In the current scenerio Practicing NCERT Maths MCQ for Class 10 CBSE with Answers Pdf is one of the best ways to prepare for the CBSE Class 10 board exam. There is no substitute for consistent practice whether one wants to understand a concept thoroughly or one wants to score better. The given questions will help to understand the chapter at minute levels. Also the new pattern of CBSE Board will more emphasis on the Multiple choice questions.So there is need also. We will came with more and more such types of problems.
Class 10-Polynomials MCQs Multiple Choice Questions with Answers
To excel in your school examinations your subject topics should be crystal clear. For this you have to practice NCERT questions thoroughly. Practicing questions will make you delve deeper into your chapter basics & help you solve advanced level of questions easily. These Assignments are specially designed for NCERT questions & therefore help students to solve questions & understand concepts more easily. Studymate provides you best Assignments designed by faculty experts with detailed coverage to NCERT textbooks as per the latest CBSE pattern.
Class 10-Polynomials MCQ Questions with Answers
Question 1.
If a polynomial p(y) is divided by y + 2, then which of the following can be the remainder:
(a)y + 1
(b)2y + 3
(c) 5
(d)y – 1
Answer
Answer: (c) 5
When p(y) is divided by y + 2, then the degree of remainder < deg of (y + 2)
Question 2.
If a polynomial p(x) is divided by b – ax; the remainder is the value of p(x) at x =
(a) a
(b) \(\frac{b}{a}\)
(c) \(\frac{- b}{a}\)
(d) \(\frac{a}{b}\)
Answer
Answer: (b) \(\frac{b}{a}\)
b – ax = 0
x = \(\frac{b}{a}\)
Question 3.
If the polynomials ax³ + 4x² + 3x – 4 and x³ – 4x + a, leave the same remainder when divided by (x – 3), then value of a is :
(a) 2b
(b) – 1
(c) 1
(d) – 2b
Answer
Answer: (b) – 1
p(x) = ax³ + 4x² + 3x – 4
q(x) = x³ – 4x + a
p(3) = q(3)
a = – 1
Question 4.
If p(x) = 2x⁴ – ax³ + 4x² + 2x + 1 is a. multiple of 1 – 2x, then find the value of a :
(a) 25
(b) \(\frac{1}{2}\)
(c)\(\frac{- 1}{2}\)
(d) 8
Answer
Answer: (a) 25
p(x) is a multiple of 1 – 2x.
1 – 2x is a factor of p(x)
Question 5.
If -2 is a zero of p(x) = (ax³ + bx² + x – 6) and p(x) leaves a remainder 4 when divided by (x – 2), then the values of a and b are (respectively):
(a)a = 2,b = 2
(b) a = 0,b = – 2
(c) a = 0, b = 2
(d) a = 0, b = 0
Answer
Answer: (c) a = 0, b = 2
If – 2 is a zero =>
p(- 2) = 0
=> – 2a + b = 2
Also, p(2) = 4
2a + b = 2=>a = 0and b = 2
Question 6.
If x101 + 1001 is divided by x + 1, then remainder is:
(a) 0
(b) 1
(c) 1490
(d) 1000
Answer
Answer: (d) 1000
p(x) is divided by x + 1
p(- 1) = (-1101) + 1001 = 1000
Question 7.
If one zero of a polynomial p(x) = ax² + bx + c(a ≠ 0) is zero, then, which of the following is correct:
(a) b = 0
(b) c = 0
(c) other zero is also zero
(d) Nothing can be said about p(x).
Answer
Answer: (b) c = 0
let ,α = 0
Product of the roots = αs = 0
= \(\frac{c}{a}\) = 0
Question 8.
If α, s are the zeroes of x² – lx + m, then
\(\frac{α}{s}\) + \(\frac{s}{α}\)
(a) \(\frac{l² – 2m}{m}\)
(b) \(\frac{l² + 2m}{m}\)
(c) \(\frac{l – 2m}{m}\)
(d) l² – 2m
Answer
Answer: (a) \(\frac{l² – 2m}{m}\)
α + s = l
αs = m
Question 9.
sum of the squares of the zeroes of the polynomial p(x) = x² + 7x – k is 25, find k.
(a) 12
(b) 49
(c) – 24
(d) – 12
Answer
Answer: (d) – 12
p(x) = x² + 7x – k
let α,s be the zeroes
α + s = – 7
αs = – k
α² + s² = 25
(α² + s) – 2αs = 25
49 + 2k = 25
k = -12
Question 10.
If one zero of 3x² – 8x + 2k + 1 is seven times the other, find k.
(a) \(\frac{2}{3}\)
(b) \(\frac{1}{3}\)
(c) \(\frac{4}{3}\)
(d) \(\frac{5}{3}\)
Answer
Answer: (a) \(\frac{2}{3}\)
α + 7α = 8α = \(\frac{8}{3}\)
α = \(\frac{1}{3}\)
k = \(\frac{2}{3}\)
Question 11.
Let, α, s, v be the zeroes of x³ + 4x² + x- 6 such that product of two of the zeroes is 6. Find the third zero.
(a) 6
(b) 2
(c) 4
(d) 1
Answer
Answer: (a) 6
α s v = 6,
αs = 61
=> v = 6
Question 12.
If a, s are the zeroes of x² – 8x + λ, such
that α – s = 2, then X =
(a) 8
(b) 22
(c) 60
(d) 15
Answer
Answer: (d) 15
α + s = 8,
αs = λ
α – s = 2
=> (α – s)2 = 4
=> α²+s²-2αs = 4
=> (α + s)² – 4as = 4
=> 64 — 4λ = 4
=> 4λ. = 60
=> X = 15
Question 13.
Find a and b so that the polynomial 6x⁴ + 8x³ – 5x² + ax + b is exactly divisible by 2x² – 5.
(a) a = 20, b = – 25
(b) a = 4, b = – 5
(c) a = 20, b = 5
(d) a = – 20, b = – 25
Answer
Answer: (d) a = – 20, b = – 25
Divide the given polynomial by 2×2 – 5 get the remainder as (20 + a)x + (b + 25) which should be zero
Question 14.
If one of the zeroes of a quadratic polynomial of the form x² + ax + b is the negative of the other, then it
(a) has no linear term and the constant term is negative.
(b) has no linear term and the constant term is positive.
(c) can have a linear term but the constant term is negative.
(d) can have a linear term but the constant term is positive.
Answer/Explanation
Answer: a
f(x) = x² + ax + b
Given: zeroes are α and – α
Sum of zeroes = α – α = 0
∴ f(x) = x² + b, which is not linear
Product of zeroes = α(-α) = – α² = \(\frac{b}{1}\)
⇒ -α² = b
It is possible when, b < 0.
Hence, it has no linear term and the constant term is negative.
Question 15.
If sum of the two zeroes of a cubic polynomial x³ – ax² + bx – c, is zero, then which of the following is true:
(a) ab = c
(b) a – b = c
(c) ab = \(\frac{c}{2}\)
(d) a = \(\frac{b}{c}\)
Answer
Answer: (a) ab = c
Let, α, s, v be the roots =α + s + v = a
v = a
now v is a zero
ab = c
Question 16.
If a, s are the zeroes of p(x) = 2x² + 5x + k such that, α²+ s²+ αs = \(\frac{21}{4}\), then k equals,
(a) 12
(b) 4
(c) 2
(d) – 12
Answer
Answer: (c) 2
α + s = – \(\frac{5}{2}\)
αs = \(\frac{k}{2}\)
α² + s² +αs = \(\frac{21}{4}\)
(α + s)² – αs = \(\frac{21}{4}\)
\(\frac{25}{4}\) – \(\frac{k}{2}\) = \(\frac{21}{4}\)
k = 2
Question 17.
If α, s are the zeroes of x² + px + q, then a polynomial having zeroes \(\frac{1}{α}\) and \(\frac{1}{s}\) is,
(a) x² + px + q
(b) x² + qx + p
(c) px² + qx + 1
(d) qx² + px +1
Answer
Answer: (d) qx² + px +1
α + s = – p
αs = q
Question 18.
Find the number of zeros in the graph given:
(a) 3
(b) 2
(c) 1
(d) 0
Answer
Answer: (b) 2
Since the graph meets X-axis at two points -2 and 1, thus it has 2 zeroes.
Question 19.
Write the zero of the polynomial p(x), whose graph is given :
(a) 1
(b) 0
(c) – 1
(d) – 2
Answer
Answer: (b) 0
Since the graph meets X-axis at x = 0
=> Zero of p(x) is ‘O’ => Correct option is (b).
Question 20.
If α, s, v are the zeros of the polynomial 2x³ – x² + 3x -1, find the value of (αsv) + (αs + sv + vα).
(a) 2
(b) \(\frac{3}{2}\)
(c) \(\frac{1}{2}\)
(d) 0
Answer
Answer: (a) 2
p(x) = 2x³ – x² + 3x -1
αsv = – d/a = \(\frac{1}{2}\)
αs + sv + vα = c/a = \(\frac{3}{2}\)
αs + sv + vα + αsv = \(\frac{3}{2}\) + \(\frac{1}{2}\) = 2
Question 21.
If the zeros of the polynomial x³ – 3x² + x +1 are p – q,p and p + q. Find the value of q.
(a) 1
(b) 0
(c) 2
(d) ±√2
Answer
Answer: (d) ±√2
x³ – 3x² + x +1
zeroes are p – q,p,p + q
sum of zeroes = (p – q) + p + (p + q)
= 3p
= 3
α + s + v = \(\frac{- b}{a}\)
further = αs + sv + vα = \(\frac{c}{a}\)
(p – q) p + p(p + q) + (p – q)(p + q)=1
q = ±√2
Question 22.
A quadratic polynomial has :
(a) at least 2 zeros
(b) exactly 2 zeros
(c) at most 2 zeros
(d) exactly 1 zero
Answer
Answer: (c) at most 2 zeros
A quadratic polynomial has atmost two zeroes.
Question 23.
If α, s are the roots of cx² – bx + a = 0 (c 0), then α + s is:
(a) \(\frac{- b}{a}\)
(b) \(\frac{b}{a}\)
(c) \(\frac{c}{a}\)
(d) \(\frac{b}{c}\)
Answer
Answer: (d) \(\frac{b}{c}\)
sum of the roots = – \(\frac{coefficient of x}{coefficient of x²}\) = \(\frac{b}{c}\)
Question 24.
If P(x) and D(r) are any two polynomials such that D(x) ≠ 0, there exists unique polynomial Q(x) and R(x) such that, P(x) = D(x). Q(x) + R(x) where :
(a) R(x) = 0 and deg R(x) > deg Q(x)
(b) R(x) = 0 or deg R(x) > deg D(x)
(c) deg R(x) < deg Q(x)
(d) R(x) = 0 or deg R(x) < deg D(x)
Answer
Answer: (b) R(x) = 0 or deg R(x) > deg D(x)
division algorithm
Question 25.
When we divide x³ + 5x + 7 by x⁴ – 7x² – 6 then quotient and remainder are (respectively):
(a) 0,x³ + 5x + 7
(b) x, 2x + 3
(c) 1,x⁴ – 7x²-6
(d) x², 4x – 9
Answer
Answer: (a) 0,x³ + 5x + 7
Degree of the divisor is more than the degree of the dividend = quotient is zero and the remainder is x³ + 5x + 7
Question 26.
The value of b, for which 2x³ + 9x² – x – b is exactly divisible by 2x + 3 is:
(a) -15
(b) 15
(c) 9
(d) – 9
Answer
Answer: (b) 15
when 2x³ + 9x² – x – b is divided by 2x + 3, remainder is – b + 15
Question 27.
If α and s are two zeros of the polynomial p(x), then which of the following is a factor of p(x):
(a) (x – α)(x – s)
(b) (x + α) (x + s)
(c) k(x – α)
(d) k(x- s)
Answer
Answer: (a) (x – α)(x – s)
if α, s are the zeros of p(x), then (x – α)(x – s) is a factor of p(x).
Question 28.
Find a cubic polynomial with the sum, sum of the product of its zeros taken two at a time and the product of its zeros as -2, +5, -3, respectively.
(a) 2x³ + 5x² + x + 3
(b) 4x³ + 5x² – 3x + 7
(c) x³ + 2x² + 5x + 3
(d) 2x³ + 5x² + 3x + 1
Answer
Answer: (c) x³ + 2x² + 5x + 3
Let the polynomial be ax³ + bx² + cx + d
– b/a = – 2
c/a = 5
– d/a = – 3
a = 1, b = 2, c = 5 and d = 3
required polynomial is x³ + 2x² + 5x + 3
Question 29.
Write a polynomial with zeros 1, – 1 and 1.
(a) x³ + x² + x + 1
(b) x³ – x² + x + 1
(c) x³ – x² – x – 1
(d) x³ – x² – x + 1
Answer
Answer: (d) x³ – x² – x + 1
zeros are 1, – 1 and 1.
required polynomial is
k(x – 1)(x +1)(x – 1)
= x³ – x² – x + 1
Question 30.
The graph of a polynomial is as shown, find the polynomial
(a) k(x² – x – 6)
(b) k(x³ + x² + 6x)
(c) k(x³ – x² – 6x)
(d) k(x³ – 6x)
Answer
Answer: (c) k(x³ – x² – 6x)
zeros are – 2,0, and 3
required polynomial = k(x – 2)(x – 0)(x – 3)
= k(x³ – x² – 6x)
Question 31.
If α, s and v are the zeroes of the polynomial 2x³ – x² + 3x – 1, find the value of => (as + sv + va + asv )²
(a) \(\frac{3}{2}\)
(b) \(\frac{5}{2}\)
(c) \(\frac{1}{2}\)
(d) 4
Answer
Answer: (d) 4
αs + sv + vα + αsv = \(\frac{3}{2}\) + \(\frac{1}{2}\) = 2
(αs + sv + vα + αsv )² = 4
Question 32.
If 2 ± √3 are the two zeros of a polynomial then the following is a factor:
(a) x² – 4x + 1
(b) x² + 4x – 1
(c) 4x² + x – 1
(d) 4x² – x + 1
Answer
Answer: (a) x² – 4x + 1
If a, s are the zeroes => (x – α) (x – s) is a factor
=> (x – (2 + √3)) (x – (2 – √3))is a factor
=> x2 – 4x + 1 is a factor.
Question 33.
If 2 is a zero of p(x) = x² + 3x + k, find k:
(a) 10
(b) 5
(c) – 3
(d) – 10
Answer
Answer: (d) – 10
p(x) = x² + 3x + k
p(2) = 0
=>4 + 6 + k = 0
=k = – 10
Question 34.
Given that two of the zeroes of the
polynomial, x³ + px² + rx + s are 0, then third zero
(a) 0
(b) \(\frac{p}{r}\)
(c) \(\frac{- p}{r}\)
(d) \(\frac{p}{q}\)
Answer
Answer: (c) \(\frac{- p}{r}\)
Two zeroes are zero, let third zero = α
=> Sum of the roots = α + 0 + 0
\(\frac{Coefficient of x²}{Coefficient of x³}\)
Question 35.
Given that one of the zeroes of the
polynomial ax³ + bx² + cx + d is zero, then the product of the other two zeroes is:
(a) \(\frac{- c}{a}\)
(b) \(\frac{c}{a}\)
(c) 0
(d) \(\frac{- b}{a}\)
Answer
Answer: (b) \(\frac{c}{a}\)
αs + sv + vα = \(\frac{c}{a}\)
now α = 0
0 + sv + 0 = \(\frac{c}{a}\)
sv = \(\frac{c}{a}\)
Question 36.
