Functions Exercise 1b Solutions
Functions Exercise 1b
The famous mathematician ” Lejeune Dirichlet” defined a function.
Function: A variable is a symbol which represents any one of a set of numbers, if two variables x and y so related that whenever a value is assigned to x there is autometically assigned by some rule or correspondence a value to y, then we say y is a function of x.
Chapter 1 Functions Exercise 1b Solutions for inter first year students, prepared by Mathematics expert of www.basicsinmaths.com
Functions Exercise 1b
Exercise 1(b) Solutions
I.
1.If f (x) = e^{x} and g(x) = , then show that f og = gof and f^{-1} = g^{-1}
Sol:
Given f (x) = e^{x} and g(x) =
(fog) (x) = f (g (x))
= f ( )
=
= x ————- (1)
(gof) (x) = g (f (x))
= g (e^{x})
=
= x
= x ————- (2)
From (1) and (2) f og = gof
let y = f(x)
x = f^{-1}(y)
y = e^{x}
^{ } x =
f^{-1} (x) =
let g(x) = z
x = g^{-1}(z)
z =
x = e^{z}
g^{-1 }(x) = e^{x}
2. If f (y) = , g (y) = then show that fog(y) = y
Sol:
Given f (y) = , g (y) =
Now
fog(y) = f(g(y))
∴ fog(y) = y
3. If f: R ⟶ R, g: R ⟶ R is defined by f(x) = 2x^{2} + 3 ang g (x) = 3x – 2 then find
(i) (fog) (x) (ii) (gof) (x) (iii) (fof) (0) (iv) go(fof)(3)
Sol:
Given f: R ⟶ R is, g: R ⟶ R is defined by f(x) = 2x^{2} + 3 ang g (x) = 3x – 2
(i) (fog) (x) = f(g(x))
= f(3x – 2)
= 2 (3x – 2)^{2} + 3
= 2 (9x^{2} – 12x + 4) + 3
= 18 x^{2} – 24x + 8 + 3
= 18 x^{2} – 24x + 11
∴ (fog) (x) = 18 x^{2} – 24x + 11
(ii) (gof) (x) = g (f (x))
= g (2x^{2} + 3)
= 3(2x^{2} + 3) – 2
= 6x^{2} + 9 – 2
= 6 x^{2} + 7
∴ (gof) (x) = 6 x^{2} + 7
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(iii) (fof) (0) = f (f (0))
= f (2(0)^{2} + 3)
= f (2(0) + 3)
= f (3)
= 2 (3)^{2} + 3
= 2 (9) + 3
= 18 + 3 = 21
(iv) go(fof) (3) = go (f (f (3)))
= go (f (2 (3)^{2} + 3))
= go (f (21))
= g (f (21))
= g (2 (21)^{2} + 3))
= g (2 (441) + 3))
= g (882 + 3)
= g (885)
= 3 (885) – 2
= 2655 – 2
= 2653
∴ go(fof) (3) = 2653
Functions Exercise 1b
4. If f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x^{2} + 1, then find
(i) (fof) (x^{2} + 1) (ii) (fog) (2) (iii) (gof) (2a – 3)
Sol:
Given f: R ⟶ R; g: R ⟶ R is defined by f(x) = 3x – 1, g (x) = x^{2} + 1
(i) (fof) (x^{2} + 1) = f (f (x^{2} + 1))
= f (3 (x^{2} + 1)– 1)
= f (3x^{2} + 3– 1)
= f (3x^{2} + 2)
= 3(3x^{2} + 2) – 1
= 9x^{2} + 6 – 1
= 9x^{2} + 5
(ii) (fog) (2) = f (g (2))
= f (2^{2} + 1)
=f (4 + 1)
= f (5)
= 3(5) – 1
= 15 – 1
= 14
(iii) (gof) (2a – 3) = g (f (2a – 3))
= g (3(2a – 3) – 1)
= g (6a – 9 – 1)
= g (6a – 10)
