# functions intermediate maths solutions

## Functions Exercise 1c Solutions

Functions Exercise 1c

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 1c Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

## Functions Exercise 1c

### I.

#### 1.Find the domains of the following real valued functions

(i)

f (x) =

given function is f (x) =

f (x) is defined when (x2 – 1) (x + 3) ≠ 0

⟹ (x2 – 1) ≠ 0 or (x + 3) ≠ 0

⟹ (x + 1) (x – 1) ≠ 0 or (x + 3) ≠ 0

⟹ x ≠ 1, x ≠ – 1 or x ≠ – 3

∴ Domain of f(x) is R – {– 1, – 3, 1}

###### (ii)

f (x) =

Given function is f (x) =

f (x) is defined when (x – 1) (x – 2) (x – 3) ≠ 0

⟹ x ≠ 1, x ≠ 2 or x ≠ 3

∴ Domain of f(x) is R – {1, 2, 3}

(iii)

f (x) =

Given function is f (x) =

f (x) is defined when 2 – x > 0 and 2 – x ≠ 1

⟹ 2 > x  and  2 – 1 ≠ x

⟹ 2 > x  and  x ≠ 1

∴ Domain of f(x) is (– ∞, 2) – {1}

(iv)

f (x) =

Given function is f (x) =

f (x) is defined when x ∈ R

∴ Domain of f(x) is R

Functions Exercise 1c

(v)

f (x) =

Given function is f (x) =

f (x) is defined when 4x – x2 ≥ 0

⟹ x (4 – x) ≥ 0

⟹ x (x – 4) ≤ 0

⟹ (x – 0) (x – 4) ≤ 0

⟹ x ∈ [0, 4]

∴ Domain of f(x) is [0, 4]

(vi)

f (x) =

Given function is f (x) =

f (x) is defined when 1 – x2 > 0

⟹ x2 – 1 < 0

⟹ (x – 1) (x + 1) < 0

⟹ x ∈ (– 1, 1)

∴ Domain of f(x) is (– 1, 1)

(vii)

f (x) =

Given function is f (x) =

f (x) is defined when x + 1≠ 0

⟹ x ≠ – 1

∴ Domain of f(x) is R – {– 1}

(viii)

f(x) =

Given function is f (x) =

f (x) is defined when x2 – 25 ≥ 0

⟹ (x – 5) (x + 5) ≥ 0

⟹ x ∈ (–∞, –5] ∪ [5, ∞)

⟹ x ∈ R – (– 5, 5)

∴ Domain of f(x) is R – (– 5, 5)

Functions Exercise 1c

(ix)

f(x) =

Given function is f (x) =

f (x) is defined when x – [x] ≥ 0

⟹ x ≥ [x]

⟹ x ∈ R

∴ Domain of f(x) is R

(x)

f(x) =

Given function is f (x) =

f (x) is defined when [x] – x ≥ 0

⟹ [x] ≥ x

⟹ x ∈ Z

∴ Domain of f(x) is Z

Functions Exercise 1c

##### 2. find the ranges of the following real valued functions

(i)

f(x) =

Given function is f (x) =

Let y =

⟹  |4 – x2| = ey

∵ ey > 0 ∀ y ∈ R

∴ Range of f(x) is R

(ii)

f(x) =

Given function is f (x) =

f (x) is defined when [x] – x ≥ 0

⟹ [x] ≥ x

⟹ x ∈ Z

Domain of f(x) is Z

Range of f = {0}

(iii)

f(x) =

Given function is f (x) =

f (x) is defined when x ∈ R

Domain of f(x) is R

For x ∈ R   [x] is an integer

Since sin nπ = 0, ∀ n ∈ z

⟹ sin π[x] = 0

∴ Range of f = {0}

(iv)

f (x) =

Given function is f (x) =

f (x) is defined when x – 2 ≠ 0

⟹ x ≠ 2

Domain of f(x) is R – {2}

Let y =

=

= x + 2

If x = 2 ⟹ y = 2 + 2 = 4

∴ Range of f(x) is R – {4}

Functions Exercise 1c

(v)

f (x) =

let y =

y2 = 9 + x2

x2 = y2 – 9

x =

it is defined when y2 – 9 ≥ 0

⟹ (y – 3) (y + 3) ≥ 0

y ∈ (– ∞, – 3] ∪ [3, ∞)

but y = ≥ 0

∴ Range of f(x) is [3, ∞)

