?

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Cevap: 6

Cevap

The value of kk is 6.
Solving g(k)=3g(k) = 3 yields 4(k⋆3)−7=34(k \star 3) - 7 = 3, so k⋆3=52k \star 3 = \frac{5}{2}. Replacing k⋆3k \star 3 with k2+93k\frac{k^2+9}{3k} results in the quadratic equation 2k2−15k+18=02k^2 - 15k + 18 = 0. The roots are 1.51.5 and 66. Because kk must be strictly greater than 33, the only valid solution is 66.

Adım Adım Çözüm

1
Isolate the custom operation term in the function equation
k⋆3=52k \star 3 = \frac{5}{2}
Setting g(k)=4(k⋆3)−7=3g(k) = 4(k \star 3) - 7 = 3 allows us to find the required value of k⋆3k \star 3 by adding 7 to both sides and dividing by 4.
2
Express the custom operation algebraically using its definition
k2+93k=52\frac{k^2 + 9}{3k} = \frac{5}{2}
By definition, x⋆y=x2+y2xyx \star y = \frac{x^2 + y^2}{xy}, so k⋆3=k2+323(k)=k2+93kk \star 3 = \frac{k^2 + 3^2}{3(k)} = \frac{k^2 + 9}{3k}.
3
Form and solve the quadratic equation
k = 1.5 \text{ or } k = 6
Cross-multiplying gives 2(k2+9)=5(3k)2(k^2 + 9) = 5(3k), which simplifies to 2k2−15k+18=02k^2 - 15k + 18 = 0. Factoring as (2k−3)(k−6)=0(2k - 3)(k - 6) = 0 yields two roots: k=1.5k = 1.5 and k=6k = 6.
4
Apply the domain constraint
k = 6
The problem specifies that k>3k > 3. Since 1.5≤31.5 \le 3, k=1.5k = 1.5 is rejected, leaving k=6k = 6 as the unique correct answer.

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Custom Binary Symbol Operations and Composite Functions
Soru 22Soru →

For all real numbers xx, the function gg is defined by g(x)=cx+5g(x) = cx + 5, where cc is a constant. The custom operation ⋄\diamond is defined for all real numbers aa and bb by a⋄b=g(a+b)−g(a−b)a \diamond b = g(a+b) - g(a-b). If 3⋄1=123 \diamond 1 = 12, what is the value of g(4)g(4)?

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Cevap: 29

Cevap

29
Applying the function definition g(x)=cx+5g(x) = cx + 5 to the custom operation yields a⋄b=[c(a+b)+5]−[c(a−b)+5]=2bca \diamond b = [c(a+b)+5] - [c(a-b)+5] = 2bc. Substituting a=3a=3 and b=1b=1 into 3⋄1=123 \diamond 1 = 12 gives 2(1)c=122(1)c = 12, so c=6c = 6. Consequently, g(x)=6x+5g(x) = 6x + 5, and evaluating at x=4x = 4 yields g(4)=6(4)+5=29g(4) = 6(4) + 5 = 29.

Adım Adım Çözüm

1
Substitute (a+b)(a+b) and (a−b)(a-b) into the function definition g(x)=cx+5g(x) = cx + 5 to simplify a⋄ba \diamond b.
a⋄b=[c(a+b)+5]−[c(a−b)+5]=2bca \diamond b = [c(a+b) + 5] - [c(a-b) + 5] = 2bc.
Applying the definition of the custom binary operation in terms of function gg eliminates the constant term 55.
2
Use the given equality 3⋄1=123 \diamond 1 = 12 to determine the constant cc.
2(1)c=12  ⟹  2c=12  ⟹  c=62(1)c = 12 \implies 2c = 12 \implies c = 6.
Plugging a=3a=3 and b=1b=1 into 2bc=122bc = 12 yields an equation in terms of cc.
3
Evaluate g(4)g(4) using c=6c = 6.
g(4)=6(4)+5=29g(4) = 6(4) + 5 = 29.
Substituting x=4x = 4 into g(x)=6x+5g(x) = 6x + 5 calculates the required numerical value.

