Introduction to Linear Polynomials Class 9 Extra Questions Maths Chapter 2

By using Extra Questions for Class 9 Maths and Ganita Manjari Class 9 Maths Chapter 2 Introduction to Linear Polynomials Extra Questions, students can improve their problem-solving skills.

Class 9 Introduction to Linear Polynomials Extra Questions

Extra Questions on Introduction to Linear Polynomials Class 9

Class 9 Ganita Manjari Chapter 2 Extra Questions

Question 1.
What will be the coefficient of y² in the polynomial 5y + 3y² – y + y² + 5 ?
Solution:
5y³ + 3y² – y + y² + 5 = 5y³ + 3y² + y² – y + 5
= 5y + 4y² – y + 5
Term containing y² = 4y²
∴ Coefficient of y² = 4

Question 2.
What is the degree of the polynomial
p(x) = 2x + \(\frac{3}{2}\)x³ – 7?
Solution:
Term with highest power of x = \(\frac{3}{2}\)x³
Exponent of x in this term = 3
∴ Degree of the polynomial = 3.

Question 3.
Write the degree of the polynomial
(x³ + 5) (4 – x ).
Solution:
(x³ + 5) (4 – x⁵)
= x³(4 – x⁵) + 5(4 – x⁵)
= 4x³ – x⁸ + 20 – 5x⁵
Term with highest power of x = – x⁸;
Exponent of x in this term = 8
Degree of the polynomial = 8.

Question 4.
If P(x) = 5x – 4x² + 3, find P(- 1)
Solution:
P(x) = 5x – 4x² + 3
P(- 1) = 5(- 1) – 4(-1)² + 3
= – 5 – 4 + 3
= -9 + 3
= -6

Introduction to Linear Polynomials Class 9 Very Short Question Answer

Question 5.
Find the value of f(x) = 3x² – 5x + 3 at x = – 2.
Solution:
f(x) = 3x² – 5x + 3
⇒ f(-2) = 3(-2)² – 5(-2) + 3
= 12 + 10 + 3
= 25

Question 6.
If P(x) = x³ + x + x + 1, find f(- 1) + p(1).
Solution:
P(x) = x³ + x + x + 1
∴ P(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 – 1 + 1
= 0
and P(1) = (1)³ + (1)² + (1) + 1
= 1 + 1 + 1 + 1
= 4
∴ p(-1) + p(1) = 0 + 4
= 4

Question 7.
Find the value of p\(\left(-\frac{2}{3}\right)\) for
p(y) = 2y³ – y² – 13y – 6.
Solution:
p(y) = 2y³ – y² – 13y – 6
⇒ p\(\left(-\frac{2}{3}\right)\) = 2\(\left(-\frac{2}{3}\right)^3\) – \(\left(\frac{-2}{3}\right)^2\) – 13\(\left(\frac{-2}{3}\right)\) – 6
= \(-\frac{16}{27}-\frac{4}{9}+\frac{26}{3}\) – 6
= \(\frac{-16-12+234-162}{27}=\frac{44}{27}\)

Question 8.
If p(x) = x² – 4x + 3, evaluate p(2) – p(- 1) + p(\(\frac{1}{2}\))
Solution:
p(x) = x² – 4x + 3
∴ p(2) = (2)² – 4(2) + 3
= 4 – 8 + 3
= – 1

p(- 1) = (- 1)² – 4(- 1) + 3
= 1 + 4 + 3
= 8

and P(\(\frac{1}{2}\)) = (\(\frac{1}{2}\))² – 4((\(\frac{1}{2}\))) + 3
= \(\frac{1}{4}\) – 2 + 4
= \(\frac{5}{4}\)
∴ p(2) – p(-1) + p(\(\frac{1}{2}\))
= -1 – 8 + \(\frac{5}{4}\)
= –\(\frac{31}{4}\)

Question 9.
If x = – 2 is the root of the equation √2 (x + p) = 0 and is also the zero of the polynomial px² + kx + 2√2, then find the value of k.
Solution:
If x = – 2 is the root of the equation √2 (x + p) = 0, then,
√2 (— 2 + p)== 0 ⇒ -2 + p = 0 ⇒ p = 2
Let f(x) = px² + kx + 2√2
Then, f(x) = 2x² + kx + 2√2
If x = – 2 is a zero of/(x), then f(- 2) = 0
⇒ 2(- 2)² + k(- 2) + 2√2 = 0
⇒ 8 – 2k + 2√2 = 0
⇒ 2k = 8 + 2√2
⇒ k = 4 + V2 .

Question 10.
Check whether the polynomial p(y) = 4y³ + 4y² – y – 1 is a multiple of (2y + 1).
Solution:
p(y) = 4y³ + 4y² – y – 1
2y + 1 = 0 ⇒ y = –\(\frac{1}{2}\)
Now,
p(-\(\frac{1}{2}\)) = 4(-\(\frac{1}{2}\))³ + 4(-\(\frac{1}{2}\))² – (-\(\frac{1}{2}\)) – 1
= –\(\frac{1}{2}\) + 1 + \(\frac{1}{2}\) – 1 = 0
∴ By factor theorem, (2y + 1) is a factor of p(y), i.e., p(y) is a multiple of (2y + 1).