##### find

Sol:

Given f and g are real valued functions f(x) = 2x – 1 ang g (x) = x2

(i)

(3f – 2g) (x) = 3 f(x) – 2g (x)

= 3 (2x – 1) – 2(x2)

= 6x – 3 – 2x2

= – 2x2 + 6x – 3

∴ (3f – 2g) (x) =– 2x2 + 6x – 3

(ii)

(fg) (x) = f (x) g (x)

= (2x – 1) (x2)

= 2x3 + x2

∴ (fg) (x) = 2x3 + x2

(iii)

##### 4. If f = {(1, 2), (2, – 3) (3, – 1)} then find (i) 2f (ii) (fog) 2 + f iii) f2 (iv)

Sol:

Given f = {(1, 2), (2, – 3) (3, – 1)}

(i)

(2f) (1) = 2 f (1) = 2 × 2 = 4

(2f) (2) = 2 f (2) = 2 × – 3 = – 6

(2f) (3) = 2 f (3) = 2 × – 1 = – 2

∴ 2f = {(1, 4), (2, – 6) (3, – 2)}

(ii)

(2 + f) (1) = 2 + f (1) = 2 + 2 = 4

(2 + f) (2) = 2 + f (2) = 2 + (– 3) = – 1

(2 + f) (3) = 2 + f (3) = 2 + (– 1) = 1

∴ 2 + f = {(1, 4), (2, – 1) (3, 1)}

(iii)

(f2) (1) = [f (1))]2 = 22 = 4

(f2) (2) = [f (2))]2 = (– 3)2 = 9

(f2) (3) = [f (1))]2 = (– 1)2 = 1

∴ f2= {(1, 4), (2, 9) (3, 1)}

(iv)

### II.

#### 1.Find the domain of the following real valued functions

(i)

f (x) =

f(x) is defined when x2 – 3x + 2 ≥ 0

x2 – 2x – x + 2 ≥ 0

x (x – 2) – 1(x – 2) ≥ 0

(x – 1) (x – 2) ≥ 0

x ∈ (– ∞, 1] ∪ [2, ∞)

∴ Domain of f(x) is R – (1, 2)

(ii)

f(x) = log (x2 – 4x + 3)

f(x) is defined when x2 – 4x + 3 > 0

x2 – 3x – x + 3 > 0

x (x – 3) – 1(x – 3) > 0

(x – 1) (x – 3) > 0

x ∈ (– ∞, 1) ∪ (3, ∞)

∴ Domain of f(x) is R – [1, 3]

(iii)

f(x) =

f(x) is defined when 2 + x ≥ 0, 2 – x ≥ 0  and x ≠ 0

x ≥ – 2, x ≤ 2 and x ≠ 0

– 2 ≤ x ≤ 2 and x ≠ 0

x ∈ [–2, 2] – {0}

∴ Domain of f(x) is [–2, 2] – {0}

(iv)

f(x) =

f(x) is defined in two cases as follows:

case (i) 4 – x2 ≥ 0 and [x] + 2 > 0

x2 – 4 ≤ 0 and [x] > –2

(x – 2) (x + 2) ≤ 0 and [x] > –2

x ∈ [– 2, 2] and x ∈ [– 1, ∞)

x ∈ [– 1, 2]

case (ii) 4 – x2 ≤ 0 and [x] + 2 < 0

x2 – 4 ≥ 0 and [x] < –2

(x – 2) (x + 2) ≥ 0 and [x] < –2

x ∈ (– ∞, –2] ∪ [2, ∞) and x ∈ (–∞, –2)

x ∈ (–∞, –2)

from case (i) and case (ii)

x ∈ (–∞, –2) ∪ [– 1, 2]

∴ Domain of f(x) is (–∞, –2) ∪ [– 1, 2]