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Functions and Custom Symbol Operations
Tahmini Süre:1m 30s
Soru 23Soru →
For all real numbers xx, the function ff is defined by f(x)=x2−kf(x) = x^2 - k, where kk is a constant. The custom operation ⋆\star is defined for all real numbers aa and bb by
a⋆b=f(a+b)−f(a)−f(b)a \star b = f(a + b) - f(a) - f(b)
If 5⋆(−3)=−125 \star (-3) = -12, what is the value of kk?
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Cevap: 18

Cevap

The value of kk is 18.
Evaluating f(2)=4−kf(2) = 4 - k, f(5)=25−kf(5) = 25 - k, and f(−3)=9−kf(-3) = 9 - k yields 5⋆(−3)=(4−k)−(34−2k)=−30+k5 \star (-3) = (4 - k) - (34 - 2k) = -30 + k. Setting −30+k=−12-30 + k = -12 correctly gives k=18k = 18.

Adım Adım Çözüm

1
Evaluate f(a+b)f(a+b) for a=5a=5 and b=−3b=-3
a+b=5+(−3)=2a + b = 5 + (-3) = 2, so f(2)=22−k=4−kf(2) = 2^2 - k = 4 - k.
Substitute the input a+b=2a+b = 2 into the function definition f(x)=x2−kf(x) = x^2 - k.
2
Evaluate f(a)f(a) and f(b)f(b) individually
f(5)=52−k=25−kf(5) = 5^2 - k = 25 - k and f(−3)=(−3)2−k=9−kf(-3) = (-3)^2 - k = 9 - k.
Apply the function rule to inputs 5 and -3.
3
Substitute the evaluated expressions into the custom operation definition
5⋆(−3)=(4−k)−[(25−k)+(9−k)]=(4−k)−(34−2k)=−30+k5 \star (-3) = (4 - k) - [(25 - k) + (9 - k)] = (4 - k) - (34 - 2k) = -30 + k.
Simplify the algebraic expression by combining like terms and distributing the negative sign.
4
Solve for kk using the given equation 5⋆(−3)=−125 \star (-3) = -12
−30+k=−12  ⟹  k=18-30 + k = -12 \implies k = 18.
Add 30 to both sides of the equation to isolate kk.

Anahtar Kavram

Evaluating custom binary operations by substituting function definitions and simplifying algebraic expressions.
Tahmini Süre:1m 30s
Soru 24Soru →

For all real numbers xx, the function ff is defined by f(x)=3−x2f(x) = 3 - x^2. For all real numbers x≠−2x \neq -2, the function gg is defined by g(x)=2x−1x+2g(x) = \frac{2x - 1}{x + 2}. What is the value of g(f(−3))g(f(-3))?

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Cevap: 134\frac{13}{4}

Cevap

The correct answer is 134\frac{13}{4}.
First, evaluate the inner expression f(−3)=3−(−3)2=3−9=−6f(-3) = 3 - (-3)^2 = 3 - 9 = -6. Next, evaluate the outer expression at this value: g(−6)=2(−6)−1−6+2=−13−4=134g(-6) = \frac{2(-6) - 1}{-6 + 2} = \frac{-13}{-4} = \frac{13}{4}.

Adım Adım Çözüm

1
Evaluate the inner function f(−3)f(-3)
f(−3)=3−(−3)2=3−9=−6f(-3) = 3 - (-3)^2 = 3 - 9 = -6
Applying the definition of f(x)f(x) where the negative base is squared to yield positive 99.
2
Substitute the result into the outer function g(x)g(x)
g(−6)=2(−6)−1−6+2=−12−1−4=−13−4=134g(-6) = \frac{2(-6) - 1}{-6 + 2} = \frac{-12 - 1}{-4} = \frac{-13}{-4} = \frac{13}{4}
Evaluating g(x)g(x) at x=−6x = -6 and simplifying the fraction.

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Nested Function Evaluation g(f(x))g(f(x))
Tahmini Süre:1m 30s
Soru 25Soru →

For all non-zero real numbers aa and bb, the custom binary operation ♢\diamondsuit is defined by a♢b=a2−b2aba \diamondsuit b = \frac{a^2 - b^2}{ab}. If the function ff is defined for all x≠0x \neq 0 by f(x)=x♢2f(x) = x \diamondsuit 2, what is the value of f(4)−f(1)f(4) - f(1)?

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Cevap: 3

Cevap

3
Evaluating f(4)=4♢2=16−48=1.5f(4) = 4 \diamondsuit 2 = \frac{16 - 4}{8} = 1.5 and f(1)=1♢2=1−42=−1.5f(1) = 1 \diamondsuit 2 = \frac{1 - 4}{2} = -1.5, the required difference is f(4)−f(1)=1.5−(−1.5)=3f(4) - f(1) = 1.5 - (-1.5) = 3.

