10th Grade > Mathematics
POLYNOMIALS MCQs
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Total Questions : 79
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Answer: Option D. -> D)
:
D
We know that, the values at whicha quadratic polynomial cuts the x-axis are its zeroes.
Option A: Polynomial cuts the x-axis at : -3, -2 ; sum of zeroes: -5
Option B: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option C: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option D: Polynomial cuts the x-axis at : -2, 2; sum of zeros: 0
Hence, the answer is D.
:
D
We know that, the values at whicha quadratic polynomial cuts the x-axis are its zeroes.
Option A: Polynomial cuts the x-axis at : -3, -2 ; sum of zeroes: -5
Option B: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option C: Polynomial cuts the x-axis at : 2, 3; sum of zeroes: 5
Option D: Polynomial cuts the x-axis at : -2, 2; sum of zeros: 0
Hence, the answer is D.
Answer: Option A. -> 0,2
:
A
Let
p(x)=ax3+bx2+x−6
By using factor theorem, (x+2) can be a factor of p(x)only when p(−2)=0
p(−2)=a(−2)3+b(−2)2+(−2)−6=0
⇒−8a+4b−8=0
∴−2a+b=2...(i)
Also when p(x) is divided by (x−2) the remainder is 4.
∴p(2)=4
⇒a(2)3+b(2)2+2−6=4
⇒8a+4b+2−6=4
⇒8a+4b=8
⇒2a+b=2...(ii)
Adding equations (i)and (ii), we get
(−2a+b)+(2a+b)=2+2
⇒2b=4⇒b=2
Putting b=2 in (i)we get
−2a+2=2
⇒−2a=0⇒a=0
Hence,
a=0andb=2
:
A
Let
p(x)=ax3+bx2+x−6
By using factor theorem, (x+2) can be a factor of p(x)only when p(−2)=0
p(−2)=a(−2)3+b(−2)2+(−2)−6=0
⇒−8a+4b−8=0
∴−2a+b=2...(i)
Also when p(x) is divided by (x−2) the remainder is 4.
∴p(2)=4
⇒a(2)3+b(2)2+2−6=4
⇒8a+4b+2−6=4
⇒8a+4b=8
⇒2a+b=2...(ii)
Adding equations (i)and (ii), we get
(−2a+b)+(2a+b)=2+2
⇒2b=4⇒b=2
Putting b=2 in (i)we get
−2a+2=2
⇒−2a=0⇒a=0
Hence,
a=0andb=2
Answer: Option A. -> −1
:
A
Given polynomials:
p(x)=ax3+4x2+3x−4
q(x)=x3−4x+a
Using remainder theorm, the remainders when p(x)and q(x)are divided by (x−3)are p(3) and q(3) respectively.
p(3)=a×(3)3+4×(3)2+3×3−4
p(3)=27a+41
q(3)=(3)3−4×3+a
q(3)=15+a
Given that the remainder is the same.
⇒p(3)=q(3)
27a+41=15+a
26a=26
a=1
:
A
Given polynomials:
p(x)=ax3+4x2+3x−4
q(x)=x3−4x+a
Using remainder theorm, the remainders when p(x)and q(x)are divided by (x−3)are p(3) and q(3) respectively.
p(3)=a×(3)3+4×(3)2+3×3−4
p(3)=27a+41
q(3)=(3)3−4×3+a
q(3)=15+a
Given that the remainder is the same.
⇒p(3)=q(3)
27a+41=15+a
26a=26
a=1
Answer: Option B. -> 0
:
B
As we know that any variable raised to the power of zero yields 1, aconstant polynomial can be represented as ax0, where, a is the constant and x is the variable of the polynomial. Thus the degree of a constant polynomial is 0.
:
B
As we know that any variable raised to the power of zero yields 1, aconstant polynomial can be represented as ax0, where, a is the constant and x is the variable of the polynomial. Thus the degree of a constant polynomial is 0.
:
The point where the polynomial cuts the x-axis is the zero of the polynomial. Since the given polynomial cuts the x-axis at 3 points, the number of zeroes is 3.
Answer: Option A. -> 119
:
A
x+x−1=11
(x+x−1)2=112 [Squaring both the sides]
⇒(x)2+(x−1)2+2(x)(x−1)=121
⇒x2+x−2+2=121 [∵(x)(x−1)=x0=1]
⇒x2+x−2=121−2=119
:
A
x+x−1=11
(x+x−1)2=112 [Squaring both the sides]
⇒(x)2+(x−1)2+2(x)(x−1)=121
⇒x2+x−2+2=121 [∵(x)(x−1)=x0=1]
⇒x2+x−2=121−2=119
Answer: Option A. -> 1
:
A
Since (x+1) is a factor, x=−1will make the given expression zero.
∴(−1)100+2×(−1)99+k=0⇒1−2+k=0⇒k=1
:
A
Since (x+1) is a factor, x=−1will make the given expression zero.
∴(−1)100+2×(−1)99+k=0⇒1−2+k=0⇒k=1
Answer: Option C. -> Only one among a, b and c is non - zero.
:
C
Since a monomial can containonly one term, only one of a, b and c can be non-zero to leave a single term.
:
C
Since a monomial can containonly one term, only one of a, b and c can be non-zero to leave a single term.
Answer: Option C. -> any negative integer
:
C
For an expression to be a polynomial, the exponents of x should be whole numbers. The above expression can be written as xg+x−a. For this to be a polynomial, −a should be a whole number. This is possible if a is a negative integer or zero.
:
C
For an expression to be a polynomial, the exponents of x should be whole numbers. The above expression can be written as xg+x−a. For this to be a polynomial, −a should be a whole number. This is possible if a is a negative integer or zero.
Answer: Option B. -> 5a−4
:
B
Let ybe the width.
Given: Area=25a2−35a+69
We know that, area of a rectangle= Length×breadth
Hence,y×(5a−3)=25a2−35a+69
y=(25a2−35a+69)÷(5a−3)
By long division method,
5a−35a−23)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯25a2−35a+6925a2−12a−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−23a+69−23a+69+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯0
∴y=5a−23Width of the rectangle =5a−23
:
B
Let ybe the width.
Given: Area=25a2−35a+69
We know that, area of a rectangle= Length×breadth
Hence,y×(5a−3)=25a2−35a+69
y=(25a2−35a+69)÷(5a−3)
By long division method,
5a−35a−23)¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯25a2−35a+6925a2−12a−+¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯−23a+69−23a+69+−¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯0
∴y=5a−23Width of the rectangle =5a−23
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