Odisha State Board CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Ex 3(c) Textbook Exercise Questions and Answers.

## CHSE Odisha Class 11 Math Solutions Chapter 3 Relations And Functions Exercise 3(c)

Question 1.

Give an example of a relation that is not a function.

Solution:

Let the relation ‘f’ be defined from

A = {1, 2, 3} to B = {1, 2} as

f = {(1, 1), (1, 2), (2, 1), (3, 2)}

f is not a function because f (1) = 1,

f(1) = 2 but 1 ± 2 i.e. f is one many.

Question 2.

If X and Y are sets containing m and n elements respectively then what is the total number of functions from X to Y?

Solution:

If |X| = m and |Y| = n then the number of functions from X to Y = |Y|^{|X|} = n^{m}

Question 3.

Find the domain of the following functions:

(i)\( \sqrt{9-x^2}\)

Solution:

y = f(x) = \(\sqrt{9-x^2}\)

Clearly, f(x) is well-defined for 9 – x^{2} ≥ 0 ⇒ x^{2} ≤ 3^{2} ⇒ – 3 ≤ x ≤ 3

∴ dom f = [ – 3, 3]

(ii) \(\frac{x}{1+x}\)

Solution:

y = f(x) = \(\frac{x}{1+x}\)

f(x) is well defined for all x ∈ R

∴ dom f = R

(iii) 1 – |x|

Solution:

y = f(x) = 1 – |x|

f(x) is well defined for all x ∈ R

∴ dom f = R

(iv) \(\frac{1}{x^2-1}\)

Solution:

y = f(x) = \(\frac{1}{x^2-1}\)

f(x) is well defined for x^{2} – 1 ≠ 0

⇒ x^{2} ≠ 1

⇒ x^{2} ≠ ± 1

∴ dom f = R – { -1, 1}

(v) \(\frac{\sin x}{1+\tan x}\)

Solution:

y – f(x) = \(\frac{\sin x}{1+\tan x}\)

f(x) is well defined for 1 + tan x

≠ 0

⇒ tan x ≠ – 1

⇒ x ≠ n π + \(\frac{3 \pi}{4}\)

∴ dom f = R – {n π + \(\frac{3 \pi}{4}\)}

(vi) \(\frac{x}{|x|}\)

Solution:

y = f(x) = \(\frac{x}{|x|}\)

f(x) is well defined for |x| ≠ 0

⇒ x ≠ 0

∴ dom f = R – {0}

(vii) \(\frac{1}{x+|x|}\)

Solution:

y = f(x) = \(\frac{1}{x+|x|}\)

f(x) is well-defined for

x + |x| ≠ 0

⇒ |x| ≠ – x

⇒ x is not negative

⇒ dom f = [0, – ∞)

(viii) \(\sqrt{\log \left(\frac{12}{x^2-x}\right)}\)

Solution:

y = f(x) = \(\sqrt{\log \left(\frac{12}{x^2-x}\right)}\)

f(x) is well defined for \(\frac{12}{x^2-x}\) ≥ 1

⇒ 12 ≥ x^{2} – x

⇒ x^{2} – x – 12 ≤ 0

⇒ (x – 4) (x + 3) ≤ 0

Let us draw the signing scheme

⇒ x ∈ [ – 3, 4]

∴ dom f =[-3, 4]

(ix) [x] – x

Solution:

y = f(x) = [x] – x

f(x) is well defined for all x ∈ R

∴ dom f = R

(x) \(\frac{1}{\sqrt{1-x^2}}\)

Solution:

y = f(x) = \(\frac{1}{\sqrt{1-x^2}}\)

f(x) is well defined for 1 – x^{2} > 0

⇒ x^{2} < 1 ⇒ – 1 < x < 1

∴ dom f = (-1, 1)

(xi) log (sin x)

Solution:

y = f(x) = log (sin x)

fix) is well-defined for sin x > 0

⇒ x ∈ (2nπ , (2n + l)π), n ∈ Z

Question 4.

Find the range of the following functions:

(i) \(\frac{x^2-1}{x^2+1}\)

Solution:

Let y = f(x) = \(\frac{x^2-1}{x^2+1}\)

⇒ x^{2} – y + y = x^{2} – 1

⇒ x^{2}y – x^{2} = – (1 + y)

⇒ x^{2} (y- 1)=-(1 + y)

⇒ x^{2} = \(\frac{\ – (1+y)}{y-1}=\frac{1+y}{1-y}\)

⇒ x = ± \(\sqrt{\frac{1+y}{1-y}}\)

Now x is well-defined for

\(\frac{1+y}{1-y}\) ≥ 0

∴ rng f = [-1,1)

(ii) \(\sqrt{x-1}\)

Solution:

y = f(x) = \(\sqrt{x-1}\)

⇒ y^{2} = x- 1

⇒ x = y^{2} + 1

x is well difierid for all y ∈ R but

y ≥ 0

∴ rng f = [0, ∞]

(iii) [x] – x

Solution:

y = f(x) = [x], – x

We have [x] ≤ x < [ x] + 1

⇒ 0 ≤ x – [x] < 1

⇒ 0 ≥ [x] – x > – 1

⇒ -1 < [x] – x ≤ 0

⇒ -1 < y ≤ 0

⇒ rng f = ( – 1 , 0]

(iv) \(\frac{x}{1-x}\)

Solution:

y = f(x) = \(\frac{x}{1-x}\)

⇒ y – xy = x

⇒ xy + x = y

⇒ x (y + 1) = y

⇒ x = \(\frac{y}{y+1}\)

x is well defined for y ≠ – 1

∴ Rng f = R – {- 1}

(v) \(\frac{x}{1+x^2}\)

Solution:

y = f(x) = \(\frac{x}{1+x^2}\)

⇒ y + x^{2}y = x

⇒ x^{2}y – x + y = 0

⇒ x = \(\frac{1 \pm \sqrt{1-4 y^2}}{2 y}\)

x is well defined for 1 – 4y^{2} ≥ 0 and y ≠ 0

⇒ y^{2} ≤ \(\frac{1}{4}\) and y ≠ 0

⇒ – \(\frac{1}{2}\) ≤ y ≤ \(\frac{1}{2}\) and y ≠ 0

but for x = 0 we have y = 0

∴ rng f = \(\left[-\frac{1}{2}, \frac{1}{2}\right]\)

(vi) \(\frac{1}{2-\cos 3 x}\)

Solution:

y = f(x) = \(\frac{1}{2-\cos 3 x}\)

we have – 1 ≤ cos 3x ≤ 1

⇒ 1 ≥ – cos 3x ≥ -1

⇒ – 1 ≤- cos 3x ≤ 1

⇒ 2 – 1 < 2 – cos 3x ≤ 3

⇒ 1 ≤ 2 – cos 3x ≤ 3

⇒ 1 ≥ \(\frac{1}{2-\cos 3 x} \geq \frac{1}{3}\)

⇒ \(\frac{1}{3}\) ≤ y ≤ 1

∴ rng f =[\(\frac{1}{3}\), 1]

(vii) log _{10} (1 – x)

Solution:

y = f(x) = log_{10} (1 – x)

Clearly y ∈ R

∴ rng f = R

(viii) \(\sqrt{1+x^2}\)

Solution:

y = f(x) = \(\sqrt{1+x^2}\)

⇒ y^{2} = 1 + x^{2}

⇒ x^{2} = y^{2} – 1

⇒ x = ± \(\sqrt{y^2-1}\)

x is well defined for y^{2} – 1 ≥ 0

⇒ y^{2} ≥ 1

⇒ – 1 ≥ y ≥ 1

But y > 0

∴ rng f = [1, ∞)

Question 5.

Find the domain and range of the following functions:

(i) \(\frac{x^2}{1+x^2}\)

Solution:

y = f(x) = \(\frac{x^2}{1+x^2}\)

Domain:

f(x) is well defined for all x ∈ R

∴ dom f = R

Range (Method – 1):

y = \(\frac{x^2}{1+x^2}\)

⇒ y + yx^{2} = x^{2}

⇒ x^{2} – yx^{2} = y

⇒ x^{2} (1 – y) = y

⇒ x^{2} = \(\frac{y}{1-y}\)

⇒ x = ± \(\sqrt{\frac{y}{1-y}}\)

x is well-defined for \(\frac{y}{1-y}\) ≥ 0

⇒ y 6 [0. 1)

∴ Rng f = [0,1)

Method – 2:

0 ≤ x^{2} < x^{2} + 1

⇒ 0 ≤ \(\frac{x^2}{x^2+1}\) < 1

⇒ 0 ≤ y ≤ 1

∴ rng f = [0, 1)

(ii) \(\sqrt{2 x-3}\)

Solution:

y = f(x) = \(\sqrt{2 x-3}\)

Domain:

f(x) is well defined for 2x- 3 ≥ 0

⇒ x ≥ 3/2

∴ dom f =[\(\frac{3}{2}\), ∞]

Range:

y = \(\sqrt{2 x-3}\) ⇒ y^{2} = 2x – 3

⇒ 2x = y^{2} + 3

⇒ x = \(\frac{y^2+3}{2}\)

x is well defined for all y ∈ R

But y is not negative.