Adım Adım Çözüm

1
Evaluate f(4)f(4) using the custom operation definition
f(4)=42−224⋅2=128=1.5f(4) = \frac{4^2 - 2^2}{4 \cdot 2} = \frac{12}{8} = 1.5
Substitute a=4a = 4 and b=2b = 2 into a♢b=a2−b2aba \diamondsuit b = \frac{a^2 - b^2}{ab}.
2
Evaluate f(1)f(1) using the custom operation definition
f(1)=12−221⋅2=−1.5f(1) = \frac{1^2 - 2^2}{1 \cdot 2} = -1.5
Substitute a=1a = 1 and b=2b = 2 into a♢b=a2−b2aba \diamondsuit b = \frac{a^2 - b^2}{ab}.
3
Compute the difference f(4)−f(1)f(4) - f(1)
3
Subtracting −1.5-1.5 from 1.51.5 yields 1.5−(−1.5)=1.5+1.5=31.5 - (-1.5) = 1.5 + 1.5 = 3.

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Custom Symbol Operations and Function Evaluation
Tahmini Süre:1m 30s
Soru 26Soru →

The function ff is defined by f(t)=2tf(t) = 2^t for all real numbers tt, and the binary operation ⊙\odot is defined for all non-zero real numbers mm and nn by m⊙n=mn−nmm \odot n = \frac{m}{n} - \frac{n}{m}. What is the value of f(2)⊙f(−2)f(2) \odot f(-2)?

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Cevap: 25516\frac{255}{16}

Cevap

25516\frac{255}{16}
Evaluating f(2)f(2) yields 22=42^2 = 4, and evaluating f(−2)f(-2) yields 2−2=142^{-2} = \frac{1}{4}. Substituting these values into the binary operation rule m⊙n=mn−nmm \odot n = \frac{m}{n} - \frac{n}{m} gives 41/4−1/44=16−116=25516\frac{4}{1/4} - \frac{1/4}{4} = 16 - \frac{1}{16} = \frac{255}{16}.

Adım Adım Çözüm

1
Evaluate the function f(t)=2tf(t) = 2^t at t=2t = 2 and t=−2t = -2.
f(2)=22=4f(2) = 2^2 = 4 and f(−2)=2−2=122=14f(-2) = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}.
Negative exponents follow the rule x−a=1xax^{-a} = \frac{1}{x^a}.
2
Substitute m=f(2)=4m = f(2) = 4 and n=f(−2)=14n = f(-2) = \frac{1}{4} into the custom operation definition m⊙n=mn−nmm \odot n = \frac{m}{n} - \frac{n}{m}.
4⊙14=414−1444 \odot \frac{1}{4} = \frac{4}{\frac{1}{4}} - \frac{\frac{1}{4}}{4}.
Apply the defined binary operation rule.
3
Simplify the complex fraction terms.
414=4×4=16\frac{4}{\frac{1}{4}} = 4 \times 4 = 16 and 144=14×4=116\frac{\frac{1}{4}}{4} = \frac{1}{4 \times 4} = \frac{1}{16}.
Dividing by a fraction is equivalent to multiplying by its reciprocal.
4
Subtract the two simplified terms.
16−116=25616−116=2551616 - \frac{1}{16} = \frac{256}{16} - \frac{1}{16} = \frac{255}{16}.
Find a common denominator to compute the final value.

Anahtar Kavram

Function evaluation with exponent rules combined with custom binary symbol operations
Soru 27Soru →

For all non-zero real numbers xx and yy, the custom binary operation ⋆\star is defined by x⋆y=xy−yxx \star y = \frac{x}{y} - \frac{y}{x}. Which of the following statements must be true for all non-zero real numbers aa, bb, and cc? Select all that apply.

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Cevap: a⋆b=−(b⋆a)a \star b = -(b \star a); a⋆a=0a \star a = 0; (a⋆b)2=a2b2+b2a2−2(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2

Cevap

The statements a⋆b=−(b⋆a)a \star b = -(b \star a), a⋆a=0a \star a = 0, and (a⋆b)2=a2b2+b2a2−2(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2 must be true for all non-zero real numbers aa, bb, and cc.
The statements showing anti-commutativity, zero self-operation, and the expanded squared expression are all mathematically valid for all non-zero real numbers upon expanding their definitions using algebra.

