**Question 1.**

If the ordered pairs (x^{2} – 3x, y^{2} + 4y) and (-2, 5) are equal, then find x and y.

Answer:

(x^{2} – 3x, y^{2} + 4y) = (-2, 5)

x^{2} – 3x = -2

x^{2} – 3x + 2 = 0

(x – 2) (x – 1) = 0

x – 2 = 0 or x – 1 = 0

x = 2 or 1

y^{2} + 4y = 5

y^{2} + 4y – 5 = 0

(y + 5) (y – 1) = 0

y + 5 = 0 or y – 1 = 0

y = -5 or y = 1

The value of x = 2, 1

and 7 = -5, 1

Question 2.

The Cartesian product A × A has 9 elements among which (-1, 0) and (0, 1) are found. Find the set A and the remaining elements of A × A.

Solution:

A = {-1, 0, 1}, B = {1, 0, -1}

A × B = {(-1, 1), (-1, 0), (-1, -1), (0, 1), (0, 0), (0, -1), (1, 1), (1, 0), (1, -1)}

Question 3.

Given that f(x) = {x−1−−−−−√4x≥1x<1.

Find

(i) f(0) (ii)f (3) (iii) f(a + 1) in terms of a.(Given that a > 0)

Answer:

f(x) = x−1−−−−−√ ; f(x) = 4

(i) f(0) = 4

(ii) f(3) = 3−1−−−−√ = 2–√

(iii) f(a + 1) = a+1−1−−−−−−−−√ = a−−√

Question 4.

Let A = {9, 10, 11, 12, 13, 14, 15, 16, 17} and let f: A → N be defined by f(n) = the highest prime factor of n ∈ A. Write f as a set of ordered pairs and find the range of f.

Solution:

A = {9, 10, 11, 12, 13, 14, 15, 16, 17}

f: A → N

f(n) = the highest prime factor of n ∈ A

f = {(9, 3), (10, 5), (11, 11), (12, 3), (13, 13), (14, 7), (15, 5), (16, 2), (17, 17)}

Range = {3, 5, 11, 13, 7, 2, 17}

= {2, 3, 5, 7, 11, 13, 17}

Question 5.

Find the domain of the function

Answer:

Domain of f(x) = {-1, 0, 1}

Question 6.

If f(x)= x^{2}, g(x) = 3x and h(x) = x – 2 Prove that (fog)oh = fo(goh).

Solution:

f(x) = x^{2}

g(x) = 3x

h(x) = x – 2

(fog)oh = x – 2

LHS = fo(goh)

fog = f(g(x)) = f(3x) = (3x)^{2} = 9x^{2}

(fog)oh = (fog) h(x) = (fog) (x – 2)

= 9(x – 2)^{2} = 9(x^{2} – 4x + 4)

= 9x^{2} – 36x + 36 ……………. (1)

RHS = fo(goh)

(goh) = g(h(x)) = g(x – 2)

= 3(x – 2) = 3x – 6

fo(goh) = f(3x – 6) = (3x – 6)^{2}

= 9x^{2} – 36x + 36 ………….. (2)

(1) = (2)

LHS = RHS

(fog)oh = fo(goh) is proved.

Question 7.

Let A= {1,2} and B = {1,2,3,4}, C = {5,6} and D = {5,6,7,8}. Verify whether A × C is a subset of B × D?

Answer:

Given A = {1, 2}

B = {1, 2, 3, 4}

C = {5,6}

D = {5,6, 7,8}

A × C = {1,2} × {5,6}

= {(1,5) (1,6) (2, 5) (2, 6)}

B × D = {1,2, 3, 4} × {5, 6, 7, 8}

= {(1,5) (1,6) (1,7) (1,8)

(2, 5) (2, 6) (2,7) (2, 8)

(3, 5) (3, 6) (3, 7) (3, 8)

(4, 5) (4, 6) (4, 7) (4, 8)}

∴ A × C ⊂ B × D

Hence it is verified

Question 8.

If f(x) = x−1x+1,x≠1 Show that

f(f(x)) = – 1x, Provided x ≠ 0.

Answer:

Question 9.

The functions f and g are defined by f{x) = 6x + 8; g(x) = x−23

(i) Calculate the value of gg [latex]\frac { 1 }{ 2 } [/latex]

(a) Write an expression for gf (x) in its simplest form.

Answer:

f(x) = 6x + 8 ; g(x) = x−23

Question 10.

Write the domain of the following real functions

(ii) if p(x) = =−54×2+1

p(x) is defined for all values of x. So domain is x ∈ R.

If n(A × B) = 6 and A= {1, 3} then n (B) is ………….

(1) 1

(2) 2

(3) 3

(4) 6

**Answer:**

(3) 3**Hint: **n(A × B) = 6

n(A) = 2

n(A × B) = n(A) × n(B)

6 = 2 × n(B)

n(B) = 62 = 3

Question 2.

A = {a, b, p}, B = {2, 3}, C = {p, q, r, s} then n[(A ∪ C) × B] is

(1) 8

(2) 20

(3) 12

(4) 16

**Answer**:

(3) 12**Hint:**

A = {a, b, p}, B = {2, 3}, C = {p, q, r, s}

n (A ∪ C) × B

A ∪ C = {a, b, p, q, r, s}

(A ∪ C) × B = {{a, 2), (a, 3), (b, 2), (b, 3), (p, 2), (p, 3), (q, 2), (q, 3), (r, 2), (r, 3), (s, 2), (s, 3)

n [(A ∪ C) × B] = 12

Question 3.

If A = {1,2}, B = {1,2, 3, 4}, C = {5,6} and D = {5, 6, 7, 8} then state which of the following statement is true ……………….

(1) (A × C) ⊂ (B × D)

(2) (B × D) ⊂ (A × C)

(3) (A × B) ⊂ (A × D)

(4) (D × A) ⊂ (B × A)

**Answer:**

(1) (A × C) ⊂ (B × D)**Hint:** n(A × B) = 2 × 4 = 8

(A × C) = 2 × 2 = 4

n(B × C) = 4 × 2 = 8

n(C × D) = 2 × 4 = 8

n(A × C) = 2 × 2 = 4

n(A × D) = 2 × 4 = 8

n(B × D) = 4 × 4 = 16

∴ (A × C) ⊂ (B × D)

Question 4.

If there are 1024 relations from a set A = {1, 2, 3, 4, 5} to a set B, then the number of elements in B is

(1) 3

(2) 2

(3) 4

(4) 6

**Answer:**

(2) 2**Hint:**

n(A) = 5

n(B) = x

n(A × B) = 1024 = 2^{10}

2^{5x} = 2^{10}

⇒ 5x = 10

⇒ x =2

Question 5.

The range of the relation R = {(x, x^{2}) a prime number less than 13} is ……………………

(1) {2, 3, 5, 7}

(2) {2, 3, 5, 7, 11}

(3) {4, 9, 25, 49, 121}

(4) {1, 4, 9, 25, 49, 121}

**Answer:**

(3) {4, 9, 25, 49, 121}**Hint:**

Prime number less than 13 = {2, 3, 5, 7, 11}

Range (R) = {(x, x^{2})}

Range = {4, 9, 25, 49, 121} (square of x)

Question 6.

If the ordered pairs (a + 2, 4) and (5, 2a + b)are equal then (a, b) is

(1) (2, -2)

(2) (5, 1)

(3) (2, 3)

(4) (3, -2)

**Answer:**

(4) (3, -2)**Hint:**

(a + 2, 4), (5, 2a + b)

a + 2 = 5

a = 3

2a + b = 4

6 + b = 4

b = -2

Question 7.

Let n(A) = m and n(B) = n then the total number of non-empty relations that can be defined from A to B is ……………..

(1) m^{n}

(2) n^{m}

(3) 2^{mn} – 1

(4) 2^{mn}

Answer:

(4) 2^{mn}

Question 8.

If {(a, 8),(6, b)}represents an identity function, then the value of a and b are respectively

(1) (8, 6)

(2) (8, 8)

(3) (6, 8)

(4) (6, 6)

Answer:

(1) (8, 6)

Hint:

{{a, 8), (6, b)}

a = 8

b = 6

Question 9.

Let A = {1, 2, 3, 4} and B = {4, 8, 9, 10}.

A function f: A → B given by f = {(1, 4), (2, 8),(3,9),(4,10)} is a ……………

(1) Many-one function

(2) Identity function

(3) One-to-one function

(4) Into function

Answer:

(3) One-to-one function

Hint:

Different elements of A has different images in B.

∴ It is one-to-one function.

Question 10.

If f (x) = 2x^{2} and g(x) = 13x, then fog is …………..

(1) 32×2

(2) 23×2

(3) 29×2

(4) 16×2

Answer:

(3) 29×2

Hint:

Question 11.

If f: A → B is a bijective function and if n(B) = 7, then n(A) is equal to

(1) 7

(2) 49

(3) 1

(4) 14

Answer:

(1) 7

Hint:

In a bijective function, n(A) = n(B)

⇒ n(A) = 7

Question 12.

Let f and g be two functions given by

f = {(0,1),(2, 0),(3-4),(4,2),(5,7)}

g = {(0,2),(1,0),(2, 4),(-4,2),(7,0)}

then the range of f o g is …………………

(1) {0,2,3,4,5}

(2) {-4,1,0,2,7}

(3) {1,2,3,4,5}

(4) {0,1,2}

**Answer**:

(4) {0,1,2}**Hint: **f = {(0, 1)(2, 0)(3, -4) (4, 2) (5, 7)}

g = {(0,2)(l,0)(2,4)(-4,2)(7,0)}

fog = f[g(x)]

f [g(0)] = f(2) = 0

f [g(1)] = f(0) = 1

f [g(2)] = f(4) = 2

f[g(-4)] = f(2) = 0

f[g(7)] = f(0) = 1

Range of fog = {0,1,2}

Question 13.

Let f(x) = 1+x2−−−−−√ then

(1) f(xy) = f(x),f(y)

(2) f(xy) ≥ f(x),f(y)

(3) f(xy) ≤ f(x).f(y)

(4) None of these

**Answer:**

(3) f(xy) ≤ f(x).f(y)**Hint:**

1+x2y2−−−−−−−√≤(1+x2)−−−−−−−√(1+y2)−−−−−−−√

⇒ f(xy) ≤ f(x) . f(y)

Question 14.