∴ Rng f = [0, ∞)

(iii) log_{e} |x – 2|

Solution:

y = f(x) = log |x – 2|

Domain:

f(x) is well defined for |x – 2| >0

⇒ x ≠ 2

∴ dom f = R – {2}

Range:

y = log |x – 2| ∈ R ⇒ rng f = R

Question 6.

Give an example of a step function on [- 1, 3] = {x ∈ R| – 1 ≤ x ≤ 3}.

Solution:

f(x) = \(\left\{\begin{array}{c}

-1,-1 \leq x<0 \\

0,0 \leq x<1 \\

1,1 \leq x<2 \\

2,2 \leq x<3 \\

3,3 \leq x<4

\end{array}\right\}\)

Question 7.

Let X ={ a, b, c}, Y = {1, 2, 3, 4}

(a) Find out which of the following relations are functions and which are not and why?

Solution:

X = [a, b, c}, Y = {1, 2, 3, 4}

(i) {(a, 1), (a, 2), (b, 3), (b, 4)}

Solution:

{(a, 1), (a, 2), (b, 3), (b, 4)} is not a function as ‘a’ has two images in Y.

(ii) {(a, 2), (b, 3), (c, 4)}

Solution:

{(a, 2), (b, 3), (c, 4)} is a function from X to Y as the elements of X has unique images in Y.

(iii) {(a, 3), (b, 1), (a, 4), (c, 2)}

Solution:

{(a, 3), (b, 1), (a, 4), (c, 2)} is not a function as ‘a’ and ‘b’ have two images also domain of the function ≠ X.

(iv) {(a, 1), (b, 1), (c, 1)}

Solution:

{(a, 1), (b, 1), (c, 1)} is a function from X to Y.

(v) {(a, 2), (b. 1), (c, 1)}

Solution:

{(a, 2), (b, 1), (c, 1)} is a function from X to Y.

(vi) {(a, a), (b, b), (c, c)}

Solution:

{(a, a), (b, b), (c, c)} is a function from X to X.

(b) Find the domain and range of those relations in a which are functions.

Solution:

(i) The domain of each function is

X = [a, b, c}

Range of functions

(ii) is {2, 3, 4}.

In (iv) is {1} and in (v) is {1, 2}. The range of function in (vi) is X.

(c) Identify the constant function if any.

Solution:

The function in (iv)

i.e. {(a, 1), ib, 1), (c, 1)} is a constant function.

(d) Identify the identity function if any.

Solution:

The function in (vi)

i.e.{(a, a), (b, b), (c, c)} is an identity function.

Question 8.

Find \(f(\sqrt{2})\) and \(\boldsymbol{f}(-\sqrt{3})\) for the function

\(f(x)=\left\{\begin{array}{l}

x^2, \text { if } x<0 \\

x, \text { if } 0 \leq x \leq 1 \\

\frac{1}{x}, \text { if } x>1

\end{array}\right.\)

Solution:

\(f(\sqrt{2})\) = \(\frac{1}{\sqrt{2}}\)

\(\boldsymbol{f}(-\sqrt{3})\) = (√3)^{2} = 3

Question 9.

Find x for which the value of f(x) = x^{2 } – 4x + 3 is

Solution:

f(x) = x^{2} – 4x + 3

(i) 0

Solution:

f(x) = 0

⇒ x^{2} – 4x + 3 = 0

⇒ (x – 3) (x – 1) = 0

⇒ x =1, 3

(ii) – 1

Solution:

f(x) = – 1

⇒ x^{2} – 4x + 3 = -1

⇒ x^{2} – 4x + 4 = 0

⇒ (x – 2)^{2} = 0

⇒ x = 2

Question 10.

