Chapter 2 Functions Ex 2.1 Solutions

Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 2 Functions Ex 2.1 Questions and Answers.

Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Functions Ex 2.1

(b) No
Reason: An element of set A has been assigned more than one element from set B.

(c) No
Reason: Not every element of set A has been assigned an image from set B.

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)} represents a function.
Reason: Every element of set A has a unique image in set B.

(iii) {(1, 3), (4, 1), (2, 2)} does not represent a function.
Reason: 3 ∈ A does not have an image in set B.

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)} represents a function
Reason: Every element of set A has been assigned a unique image in set B.

(ii) f(-3) = (-3)²– 3(-3) + 1
= 9 + 9 + 1
= 19

(iii)
=
=
=

(iv) f(x + 1) = (x + 1)²– 3(x + 1) + 1
= x²+ 2x + 1 – 3x – 3 + 1
= x²– x – 1

(v) f(-x) = (-x)²– 3(-x) + 1 = x²+ 3x + 1

(ii) g(x) =
g(x) = 0
∴ = 0
∴ 18 – 2x²= 0
∴ x²= 9
∴ x = ±3

(iii) g(x) = 6x²+ x – 2
g(x) = 0
∴ 6x²+ x – 2 = 0
∴ 6x²+ 4x – 3x – 2 = 0
∴ 2x(3x + 2) – 1(3x + 2) = 0
∴ (2x – 1)(3x + 2) = 0
∴ 2x – 1 = 0 or 3x + 2 = 0
∴ x = or x =

(ii) (f – g) (2) = f(2) – g(2)
= [3(2) + 5] – [6(2) – 1]
= 6 + 5 – 12 + 1
= 0

(iii) (fg)(3) = f(3) g(3)
= [3(3) + 5] [6(3) – 1]
= (14) (17)
= 238

(iv)
Domain = R – {}

(ii) (gof)(x) = g(f(x))
= g(2x²+ 3)
= 5(2x²+ 3) – 2
= 10x²+ 15 – 2
= 10x²+ 13

(iii) (fof)(x) = f(f(x))
= f(2x²+ 3)
= 2(2x²+ 3)²+ 3
= 2(4x⁴+ 12x²+ 9) + 3
= 8x⁴+ 24x²+ 18 + 3
= 8x⁴+ 24x²+ 21

(iv) (gog)(x) = g(g(x))
= g(5x – 2)
= 5(5x – 2) – 2
= 25x – 10 – 2
= 25x – 12