Adım Adım Çözüm

1
Evaluate a⋆b=−(b⋆a)a \star b = -(b \star a)
a⋆b=ab−baa \star b = \frac{a}{b} - \frac{b}{a} and −(b⋆a)=−(ba−ab)=ab−ba-(b \star a) = -\left(\frac{b}{a} - \frac{a}{b}\right) = \frac{a}{b} - \frac{b}{a}.
Verify anti-commutativity property by direct substitution.
2
Evaluate a⋆a=0a \star a = 0
a⋆a=aa−aa=1−1=0a \star a = \frac{a}{a} - \frac{a}{a} = 1 - 1 = 0.
Verify identity property for identical inputs.
3
Evaluate (a⋆b)2(a \star b)^2
(a⋆b)2=(ab−ba)2=(ab)2−2(ab)(ba)+(ba)2=a2b2−2(1)+b2a2=a2b2+b2a2−2(a \star b)^2 = \left(\frac{a}{b} - \frac{b}{a}\right)^2 = \left(\frac{a}{b}\right)^2 - 2\left(\frac{a}{b}\right)\left(\frac{b}{a}\right) + \left(\frac{b}{a}\right)^2 = \frac{a^2}{b^2} - 2(1) + \frac{b^2}{a^2} = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2.
Apply binomial expansion to the squared custom operation.
4
Test associativity statement a⋆(b⋆c)=(a⋆b)⋆ca \star (b \star c) = (a \star b) \star c with a counterexample
Let a=4,b=2,c=1a = 4, b = 2, c = 1. Then b⋆c=21−12=32b \star c = \frac{2}{1} - \frac{1}{2} = \frac{3}{2}. a⋆(b⋆c)=4⋆32=43/2−3/24=83−38=5524a \star (b \star c) = 4 \star \frac{3}{2} = \frac{4}{3/2} - \frac{3/2}{4} = \frac{8}{3} - \frac{3}{8} = \frac{55}{24}. Meanwhile, a⋆b=4⋆2=42−24=32a \star b = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}, and (a⋆b)⋆c=32⋆1=3/21−13/2=32−23=56(a \star b) \star c = \frac{3}{2} \star 1 = \frac{3/2}{1} - \frac{1}{3/2} = \frac{3}{2} - \frac{2}{3} = \frac{5}{6}. Since 5524≠56\frac{55}{24} \neq \frac{5}{6}, associativity fails.
A single counterexample disproves a general identity statement.
5
Test distributivity statement a⋆(b+c)=(a⋆b)+(a⋆c)a \star (b + c) = (a \star b) + (a \star c) with a counterexample
Let a=1,b=1,c=1a = 1, b = 1, c = 1. Then a⋆(b+c)=1⋆2=12−21=−32a \star (b + c) = 1 \star 2 = \frac{1}{2} - \frac{2}{1} = -\frac{3}{2}. Meanwhile, (a⋆b)+(a⋆c)=(1⋆1)+(1⋆1)=0+0=0(a \star b) + (a \star c) = (1 \star 1) + (1 \star 1) = 0 + 0 = 0. Since −32≠0-\frac{3}{2} \neq 0, distributivity fails.
A single counterexample disproves distributivity over addition.

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Evaluating algebraic properties and identity statements for custom defined binary operations.
Soru 28Soru →

For all real numbers xx and yy, the custom operation ⊙\odot is defined by x⊙y=x∣y∣−y∣x∣x \odot y = x|y| - y|x|. Which of the following statements must be true for all real numbers xx and yy? Select all such statements.

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Cevap: x⊙y=0x \odot y = 0 whenever xx and yy have the same sign; x⊙y=−(y⊙x)x \odot y = -(y \odot x); If x>0x > 0 and y<0y < 0, then x⊙y>0x \odot y > 0

Cevap

The correct statements are the statement asserting x⊙y=0x \odot y = 0 when xx and yy have the same sign, the statement asserting anti-commutativity x⊙y=−(y⊙x)x \odot y = -(y \odot x), and the statement asserting x⊙y>0x \odot y > 0 when x>0x > 0 and y<0y < 0.
The operation x⊙y=x∣y∣−y∣x∣x \odot y = x|y| - y|x| produces 0 whenever xx and yy share the same sign because terms evaluate to identical quantities. Swapping variables negates the expression, establishing anti-commutativity. When xx is positive and yy is negative, x⊙yx \odot y simplifies to −2xy-2xy, which is strictly greater than 0 since xy<0xy < 0.

