Students can track their progress and improvement through regular use of OP Malhotra Maths Class 12 Solutions Chapter 11 Inequalities Ex 11(b).
S Chand Class 11 ICSE Maths Solutions Chapter 11 Inequalities Ex 11(b)
Question 1.
x < 2 y > – 1
Solution:
Given inequalities are :
x < 2 …(1)
y > – 1 …(2)
For region x < 2; The line x = 2 is parallel to y-axis and passing through (2, 0). Since (0,0) satisfies the inequality (1). Thus region containing (0, 0) gives the solution set of inequality (1).
For region y > – 1; The line y = – 1 passing through (0, -1) and parallel to x-axis. Clearly (0, 0) satisfies y > – 1. Thus region containing (0, 0) gives the solution set of y > – 1.
Clearly shade the common region of both inequalities. So all points in the shaded region gives the solution set of given system of inequalities.
Question 2.
y > x
y ≤ 3
Solution:
Given inequalities are
y > x …(1)
y ≤ 3 …(2)
For region y > x; The line y = x passes through (0, 0) making an angle of 45° with positive direction of x-axis.
Clearly (0, 0) does not satisfies eqn. (1) ∴ region not containing (0, 0) gives the solution set of inequality (1, 1).
For region y ≤ 3; The line y = 3 passing through (0, 3) and || to x-axis. Clearly (0, 0) satisfies the inequality (2). Thus region containing (0, 0) gives the solution set of inequality (2).
Both lines y = x and y = 3 intersects at (3, 3).
Hence all points of common shaded portion gives the solution set of given system of inequalities.
Question 3.
x ≤ – y
2y < x
Solution:
Given inequalities are ;
x ≤ – y …(1)
2y < x …(2)
For region x ≤ – y; The line x = – y be a line passing through (0, 0) and making an angle of 135° with x-axis. Clearly (0, 0) satisfies x ≤ – y.
Hence the region containing origin gives the solution set of inequality (1).
For region 2y < x; The line 2y = x is a line passing through (0, 0) and (2, 1). Further origin does not satisfies 2y < x.
Hence origin not containing (0, 0) gives the solution set of inequality (2).
Thus solution set contains all the points in common shaded region of coordinate plane.
Question 4.
y > 2 – x
y > x – 2
Solution:
Given inequalities are ;
y > 2 – x …(1)
y > x – 2 …(2)
For region y > 2 – x; The line y = 2 – x meets the coordinate axes at A(2, 0) and B(0, 2). Since (0, 0) does not satisfies inequality (1). Hence region not containing (0, 0) gives the solution set of inequality (1).
For region y > x – 2; The line y = x – 2 meets coordinate axes at C(2, 0) and D(0, -2).
Clearly (0, 0) satisfies y > x – 2.
Hence region containing (0, 0) gives the solution set of inequality (2).
Both lines y = 2 – x and y = x – 2 intersect at P(2, 0).
Thus all points of common shaded region gives the solution set of system of inequalities.
Question 5.
-2y > x – 8
4y > x – 12
Solution:
Given inequalities are ;
-2y > x – 8 …(1)
4y > x – 12 …(2)
For region – 2y > x – 8; The line – 2y = x – 8 meets coordinates axes at A(8, 0) and B(0, 4). Clearly (0, 0) satisfies the inequality (1).
Hence the region containing origin gives the solution set of inequality (1).
For region 4y > x – 12; The line 4y = x – 12 meets coordinate axes at C(12, 0) and D(0, -3).
Clearly (0, 0) satisfies inequality ( 2 ).
Hence region containing (0, 0) gives the solution set of inequality (2).
Clearly both lines x + 2y = 8 and x – 4y = 12 intersects at P(\(\frac { 28 }{ 3 }\), – \(\frac { 2 }{ 3 }\))
Thus all points of common shaded region gives the solution set of given system of inequalities.
Question 6.
