Balbharati Maharashtra State Board11th Commerce Maths Solution Book PdfChapter 2 Functions Miscellaneous Exercise 2 Questions and Answers.
Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2
Question 1
Maharashtra Board Solution
Which of the following relations are functions? If it is a function determine its domain and range. (i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)} (ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)} (iii) {(1, 1), (3, 1), (5, 2)}
Solution & Step-by-Step Answer:
(i) {(2, 1), (4, 2), (6, 3), (8, 4), (10, 5) (12, 6), (14, 7)} Every element of set A has been assigned a unique element in set B. ∴ Given relation is a function. Domain = {2, 4, 6, 8, 10, 12, 14}, Range = {1, 2, 3, 4, 5, 6, 7}
(ii) {(0, 0), (1, 1), (1, -1), (4, 2), (4, -2), (9, 3), (9, -3), (16, 4), (16, -4)}
∴ (1, 1), (1, -1) ∈ the relation
∴ Given relation is not a function.
As element 1 of the domain has not been assigned a unique element of co-domain.
(iii) {(1, 1), (3, 1), (5, 2)}
Every element of set A has been assigned a unique element in set B.
∴ Given relation is a function.
Domain = {1, 3, 5}, Range = {1, 2}
Question 2
Maharashtra Board Solution
A function f: R → R defined by f(x) = + 2, x ∈ R. Show that f is one-one and onto. Hence, find f-1.
Solution & Step-by-Step Answer:
f: R → R defined by f(x) = + 2 First we have to prove that f is one-one function for that we have to prove if f(x1) = f(x2) then x1 = x2 Here f(x) = + 2 Let f(x1) = f(x2) ∴ ∴ ∴ x1 = x2 ∴ f is a one-one function. Now, we have to prove that f is an onto function. Let y ∈ R be such that y = f(x) ∴ y = + 2 ∴ y – 2 = ∴ x = ∈ R ∴ for any y ∈ co-domain R, there exist an element x = ∈ domain R such that f(x) = y ∴ f is an onto function. ∴ f is one-one onto function. ∴ f-1 exists. ∴ ∴
Question 3
Maharashtra Board Solution
A function f is defined as follows: f(x) = 4x + 5, for -4 ≤ x < 0. Find the values of f(-1), f(-2), f(0), if they exist.
Solution & Step-by-Step Answer:
f(x) = 4x + 5, -4 ≤ x < 0 f(-1) = 4(-1) + 5 = -4 + 5 = 1 f(-2) = 4(-2) + 5 = -8 + 5 = -3 x = 0 ∉ domain of f ∴ f(0) does not exist.
Question 4
Maharashtra Board Solution
A function f is defined as follows: f(x) = 5 – x for 0 ≤ x ≤ 4. Find the value of x such that f(x) = 3.
Solution & Step-by-Step Answer:
f(x) = 5 – x f(x) = 3 ∴ 5 – x = 3 ∴ x = 5 – 3 = 2
Question 5
Maharashtra Board Solution
If f(x) = 3x2 – 5x + 7, find f(x – 1).
Solution & Step-by-Step Answer:
f(x) = 3x2 – 5x + 7 ∴ f(x – 1) = 3(x – 1)2 – 5(x – 1) + 7 = 3(x2 – 2x + 1) – 5(x – 1) + 7 = 3x2 – 6x + 3 – 5x + 5 + 7 = 3x2 – 11x + 15
Question 6
Maharashtra Board Solution
If f(x) = 3x + a and f(1) = 7, find a and f(4).
Solution & Step-by-Step Answer:
f(x) = 3x + a, f(1) = 7 ∴ 3(1) + a = 7 ∴ a = 7 – 3 = 4 ∴ f(x) = 3x + 4 ∴ f(4) = 3(4) + 4 = 12 + 4 = 16
Question 7
Maharashtra Board Solution
If f(x) = ax2 + bx + 2 and f(1) = 3, f(4) = 42, find a and b.
Solution & Step-by-Step Answer:
f(x) = ax2 + bx + 2 f(1) = 3 ∴ a(1)2 + b(1) + 2 = 3 ∴ a + b = 1 …….(i) f(4) = 42 ∴ a(4)2 + b(4) + 2 = 42 ∴ 16a + 4b = 40 Dividing by 4, we get 4a + b = 10 ……….(ii) Solving (i) and (ii), we get a = 3, b = -2
Question 8
Maharashtra Board Solution
If f(x) = , verify whether (fof)(x) = x
Solution & Step-by-Step Answer:
(fof)(x) = f(f(x))
Question 9
Maharashtra Board Solution
If f(x) = , g(x) = , then verify that (fog)(x) = x.
Solution & Step-by-Step Answer:
