Balbharati Maharashtra State Board 11th Commerce Maths Solution Book Pdf Chapter 2 Functions Ex 2.1 Questions and Answers.

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

Question 1.

Check if the following relations are functions.

Solution:

(a) Yes

Reason: Every element of set A has been assigned a unique element in set B.

(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.

Question 2.

Which sets of ordered pairs represent functions from A = {1, 2, 3, 4} to B = {-1, 0, 1, 2, 3}? Justify.

(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)}

(ii) {(1, 2), (2, -1), (3, 1), (4, 3)}

(iii) {(1, 3), (4, 1), (2, 2)}

(iv) {(1, 1), (2, 1), (3, 1), (4, 1)}

Solution:

(i) {(1, 0), (3, 3), (2, -1), (4, 1), (2, 2)} does not represent a function.

Reason: (2, -1) and (2, 2) show that element 2 ∈ A has been assigned two images -1 and 2 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.

Question 3.

If f(m) = m^{2} – 3m + 1, find

(i) f(0)

(ii) f(-3)

(iii) f(\(\frac{1}{2}\))

(iv) f(x + 1)

(v) f(-x)

Solution:

f(m) = m^{2} – 3m + 1

(i) f(0) = 0^{2} – 3(0) + 1 = 1

(ii) f(-3) = (-3)^{2} – 3(-3) + 1

= 9 + 9 + 1

= 19

(iii) \(f\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2}-3\left(\frac{1}{2}\right)+1\)

= \(\frac{1}{4}-\frac{3}{2}+1\)

= \(\frac{1-6+4}{4}\)

= \(-\frac{1}{4}\)

(iv) f(x + 1) = (x + 1)^{2} – 3(x + 1) + 1

= x^{2} + 2x + 1 – 3x – 3 + 1

= x^{2} – x – 1

(v) f(-x) = (-x)^{2} – 3(-x) + 1 = x^{2} + 3x + 1

Question 4.

Find x, if g(x) = 0 where

(i) g(x) = \(\frac{5 x-6}{7}\)

(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)

(iii) g(x) = 6x^{2} + x – 2

Solution:

(i) g(x) = \(\frac{5 x-6}{7}\)

g(x) = 0

∴ \(\frac{5 x-6}{7}\) = 0

∴ 5x – 6 = 0

∴ x = \(\frac{6}{5}\)

(ii) g(x) = \(\frac{18-2 x^{2}}{7}\)

g(x) = 0

∴ \(\frac{18-2 x^{2}}{7}\) = 0

∴ 18 – 2x^{2} = 0

∴ x^{2} = 9

∴ x = ±3

(iii) g(x) = 6x^{2} + x – 2

g(x) = 0

∴ 6x^{2} + x – 2 = 0

∴ 6x^{2} + 4x – 3x – 2 = 0

∴ 2x(3x + 2) – 1(3x + 2) = 0

∴ (2x – 1)(3x + 2) = 0

∴ 2x – 1 = 0 or 3x + 2 = 0

∴ x = \(\frac{1}{2}\) or x = \(\frac{-2}{3}\)

Question 5.

Find x, if f(x) = g(x) where f(x) = x^{4} + 2x^{2}, g(x) = 11x^{2}.

Solution:

f(x) = x^{4} + 2x^{2}, g(x) = 11x^{2}

f(x) = g(x)

∴ x^{4} + 2x^{2} = 11x^{2}

∴ x^{4} – 9x^{2} = 0

∴ x^{2}(x^{2} – 9) = 0

∴ x^{2} = 0 or x^{2} – 9 = 0

∴ x = 0 or x^{2} = 9

∴ x = 0 or x = ±3

Question 6.

If f(x) = \(\begin{cases}x^{2}+3, & x \leq 2 \\ 5 x+7, & x>2\end{cases}\), then find

(i) f(3)

(ii) f(2)

(iii) f(0)

Solution:

f(x) = x^{2} + 3, x ≤ 2

= 5x + 7, x > 2

(i) f(3) = 5(3) + 7 = 15 + 7 = 22

(ii) f(2) = 2^{2} + 3 = 4 + 3 = 7

(iii) f(0) = 0^{2} + 3 = 3

Question 7.

If f(x) = \(\left\{\begin{array}{cl}

4 x-2, & x \leq-3 \\

5, & -3<x<3 \\

x^{2}, & x \geq 3

\end{array}\right.\), then fmd

(i) f(-4)

(ii) f(-3)

(iii) f(1)

(iv) f(5)

Solution:

f(x) = 4x – 2, x ≤ -3

= 5, -3 < x < 3

= x^{2}, x ≥ 3

(i) f(-4) = 4(-4) – 2 = -16 – 2 = -18

(ii) f(-3) = 4(-3) – 2 = -12 – 2 = -14

(iii) f(1) = 5

(iv) f(5) = 5^{2} = 25

Question 8.

If f(x) = 3x + 5, g(x) = 6x – 1, then find

(i) (f + g)(x)

(ii) (f – g)(2)

(iii) (fg)(3)

(iv) \(\left(\frac{\mathbf{f}}{\mathbf{g}}\right)(x)\) and its domain

Solution:

f(x) = 3x + 5, g(x) = 6x – 1

(i) (f + g)(x) = f(x) + g(x)

= 3x + 5 + 6x – 1

= 9x + 4

(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) \(\left(\frac{\mathrm{f}}{\mathrm{g}}\right) x=\frac{\mathrm{f}(x)}{\mathrm{g}(x)}=\frac{3 x+5}{6 x-1}, x \neq \frac{1}{6}\)

Domain = R – {\(\frac{1}{6}\)}

Question 9.

If f(x) = 2x^{2} + 3, g(x) = 5x – 2, then find

(i) fog

(ii) gof

(iii) fof

(iv) gog

Solution:

f(x) = 2x^{2} + 3, g(x) = 5x – 2

(i) (fog)(x) = f(g(x))

= f(5x – 2)

= 2(5x – 2)^{2} + 3

= 2(25x^{2} – 20x + 4) + 3

= 50x^{2} – 40x + 8 + 3

= 50x^{2} – 40x + 11

(ii) (gof)(x) = g(f(x))

= g(2x^{2} + 3)

= 5(2x^{2} + 3) – 2

= 10x^{2} + 15 – 2

= 10x^{2} + 13

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

= f(2x^{2} + 3)

= 2(2x^{2} + 3)^{2} + 3

= 2(4x^{4} + 12x^{2} + 9) + 3

= 8x^{4} + 24x^{2} + 18 + 3

= 8x^{4} + 24x^{2} + 21

(iv) (gog)(x) = g(g(x))

= g(5x – 2)

= 5(5x – 2) – 2

= 25x – 10 – 2

= 25x – 12