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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Ex 1.3

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?

Solution:

F = {(x, y)|x, y ∈ N and y = 2x}

x = {1, 2, 3,…}

y = {1 × 2, 2 × 2, 3 × 2, 4 × 2, 5 × 2 …}

R = {(1, 2), (2, 4), (3, 6), (4, 8), (5, 10),…}

Domain of R = {1, 2, 3, 4,…},

Co-domain = {1, 2, 3…..}

Range of R = {2, 4, 6, 8, 10,…}

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 ?

Solution:

x = {3,4, 6, 8}

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

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

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

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

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

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

R = {(3, 10), (4, 17), (6, 37), (8, 65)}

R = {(3, 10), (4, 17), (6, 37), (8, 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(2 a)

(iii) f(2)

(iv) f(x – 1)

Answer:

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 figure 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)

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

(iii) Describe the following

(i) Domain

(ii) Range.

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

Solution:

From the graph

(a) f(0) = 9

(b) f(7) = 6

(c) f(2) = 6

(d) f(10) = 0

(ii) At x = 9.5, f(x) = 1

(iii) Domain = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}

= {x |0 < x < 10, x ∈ R}

Range = {x|0 < x < 9, x ∈ R}

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

(iv) The image of 6 under f is 5.

Question 5.

Let f(x) = 2x + 5. If x ≠ 0 then find \(\frac{f(x+2)-f(2)}{x}\)

Solution:

Given f(x) = 2x + 5, x ≠ 0.

Question 6.

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

(i) find \(\frac{f(0)+f(1)}{2}\)

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

(iii) find x such that f(x) = x.

(iv) find x such that f(x) = f(1 – x).

Solution:

Given f(x) = 2x – 3

(i) find \(\frac{f(0)+f(1)}{2}\)

f(0) = 2(0) – 3 = -3

f(1) = 2(1) – 3 = -1

∴ \(\frac{f(0)+f(1)}{2}=\frac{-3-1}{2}=\frac{-4}{2}\) = -2

(ii) f(x) = 0

⇒ 2x – 3 = 0

2x = 3

x = \(\frac{3}{2}\)

(iii) f(x) = x

⇒ 2x – 3 = x ⇒ 2x – x = 3

x = 3

(iv) f(x) = f(1 – x)

2x – 3 = 2(1 – x) – 3

2x – 3 = 2x – 2x – 3

2x + 2x = 2 – 3 + 3

4x = 2

x = \(\frac{2}{4}\)

x = \(\frac{1}{2}\)

Question 7.

An open box is to be made from a square piece of material, 24 cm on a side, by cutting equal squares from the corners and turning up the sides as shown in figure. Express the volume V of the box as a function of x.

Solution:

Volume of the box = Volume of the cuboid

= l × b × h cu. units

Here l = 24 – 2x

b = 24 – 2x

h = x

∴ V = (24 – 2x) (24 – 2x) × x

= (576 – 48x – 48x + 4x^{2})x

V = 4x^{3} – 96x^{2} + 576x

Question 8.

A function f is defined bv f(x) = 3 – 2x . Find x such that f(x2) = (f(x))2.

Solution:

f(x) = 3 – 2x

f(x^{2}) = 3 – 2x^{2}

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 r in hours.

Answer:

Speed of the plane = 500 km/hr

Distance travelled in “t” hours

= 500 × t (distance = speed × time)

= 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.

Solution:

(i) Given y = ax + b …………. (1)

The ordered pairs are R = {(35, 56) (45, 65) (50, 69.5) (55, 74)}

∴ Hence this relation is a function.

Substituting a = 0.9 in (2) we get

⇒ 65 = 45(.9) + b

⇒ 65 = 40.5 + b

⇒ b = 65 – 40.5

⇒ b = 24.5

∴ a = 0.9, b = 24.5

∴ y = 0.9x + 24.5

(iii) Given x = 40 , y = ?

∴ (4) → y = 0.9 (40) + 24.5

⇒ y = 36 + 24.5

⇒ y = 60.5 inches

(iv) Given y = 53.3 inches, x = ?

(4) → 53.3 = 0.9x + 24.5

⇒ 53.3 – 24.5 = 0.9x

⇒ 28.8 = 0.9x

⇒ x = \(\frac{28.8}{0.9}\) = 32 cm

∴ When y = 53.3 inches, x = 32 cm