OP Malhotra Class 11 Maths Solutions Chapter 2 Relations and Functions Ex 2(g)

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S Chand Class 11 ICSE Maths Solutions Chapter 2 Relations and Functions Ex 2(g)

Question 1.
y = – | x |
Solution:
Given y = – | x |
Its domain is set of all real numbers and range is set all negative numbers including 0.
since | x | > 0 ⇒ – | x | < 0 ⇒j<0 When x > 0 ; y = – x [∵ | x | = x]
which is the eqn. of straight line passing through (0, 0) making a slope of – 1.
When x < 0 ; y = x [∵ | x | = x]
which is the eqn. of straight line having slope = 1.
The graph of y = – | x | is given as under ;

Question 2.
y = \(\frac{|x|-x}{2}\)
Solution:
Given y = \(\frac{|x|-x}{2}\)
Here domain of function is set of all real numbers.
When x ≥ 0 ∴ |x| = x ⇒ y = \(\frac{x-x}{2}\) = 0
which shows the x-axis.
When x < 0 ∴ | x I = – x ⇒ y = \(\frac{-x-x}{2}\) = – x
which is the eqn. of straight line having slope – 1

Question 3.
y = \(\frac{1}{|x|}\)
Solution:
Given y = \(\frac{1}{|x|}\)
Here domain of the function is the set of all non-zero real numbers.
Further the graph does not meet coordinate axes.
When x ≥ 0 ∴ |x| = x ⇒ y = \(\frac{1}{x}\)

x	1	2	3	\(\frac { 1 }{ 2 }\)	–          1
y	1	\(\frac { 1 }{ 2 }\)	\(\frac { 1 }{ 3 }\)	2	–          1

When x < 0 ⇒ | x | = – x
∴ y = – \(\frac{1}{x}\)

Question 4.
y = | 4 – x² |, – 3 ≤ x ≤ 3.
Solution:
Given y = | 4 – x² |, – 3 ≤ x ≤ 3
Here D_f = R and range (f) = [0, ∞)
When – 3 < x < – 2
⇒ x + 2 < 0 and 2 – x > 0
∴ y = | 2 – x | | 2 + x | = – (2 – x) (x + 2)
y = – (4 – x²) = x² – 4 ⇒ x² = y + 4
which represents an upward parabola with vertex (0, – 4)
When – 2 ≤ x ≤ 2
⇒ x – 2 ≤ 0 and x + 2 ≥ 0
∴ y = (2 – x) (2 + x) = 4 – x² ⇒ x² = 4 – y
⇒ x² = – (y – 4) which represents a downward parabola with vertex (0, 4) meeting x-axes at (± 2, 0),
When 2 < x ≤ 3 ⇒ 2 – x < 0, 2 + x > 0
∴ y = – (2 – x) (2 + x) = – (4 – x² )
⇒ x² – 4 = v
⇒ x² = 4 + y which is again represents a downward parabola with vertex (0, – 4).
Thus the required graph of given function is given as under :

Question 5.
y = | x | + x, – 2 ≤ x ≤ 2
Solution:
Given y = | x | + x, – 2 ≤ x ≤ 2
Here domain of function is set of all real numbers.
When – 2 ≤ x < 0 ∴ |x| = – x
⇒ y = – x + x = 0
which is the eqn. of x-axis.
When 0 ≤ x < 2 ∴ | x | = x
⇒ y = x + x = 2x
which is the eqn. of straight line passing through (0, 0) with slope 2.

Question 6.
y = |x + 2| + x
Solution:
Given y = |x + 2| + x
Here domain of the function is the set of all real numbers.
Case – I: When x ≤ – 2 ∴ x + 2 ≤ 0
⇒ | x + 2 | = – (x + 2)
∴ y = – (x + 2) + x = – 2
Hence the graph is left side of the line y = – 2 for x ≤ – 2

Case – II: When x ≥ – 2
⇒ x + 2 > 0 ⇒ | x + 2 | = x + 2
∴ y = x + 2 + x = 2(x + 1)

x	0	1	– 1	– 2
y	2	4	0	– 2

Question 7.
(i) Copy and complete this table of values :

