Inter 1st Year Maths Relations and Functions Solutions Exercise 2d

Practicing the AP Board Solutions Class 11 Maths and Chapter 2 Inter 1st Year Maths Relations and Functions Solutions Exercise 2d Pdf Download will help students to clear their doubts quickly.

Intermediate 1st Year Maths Relations and Functions Solutions Exercise 2d

Relations and Functions Exercise 2d Solutions

Relations and Functions Class 11 Exercise 2d Solutions – Relations and Functions 2d Exercise Solutions

I.

Question 1.
If f(x) = x², find \(\frac{\mathrm{f}(1.1)-\mathrm{f}(1)}{(1.1-1)}\)
Solution:
Given f(x) = x²
Hence,

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

It’s clearly seen that function f is defined for all real numbers except at x = 6 and x = 2, as the denominator becomes zero otherwise.
Therefore, the domain of f is R – (2, 6).

Question 3.
Find the domain and the range of the real function f defined by f(x) = \(\sqrt{(x-1)}\).
Solution:
Given real function f(x) = \(\sqrt{(x-1)}\)
Clearly, \(\sqrt{(x-1)}\) is defined for (x – 1) ≥ 0
So, the function f(x) = \(\sqrt{(x-1)}\) is defined for ≥ 1
Thus, the domain of f is the set of all real numbers greater than or equal to 1.
Domain of f = (1, ∞)
Now, As x ⇒ 1 ≥ (x – 1) ≥ 0
⇒ \(\sqrt{(x-1)}\) ≥ 0
Thus, the range of f is the set of all real numbers greater than or equal to 0.
Range of f = (0, ∞)

Question 4.
Find the domain and the range of the real function f defined by f(x) = |x – 1|.
Solution:
Given a real function f(x) = |x – 1|
Clearly, the function |x – 1| is defined for all real numbers.
Hence, Domain at f = R
Also, for x ∈ R, |x – 1| assumes all real numbers.
Therefore, the range of f is the set of all non-negative real numbers.

Question 5.
Let f = \(\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\) be a function from R into R. Determine the range of f.
Solution:
Given function
f = \(\left\{\left(x, \frac{x^2}{1+x^2}\right): x \in R\right\}\)
Substituting values and determining the images, we have

The range of f is the set of all second elements.
It can be observed that these elements are greater than or equal to 0 but less than 1.
[As the denominator is greater than the numerator.]
Or, we know that for x ∈ R, x² ≥ 0
Then, x² + 1 ≥ x²
1 ≥ x² (x² + 1)
Therefore, the range of f = (0, 1)

Question 6.
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.
Justify your answer in each case.
Solution:
Given relation R = {(a, b): a, b ∈ N and a = b²)}
(i) It can be seen that 2 ∈ N; however, 2 ≠ 2² = 4
Thus, the statement “(a, a) ∈ R, for all a ∈ N” is not true.

(ii) It’s clearly seen that (9, 3) ∈ N
because 9, 3 ∈ N and 9 = 3².
Now, 3 ≠ 9² = 81,
therefore (3, 9) ∈ N
Thus, the statement “(a, b) ∈ R implies (b, a) ∈ R is not true.

(iii) It’s dearly seen that (16, 4) ∈ R
(4, 2) ∈ R because 16, 4, 2 ∈ N and 16 = 4² and 4 = 2²
Now 16 ≠ 2² = 4; therefore (16, 2) ∉ N
Thus, the statement “(a, b) ∈ R (b, c) ∈ R implies (a, c) ∈ R” is not true.

Question 7.
Let A = {1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16}, and f = {(1, 5), (2, 9), (3, 1), (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:
Given A = {1, 2, 3, 4} and {1, 5, 9, 11, 15, 16}
So, A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)}
Also given that, F = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
(i) A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.
It’s clearly seen that f is a subset of A × B.

(ii) As the same first element, i.e., 2, corresponds to two different images (9 and 11), relation f is not a function.

Question 8.
Let f be the subset of Z × Z defined by f = {(ab, a + b): a, b ∈ Z}. Is f a function from Z to Z? Justify your answer.
Solution:
Given relation f is defined as f = {(ab, a + b): a, b ∈ z)}
We know that a relation from a set B is said to be a function if every element of set A has a unique image in set B.
As 2, 6, -2, -6 ∈ z (2 × 6; 2 + 6), (-2 × -6, -2 + (-6) ∈ f,
i.e., (12, 8), (12, -8) ∈ f
It’s clearly seen that the same first element 12 corresponds to two different images (8 and -8).
Therefore, the relation f is not a function.

