Gujarat Board GSEB Textbook Solutions Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Textbook Questions and Answers.

## Gujarat Board Textbook Solutions Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise

Question 1.

The function f is defined by f(x) = \(\left\{\begin{array}{l}

x^{2}, 0 \leq x \leq 3 \\

3 x, 3 \leq x \leq 10

\end{array}\right.\)

The relation g is defined by g(x) = \(\left\{\begin{array}{l}

x^{2}, 0 \leq x \leq 2 \\

3 x, 2 \leq x \leq 10

\end{array}\right.\)

Show that f is a function and g is not a function.

Solution:

(i) f(x) = x^{2} is well defined in the interval in the interval 0 ≤ x ≤ 3

Also, f(x) = 3x is well defined in the interval 3 ≤ x ≤ 10

At x = 3 from f(x) = x^{2}, f(3) = 3^{2} = 9

from g(x) = 3x, g(3) = 3 × 3 = 9

∴ f is defined at x = 3, Hence f is a function.

(ii) g(x) = x^{2} is well defined in the interval 0 ≤ x ≤ 2.

g(x) = 3x is also well defined in the interval 2 ≤ x ≤ 10.

But at x = 2, g(x) = x^{2} = 2^{2} = 4.

Also, g(x) = 3x = 3 × 2 = 6.

At x = 2, relation on g has two values.

∴ Relation g is not a function.

Question 2.

f(x) = x^{2}, find \(\frac{f(1.1)-f(1)}{1.1-1}\)

Solution:

f(x) = x^{2} is f(1.1) = 1.1^{2} = 1.21.

f(1) = 1^{2} = 1.

∴ \(\frac{f(1.1)-f(1)}{1.1-1}\) = \(\frac{1.21-1}{1.1-1}\) = \(\frac{0.21}{0.1}\) = 2.1.

Question 3.

Find the domain of the function f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}\).

Solution:

f(x) = \(\frac{x^{2}+2 x+1}{x^{2}-8 x+12}=\frac{(x+1)^{2}}{(x-2)(x-6)}\)

The function f is not defined at x = 2,6.

∴ Domain of f = {x : x ∈ R and x ≠ 2, x ≠ 6} = {x : x ∈ R – (2, 6)}.

Question 4.

Find the domain and range of real function f defined by f(x) = \(\sqrt{x-1}\).

Solution:

1. f(x) = \(\sqrt{x-1}\), f is not defined for x – 1 < 0 or x < 1

Domain (f) = {x : x ≥ 1} = [1, ∞).

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

∴ y is well defined for all values of x ≥ 1.

∴ Range : {y : y ∈ R, y ≥ 0} = [0, ∞).

Question 5.

Find the domain and range of the real function f defined by f(x) = |x – 1|.

Solution:

1. f(x) = |x – 1| is defined for all real values of x.

∴ Domain of f = {x : x ∈ R} = R.

2. f(x) = |x – 1| can acquire only non-negative values.

∴ Range = {y : y ≥ 0} = R^{+}.

Question 6.

Let f = {(x, \(\frac{x^{2}}{1+x^{2}}\)}} be a function from R into R. Determine the range of f?

Solution:

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

f(x) is positive for all values of x.

When x = 0, y = 1. Also, denominator > numerator.

∴ Range of f = {y : y ∈ R and y ∈ (0, 1)}

Question 7.

Let f, g : R → R be defined respectively by f(x) = x + 1, g(x) = 2x – 3. Find f + g, f – g and \(\frac{f}{g}\).

Solution:

f, g are defined for all x ∈ R

(i) (f + g)(x) = f(x) + g(x) = x + 1 + 2x – 3 = 3x – 2.

(ii) (f – g)(x) = f(x) – g(x) = (x + 1) – (2x – 3) = – x + 4.

(iii) (\(\frac{f}{g}\)) (x) = \(\frac{f(x)}{g(x)}\) = \(\frac{x+1}{2x-3}\), x ∈ R and x ≠ \(\frac{3}{2}\).

Question 8.

Let f= {(1, 1), (2, 3), (0, -1), (-1, -3)} be a function from Z into Z defined by f(x) = ax + b for some integers a and b. Determine a and b?

Solution:

We have:

f(x) = ax + b.

For x = 1, f(x) = 1.

∴ 1 = a × 1 + b or a + b = 1 …………………… (1)

For x = 2, f(x) = 3.

∴ 3 = a × 2 + b or 2a + b = 3 ………………….. (2)

Subtracting eq. (1) from eq. (2), we get a = 2 ⇒ 6 = -1.

∴ f(x) = 2x – 1

For x = 0, f(0) = – 1.

For x = – 1, f(x) = – 2 – 1 = – 3.

∴ (0, – 1), (- 1, – 3) also ∈ f.

Hence, f(x) = 2x – 1.

Question 9.

Let R be a relation from N to N defined by R = {(a, b) : a, b ∈ N and a = b^{2}}. Are the following true?

(i) (a, a) ∈ R for all a ∈ N

(ii) (a, b) ∈ R implies (b, a) ∈ R

(iii) (a, b) ∈ R, (b, c) ∈ R implies (a, c) ∈ R

Solution:

(i) a = a^{2} is true only when a = 0 or a = 1.

∴ It is not a relation.

(ii) a = b^{2} and b = a^{2} is not true for all a, b ∈ N; It is not a relation.

(iii) a – b^{2},b = c^{2}, ∴ a = (c^{2})^{2} = c^{4} ⇒ a ≠ c^{2}

∴ It is not a relation.

Question 10.

Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = (1, 5), (2, 9), (3, 2), (4, 5), (2, 11)}. Are the following true?

(i) f is a relation from A to B

(ii) f is a function from A to B Justify your answer in each case.

Solution:

(i) f is a subset of A × B.

∴ f is a relation from A to B.

(ii) The element 2 ∈ A, has two images 9 and 11.

∴ f is not a function from A to B.

Question 11.

Let fhe the subset of Z × Z defined by f = {(ab, a + b) : a, b ∈ Z}. Is fa function from Z to Z? Justify your answer.

Solution:

Let a = 0, b = 1,

∴ ab = 0, a + b = 1, (0, 1) ∈ f.

a = 0, b = 1,

∴ ab = 0, a + b = 2, (0, 2) ∈ f.

∴ For the same element, there are different images.

∴ f is not a function.

Question 12.

Let A = {9, 10, 11, 12, 13} and let f : A → N be defined by f(x) = the highest prime factor of n. Find the range of f?

Solution:

For n = 9; 3 is the highest prime factor of it.

n = 10; 5 is the highest prime factor of it.

n = 11; 11 is the highest prime factor of it.

n = 12; 3 is the highest prime factor of it.

n = 13; 13 is the highest prime factor of it.

∴ Range of f = (3, 5, 11, 13}.