10th Grade > Mathematics
POLYNOMIALS MCQs
1626393664750aae00754cf
Total Questions : 79
| Page 2 of 8 pages
Answer: Option A. -> 2 and 6
:
A
To find the zeroes, equate the polynomial to zero and then factorise.
∴x2−8x+12=0
⇒x2−6x−2x+12=0
⇒x(x−6)−2(x−6)=0
⇒(x−6)(x−2)=0
⇒x−6=0orx−2=0⇒x=6;2
2and6are the zeros of the quadraticpolynomial.
:
A
To find the zeroes, equate the polynomial to zero and then factorise.
∴x2−8x+12=0
⇒x2−6x−2x+12=0
⇒x(x−6)−2(x−6)=0
⇒(x−6)(x−2)=0
⇒x−6=0orx−2=0⇒x=6;2
2and6are the zeros of the quadraticpolynomial.
Answer: Option B. -> 3x2−12x+11
:
B
Letα&β be the roots.
Sum of zeroes, α+β=6+√33+6−√33 =123=4
Product of zeroes, αβ=6+√33×6−√33=62−(√3)29αβ=36−39=339=113
Required polynomial f(x)=[(x2−(α+β)x+(αβ)]=[(x2−4x+113]
Multiplying by 3
⇒f(x)=3x2−12x+11
∴ the required polynomial is 3x2−12x+11.
:
B
Letα&β be the roots.
Sum of zeroes, α+β=6+√33+6−√33 =123=4
Product of zeroes, αβ=6+√33×6−√33=62−(√3)29αβ=36−39=339=113
Required polynomial f(x)=[(x2−(α+β)x+(αβ)]=[(x2−4x+113]
Multiplying by 3
⇒f(x)=3x2−12x+11
∴ the required polynomial is 3x2−12x+11.
Answer: Option D. -> 2x2+3x+1
:
D
We know that, by divison algorithm,
Dividend = Divisor x Quotient + Remainder.
∴P(x)=(x+2)×(2x−1)+3=x(2x−1)+2(2x−1)+3=2x2−x+4x−2+3P(x)=2x2+3x+1
∴The required polynomialis2x2+3x+1.
:
D
We know that, by divison algorithm,
Dividend = Divisor x Quotient + Remainder.
∴P(x)=(x+2)×(2x−1)+3=x(2x−1)+2(2x−1)+3=2x2−x+4x−2+3P(x)=2x2+3x+1
∴The required polynomialis2x2+3x+1.
Answer: Option C. -> √2 and −√2
:
C
To find the zeroes, equate the givenpolynomial to zero.
x2−2=0
Using the identity,a2−b2=(a+b)(a−b)
⇒(x+√2)(x−√2)=0
⇒x+√2=0andx−√2=0
⇒x=−√2andx=√2
Hence, the zeroes are√2,−√2.
:
C
To find the zeroes, equate the givenpolynomial to zero.
x2−2=0
Using the identity,a2−b2=(a+b)(a−b)
⇒(x+√2)(x−√2)=0
⇒x+√2=0andx−√2=0
⇒x=−√2andx=√2
Hence, the zeroes are√2,−√2.
Answer: Option B. -> 5x2+8x+7
:
B
Given that,
Sum of zeroes =−85
Product of zeroes =75
Required quadratic polynomial is,
f(x)=[(x2−(sumofroots)x+(productofroots)]
Substituting the given values we get,
f(x)=[x2−(−8)5x+75]
f(x)=[x2+85x+75]
multiplying by 5 we get
f(x)=5x2+8x+7
∴Required polynomial is 5x2+8x+7.
:
B
Given that,
Sum of zeroes =−85
Product of zeroes =75
Required quadratic polynomial is,
f(x)=[(x2−(sumofroots)x+(productofroots)]
Substituting the given values we get,
f(x)=[x2−(−8)5x+75]
f(x)=[x2+85x+75]
multiplying by 5 we get
f(x)=5x2+8x+7
∴Required polynomial is 5x2+8x+7.
Answer: Option A. -> -3
:
A
For a cubic polynomial ax3+bx2+cx+d
Sum of zeros = −ba
Product of zeros =−da
∴ Product of zeros is−31 = -3
:
A
For a cubic polynomial ax3+bx2+cx+d
Sum of zeros = −ba
Product of zeros =−da
∴ Product of zeros is−31 = -3
Answer: Option D. -> -ba
:
D
Ifα,β and γare the zeroes of thecubic polynomialax3+bx2+cx+d, then
sum of its zeroes
= α+β+γ
=−(coefficientofx2)(coefficientofx3)=−ba
:
D
Ifα,β and γare the zeroes of thecubic polynomialax3+bx2+cx+d, then
sum of its zeroes
= α+β+γ
=−(coefficientofx2)(coefficientofx3)=−ba
Answer: Option B. -> False
:
B
Given,p(x)=x2−3√3x+1x=3√3
p(3√3)=(3√3)2−3√3(3√3)+1p(3√3)=27−27+1p(3√3)=0+1=1
Thus, the given statement is false.
:
B
Given,p(x)=x2−3√3x+1x=3√3
p(3√3)=(3√3)2−3√3(3√3)+1p(3√3)=27−27+1p(3√3)=0+1=1
Thus, the given statement is false.
Answer: Option B. -> 1
:
B
Given f(x) =3x2−7x+5
To find the remainderwhen it is divided by (x−1), we useremainder theorem.
Remainder of f(x)(x−a) is f(a).
f(1)=3−7+5=1
⇒ The remainder when 3x2−7x+5 is divided by (x−1) is 1.
:
B
Given f(x) =3x2−7x+5
To find the remainderwhen it is divided by (x−1), we useremainder theorem.
Remainder of f(x)(x−a) is f(a).
f(1)=3−7+5=1
⇒ The remainder when 3x2−7x+5 is divided by (x−1) is 1.
Latest Videos
Latest Test Papers
Login With Google
Old way of Login
Login With Email