= (6a – 10)^{2} + 1
= 36^{2} – 120a + 100 + 1
= 36^{2} – 120a + 101
5. If f(x) = and g(x) = for all x ∈ (0, ∞) then find (gof) (x)
Sol:
Given f(x) = and g(x) = for all x ∈ (0, ∞)
(gof) (x) = g (f (x))
= g ( )
=
∴ (gof) (x) =
6. If f(x) = 2x – 1 and g(x) = for all x ∈ R then find (gof) (x)
Sol:
Given f(x) = 2x – 1 and g(x) = for all x ∈ R
∴ gof(x) = x
7. If f (x) = 2, g (x) = x^{2} and h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x)
Sol:
Given f (x) = 2, g (x) = x^{2} and h(x) = 2x ∀ x ∈ R
(fo(goh)) (x) = fo (g (h (x))
= fo g(2x)
= f (g(2x))
= f((2x)^{2})
= f(4x^{2})
= 2
∴ (fo(goh)) (x) = 2
Functions Exercise 1b Solutions
8. Find the inverse of the following functions
(i) a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b, (a ≠ 0)
Given a, b ∈ R, f: R ⟶ R defined by f(x) = ax + b
Let y = f(x)
x = f^{-1}(y)
now y = ax + b
ax = y – b
x =
f^{-1}(y) =
∴ f^{-1}(x) =
(ii) f: R ⟶ (0, ∞) defined by f(x) = 5^{x}
Let y = f(x)
x = f^{-1}(y)
now y = 5^{x}
x =
f^{-1}(y) =
∴ f^{-1}(x) =
(iii) f: (0, ∞) ⟶ R defined by f(x) =
Let y = f(x)
x = f^{-1}(y)
now y =
x = 2^{y}
f^{-1}(y) = 2^{y}
∴ f^{-1}(x) = 2^{x}
9. If f(x) = 1 + x + x^{2} + … for , then show that f^{-1} (x) =
Sol:
Given, f(x) = 1 + x + x^{2} + … for
1 + x + x^{2} + … is an infinite G.P
a = 1, r = x
S_{∞} =
=
Now
f (x) =
Let y = f(x)
x = f^{-1}(y)
now y =
1 – x =
x = 1 –
=
f^{-1}(y) =
∴ f^{-1}(x) =
10 . If f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2^{x (x – 1}^{)} then find f^{-1} (x)
Sol:
Given f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2^{x (x – 1)}
Let y = f(x)
x = f^{-1}(y)
now y = 2^{x (x – 1)}
= x (x – 1)
x^{2} – x =
x^{2} – x – = 0
Functions Exercise 1b Solutions
II.
1. If f (x) = , x ≠ ± 1, then verify (fof^{-1}) (x) = x
Sol:
Given f (x) = , x ≠ ± 1
Let y = f(x)
x = f^{-1}(y)
now y =
y (x + 1) = x – 1
xy + y = x – 1
1 + y = x – xy
1 + y = x (1 – y)
x =
f^{-1}(y) =
∴ f^{-1}(x) =
Now
2. If A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}, then show that f and g are bijective functions and (gof)^{-1} = f^{-1}og^{-1}
Sol:
Given A = {1, 2, 3}, B = {α, β, γ}, C = {p, q, r} and f: A ⟶ B; g: B ⟶ C is defined by f = {(1, α), (2, γ), (3, β)}, g = {(α, p), (β, r), (γ, p)}
A = {1, 2, 3}, B = {α, β, γ}
f: A ⟶ B; and f = {(1, α), (2, γ), (3, β)}
⟹ f (1) = α; f (2) = γ; f (3) = β
⟹ Every element of set A has a unique image in set B
f is injection (one – one)
range of f = codomain of f
⟹ f is surjection (on to)
∴ f is bijective
B = {α, β, γ}, C = {p, q, r}
