$$ A \times B = \{(2,1),(2,-4),(-2,1),(-2,-4),(3,1),(3,-4)\} $$
$$ A \times A = \{(2,2),(2,-2),(2,3),(-2,2),(-2,-2),(-2,3),(3,2),(3,-2),(3,3)\} $$
$$ B \times A = \{(1,2),(1,-2),(1,3),(-4,2),(-4,-2),(-4,3)\} $$
$$ A \times B = \{(p,p),(p,q),(q,p),(q,q)\} $$
$$ A \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$
$$ B \times A = \{(p,p),(p,q),(q,p),(q,q)\} $$
$$ A \times B = \phi $$
$$ A \times A = \{(m,m),(m,n),(n,m),(n,n)\} $$
$$ B \times A = \phi $$
$$ B=\{2,3,5,7\} $$
$$ A \times B = \{ (1,2),(1,3),(1,5),(1,7), (2,2),(2,3),(2,5),(2,7), (3,2),(3,3),(3,5),(3,7) \} $$
$$ B \times A = \{ (2,1),(2,2),(2,3), (3,1),(3,2),(3,3), (5,1),(5,2),(5,3), (7,1),(7,2),(7,3) \} $$
In $B \times A$, the first component of each ordered pair belongs to $B$ and the second component belongs to $A$.
First components are $-2,0,3$, so $B=\{-2,0,3\}$.
Second components are $3,4$, so $A=\{3,4\}$.
Solution.
A × A = {(5,5),(5,6),(6,5),(6,6)}.
B × B = {(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}.
C × C = {(5,5),(5,6),(5,7),(6,5),(6,6),(6,7),(7,5),(7,6),(7,7)}.
(B × B) ∩ (C × C) = {(5,5),(5,6),(6,5),(6,6)} = A × A. Hence proved.
A ∩ C = {3}
B ∩ D = {3,5}
(A ∩ C) × (B ∩ D) = {(3,3), (3,5)}
A × B = {(1,2),(1,3),(1,5),(2,2),(2,3),(2,5),(3,2),(3,3),(3,5)}
C × D = {(3,1),(3,3),(3,5),(4,1),(4,3),(4,5)}
(A × B) ∩ (C × D) = {(3,3),(3,5)}
Hence the equality holds.
From the given definitions: A = {0,1}, B = {2,3,4}, C = {3,5}.
(i) B ∪ C = {2,3,4,5}.
A × (B ∪ C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}.
(A × B) ∪ (A × C) = {(0,2),(0,3),(0,4),(0,5),(1,2),(1,3),(1,4),(1,5)}.
Thus (i) is verified.
(ii) B ∩ C = {3}.
A × (B ∩ C) = {(0,3),(1,3)}.
(A × B) ∩ (A × C) = {(0,3),(1,3)}.
Thus (ii) is verified.
(iii) A ∪ B = {0,1,2,3,4}.
(A ∪ B) × C = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}.
(A × C) ∪ (B × C) = {(0,3),(0,5),(1,3),(1,5),(2,3),(2,5),(3,3),(3,5),(4,3),(4,5)}.
Thus (iii) is verified.
A = {1,2,3,4,5,6,7}, B = {2,3,5,7}, C = {2}.
(i) A ∩ B = {2,3,5,7}.
(A ∩ B) × C = {(2,2),(3,2),(5,2),(7,2)}.
A × C = {(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),(7,2)}.
B × C = {(2,2),(3,2),(5,2),(7,2)}.
(A × C) ∩ (B × C) = {(2,2),(3,2),(5,2),(7,2)} = (A ∩ B) × C.
(ii) B − C = {3,5,7}.
A × (B − C) = {(n,m) | n ∈ A, m ∈ {3,5,7}} — list equals {(1,3),(1,5),(1,7),(2,3),(2,5),(2,7),...,(7,3),(7,5),(7,7)}.
(A × B) − (A × C) removes pairs with second coordinate 2, yielding the same set as A × (B − C).