(v)

f(x) =

f(x) is defined when ≥ 0 and x – x2 > o

x – x2 ≥ (0.3)0   and x2 – x < 0

x – x2 ≥ 1 and x (x – 1) < 0

x2 –x + 1 ≤ 0 and (x – 0) (x – 1) < 0

it is true for all x ∈ R and x ∈ (0, 1)

∴ domain of f(x) =  R∩ (0, 1) = (0, 1)

(vi)

f(x) =

f(x) is defined when x +|x| ≠ 0

|x| ≠ – x

|x| ≠ – x

⟹|x| = x

⟹ x > 0

x ∈ (0, ∞)

∴ Domain of f(x) is (0, ∞)

#### 2. Prove that the real valued function is an even function

Sol:

Given f(x) =

#### 3. Find the domain and range of the following functions

(i)

f (x) =

Given f(x) =

Since [x] is an integer

sin π[x] = tan π[x] = 0 ∀ x ∈ R

∴ domain of f(x) is R

and

since tan π[x] = 0

Range of f(x) = {0}

(ii)

f(x) =

Given f (x) =

It is defined when 2 – 3x ≠ 0

⟹ 2 ≠ 3x

⟹ x ≠ 2/3

∴ Domain of f (x) = R – {2/3}

Let y = f(x)

y =

⟹ y (2 – 3x) = x

2y – 3xy = x

2y = x + 3xy

2y = x (1 + 3y)

⟹ x =

It is defined when 1 + 3y ≠ 0

1 ≠ –3y

y ≠ – 1/3

∴ Range of f (x) = R – {– 1/3}

(iii)

f(x) = |x| + | 1 + x|

Given function is f (x) = |x| + |1 + x|

f (x) is defined for all x ∈ R

∴ domain of f(x) = R

f (– 3) = |– 3| + |1 – 3|

=|– 3| + |– 2|

= 3 + 2 = 5

f (– 2) = |– 2| + |1 – 2|

=|– 2| + |– 1|

= 2 + 1 = 3

f (– 1) = |– 1| + |1 – 1|

=|– 1| + |0|

= 1 + 0 = 1

f (0) = |0| + |1 + 0| = 1

f (1) = |1| + |1 + 1|

= 1 + |2|

= 1 + 2 = 3

f (2) = |2| + |1 + 2|

= |2| + |3|

= 2 + 3 = 5

f (3) = |3| + |1 + 3|

= |3| + |4|

= 3 + 4 = 7

∴ Range of f(x) = [1, ∞)

## 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) = ex and g(x) = , then show that f og = gof and f-1 = g-1

Sol:

Given     f (x) = ex and g(x) =

(fog) (x) = f (g (x))

= f ( )

=

= x ————- (1)

(gof) (x) = g (f (x))

= g (ex)

=

= x

= x ————- (2)

From (1) and (2)  f og = gof

let y = f(x)

x = f-1(y)

y = ex

x =

f-1 (x) =

let g(x) = z

x = g-1(z)

z =

x = ez

g-1 (x)  = ex

##### 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

##### (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) = 2x2 + 3 ang g (x) = 3x – 2

(i)  (fog) (x) = f(g(x))

= f(3x – 2)

= 2 (3x – 2)2 + 3

= 2 (9x2 – 12x + 4) + 3

= 18 x2 – 24x + 8 + 3

= 18 x2 – 24x + 11

∴ (fog) (x) = 18 x2 – 24x + 11

(ii)  (gof) (x) = g (f (x))

= g (2x2 + 3)

= 3(2x2 + 3) – 2

= 6x2 + 9 – 2

= 6 x2 + 7

∴ (gof) (x) = 6 x2 + 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

Ts Inter Maths IA Concept

##### (i) (fof) (x2 + 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) = x2 + 1

(i) (fof) (x2 + 1) = f (f (x2 + 1))

= f (3 (x2 + 1)– 1)

= f (3x2 + 3– 1)

= f (3x2 + 2)

= 3(3x2 + 2) – 1

= 9x2 + 6 – 1

= 9x2 + 5

(ii) (fog) (2) = f (g (2))