X	-2	-1	0	1	2	3
3x	0.1	0.3	1

(ii) Draw the graph y = 3^x on squared paper, for – 2 ≤ x ≤ 3.
(iii) What features do the graphs of y = 2^x (drawn on page Ch 2-51) and y = 3^x have in common?
Solution:
(i) Giveny = 3^x
When x = 1 ⇒ y = 3
When x = 2 ⇒ y = 3² = 9
When x = 3 ⇒ y = 3³ = 27

The table of values becomes ;

X	-2	-1	0	1	2	3
y	0.1	0.3	1	3	9	27

(ii) When x → ∞ ⇒ y → ∞
When x → ∞ ⇒ 3^-∞ → 0 ⇒ y → 0

(iii)

X	-2	-1	0	1	2
y	0.25	0.5	1	2	4

Clearly both graphs pass through the point (0, 1) rise steadily from left to right. Also x-axis be the tangent to both graphs at infinity.
Further both graphs as x → ∞ ⇒ x → ∞ as x → – ∞ ⇒ y → 0

Question 8.
(i) Copy and complete this table.

X	-3	-2	-1	0	1	2	3
(\(\frac { 1 }{ 2 }\))x	8			1		0.3	0.1

(ii) Draw the graphs y = 2^x (See Solved Ex. 68) and y = (\(\frac { 1 }{ 2 }\))^x, on the same diagram, for – 3 ≤ x ≤ 3.
(iii) Which line is the axis of symmetry in the diagram?
Solution:
(i) y = (\(\frac { 1 }{ 2 }\))^x
When x = – 2 ⇒ y = (\(\frac { 1 }{ 2 }\))^-2 = 2² = 4
When x = – 1 ⇒ y = (\(\frac { 1 }{ 2 }\))^-1 = 2
When x = 1 ⇒ y = \(\frac { 1 }{ 2 }\)
The table of values is given as under:

X	-3	-2	-1	0	1	2	3
y	8	4	2	1	\(\frac { 1 }{ 2 }\)	0.3	0.1

(ii) When x → ∞ ⇒ y → 0
When x → ∞ ⇒ y → ∞
For graph of y = 2^x :
The table of values is given as under :

X	-3	-2	-1	0	1	2	3
y	\(\frac { 1 }{ 8 }\)	\(\frac { 1 }{ 4 }\)	\(\frac { 1 }{ 2 }\)	1	2	4	8

as x → ∞ ⇒ y → ∞
and as x → – ∞ ⇒ y → 0

(iii) Clearly both graphs lies in first and second quadrants and intersecting at point (0, 1).
Further y-axis is the required line of symmetry.

Question 9.
A sketch of the graph y = a log₄ (x + b) is shown in Fig. 2.70. Find the values of a and b.

Solution:
Given y = alog₄(x + b) …(1)
Clearly the graph of eqn. (1) pass through the points (- 2, 0) and (1, 5).
Thus, 0 = a log₄ (b – 2)
since a ≠ 0 ∴ 0 = log₄ (b – 2)
⇒ 4° = b – 2 ⇒ 1 = b – 2 ⇒ b = 3
Also, 5 = a log₄ (1 + b) ⇒ 5 = a log₄ 4
⇒ 5 = a x \(\frac{\log 4}{\log 4}\) ⇒ a = 5

Question 10.
Diagrams (i) shows the curve y = log_a x. What is the value of a ?

Solution:
Given eqn. of curve be
y = log_a x …(1)
Since, eqn. (1) passes through the point (81, 4).
⇒ 4 = log_a 81 ⇒ a⁴ = 81 = 3⁴ ⇒ a = 3 [if x = log_a y ⇒ a^x = y]

Question 11.
Diagram given below (ii) shows the curve y = log₁₀ (x + p). What is the value of p ?