3x – 4y < 6
2x – y < – 1
Solution:
Given inequalities are ;
3x – 4y < 6 …(1)
2x – y < – 1 …(2)
For region 3x – 4y < 6; The line 3x – 4y = 6 meets coordinate axes at A(2, 0) and B(0, \(\frac { -3 }{ 2 }\)). Clearly (0, 0) satisfies the inequality (1).
Thus region containing (0, 0) gives the solution set of inequality (1).
For region 2x – y < – 1; The line 2x -y = – 1 meets coordinate axes at C(-\(\frac { 1 }{ 2 }\), 0) and D(0, 1).
Clearly (0, 0) does not satisfies inequality (2). Thus the region not containing (0, 0) gives the solution set of 2x – y < 1.
Both lines 3x – 4y = 6 and 2x – y = – 1 intersects at P(-2, -3).
Thus all points of common shaded region gives the solution set of given system of inequalities.
Question 7.
x < y < x + 3
Solution:
Given inequalities are
x < y …(1)
y < x + 3 …(2)
For region y > x; The line y = x passing through (0, 0) and making an angle 45° with the positive direction of x-axis. Clearly (0, 0) does not satisfies inequality (1). Hence region not containing origin gives the solution set of inequality (1).
For region y < x + 3; The line y = x + 3 meets coordinates axes at A(-3, 0) and B(0, 3). Clearly (0, 0) satisfies the inequality (2). Thus region containing (0, 0) gives the solution set of inequality (2). Further both lines are || to each other. Thus all the points of common shaded region gives the solution set of system of inequalities.
Question 8.
2x + 1 ≥ y ≥ 2x – 3
Solution:
Given inequalities are
2x + 1 ≥ y …(1)
y ≥ 2x ≥ 3 …(2)
For region 2x +1 ≥ y; The line 2x + 1 = y intersects coordinate axes at A(-\(\frac { 1 }{ 2 }\), 0) and B(0, 1). Clearly (0, 0) satisfies inequality (1). Thus the region containing (0, 0) gives the solution set of inequality (1).
For region y ≥ 2x – 3; The line y = 2x – 3 meets coordinate axes at C(\(\frac { 3 }{ 2 }\), 0) and D(0, – 3). Clearly (0, 0) satisfies inequality (2). Thus region containing (0, 0) gives the solution set of inequality (2).
Clearly both lines y = 2x + 1 and y = 2x – 3 are parallel to each other.
Thus all the points of common shaded region gives the solution set of system of inequalities.
Question 9.
2 < x < 4 2y > x – 6
Solution:
Given inequalities are
2 < x < 4 …(1)
and 2y > x – 6 …(2)
For region 2 < x < 4 ; The line x = 2 be a line passing through A (2, 0) and || to y-axis and x = 4 is a line passing through B (4, 0) and || to y-axis. Thus 2 < x < 4 represents the region between the parallel lines x = 2 and x = 4. For region. 2y > x – 6 ; The line 2y = x – 6 meets coordinate axes at C (6, 0) and D (0, – 3). Clearly (0, 0) satisfies the inequality (2).
Thus region containing (0, 0) gives the solution set of inequality (2).
Clearly the line 2y = x – 6 meets x = 2 at E (2, – 2) and x = 4 at F (4, – 1).
Thus all the points of common shaded region gives the solution set of system of inequalities.
Question 10.
y ≤ x
x + y < 4
x – 2y < 1
Solution:
Given inequalities are
y ≤ x …(1)
x + y < 4 …(2)
and x – 2y < 1 …(3)
For region y ≤ x ; The line y = x passing through origin and making an angle of 45° with the direction of x-axis. Hence region containing origin gives the solution set of inequality (1).
For region x + y < 4 ; The line x + y = 4 meets coordinate axes at A (4, 0) and B (0, 4). Further (0, 0) satisfies the inequality (2). Thus region containing (0, 0) gives the solution set of inequality (2).
For region x – 2y < 1 ; The line x – 2y = 1 meets coordinate axes at C(1, 0) and D(0, – \(\frac { 1 }{ 2 }\)). Clearly (0, 0) satisfies the inequality (3).