If g= {(1,1),(2,3),(3,5),(4,7)} is a function given by g(x) = αx + β then the values of α and β are

(1) (-1,2)

(2) (2,-1)

(3) (-1,-2)

(4) (1,2)

**Answer:**

(2) (2, -1)

**Hint:** g (x) = αx + β

g(1) = α(1) + β

1 = α + β ….(1)

g (2) = α (2) + β

3 = 2α + β ….(2)

Solve the two equations we get

α = 2, β = -1

Question 15.

f(x) = (x + 1)^{3} – (x – 1)^{3} represents a function which is

(1) linear

(2) cubic

(3) reciprocal

(4) quadratic

Answer:

(4) quadratic

**Hint**:

f(x) = (x + 1)^{3} – (x – 1)^{3}

= x^{3} + 3x^{2} + 3x + 1 -[x^{3} – 3x^{2} + 3x – 1]

= x^{3} + 3x^{2} + 3x + 1 – x^{3} + 3x^{2} – 3x + 1 = 6x^{2} + 2

It is a quadratic function.

Using the functions f and g given below, find fog and gof Check whether fog = gof.

(i) f(x) = x – 6, g(x) = x^{2}

Answer:

f(x) = x – 6, g(x) = x^{2}

fog = fog (x)

= f(g(x))

fog = f(x)^{2}

= x^{2} – 6

gof = go f(x)

= g(x – 6)

= (x – 6)^{2}

= x^{2} – 12x + 36

fog ≠ gof

(ii) f(x) = 2x, g(x) = 2x^{2} – 1

Answer:

f(x) – 2x; g(x) = 2x^{2} – 1

fag = f[g (x)]

= f(2x^{2} – 1)

= 22×2−1

gof = g [f(x)]

= g (2x)

= 2 (2x)^{2} – 1

=2×4×2−1

=8×2−1

fog ≠ gof

(iii) f(x) = x+63, g(x) = 3 – x**Answer:**

f(x) = x+6x, g(x) = 3 – x

fog = f[g(x)]

= f(3 – x)

(iv) f(x) = 3 + x, g(x) = x – 4

Answer:

f(x) = 3 + x ;g(x) = x – 4

fog = f[g(x)]

= f(x – 4)

= 3 + x – 4

= x – 1

gof = g[f(x)]

= g(3 + x)

= 3 + x – 4

= x – 1

fog = gof

(v) f(x) = 4x^{2} – 1,g(x) = 1 + x

Answer:

f(x) = 4x^{2} – 1 ; g(x) = 1 + x

fog = f[g(x)]

= 4(1 + x)

= 4(1 + x)^{2} – 1

= 4[1 + x^{2} + 2x] – 1

= 4 + 4x^{2} + 8x – 1

= 4x^{2} + 8x + 3

gof = g [f(x)]

= g (4x^{2} – 1)

= 1 + 4x^{2} – 1

= 4x^{2}

fog ≠ gof

Question 2.

Find the value of k, such that fog = gof

(i) f(x) = 3x + 2, g(x) = 6x – k

(ii) f(x) = 2x – k, g(x) = 4x + 5**Solution:**

(i) f(x) = 3x + 2, g(x) = 6x – k

fog(x) = f(g(x)) = f(6x – k) = 3(6x – k) + 2

= 18x – 3k + 2 …………… (1)

gof(x) = g(f(x)) = g(3x + 2) = 6(3x + 2) – k

= 18x + 12 – k ……………. (2)

(1) = (2)

⇒ 18x – 3k + 2 = 18x + 12 – k

2k = -10

k = -5

(ii) f(x) = 2x – k, g(x) = 4x + 5

fog(x) = f(g(x)) = f(4x + 5) = 2(4x + 5) – k

= 8x + 10 – k ……………… (1)

gof(x) = g(f(x)) = g(2x – k) = 4(2x – k) + 5

= 8x – 4k + 5 ……………. (2)

(1) = (2)

⇒ 8x + 10 – k = 8x – 4k + 5

3k = -5

k = −53

Question 3.

If f(x) = 2x – 1, g(x) = x+12, show that f o g = g o f = x

Answer:

f(x) = 2x – 1 ; g(x) = x+12

fog = f[g(x)]

∴ fog = gof = x

Hence it is proved.

Question 4.

(i) If f (x) = x^{2} – 1, g(x) = x – 2 find a, if gof(a) = 1.

(ii) Find k, if f(k) = 2k – 1 and fof (k) = 5.

Solution:

(i) f(x) = x^{2} – 1, g(x) = x – 2

Given gof(a) = 1

gof(x) = g(f(x)

= g(x^{2} – 1) = x^{2} – 1 – 2

= x^{2} – 3

gof(a) ⇒ a^{2} – 3 = 1 =+ a^{2} = 4

a = ± 2

(ii) f(k) = 2k – 1

fo f(k) = 5

f(f(k)m = f(2k – 1) = 5

⇒ 2(2k – 1) – 1 = 5

4 k – 2 – 1 = 5 ⇒ 4k = 8

k = 2

Question 5.

Let A,B,C ⊂ N and a function f: A → B be defined by f(x) = 2x + 1 and g: B → C be defined by g(x) = x^{2} . Find the range of fog and gof.

Answer:

f(x) = 2x + 1 ; g(x) = x^{2}

fog = f[g(x)]

= f(x^{2})

= 2x^{2} + 1

2x^{2} + 1 ∈ N

g o f = g [f(x)]

= g (2x + 1)

g o f = (2x + 1)^{2}

(2x + 1)^{2} ∈ N

Range = {y/y = 2x^{2} + 1, x ∈ N};

{y/y = (2x + 1)^{2}, x ∈ N)

Question 6.