Solution:
Given eqn. of curve be
y = log₁₀ (x + P) … (1)
Clearly graph of eqn. (1) pass through the point (- 1, 0).
∴ 0 = log₁₀ (- 1 + p) ⇒ 10⁰ = – 1 + p
⇒ p = 1 + 1 = 2

Question 12.
(i) Sketch the graphs y = 2
and y = log₁₀ 2x on the same diagram.
(ii) Find the point of intersection of the graphs by solving the equation log10 2x = 2.
Solution:
(i) Given y = 2 which is the eqn. of straight line passing through (0, 2) and || to x-axis.
Also given y = log₁₀ 2x ⇒ 2x = 10^y
⇒ x = \(\frac { 1 }{ 2 }\).10^y
Here domain of the function D_f = (0, ∞)
The table of values is given as under:

X	\(\frac { 1 }{ 2 }\)	5	50
y	0	1	2

(ii) Given log₁₀ 2x = 2 … (1)
We know that if log_a x = y ⇒ a^y = x
∴ from (1) ; 2x = 10² ⇒ 2x = 100
⇒ x = 50
Hence both graph intersects at (50, 2).

Question 13.
The sketch shows part of the graph
y = a log₂ (x – b).
Find the values of a and b.
Solution:
Given equation of curve be
y = a log₂ (x – b) …(1)
Clearly from graph given, it is observed that eqn. (1) passes through the points (5, 0) and (6, 3).

∴ 0 = a log₂ (5 – b)
since a ≠ 0 ∴ 0 = log₂ (5 – b) ⇒ 5 – b = 2°
⇒ 5 – b = 1 ⇒ b = 4
Also, 3 = a log₂ (6 – b) ⇒ 3 = a log₂ (6 – 4)
⇒ 3 = a log₂ 2 ⇒ a = 3 [∵ log_a a = 1]
Hence, a = 3 and b = 4

Question 14.
(i) Sketch the graphs y = 4 – x and y = log₁₀ x on the same diagram.
(ii) Write down an equation to find the x-coordinate of the point of inter¬section of the graphs.
(iii) Show that x satisfies 3.4 < x < 3.5.
(iv) From your graph find x, correct to 2 decimal places.
Solution:
(i) eqn. of given curve be y = 4 – x

X	0	1	2	3	4
y	4	3	2	1	0

Hence the given curve meets x-axis at (4, 0) and y-axis at (0, 4).
Also eqn. of another curve be
y = log₁₀ x ⇒ x = 10^y

Here domain of the function is the set of all positive real numbers i.e. D_f = (0, ∞).

X	0	10	100
y	0	1	2

(ii) Both curves intersects when 4 – x = log₁₀x
which is the required equation which gives the x-coordinate of point of intersection given graphs.

(iii) Since 4 – x = log₁₀x …(1)
⇒ f(x) = log₁₀ x + x – 4
Now f(3.4) = log₁₀ (3.4) + 3.4 – 4 = 0.5314 – 0.6 = – 0.0685
f(3.5) = log₁₀ (3.5) + 3.5 – 4 = 0.044068
Thus root of eqn. (1) lies between 3.4 and 3.5
i.e. x satisfies 3.4 < x < 3.5
Also given y = log₁₀ 2x ⇒ 2x = 10^y
⇒ x = \(\frac { 1 }{ 2 }\).10^y
Here domain of the function D_fy = (0, ∞)
The table of values is given as under:

(ii) Given log₁₀2x = 2 …(1)
We know that if log_a x = y ⇒ a^y = x
∴ from (1) ; 2x = 10² ⇒ 2x = 100
⇒ x = 50
Hence both graph intersects at (50, 2).

Question 13.
The sketch shows part of the graph y = a log₂ (x – b).
Find the values of a and b.
Solution:
Given equation of curve be
y = a log₂ (x – b) …(1)
Clearly from graph given, it is observed that eqn. (1) passes through the points (5, 0) and (6, 3).

∴ 0 = a log₂ (5 – b)
since a ≠ 0 ∴ 0 = log₂ (5 -b) ⇒ 5 – b = 2°
⇒ 5 – b = 1 ⇒ b = 4
Also, 3 = a log₂ (6 – b) ⇒ 3 = a log₂ (6 – 4)
⇒ 3 = a log₂ 2 ⇒ a = 3 [∵ log_a a = 1 ]
Hence, a = 3 and b = 4

Question 14.
(i) Sketch the graphs y = 4 – x and y = log₁₀ x on the same diagram.
(ii) Write down an equation to find the x-coordinate of the point of inter-section of the graphs.
(iii) Show that x satisfies 3.4 < x < 3.5.
(iv) From your graph find x, correct to 2 decimal places.
Solution:
(i) eqn. of given curve be y = 4 – x

X	0	1	2	3	4
y	4	3	2	1	0

Hence the given curve meets x-axis at (4, 0) and y-axis at (0, 4).
Also eqn. of another curve be y = log₁₀ x ⇒ x = 10^y

Here domain of the function is the set of all positive real numbers i.e. D_f = (0, ∞).

X	1	10	100
y	0	1	2

(ii) Both curves intersects when 4 – x = log₁₀ x
which is the required equation which gives the x-coordinate of point of intersection given graphs.

(iii) Since 4 – x = log₁₀x … (1)
⇒ f(x) = log₁₀ x + x – 4
Now f(3.4) = log₁₀ (3.4) + 3.4 – 4
= 0.5314 – 0.6 = – 0.0685
f(3.5) = log₁₀ (3.5) + 3.5 – 4 = 0.044068
Thus root of eqn. (1) lies between 3.4 and 3.5
i.e. x satisfies 3.4 < x < 3.5

(iv) From graphs, it is observed that x = 3.46
(correct to 2 decimal places)

Question 15.
Sketch the graphs.
(i) y = log₂ x
(ii) y = log₂ x + 1
(iii) y = log₂ (x + 1)
Solution:
(i) Given eqn, of curve be
y = log₂ x ⇒ x = 2^y
Here domain of given function is the set of all positive real numbers.

X	1	2	4	8
y	0	1	2	3

as y → ∞ ⇒ x → ∞
as y → ∞ ⇒ x → ∞

(ii) Given eqn. of curve be
y = log₂ x + 1 ⇒ y – 1 = log₂ x
⇒ x = 2^y-1 = \(\frac { 1 }{ 2 }\)2^y
Here domain of function is set of all positive real numbers.

X	\(\frac { 1 }{ 2 }\)	1	2	4	8
y	0	1	2	3	4

as x → ∞ ⇒ y → ∞
as x → 0 ⇒ y → – ∞

(iii) Given eqn. of curve be y = log₂ (x + 1)
⇒ 2^y = x + 1 ⇒ x = 2^y – 1
For D_f : x + 1 > 0 ⇒ x > – 1
∴ domain of function f = (- 1, ∞)

X	0	–\(\frac { 1 }{ 2 }\)	1	3	7	15
y	0	-1	1	2	3	4

as x → ∞ ⇒ y → ∞

Question 16.
Sketch the graphs.
(i) log₄ x
(ii) 2 log₄ x
(iii) 3 log₄ x
Solution:
(i) Given eqn. of curve be
y = log₄ x ⇒ x = 4^y
Here domain of the function is the set of all positive real numbers.

X	0	4	2	64
y	0	1	–\(\frac { 1 }{ 2 }\)	3

as x → ∞ ⇒ y → ∞
as x → 0 ⇒ y → – ∞

(ii) Given eqn. of curve be
y = 2 log₄ x ⇒ 4^y/2 = x
Here domain is again set of all positive real numbers

X	1	2	4	8	16	\(\frac { 1 }{ 2 }\)
y	0	1	2	3	4	-1

as x → ∞ ⇒ y → ∞
as x → 0 ⇒ y → – ∞

(iii) Given eqn. of curve be
y = 3 log₄ x ⇒ x = 4^y/3
Here domain of given function = (0, ∞).

X	1	4	2	16
y	0	3	\(\frac { 3 }{ 2 }\)	6

as x → ∞ ⇒ y → ∞
as x → 0 ⇒ y → – ∞

Question 17.
For – 2 < x< 1, draw the graph of y = 2^x.
(use 1 cm = 1 unit on both axes).
Use this graph to solve 2^x = 2x.
Solution:
Given eqn. of curve be
y = 2^x, – 2 < x < 1
Here domain is set of all reals.

x	-1	0
y	\(\frac { 1 }{ 2 }\)	0

The given eqn. 2^x = 2x …(1)
be the intersection of lines y = 2^x and y = 2x
From graph it is follows that x = 1, 2 satisfies eqn. (1).

Question 18.
(i) Complete the following table for y = 4^x.
Enter the values of x and y correct to 1 decimal place.