Hence the region containing origin gives the solution set of inequality (3).
Lines y = x and x + y = 4 meets at P (2, 2). Lines y = x and x-2y= 1 intersects at Q (- 1, – 1). Both lines x + y – 4 and x – 2y = 1 intersects at R (3, 1). Thus all the points of common shaded region gives the solution set of given system of inequalities.
Question 11.
2x +y < 2 x – y > – 2
x + y > – 2
Solution:
Given inequalities are ;
2x + y < 2 …(1) x – y > – 2 …(2)
and x + y > – 2 …(3)
For region 2x + y < 2 ; The line 2x + y = 2 meet the coordinate axes at A (1, 0) and B (0, 2). Clearly (0, 0) satisfies the inequality (1). Thus the region containing origin gives the solution set of inequality (1). For region x – y > – 2 ; The line x – y = – 2 meet coordinate axes at C (- 2, 0) and D (0, 2). Clearly (0, 0) satisfies the inequality (2). Hence the region containing (0,0) gives the solution set of inequality (2).
For region x + y > – 2 ; The line x + y = – 2 meets coordinate axes at E (- 2, 0) and F (0, – 2). Clearly (0, 0) satisfies the inequality (3). Hence the region containing (0, 0) gives the solution set of inequality (3).
The lines 2x + y = 2 and x – y = – 2 intersects at P (0, 2).
The lines 2x + y = 2 and x + y = – 2 intersects at Q (4, – 6).
The lines x – y = -2 and x + y = – 2 intersects at R (- 2, 0).
Thus all the points of common shaded region gives the solution set of system of inequalities.
Question 12.
x + y ≥ 3
x – y ≤ 3
x + 5y ≥ 15
Solution:
Given inequalities are
x + y ≥ 3 …(1)
x – y ≤ 3 …(2)
and x + 5y ≥ 15 …(3)
For region x + y ≥ 3 ; The line x + y = 3 meet coordinate axes at A (3, 0) and B (0, 3). Clearly (0, 0) does not satisfies x + y ≥ 3. Thus the region not containing origin gives the solution set of inequality (1).
For region x – y ≤ 3 ; The line x – y = 3 meet coordinate axes at C (3, 0) and D (0, – 3). Clearly (0, 0) satisfies x – y ≤ 3.
Thus region containing (0, 0) gives the solution set of inequality (2).
For region x + 5y ≥ 15 ; The line x + 5y = 15 meet coordinate axes at E (15, 0) and F (0, 3). Clearly (0, 0) does not satisfies the inequality (3). Hence the region not containing origin gives the solution set of inequality (3).
All points of the common shaded region gives the solution set of system of inequalities.
Hence the region is unbounded.
Question 13.
y – 2 < 0.1 x + 3 > 0
2y + x < 2
3y + 3 < 2x
Solution:
Given system of inequalities are
y – 2 < 0 …(1) x + 3 > 0 …(2)
2y + x < 2 …(3)
3y + 3 < 2x …(4)
For region y – 2 < 0; The line y = 2 passing through (0, 2) and parallel to x-axis. Since (0, 0) satisfies eqn. (1) Thus region containing (0, 0) gives the solution set of inequality (1).
For x + 3 > 0 ; The line x + 3 = 0 passing through the point (- 3, 0) and || to y-axis. Also, (0, 0) satisfies x + 3 > 0. Thus region containing (0, 0) gives the solution set of inequality x + 3 > 0.
For region 2y + x < 2 ; The line 2y + x = 2 meets coordinate axes at A (2, 0) and B (0,1). Further (0,0) satisfies 2y + x < 2. Thus region containing origin gives the solution set of given inequality (3).
For region 3y + 3 < 2x; The line 3y + 3 = 2x meet coordinate axes at C(\(\frac { 3 }{ 2 }\), 0) and D (0, – 1).
Further (0, 0) does not satisfies 3y + 3 < 2x. Thus region not containing origin gives the solution set of given inequality (4). Thus all points of common shaded region gives the solution set of system of inequalities.