= f (22 + 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

= 362 – 120a + 100 + 1

= 362 – 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) = x2 and h(x) = 2x ∀ x ∈ R, then find (fo(goh)) (x)

Sol:

Given f (x) = 2, g (x) = x2 and h(x) = 2x ∀ x ∈ R

(fo(goh)) (x) = fo (g (h (x))

= fo g(2x)

= f (g(2x))

= f((2x)2)

= f(4x2)

= 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) = 5x

Let y = f(x)

x = f-1(y)

now y = 5x

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 = 2y

f-1(y) = 2y

∴   f-1(x) = 2x

##### 9. If f(x) = 1 + x + x2 + … for , then show that f-1 (x) =

Sol:

Given, f(x) = 1 + x + x2 + … for

1 + x + x2 + … 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) = 2x (x – 1) then find f-1 (x)

Sol:

Given f: [1, ∞) ⟶ [1, ∞) defined by f (x) = 2x (x – 1)

Let y = f(x)

x = f-1(y)

now y = 2x (x – 1)

= x (x – 1)

x2 – x =

x2 – 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-1og-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-1og-1 = {(q, 1), (p, 2), (r, 3)} ———– (2)

From (1) and (2)

(gof)-1 = f-1og-1

##### (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) = x2 + 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

= 9x2 – 30x + 25 + 1

= 9x2 – 30x + 26

∴(gof) (x – 1) = 9x2 – 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-1og-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-1og-1 = {(2, 1), (1, 2), (3, 4), (4, 3)} ———(2)

from (1) and (2)

(gof)-1 = f-1og-1

##### 5. If f: R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5 then find (fog)-1 (x)

Sol:

Given R ⟶ R; g: R ⟶ R is defined by f (x) = 2x – 3 and g(x) = x3 + 5

(fog) (x) = f (g(x))

= f (x3 + 5)

= 2 (x3 + 5) – 3

= 2 x3 + 10 – 3 = 2x3 + 7

Let y = (fog) (x)

⟹ x = (fog)-1(y)

y = 2x3 + 7

2x3 = y – 7

x3 =

x =

(fog)-1(y)=

(fog)-1(x)=

##### 6. Let f(x) = x2, g(x) = 2x then solve the equation (fog) (x) = (gof) (x)

Sol:

Given f(x) = x2, g(x) = 2x

(fog) (x) = f(g(x))

= f (2x)

= (2x)2

= 22x

(gof) (x) = g(f(x))

= g (x2)

=

Now

(fog) (x) = (gof) (x)

⟹ 22x =

2x = x2   [∵ if am = an , then m = n]

x2 – 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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## Functions Exercise 1a Solutions

Functions Exercise 1a

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 1a Solutions for inter first year  students, prepared by Mathematics expert of www.basicsinmaths.com

### Exercise 1a

#### I.

##### 1. If the function f is defined by

then find the values of (i) f (3)    (ii) f (0)      (iii) f (– 1.5)       (iv) f (2) + f (– 2)       (v) f (– 5 )

Sol:

Given

Domain of f(x) is (– 3, ∞)

(i) f (3)

3 lies in the interval x > 1

⟹ f(x) = x + 2

f(3) = 3 + 2 = 5

∴ f (3) = 5

(ii) f (0)

0 lies in interval – 1 ≤ x ≤ 1

⟹f(x) = 2

∴ f (0) = 2

(iii) f (– 1.5)

– 1.5 lies in interval – 3 < x < – 1

⟹f(x) = x – 1

f (– 1.5) = – 1.5 – 1 = – 2.5

∴ f (– 1.5) = – 2.5

(iv) f (2) + f (– 2)

2 lies in the interval x > 1

⟹ f (x) = x + 2

f (3) = 2 + 2 = 4

f (2) = 4

– 2 lies in interval – 3 < x < – 1

⟹f(x) = x – 1

f (– 2) = – 2 – 1 = – 3

f (– 2) = – 2 – 1 = – 3

now f (2) + f (– 2) = 4 – 3 = 1

∴ f (2) + f (– 2) = 1

(v) f (– 5)

since domain of f(x) is (– 3, ∞)

f (– 5) is not defined

##### 2. If f: R – {0} ⟶ R is defined by f(x) = , then show that f (x) + f (1/x) = 0

Sol:

Given f: R – {0} ⟶ R is defined by f(x) =

f (1/x)  =

Now

f (x) + f (1/x)  =

∴ f (x) + f (1/x)  = 0

##### 3. If f: R ⟶ R is defined by f(x) = , then show that f (tan θ) = cos 2θ

Sol:

Given f: R ⟶ R is defined by f(x) =

f (tan θ) =

= cos 2θ

##### 4. If f: R – {±1} ⟶ R is defined by f(x) = , then show that f  = 2 f (x)

Sol:

Given f: R – {±1} ⟶ R is defined by f(x) =

##### 5. If A = {– 2, – 1, 0, 1, 2} and f: A ⟶ B is a surjection (onto function) defined by f(x) = x2 + x + 1, then find B

Sol:

Given A = {– 2, – 1, 0, 1, 2} and   f: A ⟶ B is a surjection defined by f(x) = x2 + x + 1

f(– 2) = (–2)2 + (–2) + 1

= 4 – 2 + 1 = 3

f(– 1) = (–1)2 + (–1) + 1

= 1 – 1 + 1 = 1

f(0) = (0)2 + (0) + 1

= 0 + 0 + 1 = 1

f(1) = (1)2 + (1) + 1

=1 +1 + 1 = 3

f( 2) = (2)2 + (2) + 1

= 4 + 2 + 1 = 4

∴ B = {1, 3, 7}

TS 10th class maths concept (E/M)

##### 6. If A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) = , then find range of f.

Sol:

Given A = {1, 2, 3,4} and f: A ⟶ B is a surjection defined by f(x) =

##### 7. If f (x + y) = f (xy) ∀ x, y ∈ R, then prove that f is a constant function

Sol:

Given f (x + y) = f (xy) ∀ x, y ∈ R

Let x = 0 and y = 0

f (0 + 0) = f (0 × 0) = f (0)

f (1) = f (0 + 1)

= f (0 × 1)

= f (0)

f (2) = f (1 + 1)

= f (1 × 1)

= f (1)

= f (0)

f (3) = f (1 + 2)

= f (1 × 2)

= f (2)

= f(0)

Similarly, f(4) = 0

f(5) = 0

and so on.

∴ f is a constant function

II.

##### 1. If A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2, g(x) = x3, which of the following are surjections?

(i) f: A ⟶ A                 (ii) g: A ⟶ A

Sol:

###### (i) Given A = {x/ – 1 ≤ x ≤ 1}, f(x) = x2

A = {– 1, 0, 1}; f: A ⟶ A

f (x) = x2

f (– 1) = (– 1)2 = 1

f (0) = (0)2 = 0

f (1) = (1)2 = 1

∵ range is not equal to co domain of f

f is nor a surjection

###### (ii) A = {x/ – 1 ≤ x ≤ 1}, g (x) = x3

A = {– 1, 0, 1}; g: A ⟶ A

g (x) = x3

g (– 1) = (– 1)3 = – 1

g (0) = (0)3 = 0

g (1) = (1)3 = 1

∵ range is equal to co domain of f

f is a surjection

#### 2. Which if the following are injection, surjection or bijection? Justify your answer

##### (i) f: R ⟶ R defined by f(x) =

let x1, x2 ∈ R

f(x1) = f(x2)

2x1 + 1 = 2x2 + 1

2x1 = 2x2

x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y =

3y = 2x + 1

3y – 1 = 2x

⟹ x =  ∈ R

Now

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (ii) f: R ⟶ (0, ∞) defined by f(x) = 2x

let x1, x2 ∈ R

f(x1) = f(x2)

x1 = x2

x1, x2 ∈ R, f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y = 2x

x =   ∈ (0, ∞)

Now f(x) = 2x

=

= y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (iii) f: (0, ∞) ⟶ R defined by f(x) =

let x1, x2 ∈ (0, ∞)

f(x1) = f(x2)

x1 = x2

x1, x2 ∈ (0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y =

x = ey ∈ (0, ∞)