Hence both identities are verified.
Solution.
- (i) R1 = {(2,1),(7,1)} — Not a relation, since 1 ∉ B.
- (ii) R2 = {(-1,1)} — Not a relation, since -1 ∉ A (and 1 ∉ B).
- (iii) R3 = {(2,-1),(7,7),(1,3)} — This is a relation: all first elements are in A and all second elements are in B.
- (iv) R4 = {(7,-1),(0,3),(3,3),(0,7)} — Not a relation, since 0 ∉ A.
Solution.
Squares ≤ 45 are: 1, 4, 9, 16, 25, 36.
Hence R = {(1,1),(2,4),(3,9),(4,16),(5,25),(6,36)}.
Domain: {1,2,3,4,5,6}.
Range: {1,4,9,16,25,36}.
Compute y = x + 3 for each x:
(0,3), (1,4), (2,5), (3,6), (4,7), (5,8)
Domain: {0, 1, 2, 3, 4, 5}
Range: {3, 4, 5, 6, 7, 8}
Solution.
(i)
Valid ordered pairs: {(2,1),(4,2)}.
Arrow diagram: 2 → 1, 4 → 2.
Graph points: (2,1), (4,2).
(ii)
For natural numbers < 10, y = x+3 implies x = 1,2,3,4,5,6 (so that y < 10).
Ordered pairs: {(1,4),(2,5),(3,6),(4,7),(5,8),(6,9)}.
Arrow diagram: 1→4, 2→5, 3→6, 4→7, 5→8, 6→9.
Graph points: (1,4),(2,5),(3,6),(4,7),(5,8),(6,9).
Solution.
The relation R (as ordered pairs salary → person) is:
{(10000,A1),(10000,A2),(10000,A3),(10000,A4),(10000,A5), (25000,C1),(25000,C2),(25000,C3),(25000,C4), (50000,M1),(50000,M2),(50000,M3), (100000,E1),(100000,E2)}.
Arrow representation:
10000 → A1,A2,A3,A4,A5
25000 → C1,C2,C3,C4
50000 → M1,M2,M3
100000 → E1,E2
Solution.
Since y = 2x, f = {(1,2),(2,4),(3,6),(4,8),…}.
Domain: {1,2,3,4,…} (N).
Co-domain: N.
Range: {2,4,6,8,…} (the even natural numbers).
Each input has exactly one output, so f is a function.
Solution.
Compute values: f(3)=10, f(4)=17, f(6)=37, f(8)=65.
Thus R = {(3,10),(4,17),(6,37),(8,65)}.
Every element of X has exactly one image in N, so R is a function.
Solution.
(i) f(-1) = (-1)^2 -5(-1) +6 = 1+5+6 = 12.
(ii) f(2a) = (2a)^2 -5(2a) +6 = 4a^2 -10a +6.
(iii) f(2) = 4 -10 +6 = 0.
(iv) f(x-1) = (x-1)^2 -5(x-1) +6 = x^2 -7x +12.
It is clear that:
$$ f(9)=2 $$
$$ 9 $$
(b) $f(7)$
Answer
$$ 6 $$
(c) $f(2)$
Answer
$$ 6 $$
(d) $f(10)$
Answer
$$ 0 $$
$$ x=9.5 $$
$$ \{x\mid 0\leq x\leq10,\;x\in R\} $$
(b) Range
Answer
$$ \{y\mid 0\leq y\leq9,\;y\in R\} $$
$$ f(6)=5 $$
Solution.
(f(x)-5)/x = (2x+5-5)/x = (2x)/x = 2, for x ≠ 0.
Answer: 2 (with x ≠ 0).
Solution
Given f(x) = 2x − 3.
(i) f(0) = 2·0 − 3 = −3, f(1) = 2·1 − 3 = −1. So (f(0)+f(1))/2 = (−3 + (−1))/2 = −4/2 = −2.
(ii) Solve 2x − 3 = 0 ⇒ x = 3/2.
(iii) Solve 2x − 3 = x ⇒ x = 3.