Introduction to Linear Polynomials Class 9 Short Question Answer

Question 1.
If f(x) = 5x² – 4x + 5, find f(1) + f(- 1) + f(0).
Solution:
f(x) = 5x² – 4x + 5
∴ f(1) =5(1)² – 4(1) + 5 = 5 – 4 + 5 = 6
f(-1) = 5(-1) – 4(-1) + 5 = 5 + 4 + 5 = 14
f(0) = 5(0)2 – 4(0) + 5 = 0 – 0 + 5 = 5
∴ f(1) + f(-1) + f(0) = 6 + 14 + 5 = 25

Question 2
Find the value of the polynomial x² – 9 for x = 97.
Solution:
x² – 9 = x² – 3² = (x – 3)(x + 3)
Using Identity III
= (97 – 3)(97 + 3)
= (94)(100)
= 9400

Question 3.
If x + \(\frac{1}{x}\) = 4, find the value of x³ + \(\frac{1}{x^3}\).
Solution:
We know that
(x + \(\frac{1}{x}\))³ = x³ + \(\frac{1}{x^3}\) + 3(x)(\(\frac{1}{x}\))(x + \(\frac{1}{x}\))
⇒ (x + \(\frac{1}{x}\))³ = x³ + \(\frac{1}{x^3}\) + 3(x + \(\frac{1}{x}\))
⇒ (4)³ = x³ + \(\frac{1}{x^3}\) + 3(4)
⇒ 64 = x³ + \(\frac{1}{x^3}\) + 12
⇒ x³ + \(\frac{1}{x^3}\) = 64 – 12
⇒ x³ + \(\frac{1}{x^3}\) = 52

Introduction to Linear Polynomials Class 9 Case Based Questions

Question 1.
A= x + y, B = x – y and C = 2(y + z). Based on the above information, answer the following questions :
(i) A² + B² =
(a) (x² + y²)
(b) 2 (x² + y²)
(c) 2 (x² – y²)
(d) 3 (x² – y²)
Solution:
(b) 2 (x² + y²)

(ii) (A + B)³ – (A – B)³ =
(a) 8 (x³ – y³)
(b) 4 (x – y) (x² + xy + y²)
(c) 8 (x – y) (x² + xy + y²)
(d) 4 (x³ – y³)
Solution:
(a) 8 (x³ – y³)

(iii) (A + B)² + (A – B² =
(a) 4(x² + y²)
(b) 4(x – y)(x + y)
(c) 4(x + y)²
(d) 2(x² + y²)
Solution:
(a) 4(x² + y²)

(iv) (A + B + C)² =
(a) 8(x+y + z)²
(b) 4(x – y – z)
(c) 8(x-y-z)²
(d) 4(x + y + z)²
Solution:
(d) 4(x + y + z)²

(v) A² – B² =
(a) 4(x² – y²)
(b) -4xy
(c) (x – y)²
(d) 4xy
Solution:
(d) 4xy

Introduction to Linear Polynomials Extra Questions for Practice

Very Short Answer Type Questions

Question 1.
What is the degree of zero polynomial?
Solution:
not defined

Question 2.
What is the degree of polynomial q’?
Solution:
zero

Question 3.
What is the degree of the polynomial x² – x + x³ + 1?
Solution:
3

Question 4.
Write a binomial of degree 50.
Solution:
3x⁵⁰ – 4

Question 5.
Write the coefficient of y in the expansion of (5 – y)².
Solution:
-10

Question 6.
If p(x) = x² – 5x + 1, find p(2), p(- 1) and hence find p(2) + p(- 1).
Solution:
1

Question 7.
If f(x) = x² – 5x + 7, evaluate f(2) – f(-1) + f(\(\frac{1}{3}\))
Solution:
\(\frac{-59}{9}\)

Question 8.
Find the value of the polynomial
p(z) = 3z² – 4z + \(\sqrt{17}\) when z = 3.
Solution:
15 + \(\sqrt{17}\)

Question 9.
Find the value of the polynomial 5x – 4x² + 3 at
(i) x = 0 (ii) x = 2.
Solution:
(i) 3

Question 10.
Write the coefficient of x² in 2 – x² + x.
Solution:
-1

Short Answer Type Questions

Question 1.
Find the degree of the polynomial:
(i) 3y³ + y² – y⁴ – 7
(ii) 5 – x³
Solution:
(i) 4
(ii) 3

Question 2.
Identify the number of terms in the following polynomial:
(i) y³ + 7y² – 7
(ii) 6y² – y⁴ + 3.
Solution:
(i) Three
(ii) Three

Question 3.
Write a constant polynomial:
Solution:
5

Question 4.
Write a polynomial in z of degree 3.
Solution:
4z³ + 7z² – 2

Question 5.
What the numbers 4 and 7 in the expression y = 7x + 4 represent?
Solution:
7 represents slope and 4 represents y – intercept

Question 6.
Write the coefficient of y⁴ in the polynomial
3y³ + y² – y⁴ – 77
Solution:
-1

Question 7.
Determine the value of 4m² – 3m⁴ + 1 at (i) m = 0 (ii) m = -1
Solution:
(i) 1
(ii) 2

Question 8.
The length of each side of a square is 5 cm. Determine its (i) perimeter (ii) area.
Solution:
(i) 20 cm
(ii) 25 sq cm

Question 9.
The sum of the digits of a 2-digit number is 7. If its digits are in the ratio 2:5, then find the 2-digit number.
Solution:
25

Question 10.
Using the linear expression 2p – 1, write the number at the 15^th stage. At what stage will 47 occurs?
Solution:
29; 24^th stage