CBSE Important Questions Class 11 Maths Chapter 5 – Linear Inequalities
Class 11 Maths Chapter 5 Linear Inequalities is part of the CBSE Class 11 Maths syllabus. It explains what an inequality is and how it differs from an equation. Students learn to solve linear inequalities in one variable and show the answer on a number line, to solve a system of inequalities, and to solve linear inequalities in two variables by shading the correct half plane on a graph.
These CBSE important questions for Class 11 Maths Chapter 5 cover identifying an inequality, solving a system of inequalities, solving quadratic inequalities, reading inequalities from a graph, graphical solutions in two variables, and finding integer solutions. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.
Q1. An example of an inequality is
(a) x = 5
(b) 5x + y = 5
(c) 4x − 7 ≥ 0
(d) x − y = 5
Answer: (c) 4x − 7 ≥ 0
An inequality uses one of the signs <, >, ≤ or ≥ instead of the equality sign. Only option (c) uses such a sign. The other three are equations because they use the sign =.
Q2. The solution of the system of inequalities 3x − 7 < 5 + x and 11 − 5x ≤ 1 is
(a) 2 ≤ x < 6
(b) 2 < x < 6
(c) 3 ≤ x < 6
(d) 3 < x < 6
Answer: (a) 2 ≤ x < 6
First inequality: 3x − 7 < 5 + x
3x − x < 5 + 7
2x < 12, so x < 6
Second inequality: 11 − 5x ≤ 1
−5x ≤ −10
Dividing by −5 reverses the sign: x ≥ 2
Combining both: 2 ≤ x < 6
Q3. The solution of (x − 1)²(x + 4) < 0 is
(a) (−∞, 1)
(b) (−∞, −4)
(c) (−1, 4)
(d) (1, 4)
Answer: (b) (−∞, −4)
Note that (x − 1)² is a square, so it is always positive or zero, never negative.
For the whole product to be negative, the other factor (x + 4) must be negative, and (x − 1)² must not be zero.
x + 4 < 0 gives x < −4
So the solution is x ∈ (−∞, −4).
Q4. Solve the following system of inequalities graphically: x + y ≥ 4 and 2x − y > 0.

Answer: Draw the line x + y = 4 and shade the region above it, including the points on the line (because the sign is ≥).
On the same axes draw the line 2x − y = 0 and shade the region above it, not including the points on the line (because the sign is strictly >).
Checking with a point: for (0, 0), 2(0) − 0 = 0 which is not greater than 0, so the origin is not in the region.
The common shaded region of the two is the required solution of the system.
Q5. Solve the following system of inequalities graphically: x + y ≤ 9, y > x and x ≥ 0.

Answer: Draw the line x + y = 9 and shade the region below it, including the line (sign is ≤).
Draw the line y = x and shade the region above it, excluding the line (sign is strictly >).
Draw the line x = 0 (the y-axis) and shade the region to its right, including the line (sign is ≥).
The triple shaded region common to all three is the solution region. It comes out as a triangular region.
Q6. What are the inequalities represented by a graph in which the shaded region lies between the lines x = 2, x = −2, y = 2 and y = −2, with the boundary lines included?
(a) x ≥ 2; x ≤ −2; y ≥ 2; y ≤ −2
(b) x > 2; x < −2; y > 2; y < −2
(c) x < 2; x > −2; y < 2; y > −2
(d) x ≤ 2; x ≥ −2; y ≤ 2; y ≥ −2
Answer: (d) x ≤ 2; x ≥ −2; y ≤ 2; y ≥ −2
The region lies between the lines, so x must be less than or equal to 2 and at the same time greater than or equal to −2, and similarly for y. Since the boundary lines are included, we use ≤ and ≥ rather than the strict signs.
Q7. Solve the following system of inequalities graphically: x ≥ 3 and y ≥ 2.

Answer: Draw the vertical line x = 3 and shade everything to its right, including the line.
Draw the horizontal line y = 2 and shade everything above it, including the line.
The solution is the region common to both, which is the rectangular corner region in the first quadrant starting from the point (3, 2) and extending upwards and to the right.
Q8. If x² + 6x − 27 > 0 and x² − 3x − 4 < 0, then
(a) x > 3
(b) x < 4
(c) 3 < x < 4
(d) x = 3
Answer: (c) 3 < x < 4
First: x² + 6x − 27 > 0
(x + 9)(x − 3) > 0, which gives x < −9 or x > 3
Second: x² − 3x − 4 < 0
(x − 4)(x + 1) < 0, which gives −1 < x < 4
Taking the common part of both: 3 < x < 4
Q9. Solve the inequality y < −2 graphically in a two-dimensional plane.