(iv) Solve 2x − 3 = 1/2 ⇒ 2x = 3 + 1/2 = 7/2 ⇒ x = 7/4.
Answers: (i) −2, (ii) 3/2, (iii) 3, (iv) 7/4.
Length:
$$ 24-2x $$
Breadth:
$$ 24-2x $$
Height:
$$ x $$
Volume:
$$ V=x(24-2x)^2 $$
$$ =x(576-96x+4x^2) $$
$$ =576x-96x^2+4x^3 $$
Answer
$$ V(x)=4x^3-96x^2+576x $$
Solution.
f(x^2) = 3 - 2x^2.
(f(x))^2 = (3 - 2x)^2 = 9 - 12x + 4x^2.
Equate: 3 - 2x^2 = 9 - 12x + 4x^2 ⇒ 6x^2 - 12x + 6 = 0 ⇒ x^2 - 2x + 1 = 0 ⇒ (x-1)^2 = 0.
Answer: x = 1.
$$ \text{Distance}=\text{Speed}\times\text{Time} $$
$$ d=500t $$
Answer
$$ d(t)=500t $$
Solution
(i) Yes. The relation assigns to each forehand length x exactly one height y, so it is a function.
(ii) Use any two data points to find a and b. From (35,56) and (45,65):
slope a = (65−56)/(45−35) = 9/10 = 0.9.
Then b = 56 − 0.9×35 = 56 − 31.5 = 24.5.
So y = 0.9x + 24.5.
(iii) For x = 40: y = 0.9×40 + 24.5 = 36 + 24.5 = 60.5 inches.
(iv) For y = 53.3: 53.3 = 0.9x + 24.5 ⇒ 0.9x = 28.8 ⇒ x = 32 cm.
1. Not a function 2. Function 3. Not a function 4. Function
Reason
A graph represents a function only if every vertical line intersects the graph at most once (Vertical Line Test).
Solution
(i) Ordered pairs:
{(2,0), (4,1), (6,2), (10,4), (12,5)}
(ii) Table:
4 → 1
6 → 2
10 → 4
12 → 5
(iii) Arrow diagram: map 2↦0, 4↦1, 6↦2, 10↦4, 12↦5.
(iv) Graph: plot the points (2,0), (4,1), (6,2), (10,4), (12,5) in the Cartesian plane.
$$ f=\{(1,2),(2,2),(3,2),(4,3),(5,4)\} $$
through:
$$ 1 \to 2 $$
$$ 2 \to 2 $$
$$ 3 \to 2 $$
$$ 4 \to 3 $$
$$ 5 \to 4 $$
| x | f(x) | |---|---| | 1 | 2 | | 2 | 2 | | 3 | 2 | | 4 | 3 | | 5 | 4 |
Plot the points:
$$ (1,2),(2,2),(3,2),(4,3),(5,4) $$
Suppose
$$ f(a)=f(b) $$
Then,
$$ 2a-1=2b-1 $$
$$ 2a=2b $$
$$ a=b $$
Hence $f$ is one-one.
The range is
$$ \{1,3,5,7,\dots\} $$
Even natural numbers are not images of any element.
Hence the function is not onto.
Suppose
$$ f(a)=f(b) $$
Then,
$$ a^2+a+3=b^2+b+3 $$
$$ a^2-b^2+a-b=0 $$
$$ (a-b)(a+b+1)=0 $$
Since $a+b+1\neq0$,
$$ a-b=0 $$
$$ a=b $$
Hence the function is one-one.
Solution
(i) Compute images: f(1)=1, f(2)=8, f(3)=27, f(4)=64. So the range is {1, 8, 27, 64}.
(ii) The function is one-one (distinct inputs give distinct outputs). Since the codomain B is N and the range is a proper subset of N, f is not onto; it is an into function.
The function is linear and every real number has a unique pre-image.
Hence it is both one-one and onto.
Answer
Bijective function.
$$ $$ f(1)=f(-1) $$
Hence it is not one-one.