g: B ⟶ C is defined by g = {(α, q), (β, r), (γ, p)}
⟹ g (α) =q; g (β) = r; g (γ) = p
⟹ Every element of set B has a unique image in set C
g is injection (one – one)
range of g = codomain of g
⟹ g is surjection (on to)
∴ g is bijective
Now
f = {(1, α), (2, γ), (3, β)}, g = {(α, q), (β, r), (γ, p)}
gof = {(1, q), (2, p), (3, r)
(gof)^{-1} = {(q, 1), (p, 2), (r, 3)} ———– (1)
f^{-1} = {(α, 1), (γ, 2), (β, 3)
g^{-1} = {(q, α), (r, β), (p, γ)}
f^{-1}og^{-1} = {(q, 1), (p, 2), (r, 3)} ———– (2)
From (1) and (2)
(gof)^{-1} = f^{-1}og^{-1}
3. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x^{2} + 1 then find
(i) (gof^{-1}) (2) (ii) (gof) (x – 1)
Sol:
Given f: R ⟶ R; g: R ⟶ R is defined by f (x) = 3x – 2 and g(x) = x^{2} + 1
Let y = f(x)
x = f^{-1}(y)
now y = 3x – 2
3x = y + 2
x =
f^{-1}(y) =
f^{-1}(x) =
∴ (gof^{-1}) (2) =
(ii) (gof) (x – 1) = g (f (x – 1))
= g (3(x – 1) – 2)
= g (3x – 3 – 2)
= g (3x – 5)
= (3x – 5)^{2} + 1
= 9x^{2} – 30x + 25 + 1
= 9x^{2} – 30x + 26
∴(gof) (x – 1) = 9x^{2} – 30x + 26
4. Let f = {(1, a), (2, c), (4, d), (3, b)} and g^{-1} = {(2, a), (4, b), (1, c), (3, d)}, then show that (gof)^{-1} = f^{-1}og^{-1}
Sol:
Given f = {(1, a), (2, c), (4, d), (3, b)} and g^{-1} = {(2, a), (4, b), (1, c), (3, d)}
⟹ f ^{-1} = {(a, 1), (c, 2), (d, 4), (b, 3)} and g = {(a, 2), (b, 4), (c, 1), (d, 3)}
Now gof = {(1, 2), (2, 1), (4, 3), (3, 4)}
(gof)^{-1} = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(1)
g^{-1} = {(2, a), (4, b), (1, c), (3, d)} ; f ^{-1} = {(a, 1), (c, 2), (d, 4), (b, 3)}
f^{-1}og^{-1} = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(2)
from (1) and (2)
(gof)^{-1} = f^{-1}og^{-1 }
5. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x^{3} + 5 then find (fog)^{-1} (x)
Sol:
Given R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x^{3} + 5
(fog) (x) = f (g(x))
= f (x^{3} + 5)
= 2 (x^{3} + 5) – 3
= 2 x^{3} + 10 – 3 = 2x^{3} + 7
Let y = (fog) (x)
⟹ x = (fog)^{-1}(y)
y = 2x^{3} + 7
2x^{3} = y – 7
x^{3} =
x =
(fog)^{-1}(y)=
(fog)^{-1}(x)=
6. Let f(x) = x^{2}, g(x) = 2^{x} then solve the equation (fog) (x) = (gof) (x)
Sol:
Given f(x) = x^{2}, g(x) = 2^{x}
(fog) (x) = f(g(x))
= f (2^{x})
= (2^{x})^{2}
= 2^{2x}
(gof) (x) = g(f(x))
= g (x^{2})
=
Now
(fog) (x) = (gof) (x)
⟹ 2^{2x} =
2x = x^{2} [∵ if a^{m} = a^{n} , then m = n]
x^{2} – 2x = 0
⟹ x (x – 2) = 0
⟹x = 0 or x – 2 = 0
∴ x = 0 or x = 2
7. If f (x) = , x ≠ ±1 then find (fofof) (x) and (fofofof) (x)
Sol:
Given f (x) = , x ≠ ±1
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