If f(x) = x^{2} – 1. Find (i)f(x) = x^{2} – 1, (ii)fofof

Solution:

(i) f(x) = x^{2} – 1

fof(x) = f(fx)) = f(x^{2} – 1)

= (x^{2} – 1 )^{2} – 1;

= x^{4} – 2x^{2} + 1 – 1

= x^{4} – 2x^{2}

(ii) fofof = f o f(f(x))

= f o f (x^{4} – 2x^{2})

= f(f(x^{4} – 2x^{2}))

= (x^{4} – 2x^{2})^{2} – 1

= x^{8} – 4x^{6} + 4x^{4} – 1

Question 7.

If f : R → R and g : R → R are defined by f(x) = x^{5} and g(x) = x^{4} then check if f, g are one – one and fog is one – one?

Answer:

f(x) = x^{5} – It is one – one function

g(x) = x^{4} – It is one – one function

fog = f[g(x)]

= f(x^{4})

= (x^{4})^{5}

fag = x^{20}

It is also one-one function.

Question 8.

Consider the functions f(x), g(x), h(x) as given below. Show that (fog)oh = fo(goh) in each case.

(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x^{2}

(ii) f(x) = x^{2}, g(x) = 2x and h(x) = x + 4

(iii) f(x) = x – 4, g(x) = x^{2} and h(x) = 3x – 5

Solution:

(i) f(x) = x – 1, g(x) = 3x + 1 and h(x) = x^{2}

f(x) = x – 1

g(x) = 3x + 1

f(x) = x^{2}

(fog)oh = fo(goh)

LHS = (fog)oh

fog = f(g(x)) = f(3x + 1) = 3x + 1 – 1 = 3x

(fog)oh = (fog)(h(x)) = (fog)(x^{2 }) = 3^{2 }……………. (1)

RHS = fo(goh)

goh = g(h(x)) = g(x^{2}) = 3x^{2} + 1

fo(goh) = f(3x^{2} + 1) = 3x^{2} + 1 – 1= 3x^{2 }………… (2)

LHS = RHS Hence it is verified.

(ii) f(x) = x^{2}, g(x) = 2x, h(x) = x + 4

(fog)oh = fo(goh)

LHS = (fog)oh

fog = f(g(x)) = f(2x) = (2x)^{2} = 4x^{2}

(fog)oh = (fog) h(x) = (fog) (x + 4)

= 4(x + 4)^{2} = 4(x^{2} + 8x+16)

= 4x^{2} + 32x + 64 ………….. (1)

RHS = fo(goh) goh = g(h(x)) = g(x + 4)

= 2(x + 4) = (2x + 8)

fo(goh) = f(goh) = f(2x + 8) = (2x + 8)^{2}

= 4x^{2} + 32x + 64 ……………… (2)

(1) = (2)

LHS = RHS

∴ (fog)oh = fo(goh) It is proved.

(iii) f(x) = x – 4, g(x) = x^{2}, h(x) = 3x – 5

(fog)oh = fo(goh)

LHS = (fog)oh

fog = f(g(x)) = f(x^{2}) = x^{2} – 4

(fog)oh = (fog)(3x – 5) = (3x – 5)^{2} – 4

= 9x^{2} – 30x + 25 -4

= 9x^{2} – 30x + 21 ………….. (1)

∴ RHS = fo(goh)

(goh) = g(h(x)) = g(3x – 5) = (3x – 5)^{2}

= 9x^{2} – 30x + 25

fo(goh) = f(9x^{2} – 30 x + 25)

= 9x^{2} – 30x + 25 – 4

= 9x^{2} – 30x + 21 …………… (2)

(1) = (2)

LHS = RHS

∴ (fog)oh = fo(goh)

It is proved.

Question 9.

Let f = {(-1, 3), (0, -1), (2, -9)} be a linear function from Z into Z. Find f(x).**Answer:**

The linear equation is f(x) = ax + b

f(-1) = 3

a(-1) + b = 3

-a + b = 3 ….(1)

f(0) = -1

a(0) + b = -1

0 + b = -1

b = -1

Substitute the value of b = -1 in (1)

-a – 1 = 3

-a = 3 + 1

-a = 4

a = -4

∴ The linear equation is -4(x) -1 = -4x – 1 (or) – (4x + 1)

Question 10.

In electrical circuit theory, a circuit C(t) is called a linear circuit if it satisfies the superposition principle given by C(at_{1} + bt_{2}) = aC(t_{1}) + bC(t_{2}), where a,b are constants. Show that the circuit C(t) = 31 is linear.**Solution:**

Given C(t) = 3t. To prove that the function is linear

C(at_{1}) = 3a(t_{1})

C(bt_{2}) = 3 b(t_{2})

C(at_{1} + bt_{2}) = 3 [at_{1} + bt_{2}] = 3at_{1} + 3bt_{2}

= a(3t_{1}) + b(3t_{2}) = a[C(t_{1}) + b(Ct_{2})]

∴ Superposition principle is satisfied.

Hence C(t) = 3t is linear function.

Question 1.

Let f = {(x, y) |x,y; ∈ N and y = 2x} be a relation on N. Find the domain, co-domain and range. Is this relation a function?

Answer:

X = {1,2,3,….}

Y = {1,2,3,….}

f = {(1,2) (2, 4) (3, 6) (4, 8) ….}

Domain = {1, 2, 3, 4 ….}

Co – Domain = {1, 2, 3, 4 ….}

Range = {2, 4, 6, 8 }

Yes this relation is a function.

Question 2.

