Mathematics solution NCERT
Class 9 - Chapter 2: Introduction to Linear Polynomials
Q1: Write a polynomial of degree 3 in the variable x, in which the coefficient of the x² term is −7.
Solution:
A polynomial of degree 3 must have the highest power of x equal to 3.
The coefficient of x² must be −7.
One such polynomial is:
p(x) = x³ − 7x² + 4x + 2
Here,
- Degree = 3
- Coefficient of x² = −7
Answer: x³ − 7x² + 4x + 2
Q2: Find the values of the following polynomials at the indicated values of the variables.
(i) Find the value of 5x² − 3x + 7 if x = 1
Substituting x = 1:
= 5(1)² − 3(1) + 7
= 5 − 3 + 7
= 9
Answer: 9
(ii) Find the value of 4t³ − t² + 6 if t = a
Substituting t = a:
= 4(a)³ − (a)² + 6
= 4a³ − a² + 6
Answer: 4a³ − a² + 6
Q3: If we multiply a number by 5/2 and add 2/3 to the product, we get −7/12. Find the number.
Solution:
Let the required number be x.
According to the question:
(5/2)x + 2/3 = −7/12
Subtract 2/3 from both sides:
(5/2)x = −7/12 − 2/3
(5/2)x = −7/12 − 8/12
(5/2)x = −15/12
(5/2)x = −5/4
Multiply both sides by 2/5:
x = (−5/4) × (2/5)
x = −1/2
Answer: The number is −1/2.
Q4: A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers?
Solution:
Let the smaller number be x.
Then the larger number is 5x.
After adding 21:
New numbers become:
x + 21 and 5x + 21
According to the question:
5x + 21 = 2(x + 21)
5x + 21 = 2x + 42
5x − 2x = 42 − 21
3x = 21
x = 7
Larger number = 5 × 7 = 35
Answer:
- First number = 7
- Second number = 35
Verification:
7 + 21 = 28
35 + 21 = 56
56 = 2 × 28 ✔
Q5: If you have ₹800 and you save ₹250 every month, find the amount you have after (i) 6 months (ii) 2 years. Express this as a linear pattern.
Solution:
Initial amount = ₹800
Monthly saving = ₹250
The amount increases by ₹250 every month.
(i) Amount after 6 months
Amount = 800 + (250 × 6)
= 800 + 1500
= ₹2300
Answer: ₹2300
(ii) Amount after 2 years
2 years = 24 months
Amount = 800 + (250 × 24)
= 800 + 6000
= ₹6800
Answer: ₹6800
Linear Pattern
| Months (m) | Amount (₹) |
|---|---|
| 0 | 800 |
| 1 | 1050 |
| 2 | 1300 |
| 3 | 1550 |
| 4 | 1800 |
| 5 | 2050 |
| 6 | 2300 |
If m represents the number of months, then
A = 800 + 250m
Linear Pattern: A = 800 + 250m
Q6: The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. Find both the numbers.
Solution:
Let the tens digit be x.
Then the units digit is x − 3.
Original number = 10x + (x − 3)
= 11x − 3
Interchanged number = 10(x − 3) + x
= 11x − 30
According to the question:
(11x − 3) + (11x − 30) = 143
22x − 33 = 143
22x = 176
x = 8
Units digit = 8 − 3 = 5
Original number = 85
Interchanged number = 58
Verification:
85 + 58 = 143 ✔
Answer: The two numbers are 85 and 58.
Q7: Draw the graph of the following equations, and identify their slopes and y-intercepts. Also, find the coordinates of the points where these lines cut the y-axis.
Given Equations:
- (i) y = -3x + 4
- (ii) 2y = 4x + 7
- (iii) 5y = 6x - 10
- (iv) 3y = 6x - 11
First convert each equation into slope-intercept form y = mx + c.
| Equation | Slope (m) | Y-Intercept (c) | Point on Y-Axis |
|---|---|---|---|
| y = -3x + 4 | -3 | 4 | (0,4) |
| 2y = 4x + 7 y = 2x + 3.5 |
2 | 3.5 | (0,3.5) |
| 5y = 6x - 10 y = 1.2x - 2 |
1.2 | -2 | (0,-2) |
| 3y = 6x - 11 y = 2x - 11/3 |
2 | -11/3 | (0,-11/3) |
Graph
Are any lines parallel?
Yes. The equations y = 2x + 3.5 and y = 2x - 11/3 have the same slope (m = 2), so they are parallel.
Q8: If the temperature of a liquid can be measured in Kelvin units as x K and in Fahrenheit units as y °F, the relation between the two systems is given by
y = (9/5)(x - 273) + 32
(i) Find the temperature in Fahrenheit if the temperature is 313 K.
Substitute x = 313:
y = (9/5)(313 - 273) + 32
y = (9/5)(40) + 32
y = 72 + 32
y = 104°F
Answer: 104°F
(ii) If the temperature is 158°F, find the temperature in Kelvin.
158 = (9/5)(x - 273) + 32
126 = (9/5)(x - 273)
70 = x - 273
x = 343
Answer: 343 K
Q9: The work done by a body on the application of a constant force is the product of the constant force and the distance travelled by the body in the direction of the force.
Given:
Work Done = Force × Distance
Let:
- w = work done
- d = distance travelled
- Force = 3 units
Therefore,
w = 3d
This is a linear equation in two variables.