Also range does not cover all real numbers.
Answer
Not bijective.
Solution
Onto means both elements of B are attained. Use f(−1)=0 and f(1)=2.
f(−1)= −a + b = 0
f(1)= a + b = 2
Add: 2b = 2 ⇒ b = 1. Then −a + 1 = 0 ⇒ a = 1.
Answer: a = 1, b = 1.
Solution
(i) f(3) = 3 + 2 = 5.
(ii) f(0) = 0 + 2 = 2.
(iii) For x = −1.5 (<0): f(−1.5) = 2(−1.5) + 1 = −3 + 1 = −2.
(iv) f(2) = 2 + 2 = 4; f(−2) = 2(−2) + 1 = −4 + 1 = −3. So f(2) + f(−2) = 4 + (−3) = 1.
1. $f(-3)+f(2)=2$
2. $f(7)-f(1)=10$
3. $2f(4)+f(8)=178$
4.
$$ -\frac{9}{17} $$
Solution
Consider t ≥ 0 (time is nonnegative). S'(t) = g t + a. Since g > 0 and a > 0, we have S'(t) = g t + a ≥ a > 0 for all t ≥ 0. Thus S is strictly increasing on t ≥ 0, so S is one-one on the practical domain of time (t ≥ 0).
F = (9/5)C + 32
- (i) F(0) = (9/5)·0 + 32 = 32 °F
- (ii) F(28) = (9/5)·28 + 32 = 50.4 + 32 = 82.4 °F
- (iii) F(−10) = (9/5)·(−10) + 32 = −18 + 32 = 14 °F
- (iv) For C = 100 °C: F = (9/5)·100 + 32 = 180 + 32 = 212 °F
- (v) Solve (9/5)C + 32 = C ⇒ (9/5)C − C = −32 ⇒ (4/5)C = −32 ⇒ C = −32·(5/4) = −40. Thus Celsius = Fahrenheit = −40°.
$$ f\circ g=g\circ f $$
$$ $$ (f\circ g)(x)=f(g(x)) $$
$$ =f(x^2) $$
$$ =x^2-6 $$
Thus,
$$ (f\circ g)(x)=x^2-6 $$
Now,
$$ (g\circ f)(x)=g(f(x)) $$
$$ =g(x-6) $$
$$ =(x-6)^2 $$
$$ =x^2-12x+36 $$
Answer
$$ (f\circ g)(x)=x^2-6 $$
$$ (g\circ f)(x)=x^2-12x+36 $$
Hence,
$$ f\circ g\neq g\circ f $$
$$ $$ (f\circ g)(x)=f(2x^2-1) $$
$$ =\frac2{2x^2-1} $$
Now,
$$ (g\circ f)(x)=g\left(\frac2x\right) $$
$$ =2\left(\frac2x\right)^2-1 $$
$$ =\frac8{x^2}-1 $$
Answer
$$ (f\circ g)(x)=\frac2{2x^2-1} $$
$$ (g\circ f)(x)=\frac8{x^2}-1 $$
Hence,
$$ f\circ g\neq g\circ f $$
$$ $$ (f\circ g)(x)=f(3-x) $$
$$ =\frac1{3-x} $$
Now,
$$ (g\circ f)(x)=g\left(\frac1x\right) $$
$$ =3-\frac1x $$
Answer
$$ (f\circ g)(x)=\frac1{3-x} $$
$$ (g\circ f)(x)=3-\frac1x $$
Hence,
$$ f\circ g\neq g\circ f $$
$$ $$ (f\circ g)(x)=f(x-4) $$
$$ =(x-4)+3=x-1 $$
Now,
$$ (g\circ f)(x)=g(x+3) $$
$$ =(x+3)-4=x-1 $$
Answer
$$ (f\circ g)(x)=x-1 $$
$$ (g\circ f)(x)=x-1 $$