Let X = {3, 4, 6, 8}. Determine whether the relation

R = {(x,f(x)) |x ∈ X, f(x) = x^{2} + 1}

is a function from X to N?

Answer:

f(x) = x^{2} + 1

f(3) = 3^{2} + 1 = 9 + 1 = 10

f(4) = 4^{2} + 1 = 16 + 1 = 17

f(6) = 6^{2} + 1 = 36 + 1 = 37

f(8) = 8^{2} + 1 = 64 + 1 = 65

yes, R is a function from X to N

Question 3.

Given the function f: x → x^{2} – 5x + 6, evaluate

(i) f(-1)

(ii) f(2a)

(iii) f(2)

(iv) f(x – 1)

Solution:

Give the function f: x → x^{2} – 5x + 6.

(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) = 2^{2} – 5(2) + 6 = 4 – 10 + 6 = 0

(iv) f(x – 1) = (x – 1)^{2} – 5(x – 1) + 6

= x^{2} – 2x + 1 – 5x + 5 + 6

= x^{2} – 7x + 12

Question 4.

A graph representing the function f(x) is given in it is clear that f(9) = 2.

**(i) Find the following values of the function**

(a) f(0)

(b) f(7)

(c) f(2)

(d) f(10)**Answer:**

(a) f (0) = 9

(b) f (7) = 6

(c) f (2) = 6

(d) f(10) = 0

(ii) For what value of x is f(x) = 1 ?

Answer:

When f(x) = 1 the value of x is 9.5

(iii) Describe the following

(i) Domain

(ii) Range.

Answer:

Domain = {0, 1, 2, 3,… .10}

= {x / 0 < x < 10, x ∈ R}

Range = {0,1,2,3,4,5,6,7,8,9}

= {x / 0 < x < 9, x ∈ R}

(iv) What is the image of 6 under f?**Answer:**

The image of 6 under f is 5.

Question 5.

Let f (x) = 2x + 5. If x ≠ 0 then find

f(x+2)−f(2)x**Answer:**

f(x) = 2x + 5

f(x + 2) = 2(x + 2) + 5

= 2x + 4 + 5

= 2x + 9

**Question 6.**

A function/is defined by f(x) = 2x – 3

(i) find f(0)+f(1)2

(ii) find x such that f(x) = 0.

(iii) find x such that/ (A:) = x.

(iv) find x such that fix) =/(l – x).

Answer:

(i) f(x) = 2x – 3

f(0) = 2(0) – 3 = -3

f(1) = 2(1) – 3 = 2 – 3 = -1

(ii) f(x) = 0

2x – 3 = 0

2x = 3

x = 32

(iii) f(x) = x

2x – 3 = x

2x – x = 3

x = 3

(iv) f(1 – x) = 2(1 – x) – 3

= 2 – 2x – 3

= – 2x – 1

f(x) = f(1 – x)

2x – 3 = – 2x – 1

2x + 2x = 3 – 1

4x = 2

x = 24 = 12

Question 7.

square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown. Express the volume V of the box as a function of x.

Solution:

After cutting squares we will get a cuboid,

length of the cuboid (l) = 24 – 2x

breadth of the cuboid (b) = 24 – 2x

height of the cuboid (h) = 2x

Volume of the box = Volume of the cuboid

V = (24 – 2x)(24 – 2x) (x)

= (24 – 2x)^{2} (x)

= (576 + 4x^{2} – 96x) x

= 576x + 4x^{3} – 96x^{2}

V = 4x^{3} – 96x^{2} + 576x

V(x) = 4x^{3} – 96x^{2} + 576x

Question 8.

A function f is defined by f(x) = 3 – 2x. Find x such that f(x^{2}) = (f (x))^{2}.

Answer:

f(x) = 3 – 2x

f(x^{2}) = 3 – 2 (x^{2})

= 3 – 2x^{2}

(f (x))^{2} = (3 – 2x)^{2}

= 9 + 4x^{2} – 12x

But f(x^{2}) = (f(x))^{2}

3 – 2 x^{2} = 9 + 4x^{2} – 12x

-2x^{2} – 4x^{2 }+ 12x + 3 – 9 = 0

-6x^{2} + 12x – 6 = 0

(÷ by – 6) ⇒ x^{2} – 2x + 1 = 0

(x – 1) (x – 1) = 0

x – 1 = 0 or x – 1 = 0

x = 1

The value of x = 1

Question 9.

A plane is flying at a speed of 500 km per hour. Express the distance d travelled by the plane as function of time t in hours.

Solution:

Speed = distance covered time taken

⇒ distance = Speed × time

⇒ d = 500 × t [ ∵ time = t hrs]

⇒ d = 500 t

Question 10.

The data in the adjacent table depicts the length of a woman’s forehand and her corresponding height. Based on this data, a student finds a relationship between the height (y) and the forehand length (x) as y = ax + b , where a, b are constants.

(i) Check if this relation is a function.

(ii) Find a and b.

(iii) Find the height of a woman whose forehand length is 40 cm.

(iv) Find the length of forehand of a woman if her height is 53.3 inches.

Length ‘x’ of forehand (in cm) | Height y (in inches) |

35 | 56 |

45 | 65 |

50 | 69.5 |

55 | 74 |

Answer:

The relation is y = 0.9x + 24.5

(i) Yes the relation is a function.

(ii) When compare with y = ax + b

a = 0.9, b = 24.5

(iii) When the forehand length is 40 cm, then height is 60.5 inches.