Now f(x) =

=

= y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (iv) f: [0, ∞) ⟶ [0, ∞) defined by f(x) = x2

let x1, x2 ∈ [0, ∞)

f(x1) = f(x2)

x1 = x2 [∵ x1, x2 ∈ [0, ∞)]

x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 = x2

∴ f is an injection

Let y = f(x)

y = x2

x =  ∈ [0, ∞)

Now f(x) = x2

=

= y

∴ f is surjection

f is injection and surjection

∴ f is a bijection

##### (v) f: R ⟶ [0, ∞) defined by f(x) = x2

let x1, x2 ∈ R

f(x1) = f(x2)

x1 = ± x2 [∵ x1, x2 ∈ R]

x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

Let y = f(x)

y = x2

x = ∈ R

Now f(x) = x2

=

= y

∴ f is surjection

f is not an injection but surjection

∴ f is not a bijection

##### (vi) f: R ⟶ R defined by f(x) = x2

let x1, x2 ∈ R

f(x1) = f(x2)

x1 = ± x2 [∵ x1, x2 ∈ R]

x1, x2 ∈ [0, ∞), f(x1) = f(x2) ⟹ x1 ≠ x2

∴ f is not an injection

f (1) = 12 = 1

f (– 1) = (–1)2 = 1

here ‘– 1’ has no pre image

∴ f is not a surjection

f is not an injection and not surjection

∴ f is not a bijection

Ts Inter Maths IA Concept

hai

#### 3. Is g = {(1, 1), (2, 3), (3, 5), (4, 7)} is a function from A = {1, 2, 3, 4} to B = {1, 3, 5, 7}? If this is given by the formula g(x) = ax + b, then find a and b

Sol:

Given A = {1, 2, 3, 4}, B = {1, 3, 5, 7} and g = {(1, 1), (2, 3), (3, 5), (4, 7)}

g (1) = 1; g(2) = 3; g(3) = 5 ; g(4) = 7

here, every element of set A has a unique image in set B

∴ g: A ⟶ B is a function

And also given g(x) = ax + b

g (1) = 1

⟹ a (1) + b = 1

a + b = 1

b = 1 – a _______________   (1)

g (2) = 3

a (2) + b = 3

2a + b = 3

2a + 1 – a = 3 (from (1))

a+ 1 = 3

a = 2

b = 1 – 2

b = – 1

∴ a = 2, b = – 1

#### 4. If the function f: R ⟶ R defined by f(x) = , then show that f (x + y) + f (x – y) = 2 f (x) f (y).

Sol:

Given the function f: R ⟶ R defined by f(x) =

= 2 f (x) f (y)

5. If the function f: R ⟶ R defined by f(x) = , then show that f (1 – x) = 1 – f (x)

and hence reduce the value of

Sol:

Given the function f: R ⟶ R defined by f(x) =

∴ f (1 – x) = 1 – f(x)

#### 6. If the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection, then find a and b

Sol:

Given the function f: {– 1, 1} ⟶ {0, 2}, defined by f(x) = ax + b is a surjection

Case(i)

If f(– 1) = 0 and f(1) = 2

a (– 1) + b = 0

– a + b = 0

b = a ————(1)

and

a (1) + b = 2

a + b = 2

a + a = 2         [ from (1)]

2a = 2

a = 1

b = 1

Case (ii)

If f(– 1) = 2 and f(1) = 0

a (– 1) + b = 2

– a + b =

b = 2 + a ————(2)

and

a (1) + b = 0

a + b = 0

a + 2 + a = 0         [ from (2)]

2a + 2 = 0

2a = – 2

a = – 1

b = 1

From Case(i) and  Case (ii) a = ±1 and b = 1

##### 7. If f(x) = cos (log x), then show that = 0

Sol

Given f(x) = cos (log x)

##### =

cos (log x) cos (log y) – [cos (log x) cos (log y) – sin (log x) sin (log x)

+ cos (log x) cos (log y) + sin (log x) sin (log x)]

= cos (log x) cos (log y) – [2cos (log x) cos (log y)]

= cos (log x) cos (log y) – cos (log x) cos (log y)

∴    = 0