Hence,
$$ f\circ g=g\circ f $$
$$ $$ (f\circ g)(x)=f(x+1) $$
$$ =4(x+1)^2-1 $$
$$ =4(x^2+2x+1)-1 $$
$$ =4x^2+8x+3 $$
Now,
$$ (g\circ f)(x)=g(4x^2-1) $$
$$ =(4x^2-1)+1 $$
$$ =4x^2 $$
Answer
$$ (f\circ g)(x)=4x^2+8x+3 $$
$$ (g\circ f)(x)=4x^2 $$
Hence,
$$ f\circ g\neq g\circ f $$
$$ f\circ g=g\circ f $$
$$ $$ (f\circ g)(x)=3(6x-k)+2 $$
$$ =18x-3k+2 $$
Now,
$$ (g\circ f)(x)=6(3x+2)-k $$
$$ =18x+12-k $$
Equating:
$$ 18x-3k+2=18x+12-k $$
$$ -3k+2=12-k $$
$$ -2k=10 $$
$$ k=-5 $$
Answer
$$ k=-5 $$
$$ $$ (f\circ g)(x)=2(4x+5)-k $$
$$ =8x+10-k $$
Now,
$$ (g\circ f)(x)=4(2x-k)+5 $$
$$ =8x-4k+5 $$
Equating:
$$ 8x+10-k=8x-4k+5 $$
$$ 10-k=-4k+5 $$
$$ 3k=-5 $$
$$ k=-\frac53 $$
Answer
$$ k=-\frac53 $$
Solution
(f ∘ g)(x) = f(g(x)) = f((x+1)/2) = 2((x+1)/2) − 1 = x + 1 − 1 = x.
(g ∘ f)(x) = g(f(x)) = g(2x − 1) = ((2x − 1) + 1)/2 = 2x/2 = x.
Hence (f ∘ g)(x) = (g ∘ f)(x) = x.
If
$$ f(x)=x^2-1,\quad g(x)=x-2 $$
find $a$, if
$$ (g\circ f)(a)=1 $$
Solution
$$ (g\circ f)(a)=g(a^2-1) $$
$$ =(a^2-1)-2 $$
$$ =a^2-3 $$
Given:
$$ a^2-3=1 $$
$$ a^2=4 $$
$$ a=\pm2 $$
Answer
$$ a=2,-2 $$
$$ f(f(k))=f(2k-1) $$
Since $f(x)=2x-1$,
$$ f(2k-1)=2(2k-1)-1 $$
$$ =4k-3 $$
Given:
$$ 4k-3=5 $$
$$ 4k=8 $$
$$ k=2 $$
Answer
$$ k=2 $$
Solution
(f ∘ g)(x) = f(g(x)) = 2x^2 + 1. For x ∈ ℝ, x^2 ≥ 0, so 2x^2 ≥ 0 and 2x^2 + 1 ≥ 1. Range: [1, ∞).
(g ∘ f)(x) = g(f(x)) = (2x + 1)^2 = 4x^2 + 4x + 1 = 4(x + 1/2)^2. This expression ≥ 0 for all real x and attains 0 at x = −1/2. Range: [0, ∞).
Solution
(i) (f ∘ f)(x) = f(f(x)) = f(x^2 − 1) = (x^2 − 1)^2 − 1 = x^4 − 2x^2.
(ii) (f ∘ f ∘ f)(x) = f((f ∘ f)(x)) = f(x^4 − 2x^2) = (x^4 − 2x^2)^2 − 1 = x^8 − 4x^6 + 4x^4 − 1.
Solution
f(x)=x^5 is strictly increasing on ℝ, so f is one-one.
g(x)=x^4 is not one-one on ℝ because g(2)=16 and g(−2)=16 (distinct inputs give the same output), so g is not one-one.
(f ∘ g)(x) = f(g(x)) = f(x^4) = (x^4)^5 = x^20. Since x^20 = (−x)^20, (f ∘ g)(2) = (f ∘ g)(−2), so f ∘ g is not one-one on ℝ.
For any x in the domain,
( (f ∘ g) ∘ h )(x) = (f ∘ g)(h(x)) = f(g(h(x))).