Hint: y = 0.9x + 24.5

= 0.9 × 40 + 24.5

= 36 + 24.5

= 60.5 feet

(iv) When the height is 53.3 inches, her forehand length is 32 cm

Hint: y = 0.9x + 24.5

53.3 = 0.9x + 24.5

53.3 – 24.5 = 0.9 x

28.8 = 0.9 x

x = 28.80.9

x = 32 cm

A function may be represented by

(a) Set of ordered pairs

(b) Table form

(c) Arrow diagram

(d) Graphical form

Vertical line test

A curve drawn in a graph represents a function, if every vertical line intersects the curve in at most one point.

1. One – One function (injection)

A function f: A → B is called one-one function if distinct elements of A have a distinct image in B.

2. Many – One function

,A function f: A → B is called many-on function if two or more elements of A have same image in B.

3. Onto function (surjection)

A function f: A → B is said to be on to function if the range of f is equal to the co-domain of f.

4. Into function

A function f: A → B is called an into function if there exists at least one element in B which is not the image of any element of A.

5. Bijection

A function f: A → B is both one – one and onto, then f is called a bijection from A to B.

**Horizontal line test**

A function represented in a graph is one – one, if every horizontal line intersects the curve in at most one point.

**Special cases of function**

1. Constant function

A function f: A → B is called a constant function if the range of f contains only one element.

2. Identity function

A function f : A → A defined by f(x) = x for all x ∈ A is called an identity function on A and is denoted by I_{A}.

3. Real valued function

A function f: A → B is called a real valued function if the range of f is a subset of the set of all real numbers R.

**Question 1.**

Let A = {1, 2, 3, 7} and B = {3, 0, -1, 7}, which of the following are relation from A to B?

(i) R_{1} = {(2,1), (7,1)}

(ii) R_{2} = {(-1,1)}

(iii) R_{3} = {(2,-1), (7, 7), (1,3)}

(iv) R_{4} = {(7, -1), (0, 3), (3, 3), (0, 7)}

Answer:

A = {1,2,3,7} B = {3,0,-1, 7}

A × B = {1,2,3} × {3, 0,-1, 7}

A × B = {(1,3) (1,0) (1,-1) (1,7) (2,3) (2, 0)

(2, -1) (2, 7) (3, 3) (3,0) (3,-1)

(3, 7) (7, 3) (7, 0) (7,-1) (7, 7)}

(i) R_{1} = {(2, 1)} (7, 1)

It is not a relation, there is no element of (2, 1) and (7, 1) in A × B

(ii) R_{2} = {(-1),1)}

It is not a relation, there is no element of

(-1, 1) in A × B

(iii) R_{3} = {(2,-1) (7, 7) (1,3)}

Yes, It is a relation

(iv) R_{4} = {(7,-1) (0,3) (3, 3) (0,7)}

It is not a relation, there is no element of (0, 3) and (0, 7) in A × B

Question 2.

Let A = {1, 2, 3, 4,…,45} and R be the relation defined as “is square of ” on A. Write R as a subset of A × A. Also, find the domain and range of R.

Solution:

A = {1, 2, 3, 4, . . . 45}, A × A = {(1, 1), (2, 2) ….. (45, 45)}

R – is square of’

R = {(1, 1), (2, 4), (3, 9), (4, 16), (5, 25), (6, 36)}

R ⊂ (A × A)

Domain of R = {1, 2, 3, 4, 5, 6}

Range of R = {1, 4, 9, 16, 25, 36}

Question 3.

A Relation R is given by the set {(x, y)/y = x + 3, x ∈ {0, 1, 2, 3, 4, 5}}. Determine its domain and range.

Answer:

x = {0, 1, 2, 3, 4, 5}

y = x + 3

when x = 0 ⇒ y = 0 + 3 = 3

when x = 1 ⇒ y = 1 + 3 = 4

when x = 2 ⇒ y = 2 + 3 = 5

when x = 3 ⇒ y = 3 + 3 = 6

when x = 4 ⇒ y = 4 + 3 = 7

when x = 5 y = 5 + 3 = 8

R = {(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}

Question 4.

Represent each of the given relations by

(a) an arrow diagram

(b) a graph and

(c) a set in roster form, wherever possible.

(i) {(x,y) | x = 2y,x ∈ {2, 3, 4, 5}, y ∈ {1, 2, 3, 4}

(ii) {(x, y) | y = x + 3, x, y are natural numbers < 10}**Answer:**

(i) x = {2, 3, 4, 5} y = {1, 2, 3, 4}

x = 2y

wheny y = 1 ⇒ x = 2 × 1 = 2

when y = 2 ⇒ x = 2 × 2 = 4

when y = 3 ⇒ r = 2 × 3 = 6

when y = 4 ⇒ x = 2 × 4 = 8

(a) Arrow diagram

(ii) x = {1, 2, 3, 4, 5, 6, 7, 8, 9}

y = {1,2, 3, 4, 5, 6, 7, 8,9}

y = x + 3

when x = 1 ⇒ y = 1 + 3 = 4

when x = 2 ⇒ y = 2 + 3 = 5

when x = 3 ⇒ y = 3 + 3 = 6

when x = 4 ⇒ y = 4 + 3 = 7

when x = 5 ⇒ y = 5 + 3 = 8

when x = 6 ⇒ y = 6 + 3 = 9

when x = 7 ⇒ y = 7 + 3 = 10

when x = 8 ⇒ y = 8 + 3 = 11

when x = 9 ⇒ y = 9 + 3 = 12

R = {(1,4) (2, 5) (3,6) (4, 7) (5, 8) (6, 9)}

B. Solution.

(c) Roster form: R = {(1, 4) (2, 5) (3, 6) (4, 7) (5, 8) (6, 9)}

Question 5.