Question 14.
2y – 6 ≤ x
2y + 3x ≥ – 6
5y + 15 ≥ 2X
3y + 5x ≤ 15
Solution:
Given system of inequalities are ;
2y – 6 ≤ x …(1)
2y + 3x ≥ – 6 …(2)
5y + 15 ≥ 2x …(3)
and 3y + 5x ≤ 15 …(4)
For region 2y – 6 ≤ x ; The line 2y – 6 = x meets coordinates axes at A (- 6, 0) and B (0, 3). Clearly (0, 0) satisfies the inequality 2y – 6 < x.
Thus region containing origin gives the solution set of inequality (1).
For region 2y + 3x ≥ – 6 ;
The line 2y + 3x ≥ – 6 meets coordinate axes at C (- 2, 0) and D (0, – 3). Clearly (0, 0) satisfies the inequality (2). The region containing origin gives the solution set of inequality (2).
For region 5y + 15 ≥ 2x ; The line 5y + 15 = 2x meets coordinate axes at E (\(\frac { 15 }{ 2 }\), 0) and F (0, – 3). Clearly (0, 0) satisfies the inequality (3). Thus the region containing (0, 0) gives the solution set of inequality (3).
For region 3y + 5x ≤ 15 ; The line 3y + 5x = 15 meets coordinate axes at G (3, 0) and M (0, 5). Clearly (0,0) satisfies the inequality (4). Hence the region containing (0, 0) gives the solution set of inequality (4).
Thus all the points of common shaded region gives the solution set of system of inequalities.
Question 15.
(i) Find the linear constraints for which the shaded area in the figure given below is the solution set.
Solution:
The shaded area is bounded by the lines
-7x + 4y = 14 …(1)
3x + 4y = 18 …(2)
2x + 3y = 3 …(3)
x – 6y = 3 …(4)
For line (1); Here the shaded area and (0, 0) are on the same side of line (1). Thus corresponding constraint is – 7x + 4y ≤ 14
(∵ – 7 × 0 + 4 × 0 = 0 < 14 i.e. (0, 0) satisfies the constraint)
For line (2); Here the shaded area and origin on the same side of line (2).
∴ the corresponding constraint is 3x + 4y ≤ 18
For line (3); Here the shaded area and origin are on the opposite of line (3)
Thus the corresponding constraint is 2x + 3y ≥ 3
For line (4); Here, the origin and shaded area on the same side of line (4). Thus the corresponding constraint is x – 6y ≤ 3 [∵ 0 – 6 × 0 = 0 < 3]
Also the shaded area lies in first quadrant of XOY plane.
∴ x ≥ 0 ; y ≥ 0 Thus the required system of constraints are ;
– 7x + 4y ≤ 14; 3x + 4y ≤ 18 ;
2x + 3y ≥ 3; x – 6y ≤ 3
and x ≥ 0; y ≥ 0
(ii) Here the shaded area bounded by the lines
3x + y – 6 = 0
4x + 9y – 36 = 0
x + 3y – 6 = 0
4x – 3y – 12 = 0
For line (1); The shaded area and origin lies on opposite side of line (1).
Thus corresponding constraint is
3x + y ≥ 6
For line (2); The shaded area and origin lies on same side of line (2).
Thus corresponding constraint is
4x + 9y – 36 ≤ 0
[∵ 4 × 0 + 9 × 0 – 36 = – 36 ≤ 0]
For line (3); The shaded area and origin lies on opposite side of line (3).
Thus corresponding constraint is
4x + 3y – 6 ≥ 0
For line (4); The shaded area and origin lies on same side of line (4).
Thus corresponding constraint is
4x – 3y – 12 ≤ 0
Also shaded area lies in first quadrant of XOY plane. Thus corresponding constraint are
x ≥ 0; y ≥ 0.
