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CBSE Questions for Class 9 Maths Polynomials Quiz 3 - MCQExams.com
CBSE
Class 9 Maths
Polynomials
Quiz 3
A term of the expression having no literal factors is called a constant term.
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True
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False
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Ambiguous
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Data insufficient
Explanation
Let, A term 3x which is a product of 3 and x. 3 is a numerical co-efficient of 3x and x is called literal factor of 3x.
So A term of the expression having no literal factors is called a constant term.
example: 4
Option A is correct.
Substituting $$x= -3$$ in
$$x^2-5x + 4$$
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$$-2$$
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$$28$$
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$$2$$
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$$-1$$
Explanation
For $$x=-3$$
$$x^2-5x+4$$
$$=(-3)^2-5(-3)+4$$
$$=9+15+4$$
$$=28 \neq R.H.S$$
The correct statement is $$x^2+5x+4=28$$
Substituting $$x = -3$$ in
$$x^2+5x$$
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$$x^2+5x$$ = -24
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$$x^2+5x$$ = -30
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$$x^2+5x$$ = -10
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$$x^2+5x$$ = -6
Explanation
For $$x=-3$$
$$x^2+5x$$
$$=(-3)^2+5(-3)$$
$$=9-15$$
$$=-6 $$
The correct statement is $$x^2+5x=-6$$
$$(a + b)^2-(a -b)^2$$ will be equal to $$2ab$$:
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True
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False
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Ambiguous
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Data insufficient
Explanation
$$(a+b)^2 -(a-b)^2=(a^2+2ab+b^2)-(a^2-2ab+b^2)$$
$$=a^2+2ab+b^2-a^2+2ab-b^2$$
$$=4ab$$.
Hence, answer is $$4ab$$ but not $$2ab$$.
Option $$B$$ is correct.
State true or false:
$$x = 2$$ is a root of $$2x^3+x^2-7x-6$$.
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True
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False
Explanation
$$p(x) = 2x^3+ x^2-7x-6.$$
Then, $$p(2)=2(2)^3+(2)^2-7(2)-6 = 16+4-14-6=0$$.
Hence, $$x = 2$$ is a root of $$p(x)$$.
Therefore, it is true and option $$A$$ is correct.
Simplify $$a (a^2 + a + 1) + 5$$ and find its value for a = 1
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6
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5
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8
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7
Explanation
$$a (a^2 + a + 1) + 5$$
$$=a^3+a^2+a+5$$
At $$a=1$$
$$=1^3+1^2+1+5$$
$$=1+1+1+5$$
$$=8$$
The degree of a polynomial $$x^{3} - 27$$ is
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$$3$$
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$$1$$
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$$26$$
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$$27$$
Explanation
The highest power of the variable $$x$$ in the given polynomial is $$3$$.
Hence, its degree is $$3$$.
The degree of the polynomial $$4x^{2} + 3x - 7$$ is:
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$$1$$
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$$3$$
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$$7$$
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$$2$$
Explanation
The highest power of the variable $$x$$ in the given polynomial is $$2$$.
Hence, its degree is $$2$$.
The degree of the polynomial $$2x^{4} - 3x^{2} + 9$$ is
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$$2$$
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$$3$$
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$$4$$
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None of the above
Explanation
The highest power of the variable $$x$$ in the given polynomial is $$4$$.
Hence, its degree is $$4$$.
In a polynomial, the exponents of the variables are always
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integers
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positive integers
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non-negative integers
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non-positive integers
Explanation
A polynomial is an algebraic expression that contains variables having whole number power.
The degree of a polynomial $$2x^{5} - 5x^{3} - 10x + 9$$ is
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$$5$$
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$$3$$
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$$1$$
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$$9$$
Explanation
Given polynomial is $$2x^5-5x^3-10x+9$$.
We know that, the degree of a polynomial is the highest power of the variable.
The highest power of the variable $$x$$ in the given polynomial is $$5$$.
Hence, the degree of the given polynomial is $$5$$.
The value of the polynomial $$z^3+2z^2+5z+1$$
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is equal when $$z=1$$ and $$z=-1$$
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is equal when $$z=1$$ and $$z=0$$
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is equal when $$z=-1$$ and $$z=0$$
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None of the above
Explanation
Let $$ f(z) = z^3+2z^2+5z+1 $$
At $$ z = 0 $$ value of the given polynomial = $$ f(0) = 1 $$
Similarly, $$ f(1) = 9 $$ &
$$ f(-1) = -3 $$
None of the three values are equal.
What is the output of $$x^2+3x+5$$, where $$x = 3?$$
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$$21$$
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$$22$$
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$$23$$
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None of the above
Explanation
Given that:
$$ x^2 + 3x + 5 :$$
Substituting, $$x = 3 \:we\: get ,$$
or $$ (3)^2 + 3\times 3 + 5 = 9+9+5 = 23$$
$$\therefore x^2 + 3x +5$$ is $$23$$
Given $${y= 3 ^x}$$, evaluate $$y$$ when $$x=3$$
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$$3$$
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$$9$$
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$$27$$
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$$81$$
Explanation
We have, $$y=3^x$$
At $$x=3$$
$$y =3^3=3\times 3\times 3=27$$
Hence, option C is correct choice.
