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

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) = \(\left\{\begin{array}{rl}

{\sqrt{x-1}} & {x \geq 1} \\

{4} & {x<1}

\end{array}\right.\).

Find

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

Answer:

f(x) = \(\sqrt { x-1 }\) ; f(x) = 4

(i) f(0) = 4

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

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

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) = \(\frac{x-1}{x+1}, x \neq 1\) Show that

f(f(x)) = – \(\frac { 1 }{ x } \), Provided x ≠ 0.

Answer:

Question 9.

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

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

Question 10.

Write the domain of the following real functions

Answer:

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

If the denominator vanishes when x = 9

So f(x) is not defined at x = 9

∴ Domain is x ∈ [R – {9}]

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

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

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

When x < 2 g(x) becomes complex. But given “g” is real valued function.

So x > 2

Domain x ∈ (2, α)

(iv) h (x) = x + 6

For all values of x, h(x) is defined. Hence domain is x ∈ R.