Find the values of x for which the following functions are not defined:

(i) \(\frac{x^2-4}{x-2}\)

Solution:

y = f(x) = \( \frac{x^2-4}{x-2}\)

Clearly f(x) is not defined for x = 2

(ii) \(\frac{\sin x}{x}\)

Solution:

y = f(x) = \(\frac{\sin x}{x}\)

f(x) is not defined for x = 0

(iii) \(\frac{\log \cos x}{\sec x}\)

Solution:

y = f(x) = \(\frac{\log \cos x}{\sec x}\)

f(x) is not defined for cos x ≤ 0

⇒ for x ∈ x \(\left[\frac{\pi}{2}, \frac{3 \pi}{2}\right]\) or \(\left[\frac{5 \pi}{2}, \frac{7 \pi}{2}\right]\)

In general x ∈ \(\left[(4 n+1) \frac{\pi}{2},(4 n+3) \frac{\pi}{2}\right]\), n ∈ Z

Question 11.

Let f(x) = \(\sqrt{1+x}\), g(x) = \(\sqrt{1-x}\)

Find & Also, find the domain in each case.

Solution:

f(x) = \(\sqrt{1+x}\), g(x) = \(\sqrt{1-x}\)

f(x) is well defined for x + 1 ≥ 0

⇒ x ≥ – 1

∴ dom f = [- 1 , ∞) = D_{1}

g(x) is well defined for 1 – x ≥ 0

⇒ x ≤ 1

⇒ x ∈ (-∞, 1]

∴ dom g (-∞, 1] = D_{2}

∴ D_{1} ∩ D_{2} = [-1, 1]

(i) f + g

Solution:

f + g: [-1, 1] → R

defined as (f + g) (x)

= f(x) + g(x) = \(\sqrt{1-x}\) + \(\sqrt{1-x}\)

(ii) f – g

Solution:

f – g :[-1, 1] → R defined as

(f – g) (x) = \(\sqrt{1-x}\) – \(\sqrt{1-x}\)

(iii) fg

Solution:

fg : [ – 1, 1] → R defined as

(fg) (x) > f(x) g(x)

\(\sqrt{1-x}\)\(\sqrt{1-x}\) = \(\sqrt{1-x^2}\)

(iv) f/g

Solution:

f/g : [- 1,1) → R defined as

\(\left(\frac{f}{g}\right)\)(x) = \(\frac{f(x)}{g(x)}=\frac{\sqrt{1+x}}{\sqrt{1-x}}\)

for (i) to (iii) domain = [- 1, 1]

but for (iv) domain = [- 1, 1)

Question 12.

If(x) = log_{e} \(\left(\frac{1-x}{1+x}\right)\), then prove that f(x) + f(y) = \(\left(\frac{x+y}{1+x y}\right)\)

Solution:

f(x) = log\(\left(\frac{1-x}{1+x}\right)\)

f(x) + f(y) = log\(\left(\frac{1-x}{1+x}\right)\) + \(\left(\frac{1-y}{1+y}\right)\)

Question 13.

Let f = {(-1, 4), (2, 7), (-2, 11}. (0,1), (1, 2) be a quadratic polynomial from Z to Z, find f(x).

Solution:

Let f(x) = ax^{2} + by^{2} + C is the quadratic polynomial

f(-1) = 4 ⇒ a – b + c = 4 …..(1)

f(0) = 1⇒ c = 1 ….(2)

f(1) = 2 ⇒ a + + c = 2 ….(3)

∴ a – b + 1 = 4 and a + b + 1 = 2

⇒ a – b = 3 and a + b = 1

⇒ a = 2, b = -1

f(x) = 2x^{2} – x + 1

f(2) = 8 – 2 + 1 = 7

f(- 2) = 8 + 2 + 1 = 1 1

Thus the required function is f(x) = 2x^{2} – x + 1

Question 14.

Sketch the graphs of the following functions.

(i) f(x) = x^{3}

(ii) f(x) = 1 + \(\frac{1}{x^2}\)

(iii) f(x) = (x-1)^{2}

Solution:

(i) f(x) = x^{3}

x | 0 | 1 | 2 | – 1 | -2 | 3 |

y | 1 | 1 | 8 | -1 | -8 | 27 |

(ii) f(x) = 1 + \(\frac{1}{x^2}\)

x | 1 | -1 | 2 | – 2 | 3 | -3 |

y | 2 | 2 | 5/4 | 5/4 | 10/9 | 10/9 |

(iii) f(x) = (x-1)^{2}

x | -1 | 0 | 1 | 2 | 3 | 4 |

f(x) | 4 | 1 | 0 | 1 | 4 | 9 |