Answer: First draw the line y = −2. This line divides the plane into two half planes.
Take a test point that is not on the line, say the origin (0, 0).
Put it in the inequality: 0 < −2, which is false.
So the origin is not in the solution region, and the solution is the other half plane, the one below the line.
Since the inequality is strict (<), the points on the line itself are not included, so the line is drawn as a dotted line.
Q10. The solutions of 2x − 3 ≤ −13 when x is an integer are
(a) ... −7, −6, −5
(b) −4, −3, −2, −1
(c) 2, 4, 6, ...
(d) 5, 6, 7, 8, ...
Answer: (a) ... −7, −6, −5
2x − 3 ≤ −13
2x ≤ −10
x ≤ −5
Since x is an integer, the solutions are all integers less than or equal to −5, that is ..., −7, −6, −5.
Q11. Solve 5x − 3 < 7 when (i) x is a natural number and (ii) x is an integer.
Answer: First solve the inequality:
5x − 3 < 7
5x < 10
x < 2
(i) When x is a natural number: The natural numbers less than 2 are only 1.
Solution set = {1}
(ii) When x is an integer: All integers less than 2 qualify.
Solution set = {..., −3, −2, −1, 0, 1}
Q12. Ravi obtained 70 and 75 marks in the first two unit tests. Find the minimum marks he should get in the third test to have an average of at least 60 marks.
Answer: Let the marks in the third test be x.
The average of the three tests must be at least 60, so:
(70 + 75 + x) / 3 ≥ 60
145 + x ≥ 180
x ≥ 35
So Ravi must score at least 35 marks in the third test.
Related Study Material on Extramarks
- CBSE Important Questions Class 11 Maths Chapter 1
- CBSE Important Questions Class 11 Maths Chapter 2
- CBSE Important Questions Class 11 Maths Chapter 3
- CBSE Important Questions Class 11 Maths Chapter 4
- CBSE Important Questions Class 11 Maths Chapter 6
- CBSE Important Questions Class 11 Maths Chapter 7
- CBSE Important Questions Class 11 Maths Chapter 8
- CBSE Important Questions Class 11 Maths Chapter 9
- CBSE Important Questions Class 11 Maths Chapter 10
- CBSE Important Questions Class 11 Maths Chapter 11
- CBSE Important Questions Class 11 Maths Chapter 12
- CBSE Important Questions Class 11 Maths Chapter 13
- CBSE Important Questions Class 11 Maths Chapter 14
Q.1 If the complex numbers z1, z2 and z3 represent the vertices of an equilateral triangle and |z1| = |z2| = |z3|, then
Marks:1
Ans
z1+z2+z3 = 0.
Q.2![]()
Marks:1
Ans

Q.3
If z1=2+i,z2=23i and z3=4+5i, evaluate (i)?Re?(z1?z2z3) (ii)?Im?(z1?z2z3)
Marks:4
Ans
z1‹…z2 z3=(2+i) (23i ) 4+5i =(2+i) (2+3i) (45i)16+25 =4441+2741iRez1‹…z2 z3 =4441iImz1‹…z2 z3 =2741
Q.4
Find the modulus and the argument of the following: (i)?(2?i)2 (ii)1+i1?i
Marks:4
Ans
i2i2=414i=34i4i|=9+16=5modulus = 5Since (3,4i) lies in third quadrant, therefore its argument is given by¸
=+tan143ii1+i1i =1+i1i—1+i1+i=11+2i1+1 =i=0+0+i =1=1„modulus =1and argument is given by=tan110
Q.5
Convert the complex number ?3+i in the polar form and represent it in Argand Plane.
Marks:6
Ans
Let z=3+i=r(cos¸+isin¸)Comparing real part and imaginary part, we getrcos¸=3 and rsin¸=1On squaring and adding, we getr2cos2¸+sin2¸=32+12r2=4r=2cos¸=32 and sin¸=12The value of ¸ satistying both the equation is by, ¸=56Hence Z=2cos56+isin56