ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.5

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ML Aggarwal Class 11 Maths Solutions Section A Chapter 2 Relations and Functions Ex 2.5

Draw the graphs of the following (1 to 4) real functions. Hence, find their range.

Question 1.
(i) f (x) = x + 10
(ii) f (x) = 1 – 2x
Solutions:
(i) Given f(x) = x + 10
⇒ D_f = R
Let y = f(x) = x + 10
which is linear equation in x, y and hence represents a straight line.
So two points are sufficient to represent this straight line uniquely.

x	0	– 5
y	10	5

Clearly y can take any real values.
∴ R_f = R

(ii) Given f (x) = 1 – 2x
∴ D_f = R
Let y = f(x) = 1 – 2x, which is a first degree eqn. in x, y and hence represents a straight line.
So two points are sufficient to represents this line uniquely.

x	0	1
y	1	– 1

Clearly from graph, y can takes any real values.
∴ R_f = R.

Question 2.
(i) f (x) = \(\frac{x^2-4}{x-2}\)
(ii) f (x) = \(\left\{\begin{array}{cc}
x+3 & , x<2 \\
2 x+1 & , x \geq 2
\end{array}\right.\)
Solution:
(i) Given f(x) = \(\frac{x^2-4}{x-2}\)
For D_f : f (x) must be a real number
⇒ \(\frac{x^2-4}{x-2}\) ∈ R
⇒ x – 2 ≠ 0
⇒ x ≠ 2
D_f = R – {2}
Let y = f (x)
= \(\frac{x^2-4}{x-2}\)
= \(\frac{(x-2)(x+2)}{x-2}\)
⇒ y = x + 2 [∵ x ≠ 2]
which represents a straight line as it is a linear eqn. in x and y.
So two points are sufficient to represents the line uniquely.

x	0	1
y	2	3

Clearly from graph, y takes all values except y = 4
∴ R = R – {4}.

(ii) Given f (x) = \(\left\{\begin{array}{cc}
x+3 & , x<2 \\
2 x+1 & , x \geq 2
\end{array}\right.\)
Clearly D_f = R
its graph consists of two parts

Case – I :
When x < 2
∴ f(x) = x + 3
which is a linear in x, y and hence represents a straight line and two points are sufficient to represents the line uniquely.

Table of values is given as under:

x	0	1
y	3	4

Case – II :
When x ≥ 2, y = 2x + 1
which is a linear in x and y
and represents a straight line.

x	2	3
y	5	7

from graph, it is clear that y can take any value.
∴ R_f = R.

Question 3.
(i) f (x) = \(\left\{\begin{array}{cc}
x-1 & , x<2 \\
2 x+3 & , x \geq 2
\end{array}\right.\)
(ii) f (x) = \(\left\{\begin{aligned}
-1 & , x \leq-1 \\
x & ,-1<x<1 \\
1 & , x \geq 1
\end{aligned}\right.\)
Solution:
(i) Given y = f (x)
= \(\left\{\begin{array}{cc}
x-1 & , x<2 \\
2 x+3 & , x \geq 2
\end{array}\right.\)
Clearly D_f = R
its graph consists of two parts

Case – I :
When x < 2, y = x – 1
which is linear in x and y and hence represents a straight line.
Table of values is given as under:

x	0	1
y	– 1	0

From graph, it is clear that
R_f = (- ∞, 1] ∪ [7, ∞)

Case – II :
When x ≥ 2, y = 2x + 3, which is linear in x and y and hence represents a straight line.

x	2	3
y	7	9

(ii) Given y = f(x)
= \(\left\{\begin{aligned}
-1, & x \leq-1 \\
x & ,-1<x<1 \\
1 & , x \geq 1
\end{aligned}\right.\)
Clearly D_f = R
The graph consists of three parts :

Case – I :
When x ≤ – 1, y = – 1
Thus the graph is a straight line.

Table of values is given as under:

x	– 2	– 1
y	– 1	– 1

Case – II :
When – 1 < x < 1,
y = f (x) = x
which represents a straight line.

x	0	1/2
y	0	1/2

Case – III:
When x ≥ 1, y = 1
which represents a straight line.

x	1	2
y	1	1

From graph, it is clear that, R_f = [- 1, 1].

Question 4.
(i) f (x) = \(\frac{1}{2}\) x²
(ii) f (x) = 1 – x²
Solution:
(i) Let y = f(x) = \(\frac{1}{2}\) x²
⇒ D_f = R
Table of values is given as under:

Now plot the points (± 3, \(\frac{9}{2}\)), (± 2, 2), (± 1, \(\frac{1}{2}\)), (0, 0) …. and join these points by free hand curve to give the required graph.
From graph, it is clear that R_f = [0, ∞).

(ii) Given f (x) = 1 – x²
⇒ D_f = R
Let y = f (x)
= 1 – x_f
Table of values are given as under :

Plot all three points and join then by free hand drawing.
From graph, it is clear that
R = (- ∞, 1]

Question 5.
Draw the graph of the function f (x) = |x| + |x – 2|. Hence, find its range.
Solution:
Let y = f (x)
= |x| + |x – 2|
∴ D_f = R
The graph consists of three parts :
Case – I :
When x < 0
⇒ |x| = x ;
x < 2 ⇒ x – 2 < 0
∴ |x – 2| = – (x – 2)
∴ y = – x – (x – 2)
= – 2x + 2
Thus the graph is a part of straight line.

x	– 1	– 2
y	4	6

Case – II :
When 0 ≤ x < 2
⇒ x – 2 < 0 and x ≥ 0
∴ y = x – (x – 2) = 2
∴ The graph is a part of straight line.

x	0	1
y	2	2

Case – III :
When x ≥ 2 ⇒ x – 2 ≥ 0, x > 0
∴ y = x + x – 2
= 2x – 2
Thus, the graph is a part of straight line.
From graph, it is clear that R_f = [2, ∞).

x	2	3
y	2	4