Thus, the required system of constraints are ;
3x + y – 6 ≥ 0;
4x + 9y – 36 ≤ 0 ;
x + 3y – 6 ≥ 0
4x – 3y – 12 ≤ 0;
x ≥ 0; y ≥ 0
Question 16.
x + y ≥ 3,
7x + 6y ≤ 42,
x ≤ 5,
y ≤ 4,
x,y > 0.
Solution:
Given system of inequalities are
x + y ≥ 3 …(1)
7x + 6y ≤ 42 …(2)
x ≤ 5 …(3)
y ≤ 4 …(4)
x, y > 0 …(5)
For region x + y ≥ 3 ; The line x + y = 3 intersects coordinate axes at A (3, 0) and B (0, 3). Also, (0, 0) does not satisfies x + y ≥ 3. Thus region not containing gives the solution set of inequality (1).
For region 7x + 6y ≤ 42 ;
The line 7x + 6y = 42 meet coordinate axes at C (3, 0) and (0, 7).
Clearly (0, 0) satisfies inequality (2).
Hence region containing (0, 0) gives the solution set of given inequality (2).
For region x ≤ 5 ; The line x = 5 passing through (5, 0) and parallel to y-axis. Further (0, 0) satisfies x ≤ 5. Hence region containing (0, 0) gives the solution set of x ≤ 5.
For region y ≤ 4 ; The line y = 4 passing through (0, 4) and parallel to x-axis. Since (0, 0) satisfies y ≤ 4. Hence the region containing (0, 0) gives the solution set of given inequality.
x, y > 0 represent the first quadrant of XOY plane. All points of common shaded region gives the solution set of system of inequalities.
Question 17.
Find the region where all the inequations x + y ≥ 0, 2x + y ≤ 4, x ≥ 0 and y ≤ 2 hold good. Find the coordinates of the vertices of the region.
Solution:
Given system of inequalities ;
x + y ≥ 0 …(1)
2x + y ≤ 4 …(2)
x ≥ 0 …(3)
and y ≥ 2 …(4)
For region x + y ≥ 0; The line x + y = 0 meets coordinate axes at (0, 0). Thus region containing origin gives the solution of inequality (1).
For region 2x + y ≤ 4 ; The line 2x + y = 4 meets coordinate axes at A (2, 0) and B (0,4). Clearly (0, 0) satisfies the inequality (2). Thus the region containing (0, 0) gives the solution set of inequality (2).
For region x ≥ 0 ; Clearly x ≥ 0 represents the right side of y-axis.
For region y ≤ 2 ; The line y = 2 passing through the point (0, 2) and parallel to x-axis. Since (0,0) satisfies the inequality (4). Hence the region containing (0, 0) gives the solution set of y ≤ 2.
The line y = 2 meet line x + y = 0 at Q (- 2, 2) and meets line 2x + y = 4 at P (1, 2).
Thus the points of common shaded region OAPQ gives the solution set of given system of eqns. and comer points of region are O (0, 0); A (2, 0); P (1, 2) and Q (- 2, 2).
Question 18.
Find the region where all the inequations x + y ≤ 6, x ≥ y, x ≥ 0, y ≥ 0 hold good. Find the ordered pairs of the vertices of the region so formed.
Solution:
Given system of inequalities are
x + y ≤ 6 …(1)
x ≥ y …(2)
x ≥ 0, y ≥ 0 …(3)
For region x + y ≤ 6 ; The line x + y = 6 meets coordinate axes at A (6, 0) and B (0, 6). (0,6).
Clearly (0, 0) satisfies inequality (1). Thus region containing (0, 0) gives the solution set of inequality (1).
For region x ≥ y; The line x = y passing through origin making an angle of 45° with the direction of x-axis. Since (0, 0) satisfies the inequality (2).
Thus the region containing origin gives the solution set of x ≥ y.
Further x ≥ 0, y ≥ 0 represents the first quadrant of XOY plane.
The lines x = y and x + y = 6 intersects at P (3, 3).
Hence all the points of common shaded region OAP gives the solution set of system of inequalities and corner points are O (0, 0); A (6, 0) and P (3, 3).