Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Functions Miscellaneous Exercise 2 Questions and Answers.

## Maharashtra State Board 11th Commerce Maths Solutions Chapter 2 Functions Miscellaneous Exercise 2

Question 1.

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:

(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.

A function f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2, x ∈ R. Show that f is one-one and onto. Hence, find f^{-1}.

Solution:

f: R → R defined by f(x) = \(\frac{3 x}{5}\) + 2

First we have to prove that f is one-one function for that we have to prove if

f(x_{1}) = f(x_{2}) then x_{1} = x_{2}

Here f(x) = \(\frac{3 x}{5}\) + 2

Let f(x_{1}) = f(x_{2})

∴ \(\frac{3 x_{1}}{5}+2=\frac{3 x_{2}}{5}+2\)

∴ \(\frac{3 x_{1}}{5}=\frac{3 x_{2}}{5}\)

∴ x_{1} = x_{2}

∴ 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 = \(\frac{3 x}{5}\) + 2

∴ y – 2 = \(\frac{3 x}{5}\)

∴ x = \(\frac{5(y-2)}{3}\) ∈ R

∴ for any y ∈ co-domain R, there exist an element x = \(\frac{5(y-2)}{3}\) ∈ domain R such that f(x) = y

∴ f is an onto function.

∴ f is one-one onto function.

∴ f^{-1} exists.

∴ \(\mathrm{f}^{-1}(y)=\frac{5(y-2)}{3}\)

∴ \(f^{-1}(x)=\frac{5(x-2)}{3}\)

Question 3.

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:

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.

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:

f(x) = 5 – x

f(x) = 3

∴ 5 – x = 3

∴ x = 5 – 3 = 2

Question 5.

If f(x) = 3x^{2} – 5x + 7, find f(x – 1).

Solution:

f(x) = 3x^{2} – 5x + 7

∴ f(x – 1) = 3(x – 1)^{2} – 5(x – 1) + 7

= 3(x^{2} – 2x + 1) – 5(x – 1) + 7

= 3x^{2} – 6x + 3 – 5x + 5 + 7

= 3x^{2} – 11x + 15

Question 6.

If f(x) = 3x + a and f(1) = 7, find a and f(4).

Solution:

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.

If f(x) = ax^{2} + bx + 2 and f(1) = 3, f(4) = 42, find a and b.

Solution:

f(x) = ax^{2} + 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.

If f(x) = \(\frac{2 x-1}{5 x-2}, x \neq \frac{2}{5}\), verify whether (fof)(x) = x

Solution:

(fof)(x) = f(f(x))

Question 9.

If f(x) = \(\frac{x+3}{4 x-5}\), g(x) = \(\frac{3+5 x}{4 x-1}\), then verify that (fog)(x) = x.

Solution: