**Samacheer Kalvi 10th Maths Book Back Solution:**

Tamil Nadu 10th Maths Book Back Answers **Unit 1 – Relations and Functions** Ex 1.3 are provided on this page. Samacheer Kalvi Maths Book Back Solutions/ Guide available for all Units. TN Samacheer Kalvi 10th Maths Book consists of 8 Units and each unit book back solutions given below topics wise with Questions and Answers. The complete Samacheer Kalvi Books Back Answers/Solutions are available on our site.

The Samacheer kalvi 10th Maths solutions are useful to enhance your skills. Candidates who prepared for the Competitive and board exams 10th Maths Book Back Answers in English and Tamil Medium. The 10th Maths Unit 1 Relation and Functions consist of 6 units. Each Unit Book Back Answers provide topic-wise on this page. 10th Maths Book Back Answers are prepared according to the latest syllabus. The 10th Maths Book Back Relations and Functions Ex 1.3 Answers in English.

**10th Maths Book Back Answers/Solutions:**

TN Samacheer Kalvi 10th Maths Chapter 1 Book Back Exercise given below. The 10th Maths Book Back Solutions Guide is uploaded below:

**Chapter 1**

**Exercise 1.3 Relations and Functions**

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.

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 ?

**Solu.:**

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.

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)

**Solu.:**

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

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?

**Solu.:**

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.

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

**Solu.:
**Given f(x) = 2x + 5, x ≠ 0.

6. A function fis 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 f(x) = x.

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

**Solu.:**

Given f(x) = 2x – 3

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

f(0) = 2(0) – 3 = -3

f(1) = 2(1) – 3 = -1

∴ f(0)+f(1)2=−3−12=−42 = -2

(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(x) = f(1 – x)

2x – 3 = 2(1 – x) – 3

2x – 3 = 2x – 2x – 3

2x + 2x = 2 – 3 + 3

4x = 2

x = 24

x = 12

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 the figure. Express the volume V of the box as a function of x.

**Solu.:
**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

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}

9. A plane is flying at a speed of 500 km per hour. Express the distance traveled by plane as a function of time r in hours.

**Solu.:**

Speed of the plane = 500 km/hr

Distance traveled in “t” hours

= 500 × t (distance = speed × time)

= 500 t

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 the forehand of a woman if her height is 53.3 inches.

**Solu.:**

(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 = 28.80.9 = 32 cm

∴ When y = 53.3 inches, x = 32 cm

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