A company has four categories of employees given by Assistants (A), Clerks (C), Managers (M) and an Executive Officer (E). The company provide ₹10,000, ₹25,000, ₹50,000 and ₹1,00,000 as salaries to the people who work in the categories A, C, M and E respectively. If A_{1}, A_{2}, A_{3}, A_{4} and A_{5} were Assistants; C_{1}, C_{2}, C_{3}, C_{4} were Clerks; M_{1}, M_{2}, M_{3} were managers and E_{1}, E_{2} were Executive officers and if the relation R is defined by xRy, where x is the salary given to person y, express the relation R through an ordered pair and an arrow diagram.

Answer:

Assistants → A_{1}, A_{2}, A_{3}, A_{4}, A_{5}

Clerks → C_{1}, C_{2}, C_{3}, C_{4}

Managers → M_{1}, M_{2}, M_{3}

Executive officers → E_{1}, E_{2}

R = {00000, A_{1}) (10000, A_{2}) (10000, A_{3}) (10000, A_{4}) (10000, A_{5})

(25000, C_{1}) (25000, C_{2}) (25000, C_{3}) (25000, C_{4})

(50000, M_{1}) (50000, M_{2}) (50000, M_{3}) (100000, E_{1}) (100000, E_{2})}

(a) Arrow diagram

**Functions Definition**

A relation f between two non – empty sets X and Y is called a function from X to Y if for each x ∈ X there exists only one Y ∈ Y such that (x, y) ∈ f

f = {(x, y) / for all x ∈ X, y ∈ f}

Note: The range of a function is a subset of its co-domain

1. Find A × B, A × A and B × A

(i) A = {2, -2, 3} and B = {1, -4}

(ii) A = B = {p, q}

(iii) A – {m, n} ; B = Φ

Answer:

(i) A = {2, -2, 3} and B = {1, -4}

A × B = {2,-2, 3} × {1,-4}

= {(2, 1), (2, -4)(-2, 1) (-2, -4) (3, 1) (3,-4)}

A × A = {2,-2, 3} × {2,-2, 3}

= {(2, 2)(2, -2)(2, 3)(-2, 2)

(-2, -2)(-2, 3X3,2) (3,-2) (3,3)}

B × A = {1,-4} × {2,-2, 3}

= {(1, 2)(1, -2)( 1, 3)(-4, 2) (-4,-2)(-4, 3).

(ii) A = B = {p, q}

A × B = {p, q) × {p, q}

= {(p,p),(p,q)(q,p)(q,q)}

A × A = {p,q) × (p,q)

= {(p,p)(p,q)(q,p)(q,q)

B × A = {p,q} × {p,q}

= {(p,p)(p,q)(q,p)(q,q)

(iii) A = {m, n} × B = Φ

Note: B = Φ or {}

A × B = {m, n) × { }

= { )

A × A = {m, n) × (m, n)}

= {(m, m)(w, w)(n, m)(n, n)}

B × A = { } × {w, n}

= { }

Question 2.

Let A = {1, 2, 3} and B = {x | x is a prime number less than 10}. Find A × B and B × A.

Solution:

A = {1, 2, 3}, B = {2, 3, 5, 7}

A × 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 × 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)}

Question 3.

If B × A = {(-2, 3),(-2, 4),(0, 3),(0, 4), (3,3) ,(3, 4)} find A and B.

Answer:

B × A = {(-2, 3)(-2, 4) (0, 3) (0, 4) (3, 3) (3,4)}

A = {3,4}

B = {-2,0,3}

Question 4.

If A ={5, 6}, B = {4, 5, 6} , C = {5, 6, 7}, Show that A × A = (B × B) ∩ (C × C).

Solution:

A = {5,6}, B = {4, 5, 6},C = {5, 6, 7}

A × A = {(5, 5), (5, 6), (6, 5), (6, 6)} ……….. (1)

B × B = {(4, 4), (4, 5), (4, 6), (5, 4),

(5, 5), (5, 6), (6, 4), (6, 5), (6, 6)} …(2)

C × C = {(5, 5), (5, 6), (5, 7), (6, 5), (6, 6),

(6, 7), (7, 5), (7, 6), (7, 7)} …(3)

(B × B) ∩ (C × C) = {(5, 5), (5,6), (6, 5), (6,6)} …(4)

(1) = (4)

A × A = (B × B) ∩ (C × C)

It is proved.

Question 5.

Given A = {1,2,3}, B = {2,3,5}, C = {3,4} and D = {1,3,5}, check if

(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D) is true?

Answer:

A = {1,2, 3}, B = {2, 3, 5}, C = {3,4} D = {1,3,5}

A ∩ c = {1,2,3} ∩ {3,4}

= (3}

B ∩ D = {2,3, 5} ∩ {1,3,5}

= {3,5}

(A ∩ C) × (B ∩ D) = {3} × {3,5}

= {(3, 3)(3, 5)} ….(1)

A × B = {1,2,3} × {2,3,5}

= {(1,2) (1,3) (1,5) (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5)}

C × D = {3,4} × {1,3,5}

= {(3,1) (3, 3) (3, 5) (4,1) (4, 3) (4, 5)}

(A × B) ∩ (C × D) = {(3, 3) (3, 5)} ….(2)

From (1) and (2) we get

(A ∩ C) × (B ∩ D) = (A × B) ∩ (C × D)

This is true.