X	-2	-1	…	0.5	0.75
y	…	…	1	…	…

Copy the table on your answer book and enter the values there.
(ii) Taking 4 cm = 1 unit on both axes, draw the graph of y = 4^x for 0.75 ≥ x ≥ – 2.
(iii) From your graph estimate log₄ 1.25.
Solution:
Given equation of curve be y = 4^x

Further x → ∞ ⇒ y → ∞
and x → – ∞ ⇒ y → 0
since y = 4^x ⇒ x = log₄ y
i.e. when y = 1.25 it is observed from graph that x = 0.1609

Question 19.
Copy and complete the table, for the function y = \(\frac { 3 }{ x }\), giving your answer correct to 1 d.p. Then draw and graph.

x	-3	-2.5	-2	-1.5	-1	-0.5	0.5	1	1.5	2	2.5	3
y	-1	-1.2	…	…	-3	…	6	…	…	1.5	…	…

Solution:

Question 20.
Sketch the graphs of the following rational functions
(i) y = \(\frac{x+3}{x-2}\)
(ii) y = \(\frac{6}{x-6}\)
(iii) y = \(\frac{5}{2 x+1}\)
(iv) y = \(\frac{2 x+1}{x-3}\)
(v) y = \(\frac{7-2 x}{3 x+5}\)
Solution:
(i) Given eqn. of curve be y = \(\frac{x+3}{x-2}\)
∴ the function exists for x ≠ 2.
⇒ (x – 2) y = x + 3 ⇒ x (y – 1) = 3 + 2y ⇒ x = \(\frac{3+2 y}{y-1}\)
∴ given function exists for y ≠ 1.
Thus x = 2 be the vertical asymptote and y = 1 be the horizontal asymptote. Also given curve meets x-axis at (-3, 0) and y-axis at (0, – \(\frac { 3 }{ 2 }\)).
The table of values is given as under:

(ii) Given eqn. of curve be y = \(\frac { 6 }{ x-6 }\). Thus the given function does exists for x + 6 ∴ x = 6 be the vertical asymptote.
Further xy – 6y = 6 ⇒ xy = 6y + 6 ⇒ x = \(\frac{6 y+6}{y}\)
∴ given function exists for y ≠ 0
∴ y = 0 be the horizontal asymptote.
The table of values is given as under :

(iii) Given y = \(\frac{5}{2 x+1}\) ⇒ (2x + 1)y = 5
⇒ 2xy = 5 – y ⇒ x = \(\frac{5-y}{x y}\)
Thus given function exists for 2x + 1 ≠ 0 i.e. x ≠ – \(\frac { 1 }{ 2 }\)
∴ x = – \(\frac { 1 }{ 2 }\) be the vertical asymptote to given curve.
Also function exists for y ≠ 0
∴ y = 0 i. e. x-axis be the horizontal asymptote. Clearly the curve meets y-axis at (0, 5) and curve does not meet x-axis.

(iv) Given eqn. of curve be y = \(\frac{2 x+1}{x-3}\). Thus function exists for x ≠ 3
Also, xy – 3y = 2x + 1 ⇒ x (y – 2) = 1 + 3y ⇒ x = \(\frac{1+3 y}{y-2}\)
∴ given function exists for y ≠ 2
∴ x = 3 be the vertical and y = 2 be the horizontal asymptote. Further given curve meets coordinate axis at (- \(\frac { 1 }{ 2 }\), 0) and (0, – \(\frac { 1 }{ 3 }\)).

(v) Given eqn. of curve be y = \(\frac{7-2 x}{3 x+5}\)
Thus, given function exists for x ≠ \(\frac { – 5 }{ 3 }\)
∴ x = – \(\frac { 5 }{ 3 }\) be the vertical asymptote.
Also 3xy + 5y = 7 – 2x ⇒ x = \(\frac{7-5 y}{3 y+2}\)
∴ given function exists for y ≠ – \(\frac { 2 }{ 3 }\)
∴ y = \(\frac { 2 }{ 3 }\) be the horizontal asymptote. Further, given curve
meets coordinate axes at (\(\frac { 7 }{ 2 }\), 0) and (0, \(\frac { 7 }{ 5 }\)).