KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Additional Questions
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Additional Questions
Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Additional Questions
I. Multiple Choice Questions:
Question 1.
The degree of polynomial is x + 2
a. 2
b. 1
c. 3
d. 4
Answer:
b. 1
Question 2.
The degree of a quadratic polynomial is
a. 0
b. 1
c. 2
d. 3
Answer:
c. 2
Question 3.
The degree polynomial of a bi-quadratic
a. 0
b. 2
c. 4
d. 1
Answer:
c. 4
Question 4.
The standard polynomial. form of a linear
a. ax + c
b. ax2 + c
c. ax2 + bx + c = 0
d. ax3 + bx2 + cx + d = 0
Answer:
a. ax + c
Question 5.
The standard form of a quadratic equation.
a. ax2 + bx + c = 0
b. ax + c = 0
c. ax2 + c = 0
d. ax3 + bx + c = 0
Answer:
a. ax2 + bx + c = 0
Question 6.
A polynomial of degree 3 is called
a. a linear polynomial
b. a quadratic polynomial
c. a cubic polynomial
d. a biquadratic polynomial
Answer:
c. a cubic polynomial
Question 7.
The value of px. = x2 – 3x – 4, x = -1
a. 1
b. – 4
c. 0
d. – 3
Answer:
c. 0
Question 8.
The degree of the polynomial x4 + x3 is
a. 2
b. 3
c. 5
d. 4
Answer:
d. 4
Question 9.
The number of zeroes of linear polynomial at most is
a. 0
b. 1
c. 2
d. 3
Answer:
b. 1
Question 10.
The degree of polynomial x + 2. x + 1. is
a. 1
b. 3
c. 4
d. 2
Answer:
d. 2
Question 11.
The degree of a zero polynomial is
a. not defined
b. 1
c. 2
d. 3
Answer:
a. not defined
Question 12.
The zeroes of the polynomial x3 -4x are
a. 0, ± 2
b. 0, ± 1
c. 0, ± 3
d. 0, 0
Answer:
a. 0, ± 2
Question 13.
The zeroes of the polynomials, t2 – 15 are.
a. ± √15
b. ± √5
c. ± √3
d. ± 3
Answer:
a. ± √15
Question 14.
Dividend is equal to.
a. divisor × quotient + remainder
b. divisor × quotient
c. divisor × quotient – remainder
d. divisor × quotient × remainder
Answer:
a. divisor × quotient + remainder
Question 15.
If the divisor is x2 and quotient is x while the number remainder is 1, then the dividend is.
a. x2
b. x
c. x3
d. x3 + 1
Answer:
d. x3 + 1
Question 16.
What is the co-efficient of the first term of the quotient when 2x2 + 2x + 1 is divided by x + 2?
a. 1
b. 2
c. 3
d. – 2
Answer:
b. 2
Question 17.
The zeroes of the polynomial x2 – 3x – 4 are
a. 4, – 1
b. 4, 1
c. – 4, 1
d.- 4, – 1
Answer:
a. 4, – 1
Question 18.
If f(x) = x2 – 1 the value of f(2) is
a. 1
b. 3
c. 4
d. 0
Answer:
b. 3
Question 19.
If f(x) = 8 f(x) is called
a. Constant polynomials
b. linear polynomials
c. quadratic polynomials
d. Cubic polynomials
Answer:
a. Constant polynomials
Question 20.
A cubic polynomial has at most
a. 1 zeroes
b. 2 zeroes
c. 3 zeroes
d. 4 zeroes.
Answer:
c. 3 zeroes
II. Short Answer Questions:
Question 1.
Find the value of P(x) = x2 + 2x – 5 at x = 1
Answer:
P(x) = x2 + 2x – 5
P (1) = (1)2 + 2(1) – 5 = 1 + 2 – 5
= 3 – 5
P(1) = – 2
Question 2.
If x = 1 is a zero of the Polynomial f(x) = x3 – 2x2 + 4x + K, write the value of K.
Answer:
f(x) = x3 – 2x2 + 4x + K x = 1
f(1) = (1)3 – 2(1)2 + 4(1) + K
f(1) = 0 zero of the polynomial
0 = 1 – 2 + 4 + K
0 = 3 + K
K = – 3
Question 5.
If 1 is a zero of the Polynomial P(x) = ax2 – 3(a – 1) x – 1, then find the value of a.
Answer:
P(x) = ax2 – 3(a – 1) x – 1
∴ x = 1
P(1) = a(1)2 – 3(a – 1) 1 – 1
P(1) = a – 3a + 3 – 1
0 = – 2a + 2
2a = 2
a = 2/2 = 1
a = 1
Question 6.
Write the degree of the Polynomial 3x3 – x4 + 5x + 3
Answer:
3x3 – x4 + 5x + 3
– x4 + 3x3 + 5x + 3
∴ degree of the Polynomial is 4
Question 7.
Write the standard form of a cubic Polynomial.
Answer:
ax3 + bx3 + cx + d.
III. Long Answer Questions:
Question 1.
Find the zeroes of the following quadratic polynomials x2 + 4x + 4.
Answer:
f(x) = x2 + 4x + 4
0 = x2 + 2x + 2x + 4
0 = x(x + 2) + 2(x + 2)
(x + 2) (or) (x + 2) = 0
x + 2 = 0 (or) x + 2 = 0
x = – 2 x = – 2.
Zeroes of the Polynomials x2 + 4x + 4 are – 2 and – 2
Question 2.
If x = 1 is a zero of the polynomial f(x) = x3 – 2x2 + 4x + K, find the value of K.
Answer:
f(x) = x3 – 2x2 + 4x + K
∴ x = 1
f(x) = (1)3 – 2(1)2 + 4(1) + K
0 = 1 – 2(1) + 4 + K [f(1) = 0]
0 = 1 – 2 + 4 + K.
K + 3 = 0
K = – 3
Question 3.
For what value of K, – 4 is a zero of the polynomial x2 – x – (2K + 2)?
Answer:
f(x) = x2 – x – (2K + 2) and f(x) = 0
x = – 4
f(- 4) = (- 4)2 (- 4) – (2K + 2)
0 = 16 + 4 – 2K – 2
20 – 2 – 2K = 0 ⇒ 18 = 2K
K = 18/2 ⇒ K = 9
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