Question 6.

Let A = {x ∈ W | x < 2}, B = {x ∈ N |1 < x < 4} and C = {3, 5}. Verify that

(i) A × (B ∪ C) = (A × B) ∪ (A × C)

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

(iv) A × (B ∪ C) = (A × B) ∪ (A × C)**Solution:**

A = {x ∈ W|x < 2} = {0,1}

B = {x ∈ N |1 < x < 4} = {2,3,4}

C = {3,5}

LHS =A × (B ∪ C)

B ∪ C = {2, 3, 4} ∪ {3, 5}

= {2, 3, 4, 5}

A × (B ∪ C) = {(0, 2), (0, 3), (0,4), (0, 5), (1, 2) , (1, 3), (1, 4),(1, 5)} ………. (1)

RHS = (A × B) ∪ (A × C)

(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}

(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}

(A × B) ∪ (A × C)= {(0, 2), (0, 3), (0,4), (1, 2), (1, 3), (1, 4), (0, 5), (1, 5)} ….(2)

(1) = (2),

LHS = RHS

Hence it is proved.

(ii) A × (B ∩ C) = (A × B) ∩ (A × C)

LHS = A × (B ∩ C)

(B ∩ C) = {3}

A × (B ∩ C) = {(0, 3), (1, 3)} …(1)

RHS = (A × B) ∩ (A × C)

(A × B) = {(0, 2), (0, 3), (0, 4), (1, 2), (1, 3), (1, 4)}

(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}

(A × B) ∩ (A × C) = {(0, 3), (1, 3)} ……….. (2)

(1) = (2) ⇒ LHS = RHS.

Hence it is verified.

(iii) (A ∪ B) × C = (A × C) ∪ (B × C)

LHS = (A ∪ B) × C

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)} …………. (1)

RHS = (A × C) ∪ (B × C)

(A × C) = {(0, 3), (0, 5), (1, 3), (1, 5)}

(B × C) = {(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)} ………… (2)

(1) = (2)

∴ LHS = RHS.

Hence it is verified.

**Question 7.**

Let A = The set of all natural numbers less than 8, B = The set of all prime numbers less than 8, C = The set of even prime number. Verify that

(i) (A ∩ B) × C = (A × c) ∩ (B × C)

(ii) A × (B – C) = (A × B) – (A × C)**Answer:**

A = {1,2, 3, 4, 5,6, 7}

B = {2, 3, 5,7}

C = {2}

(i) (A ∩ B) × C = (A × C) ∩ (B × C)

A ∩ B = {1, 2, 3, 4, 5, 6, 7} ∩ {2, 3, 5, 7}

= {2, 3, 5, 7}

(A ∩ B) × C = {2, 3, 5, 7} × {2}

= {(2, 2) (3, 2) (5, 2) (7, 2)} ….(1)

A × C = {1,2, 3, 4, 5, 6, 7} × {2}

= {(1,2) (2, 2) (3, 2) (4, 2)

(5.2) (6, 2) (7, 2)}

B × C = {2, 3, 5, 7} × {2}

= {(2, 2) (3, 2) (5, 2) (7, 2)}

(A × C) ∩ (B × C) = {(2, 2) (3, 2) (5, 2) (7, 2)} ….(2)

From (1) and (2) we get

(A ∩ B) × C = (A × C) ∩ (B × C)

(ii) A × (B – C) = (A × B) – (A × C)

B – C = {2, 3, 5, 7} – {2}

= {3,5,7}

A × (B – C) = {1,2, 3, 4, 5, 6,7} × {3,5,7}

= {(1, 3) (1, 5) (1, 7) (2, 3) (2, 5)

(2, 7) (3, 3) (3, 5) (3, 7) (4, 3)

(4, 5) (4, 7) (5, 3) (5, 5) (5, 7)

(6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ………….(1)

A × B = {1,2, 3, 4, 5, 6, 7} × {2, 3, 5,7}

= {(1, 2) (1, 3) (1, 5) (1, 7) (2, 2) (2, 3)

(2, 5) (2, 7) (3, 2) (3, 3) (3, 5) (3, 7)

(4, 2) (4, 3) (4, 5) (4, 7) (5, 2) (5, 3) (5, 5)

(5, 7) (6, 2) (6, 3) (6, 5) (6, 7)

(7, 2) (7, 3) (7, 5) (7, 7)}

A × C = {1,2, 3,4, 5, 6, 7} × {2}

= {(1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6.2) (7,2)}

(A × B) – (A × C) = {(1, 3) (1, 5) (1, 7)

(2, 3) (2, 5) (2, 7) (3, 3) (3, 5)

(3, 7) (4, 3) (4, 5) (4, 7) (5, 3) (5, 5)

(5, 7) (6, 3) (6, 5) (6, 7) (7, 3) (7, 5) (7, 7)} ….(2)

From (1) and (2) we get

A × (B – C) = (A × B) – (A × C)

Relations

Let A and B be any two non-empty sets. A “relation” R from A to B is a subset of A × B satisfying some specified conditions.

**Note:**

- The domain of the relations R = {x ∈ A/xRy, for some y ∈ B}
- The co-domain of the relation R is B
- The range of the ralation

R = (y ∈ B/xRy for some x ∈ A}