Question 9.
Let A = {9, 10, 11, 12, 13} and let f: A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
Solution:
Given, A = (9, 10, 11, 12, 13)
Now, f: A → N is defined as
f(n) = The highest prime factor of n.
So, the Prime factor of 9 = 3
Prime factors of 10 = 2, 5
prime factor of 11 = 11
prime factor of 12, 2, 3
prime factor of 13 = 13
Thus, it can be expressed as
f(9) = The highest prime factor of 9 = 3
f(10) = The highest prime factor of 10 = 5
f(11) = The highest prime factor of 11 = 11
f(12) = The highest prime factor of 12 = 3
f(13) = The highest prime factor of 13 = 13
The range of the set of all f(n) where n ∈ A
Therefore, the Range of f = (3, 5, 11, 13)

II.

Question 1.
The relation 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:
The given relation f is defined as
g(x) = \(\left\{\begin{array}{l}
x^2, 0 \leq x \leq 2 \\
3 x, 2 \leq x \leq 10
\end{array}\right.\)
It is seen that for 0 ≤ x ≤ 3
f(x) = x² and for 3 ≤ x ≤ 10,
f(x) = 3x
Also, at x = 3
f(x) = 3² = 9 or f(x) = 3 × 3 = 9
i.e., at x = 3, f(x) = 9 (single image)
Hence, for 0 ≤ x ≤ 10, the images of f(x) are unique.
Therefore, the given relation is a function.
Now, in the given relation, g is defined as
g(x) = \(\left\{\begin{array}{l}
x^2, 0 \leq x \leq 2 \\
3 x, 2 \leq x \leq 10
\end{array}\right.\)
It is seen that, for x = 2
g(x) = 2² = 4 and g(x) = 3 × 2 = 6
Thus, element 2 of the domain of the relation g corresponds to two different images, i.e., 4 and 6.
Therefore, this relation is not a function.

Question 2.
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:
Given the functions f, g: R → R, defined as
f(x) = x + 1, g(x) = 2x – 3
Now, (f + g) (x) = f(x)
= (x + 1) + (2x – 3)
= 3x – 2
Thus, (f + g)(x) = 3x – 2
(f-g) (x) = f(x) – g(x)
= (x + 1) – (2x – 3)
= x + 1 – 2x + 3
= -x + 4
Thus, (f-g) (x) = -x + 4
\(\frac{f}{g}(x)=\frac{f(x)}{g(x)}\) = g(x) ≠ 0, x ∈ R.
\(\frac{f}{g}(x)=x+\frac{1}{2} x-3\), 2x – 3 ≠ 0
Thus \(\frac{f}{g}(x)=x+\frac{1}{2} x-3 x \neq \frac{3}{2}\)

Question 3.
Let f = {(1, 1), (2, 3), (0, -1), (-1, -3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.
Solution:
Given f = {(1, 1), (2, 3), (0, -1), (-1, -3)}
And the function defined as f(x) = ax + b
For (1, 1) ∈ f
We have, f(1) = 1
So, a × 1 + b = 1
⇒ a + b = 1 …………(1)
And for (0, -1) ∈ f
We have f(0) = -1
a × 0 + b = -1
⇒ b = -1
On substituting b = -1 in (i), we get
a + (-1) = 1
⇒ a = 1 + 1
⇒ a = 2
Therefore, the values of a and b are 2 and -1, respectively.

III.

Question 1.
If f = {(4, 5), (5, 6), (6, -4)} and g = {(4, -4), (6, 5), (8, 5)}, then find
(i) f + g
(ii) f – g
(iii) 2f + 4g
(iv) f + 4
(v) fg
(vi) \(\frac {f}{g}\)
Solution:
Given f = {(4, 5), (5, 6), (6, -4)} and g = {(4, -4), (6, 5), (8, 5)}
(i) f + g = {(4, 5 – 4), (6, -4 + 5)} = {(4, 1), (6, 1)}
(ii) f – g = {(4, 5 + 4), (6, -4 – 5)} = {(4, 9), (6, -9)}
(iii) 2f + 4g = {(4, 10 – 16), (6, -8 + 20)} = {(4, -6), (6, 12)}
(iv) f + 4 = {(4, 5 + 4), (5, 6 + 4), (6, -4 + 4)} = {(4, 9), (5, 10), (6, 0)}
(v) fg = {(4, (5) (-4), (6, (-4) (-5))} = {(4, -20), (6, 20)}
(vi) \(\frac{f}{g}=\left\{\left(4, \frac{-5}{4}\right),\left(6, \frac{-4}{5}\right)\right\}\)

