Functions and Custom Symbol Operations

29 soru

Soru 21Soru

For all positive real numbers xx and yy, the custom operation \star is defined by xy=x2+y2xyx \star y = \frac{x^2 + y^2}{xy}. The function gg is defined by g(t)=4(t3)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(k3)7=34(k \star 3) - 7 = 3, so k3=52k \star 3 = \frac{5}{2}. Replacing k3k \star 3 with k2+93k\frac{k^2+9}{3k} results in the quadratic equation 2k215k+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
k3=52k \star 3 = \frac{5}{2}
Setting g(k)=4(k3)7=3g(k) = 4(k \star 3) - 7 = 3 allows us to find the required value of k3k \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, xy=x2+y2xyx \star y = \frac{x^2 + y^2}{xy}, so k3=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 2k215k+18=02k^2 - 15k + 18 = 0. Factoring as (2k3)(k6)=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 1.531.5 \le 3, k=1.5k = 1.5 is rejected, leaving k=6k = 6 as the unique correct answer.

Anahtar Kavram

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 ab=g(a+b)g(ab)a \diamond b = g(a+b) - g(a-b). If 31=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 ab=[c(a+b)+5][c(ab)+5]=2bca \diamond b = [c(a+b)+5] - [c(a-b)+5] = 2bc. Substituting a=3a=3 and b=1b=1 into 31=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 (ab)(a-b) into the function definition g(x)=cx+5g(x) = cx + 5 to simplify aba \diamond b.
ab=[c(a+b)+5][c(ab)+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 31=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.

Anahtar Kavram

Functions and Custom Symbol Operations
Tahmini Süre:1m 30s
Soru 23Soru
For all real numbers xx, the function ff is defined by f(x)=x2kf(x) = x^2 - k, where kk is a constant. The custom operation \star is defined for all real numbers aa and bb by
ab=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)=4kf(2) = 4 - k, f(5)=25kf(5) = 25 - k, and f(3)=9kf(-3) = 9 - k yields 5(3)=(4k)(342k)=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)=22k=4kf(2) = 2^2 - k = 4 - k.
Substitute the input a+b=2a+b = 2 into the function definition f(x)=x2kf(x) = x^2 - k.
2
Evaluate f(a)f(a) and f(b)f(b) individually
f(5)=52k=25kf(5) = 5^2 - k = 25 - k and f(3)=(3)2k=9kf(-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)=(4k)[(25k)+(9k)]=(4k)(342k)=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)=3x2f(x) = 3 - x^2. For all real numbers x2x \neq -2, the function gg is defined by g(x)=2x1x+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=39=6f(-3) = 3 - (-3)^2 = 3 - 9 = -6. Next, evaluate the outer expression at this value: g(6)=2(6)16+2=134=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=39=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)16+2=1214=134=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.

Anahtar Kavram

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 ab=a2b2aba \diamondsuit b = \frac{a^2 - b^2}{ab}. If the function ff is defined for all x0x \neq 0 by f(x)=x2f(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)=42=1648=1.5f(4) = 4 \diamondsuit 2 = \frac{16 - 4}{8} = 1.5 and f(1)=12=142=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)=422242=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 ab=a2b2aba \diamondsuit b = \frac{a^2 - b^2}{ab}.
2
Evaluate f(1)f(1) using the custom operation definition
f(1)=122212=1.5f(1) = \frac{1^2 - 2^2}{1 \cdot 2} = -1.5
Substitute a=1a = 1 and b=2b = 2 into ab=a2b2aba \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.

Anahtar Kavram

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 mn=mnnmm \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 22=142^{-2} = \frac{1}{4}. Substituting these values into the binary operation rule mn=mnnmm \odot n = \frac{m}{n} - \frac{n}{m} gives 41/41/44=16116=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)=22=122=14f(-2) = 2^{-2} = \frac{1}{2^2} = \frac{1}{4}.
Negative exponents follow the rule xa=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 mn=mnnmm \odot n = \frac{m}{n} - \frac{n}{m}.
414=4141444 \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.
16116=25616116=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 xy=xyyxx \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: ab=(ba)a \star b = -(b \star a); aa=0a \star a = 0; (ab)2=a2b2+b2a22(a \star b)^2 = \frac{a^2}{b^2} + \frac{b^2}{a^2} - 2

Cevap

The statements ab=(ba)a \star b = -(b \star a), aa=0a \star a = 0, and (ab)2=a2b2+b2a22(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.

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1
Evaluate ab=(ba)a \star b = -(b \star a)
ab=abbaa \star b = \frac{a}{b} - \frac{b}{a} and (ba)=(baab)=abba-(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 aa=0a \star a = 0
aa=aaaa=11=0a \star a = \frac{a}{a} - \frac{a}{a} = 1 - 1 = 0.
Verify identity property for identical inputs.
3
Evaluate (ab)2(a \star b)^2
(ab)2=(abba)2=(ab)22(ab)(ba)+(ba)2=a2b22(1)+b2a2=a2b2+b2a22(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(bc)=(ab)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 bc=2112=32b \star c = \frac{2}{1} - \frac{1}{2} = \frac{3}{2}. a(bc)=432=43/23/24=8338=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, ab=42=4224=32a \star b = 4 \star 2 = \frac{4}{2} - \frac{2}{4} = \frac{3}{2}, and (ab)c=321=3/2113/2=3223=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 552456\frac{55}{24} \neq \frac{5}{6}, associativity fails.
A single counterexample disproves a general identity statement.
5
Test distributivity statement a(b+c)=(ab)+(ac)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)=12=1221=32a \star (b + c) = 1 \star 2 = \frac{1}{2} - \frac{2}{1} = -\frac{3}{2}. Meanwhile, (ab)+(ac)=(11)+(11)=0+0=0(a \star b) + (a \star c) = (1 \star 1) + (1 \star 1) = 0 + 0 = 0. Since 320-\frac{3}{2} \neq 0, distributivity fails.
A single counterexample disproves distributivity over addition.

Anahtar Kavram

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

Cevap

The correct statements are the statement asserting xy=0x \odot y = 0 when xx and yy have the same sign, the statement asserting anti-commutativity xy=(yx)x \odot y = -(y \odot x), and the statement asserting xy>0x \odot y > 0 when x>0x > 0 and y<0y < 0.
The operation xy=xyyxx \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, xyx \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 xy=xyyx=0x \odot y = xy - yx = 0. If x<0x < 0 and y<0y < 0, x=x|x|=-x and y=y|y|=-y, so xy=x(y)y(x)=xy+xy=0x \odot y = x(-y) - y(-x) = -xy + xy = 0. Thus, xy=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
yx=yxxy=(xyyx)=(xy)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)=xyxy=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)=11(1)1=1(1)=201 \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: (21)1=41=8(2 \odot -1) \odot -1 = 4 \odot -1 = 8, but 2(11)=20=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 ab=a2+b2aba \diamondsuit b = \frac{a^2 + b^2}{ab}. The function ff is defined for all x>0x > 0 by f(x)=x4f(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 x210x+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 ab=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.
x210x+16=0    (x2)(x8)=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