GSEB Solutions Class 11 Maths Chapter 2 Relations and Functions Miscellaneous Exercise

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) = x² 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) = x², f(3) = 3² = 9
from g(x) = 3x, g(3) = 3 × 3 = 9
∴ f is defined at x = 3, Hence f is a function.

(ii) g(x) = x² 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) = x² = 2² = 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) = x², find \(\frac{f(1.1)-f(1)}{1.1-1}\)
Solution:
f(x) = x² is f(1.1) = 1.1² = 1.21.
f(1) = 1² = 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 = b²}. 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 = a² is true only when a = 0 or a = 1.
∴ It is not a relation.
(ii) a = b² and b = a² is not true for all a, b ∈ N; It is not a relation.
(iii) a – b²,b = c², ∴ a = (c²)² = c⁴ ⇒ a ≠ c²
∴ 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}.