Adım Adım Çözüm

1
Analyze the first statement regarding same-sign inputs
If x>0x > 0 and y>0y > 0, ∣x∣=x|x|=x and ∣y∣=y|y|=y, so x⊙y=xy−yx=0x \odot y = xy - yx = 0. If x<0x < 0 and y<0y < 0, ∣x∣=−x|x|=-x and ∣y∣=−y|y|=-y, so x⊙y=x(−y)−y(−x)=−xy+xy=0x \odot y = x(-y) - y(-x) = -xy + xy = 0. Thus, x⊙y=0x \odot y = 0 when xx and yy have the same sign.
Verifying the definition under both positive and negative cases of identical sign.
2
Analyze the second statement regarding operand order reversal
y⊙x=y∣x∣−x∣y∣=−(x∣y∣−y∣x∣)=−(x⊙y)y \odot x = y|x| - x|y| = -(x|y| - y|x|) = -(x \odot y), which holds universally for all real numbers.
Testing anti-commutativity by algebraic substitution into the custom operation.
3
Analyze the third statement for opposite signs (x>0x > 0 and y<0y < 0)
Since x>0x > 0, ∣x∣=x|x|=x. Since y<0y < 0, ∣y∣=−y|y|=-y. Substituting yields x(−y)−y(x)=−xy−xy=−2xyx(-y) - y(x) = -xy - xy = -2xy. Because x>0x > 0 and y<0y < 0, the product xyxy is negative, making −2xy-2xy strictly positive.
Determining the overall algebraic sign of the expression when variables have opposite signs.
4
Counter-test the remaining statements to verify incorrectness
For x⊙(−x)x \odot (-x) with x=1x=1: 1⊙(−1)=1∣−1∣−(−1)∣1∣=1−(−1)=2≠01 \odot (-1) = 1|-1| - (-1)|1| = 1 - (-1) = 2 \neq 0. For associativity with x=2,y=−1,z=−1x=2, y=-1, z=-1: (2⊙−1)⊙−1=4⊙−1=8(2 \odot -1) \odot -1 = 4 \odot -1 = 8, but 2⊙(−1⊙−1)=2⊙0=02 \odot (-1 \odot -1) = 2 \odot 0 = 0.
Demonstrating specific counterexamples for false generalizations.

Anahtar Kavram

Custom Binary Operations and Absolute Value Properties
Tahmini Süre:1m 30s
Soru 29Soru →

For all positive real numbers aa and bb, the custom operation ♢\diamondsuit is defined by a♢b=a2+b2aba \diamondsuit b = \frac{a^2 + b^2}{ab}. The function ff is defined for all x>0x > 0 by f(x)=x♢4f(x) = x \diamondsuit 4. If f(x)=2.5f(x) = 2.5, what is the value of xx that is greater than 44?

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Cevap: 8

Cevap

The value of xx greater than 44 is 88.
Applying the custom operator gives f(x)=x2+164xf(x) = \frac{x^2 + 16}{4x}. Setting this equal to 2.52.5 yields x2+164x=52\frac{x^2 + 16}{4x} = \frac{5}{2}, which simplifies to x2−10x+16=0x^2 - 10x + 16 = 0. The roots are x=2x = 2 and x=8x = 8. Since xx must be greater than 44, the only valid answer is 88.

Adım Adım Çözüm

1
Substitute a=xa = x and b=4b = 4 into the custom operation definition a♢b=a2+b2aba \diamondsuit b = \frac{a^2 + b^2}{ab}.
f(x)=x2+164xf(x) = \frac{x^2 + 16}{4x}
This establishes the explicit algebraic rule for the function f(x)f(x).
2
Set f(x)f(x) equal to 2.52.5 and clear the fraction.
x2+164x=2.5  ⟹  x2+16=10x\frac{x^2 + 16}{4x} = 2.5 \implies x^2 + 16 = 10x
Multiplying both sides by 4x4x converts the rational equation into a standard polynomial equation.
3
Rearrange into standard quadratic form and solve by factoring.
x2−10x+16=0  ⟹  (x−2)(x−8)=0  ⟹  x=2x^2 - 10x + 16 = 0 \implies (x - 2)(x - 8) = 0 \implies x = 2 or x=8x = 8
Factoring determines all potential positive real solutions for xx.
4
Select the solution satisfying the constraint x>4x > 4.
x=8x = 8
The question explicitly specifies that xx must be greater than 44, eliminating x=2x = 2.

Anahtar Kavram

Evaluating custom binary operations and solving algebraic function equations involving quadratic constraints.
ÖncekiSayfa 2 / 2
Functions and Custom Symbol Operations Alıştırma Soruları — GRE General Test — Sayfa 2 | Examkin