Also, ( f ∘ (g ∘ h) )(x) = f((g ∘ h)(x)) = f(g(h(x))).
Since both expressions equal f(g(h(x))) for every x, (f ∘ g) ∘ h = f ∘ (g ∘ h).
Solution. Let f(x) = ax + b. From f(0) = −1 we get b = −1. From f(−1) = 3 we get −a + b = 3 ⇒ −a − 1 = 3 ⇒ a = −4. Thus f(x) = −4x − 1. Check: f(2) = −8 − 1 = −9.
Answer: f(x) = −4x − 1.
Compute:
C(a t1 + b t2) = 3(a t1 + b t2) = 3a t1 + 3b t2 = a(3 t1) + b(3 t2) = a C(t1) + b C(t2).
Hence C(t) = 3t satisfies the superposition principle and is linear.
$$ (x,y)=(1,-5),(1,1),(2,-5),(2,1) $$
$$ A=\{-1,0,1\} $$
Remaining elements:
$$ \{ (-1,-1),(-1,1), (0,-1),(0,0), (1,-1),(1,0),(1,1) \} $$
Solution
f(0) = √(0 + 4) = 2
f(3) = √(3 + 4) = √7
f(a + 1) = √(a + 1 + 4) = √(a + 5)
Solution
The set of ordered pairs is
{ (9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17) }.
The range is {2, 3, 5, 7, 11, 13, 17}.
$$ [-1,1] $$
Solution
Compute g(h(x)) = 3·(x − 2) = 3(x − 2). Then f(g(h(x))) = [3(x − 2)]^2 = 9(x − 2)^2.
Compute (f ∘ g)(x) = f(g(x)) = [3x]^2 = 9x^2. Then ((f ∘ g) ∘ h)(x) = (f ∘ g)(h(x)) = 9·(h(x))^2 = 9(x − 2)^2.
Hence (f ∘ g) ∘ h = f ∘ (g ∘ h), both equal to 9(x − 2)^2.
Solution
A×C = {(1,5), (1,6), (2,5), (2,6)}. Each first component (1 or 2) belongs to B and each second component (5 or 6) belongs to D, so every pair of A×C is also in B×D. Therefore A×C ⊆ B×D.
Solution
f(x) = (x − 1)/(x + 1). Compute f(f(x)) = ( f(x) − 1 ) / ( f(x) + 1 ).
Substitute f(x):
f(f(x)) = ( (x − 1)/(x + 1) − 1 ) / ( (x − 1)/(x + 1) + 1 )
= ( (x − 1 − x − 1)/(x + 1) ) / ( (x − 1 + x + 1)/(x + 1) )
= ( −2/(x + 1) ) / ( 2x/(x + 1) ) = −2/(2x) = −1/x.
Thus f(f(x)) = −1/x, for x ≠ 0 (and x ≠ −1 for domain of f).
Solution
(i) g(1/2) = (1/2 − 2)/3 = (−3/2)/3 = −1/2. Then g(g(1/2)) = g(−1/2) = (−1/2 − 2)/3 = (−5/2)/3 = −5/6.
(ii) g(f(x)) = (f(x) − 2)/3 = (6x + 8 − 2)/3 = (6x + 6)/3 = 2x + 2 = 2(x + 1).
- (i) R \ {9}
- (ii) R
- (iii) [2, ∞)
- (iv) R
Equate corresponding coordinates: 2x − y = −2 and 3x + 4 = 5. From 3x + 4 = 5, 3x = 1 ⇒ x = 1/3. Substitute into 2x − y = −2: 2(1/3) − y = −2 ⇒ 2/3 − y = −2 ⇒ −y = −2 − 2/3 = −8/3 ⇒ y = 8/3.
Answer: x = 1/3, y = 8/3.
If A × A has 9 elements then n(A) = 3, so A has three elements. The given ordered pairs show that −1, 0 and 1 must be elements of A. Thus A = {−1, 0, 1}.
