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## Tamilnadu Samacheer Kalvi 10th Maths Solutions Chapter 1 Relations and Functions Unit Exercise 1

**10th Maths Unit Exercise 1 Question 1.**

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

Solution:

(x^{2} – 3x, y^{2 }+ 4y) = (-2, 5)

x^{2} – 3x = -2

x^{2} – 3x + 2 = 0

**10th Maths Chapter 1 Unit Exercise 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.

Answer:

n(A × A) = 9

n(A) = 3

A = {-1,0,1}

A × A = {-1, 0, 1} × {-1, 0, 1}

A × A = {(-1,-1)(-1, 0) (-1, 1)

(0, -1) (0, 0) (0, 1)

(1,-1) (1, 0) (1, 1)}

The remaining elements of A × A =

{(-1, -1) (-1, 1) (0, -1) (0, 0) (1,-1) (1,0) (1,1)}

**10th Maths Unit Exercise Solutions Question 3.**

Given that

(i) f(0)

(ii) f(3)

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

Solution:

(i) f(0) = 4

(ii) f(3) = \(\sqrt{3-1}=\sqrt{2}\)

(iii) f(a + 1) = \(\sqrt{a+1-1}=\sqrt{a}\)

**10th Maths Unit 1 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.

Answer:

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

f: A → N

f(x) = the highest prime factor n ∈ A

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

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

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

**10th Maths Unit Exercise Question 5.**

Find the domain of the function f(x) = \(\sqrt{1+\sqrt{1-\sqrt{1-x^{2}}}}\)

Solution:

f(x) = \(\sqrt{1+\sqrt{1-\sqrt{1-x^{2}}}}\)

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

(x^{2} = 1, -1, 0, because \(\sqrt{1-x^{2}}\) should be +ve, or 0)

**10th Maths Exercise 1.1 Samacheer Kalvi Question 6.**

If f (x) = x^{2}, g(x) = 3x and h(x) = x – 2, Prove that (f o g)o h = f o(g o h).

Answer:

f(x) = x^{2} ; g(x) = 3x and h(x) = x – 2

L.H.S. = (fog) oh

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)

R.H.S. = fo(goh)

goh = g [h(x)]

= g(x – 2)

= 3(x – 2)

= 3x – 6

fo(goh) = fo [goh (x)]

= f(3x – 6)

= (3x – 6)^{2}

= 9x^{2} – 36x + 36 ….(2)

From (1) and (2) we get

L.H.S. = R.H.S.

(fog) oh = fo {goh)

**10th Maths Chapter 1 Samacheer Kalvi Question 7.**

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?

Solution:

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

C = {5, 6}, D = {5, 6, 7, 8)

A × C = {(1, 5), (1, 6), (2, 5), (2, 6)}

B × D = {(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) It is proved.

**Samacheer Kalvi 10th Maths Exercise 1.1 Question 8.**

If f(x) = \(\frac{x-1}{x+1}\), x ≠ 1 show that f(f(x)) = \(-\frac{1}{x}\), Provided x ≠ 0.

Solution:

Hence it is proved.

**Ex 1.1 Class 10 Samacheer Question 9.**

The function/and g are defined by f(x) = 6x + 8; g(x) = \(\frac{x-2}{3}\).

(i) Calculate the value of \(gg\left(\frac{1}{2}\right)\)

(ii) Write an expression for g f(x) in its simplest form.

Solution:

**10th Maths Unit 1 Question Paper Question 10.**

Write the domain of the following real functions

Solution:

(i) f(x) = \(\frac{2x+1}{x-9}\)

The denominator should not be zero as the function is a real function.

∴ The domain = R – {9}

(ii) p(x) = \(\frac{-5}{4 x^{2}+1}\)

The domain is R.

(iii) g(x) = \(\sqrt{x-2}\)

The domain = [2, ∝)

(iv) h(x) = x + 6

The domain is R.