Question 2.
If f and g are real valued functions defined by f(x) = 2x – 1 and g(x) = x², then find
(i) (3f – 2g) (x)
(ii) (fg) (x)
(iii) \(\left(\frac{f}{g}\right)(x)\)
(iv) (f + g + 2) (x)
(v) 2f(x)
(vi) 2 + f(x)
Solution:
Given f(x) = 2x – 1, g(x) = x²
(i) (3f – 2g)(x) = 3f(x) – 2g(x)
= 3(2x – 1) – 2x²
= 6x – 3 – 2x²

(ii) fg(x) = f(x) . g(x)
= (2x – 1) (x²)
= 2×3 – x²

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

(iv) (f + g + 2)(x) = f(x) + g(x) + 2
= x² + 2x – 1 + 2
= x² + 2x + 1
= (x + 1)²

(v) 2f(x) = 2(2x – 1) = 4x – 2

(vi) 2 + f(x) = 2 + 2x – 1 = 2x + 1

Question 3.
If f(x) = x² and g(x) = |x|, find the following functions.
(i) f + g
(ii) f – g
(iii) f . g
(iv) 2f
(v) f + 3
(vi) \(\frac {f}{g}\) (for x ≠ 0)
Solution:
Given f(x) = x², g(x) = |x|
(i) (f + g)(x) = f(x) + g(x)
= x² + |x|
= \(\left\{\begin{array}{l}
\mathrm{x}^2+\mathrm{x}, \mathrm{x} \geq 0 \\
\mathrm{x}^2-\mathrm{x}, \mathrm{x}<0
\end{array}\right.\)

(ii) (f – g) (x) = f(x) – g(x)
= x² – |x|
= \(\left\{\begin{array}{c}
x^2-x, x \geq 0 \\
x^2+x, x<0
\end{array}\right.\)

(iii) fg(x) = f(x) . g(x)
= (x)² |x|
= \(\left\{\begin{array}{c}
x^3, x \geq 0 \\
-x^3, x<0 \end{array}\right.\)

(iv) 2f(x) = 2x²

(v) (f + 3) (x) = f(x) + 3 = x² + 3

(vi) \(\frac{f}{g}(x)=\frac{f(x)}{g(x)}=\frac{x^2}{|x|}=\left\{\begin{array}{c} x, x>0 \\
-x, x<0
\end{array}\right.\)

Question 4.
If the function f is defined by f(x) = \(\begin{cases}3 x-2, & x>3 \\ x^2-2, & -2 \leq x \leq 2 \\ 2 x+1, & x<-3\end{cases}\) then find the values if exists of f(4), f(2.5), f(-2), f(-4), f(0), f(-7), f(1), f(9).
Solution:
Given f(x) = \(\begin{cases}3 x-2, & x>3 \\ x^2-2, & -2 \leq x \leq 2 \\ 2 x+1, & x<-3\end{cases}\)
f(4) = 3(4) – 2 = 12 – 2 = 10
f(2.5) is not defined
f(-4) = 2(-4) + 1 = -8 + 1 = -7
f(0) = (0)² – 2 = -2
f(-7) = 2(-7) + 1 = -14 + 1 = -13
f(1) = (1)² – 2 = -1
f(9) = 3(9) – 2 = 27 – 2 = 25

Question 5.
Determine a Quadratic function f is defined by f(x) = ax² + bx + c, if f(0) = 6, f(2) = 1, f(-3) = 6.
Solution:
f(x) = ax² + bx + c;
Given f(0) = 6, f(2) = 1, f(-3) = 6
f(0) = a(0)² + b(0) + c
⇒ 6 = c
f(2) = a(2)² + b(2) + 6
⇒ 4a + 2b + 6 = 1
⇒ 4a + 2b = -5 ………(1)
f(-3) = a(-3)² + b(-3) + 6
⇒ 9a – 3b + 6 = 6
⇒ 9a – 3b = 0 ………(2)
Solving (1) and (2), we get,