The Cartesian product A × A is the set of all ordered pairs with first and second entries from A, so
{(−1,−1), (−1,0), (−1,1), (0,−1), (0,0), (0,1), (1,−1), (1,0), (1,1)}.
A × C = {(1,5), (1,6), (2,5), (2,6)}. Each of these ordered pairs has first coordinate in B and second coordinate in D, so every element of A × C is in B × D. Therefore A × C ⊆ B × D.
Number of relations:
$$ 2^{mn}=1024 $$
$$ 2^{5n}=2^{10} $$
$$ 5n=10 $$
$$ n=2 $$
Correct option: (2)
Primes less than 13:
$$ 2,3,5,7,11 $$
Squares:
$$ 4,9,25,49,121 $$
Correct option: (3)
Solution
From equality of ordered pairs: a + 2 = 5 ⇒ a = 3.
Also 2a + b = 4 ⇒ 2·3 + b = 4 ⇒ 6 + b = 4 ⇒ b = −2.
Hence a = 3, b = −2.
Compute A × C = {(1,5), (1,6), (2,5), (2,6)}.
Since every first component (1 or 2) belongs to B and every second component (5 or 6) belongs to D, each ordered pair of A × C is also an element of B × D. Hence A × C ⊆ B × D.
Let y = f(x) = (x − 1)/(x + 1). Then
f(f(x)) = f(y) = (y − 1)/(y + 1) = ( (x−1)/(x+1) − 1 ) / ( (x−1)/(x+1) + 1 ).
Simplify numerator: (x−1) − (x+1) = −2, denominator: (x−1) + (x+1) = 2x, both over (x+1), so
f(f(x)) = [ (−2)/(x+1) ] / [ (2x)/(x+1) ] = (−2)/(2x) = −1/x.
Thus f(f(x)) = −1/x, provided x ≠ 0 and x ≠ −1.
(i) g(1/2) = (1/2 − 2)/3 = (−3/2)/3 = −1/2. Then g(g(1/2)) = g(−1/2) = (−1/2 − 2)/3 = (−5/2)/3 = −5/6.
(ii) (g ∘ f)(x) = g(f(x)) = (f(x) − 2)/3 = ((x + 6)/8 − 2)/3 = ((x + 6 − 16)/8)/3 = (x − 10)/24.
$$ (f\circ g)(x)=2\left(\frac{x}{3}\right)^2 $$
$$ =\frac{2x^2}{9} $$
Correct option: (3)
Bijective functions have equal cardinalities.
$$ n(A)=n(B)=7 $$
Correct option: (1)
Since f = g^{-1}, each ordered pair (a,b) in g gives (b,a) in f. From g:
- g(0)=2 ⇒ f(2)=0
- g(1)=0 ⇒ f(0)=1
- g(2)=4 ⇒ f(4)=2
- g(-4)=2 ⇒ f(2)=0 (repeat)
- g(7)=0 ⇒ f(0)=1 (repeat)
Thus the values (images) of f are {0,1,2}. Hence the range of f is {0, 1, 2}.
Answer: (D) None of these.
Explanation: f(x+y)=x+y+1/2, f(x)+f(y)=x+y+1, and f(x)·f(y)=(x+1/2)(y+1/2)=xy+½(x+y)+1/4. None of the equalities/inequalities hold for all x,y (for example x=1,y=1 gives f(2)=2.5, f(1)+f(1)=3, f(1)·f(1)=2.25).
Answer: α = 2, β = −1 (option (B)).
Work: g(1)=1 ⇒ α+β=1. g(2)=3 ⇒ 2α+β=3. Subtracting gives α=2, then β=1−α=−1.
- (a) linear
- (b) cubic
- (c) reciprocal
- (d) quadratic
Answer: (D) Quadratic.
Work: (x+1)^3 = x^3+3x^2+3x+1 and (x−1)^3 = x^3−3x^2+3x−1. Subtracting gives f(x)=6x^2+2, which is a quadratic function.
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