Table of Values
| Distance (d) | Work (w) |
|---|---|
| 0 | 0 |
| 1 | 3 |
| 2 | 6 |
| 3 | 9 |
| 4 | 12 |
| 5 | 15 |
Graph of w = 3d
Work done when distance = 2 units:
w = 3 × 2 = 6
Answer: 6 units
On the graph, the point (2,6) lies exactly on the line, verifying the answer.
Q10: The graph of a linear polynomial p(x) passes through the points (1,5) and (3,11).
(i) Find the polynomial p(x).
Let the polynomial be:
p(x) = mx + c
Since the graph passes through (1,5) and (3,11), first find the slope.
m = (11 - 5)/(3 - 1)
m = 6/2
m = 3
Substitute m = 3 and point (1,5) into y = mx + c:
5 = 3(1) + c
5 = 3 + c
c = 2
Therefore,
p(x) = 3x + 2
Answer: p(x) = 3x + 2
(ii) Find the coordinates where the graph cuts the axes.
Y-axis:
Put x = 0
p(0) = 3(0) + 2 = 2
Therefore, the graph cuts the y-axis at:
(0,2)
X-axis:
Put y = 0
3x + 2 = 0
3x = -2
x = -2/3
Therefore, the graph cuts the x-axis at:
(-2/3, 0)
Answer:
- X-intercept = (-2/3, 0)
- Y-intercept = (0, 2)
(iii) Graph of p(x) = 3x + 2
The graph passes through (1,5) and (3,11) and cuts the axes at (-2/3,0) and (0,2), verifying our answers.
Q11: Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
- p(0) = 5
- p(x) − q(x) cuts the x-axis at (3,0)
- p(x) + q(x) = 6x + 4 for all real x
Solution:
Given:
p(x) = ax + b
q(x) = cx + d
Step 1: Use p(0) = 5
a(0) + b = 5
b = 5
Therefore,
p(x) = ax + 5
Step 2: Use p(x) + q(x) = 6x + 4
(ax + 5) + (cx + d) = 6x + 4
(a + c)x + (5 + d) = 6x + 4
Comparing coefficients:
a + c = 6 .......... (1)
5 + d = 4
d = -1
Step 3: Use p(x) − q(x) cuts the x-axis at (3,0)
(ax + 5) - (cx - 1)
(a - c)x + 6
Since (3,0) lies on this graph:
(a - c)(3) + 6 = 0
3(a - c) = -6
a - c = -2 .......... (2)
Now solve equations (1) and (2):
a + c = 6
a - c = -2
Add both equations:
2a = 4
a = 2
c = 4
Substitute values:
p(x) = 2x + 5
q(x) = 4x - 1
Verification:
p(x)+q(x)
=(2x+5)+(4x-1)
=6x+4 ✔
p(x)-q(x)
=(2x+5)-(4x-1)
=-2x+6
At x = 3,
-2(3)+6=0 ✔
Answer:
p(x) = 2x + 5
q(x) = 4x - 1
Q13: Let p(x) = ax + b and q(x) = cx + d be two linear polynomials such that:
- The graph of p(x) passes through the points (2,3) and (6,11).
- The graph of q(x) passes through the point (4,-1).
- The graph of q(x) is parallel to the graph of p(x).
Solution:
Step 1: Find p(x)
The graph of p(x) passes through the points (2,3) and (6,11).
First find the slope:
m = (11 - 3)/(6 - 2)
m = 8/4
m = 2
Therefore,
p(x) = 2x + b
Using point (2,3):
3 = 2(2) + b
3 = 4 + b
b = -1
Hence,
p(x) = 2x - 1
Step 2: Find q(x)
Since q(x) is parallel to p(x), both lines have the same slope.
Therefore,
q(x) = 2x + d
Given that q(x) passes through (4,-1).
-1 = 2(4) + d
-1 = 8 + d
d = -9
Hence,
q(x) = 2x - 9
Step 3: Find where the lines meet the x-axis.
For p(x):
2x - 1 = 0
2x = 1
x = 1/2
Therefore, p(x) cuts the x-axis at
(1/2, 0)
For q(x):
2x - 9 = 0
2x = 9
x = 9/2
Therefore, q(x) cuts the x-axis at
(9/2, 0)
Answer:
| Polynomial | Equation | X-Intercept |
|---|---|---|
| p(x) | 2x - 1 | (1/2, 0) |
| q(x) | 2x - 9 | (9/2, 0) |
Graph of p(x) and q(x)
The graph shows that both lines are parallel because they have the same slope (2).
Q14: What do all linear functions of the form f(x) = ax + a, a > 0, have in common?
Solution:
Consider the family of linear functions:
f(x) = ax + a
where a > 0.
We can factor out a:
f(x) = a(x + 1)
To find the x-intercept, put f(x) = 0.
a(x + 1) = 0
Since a > 0, a cannot be zero.
Therefore,
x + 1 = 0
x = -1
Thus, every graph in this family cuts the x-axis at the same point:
(-1, 0)
Examples:
| Function | X-Intercept |
|---|---|
| f(x)=x+1 | (-1,0) |
| f(x)=2x+2 | (-1,0) |
| f(x)=5x+5 | (-1,0) |
| f(x)=10x+10 | (-1,0) |
Common Property:
All linear functions of the form f(x)=ax+a pass through the fixed point (-1,0).
They all have positive slopes because a > 0, so every graph rises from left to right.