The number of polynomials having zeroes – 1 and – 5 is :
(a) 2
(b) 3
(c) 1
(d) More than 3.
Answer
Answer: (d) More than 3.
n – number of polynomials can have zeroes -1 and -5.
Question 37.
The graph of the polynomial f(x) = 2x – 5 intersects the x – axis at
(a) (\(\frac { 5 }{ 2 }\), 0)
(b) (\(\frac { -5 }{ 2 }\), 0)
(c) (\(\frac { -5 }{ 2 }\), \(\frac { 5 }{ 2 }\))
(d) (\(\frac { 5 }{ 2 }\), \(\frac { -5 }{ 2 }\))
Answer
Answer: (a) (\(\frac { 5 }{ 2 }\), 0)
Question 38.
If the zeroes of the quadratic polynomial Ax2 + Bx + C, C # 0 are equal, then
(a) A and B have the same sign
(b) A and C have the same sign
(c) B and C have the same sign
(d) A and C have opposite signs
Answer
Answer: (b) A and C have the same sign
Question 39.
The number of polynomials having zeroes as 4 and 7 is
(a) 2
(b) 3
(c) 4
(d) more than 4
Answer
Answer: (d) more than 4
Question 40.
If one of the zeroes of the cubic polynomial x3 + ax2 + bx + c is -1, then the product of the
other two zeroes is
(a) b – a + 1
(b) b – a – 1
(c) a – b + 1
(d) a – b – 1
Answer
Answer: (a) b – a + 1
Question 41.
The number of zeros of a cubic polynomial is
(a) 3
(b) at least 3
(c) 2
(d) at most 3
Answer
Answer: (d) at most 3
Question 42.
Find the quadratic polynomial whose zeros are 2 and -6
(a) x2 + 4x + 12
(b) x2 – 4x – 12
(c) x2 + 4x – 12
(d) x2 – 4x + 12
Answer
Answer: (c) x2 + 4x – 12
Question 43.
If 5 is a zero of the quadratic polynomial, x2 – kx – 15 then the value of k is
(a) 2
(b) -2
(c) 4
(d) – 4
Answer
Answer: (a) 2
Question 44.
The number of polynomials having zeroes as -2 and 5 is
(a) 1
(b) 2
(c) 3
(d) more than 3
Answer
Answer: (d) more than 3
Question 45.
The zeroes of the quadratic polynomial x2 + 1750x + 175000 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal
Answer
Answer: (a) both negative
Question 46.
If the zeroes of the quadratic polynomial x2 + (a + 1) x + b are 2 and -3, then
(a) a = -7, b = -1
(b) a = 5, b = -1
(c) a = 2, b = -6
(d) a – 0, b = -6
Answer
Answer: (d) a – 0, b = -6
Question 47.
Sum and the product of zeroes of the polynomial x2 +7x +10 is
(a) \(\frac { 10 }{ 7 }\) and \(\frac { -10 }{ 7 }\)
(b) \(\frac { 7 }{ 10 }\) and \(\frac { -7 }{ 10 }\)
(c) -7 and 10
(d) 7 and -10
Answer
Answer: (c) -7 and 10
Question 48.
If x = 2 and x = 3 are zeros of the quadratic polynomial x2 + ax + b, the values of a and b respectively are :
(a) 5, 6
(b) – 5, – 6
(c) – 5, 6
(d) 5, – 6
Answer
Answer: (c) – 5, 6
Question 49.
The zeroes of the quadratic polynomial 3x2 – 48 are
(a) both negative
(b) one positive and one negative
(c) both positive
(d) both equal
Answer
Answer: (b) one positive and one negative
Question 50.
The sum and product of the zeroes of the polynomial x2-6x+8 are respectively
(a) \(\frac { -3 }{ 2 }\) and – 1
(b) 6 and 8
(c) \(\frac { -3 }{ 2 }\) and 1
(d) \(\frac { 3 }{ 2 }\) and 1
Answer
Answer: (b) 6 and 8