The degree of a polynomial $$7x^{4} + 6x^{2} + x$$ is
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$$7$$
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$$2$$
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$$4$$
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$$1$$
Explanation
The highest power of the variable $$x$$ in the given polynomial is $$4$$.
Hence, its degree is $$4$$.
The value of a polynomial
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changes with the change in variable
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doesn"t change with the change in variable
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many or may not change with the change in variable
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all of the above
Identify zero polynomial among the following.
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$$0$$
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$$x$$
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$$x^2$$
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None of the above
Explanation
Consider the polynomial, $$p(x) = ax^2 + bx +c$$ , if $$a=b=c= 0$$ then the expression becomes zero polynomial.
Therefore
, the constant polynomial whose coefficients are all equal to $$0$$ is called a zero polynomial.
Hence, zero polynomial can be written as $$p(x) = 0$$.
The highest power in the polynomial $$x^2+6x+5$$
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$$2$$
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$$1$$
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$$0$$
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None of the above
Explanation
The highest power in the polynomial $$x^2+6x+5$$ is 2.
Always check the highest exponential term of the polynomial.
Find the degree of the polynomial $$5t+\sqrt{7}$$.
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$$5$$
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$$1$$
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$$\dfrac{1}{2}$$
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None of the above
Explanation
The given polynomial is
$$5t+\sqrt{7}$$
The degree is the highest power of the variable in the polynomial.
Hence, the degree of the given polynomial is $$1$$.
The zero of the linear polynomial $$x+5$$ is:
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$$5$$
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$$-5$$
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$$4$$
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$$-4$$
Explanation
A $$zero$$
or root of a $$polynomial$$
function is a number that, when plugged in for the variable, makes the function equal to
$$zero.$$
Hence lets by substituting $$ x= -5$$ we get
Equation : $$x + 5 = -5 + 5 = 0 $$.
$$\therefore$$ zero of the linear polynomial $$x + 5$$ is $$ - 5 $$.
Therefore, option $$B$$ is correct.
Is $$x=-1$$ a solution of $$\displaystyle { x }^{ 2 }+x+1=0$$?
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Yes
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No
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Can't say
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None
Explanation
Put $$x=-1$$ in equation $$\displaystyle { x }^{ 2 }+x+1=0$$.
$$\displaystyle f(-1)= { \left( -1 \right) }^{ 2 }+\left( -1 \right) +1$$
$$\displaystyle =1-1+1\ne0$$.
$$\displaystyle \therefore \quad x=-1$$ is not a solution of
$$\displaystyle { x }^{ 2 }+x+1=0$$.
Therefore, option $$B$$ is correct.
Identify the zero polynomial from the following.
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$$0$$
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$$5$$
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$$4$$
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$$3$$
Explanation
The degree of the zero polynomial is undefined.
Example:
$$p(x) = ax^2+bx+c$$, when the constants $$a = b = c = 0$$ is considered as zero polynomial.
Find the degree of $$2-y^2-y^3+2y^8$$
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$$2$$
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$$-1$$
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$$3$$
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$$8$$
Explanation
Put the polynomial $$2-y^2-y^3+2y^8$$ in standard form. The term with the highest exponent should be first, and the term with the lowest exponent should be last. This will help us see which term has the exponent with the largest value. Therefore, the standard form is:
$$2y^8-y^3-y^2+2$$
The power is simply number in the exponent. In the polynomial,
$$2y^8-y^3-y^2+2$$
, the power of the first term is $$8$$which is the largest exponent of $$y$$ and hence it is the degree of given polynomial.
Find the degree of the polynomial $$x^3+x+2$$.
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$$3$$
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$$2$$
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$$1$$
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None of the above
Explanation
The degree of the polynomial $$x^3+x+2$$ is $$3$$.
Find the degree of the polynomial $$4-y^2$$.
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$$4$$
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$$-1$$
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$$2$$
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None of the above
Explanation
The $$degree$$
of a $$polynomial$$
is the highest degree of its $$terms$$
Here $$y^2$$ has degree of $$2$$.
Hence, the degree of the polynomial $$4-y^2$$ is $$2$$
A linear polynomial has _____ zero(s).
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Explanation
Consider the polynomial, $$p(x) = bx +c$$.
Here, since the highest power of $$x$$ is $$1$$,
the degree of the given polynomial is $$1$$, i.e. the given polynomial is a linear polynomial
.
To find the zero of a polynomial, we write $$p(x) = 0$$.
That is,
$$ bx +c=0$$ $$\implies$$
$$ bx =-c$$
$$\implies$$
$$ x =\dfrac{-c}{b}$$
$$\implies$$
$$ x =-\dfrac{c}{b}$$.
That is, a linear polynomial can have one and only one zero.
Therefore, option $$A$$ is correct.
Find roots of equation
$$\displaystyle { a }^{ 4 }{ x }^{ 2 }-{ b }^{ 4 }=0$$
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$$\displaystyle { { b }^{ 2 } }/{ { a }^{ 2 } },-{ { b }^{ 2 } }/{ { a }^{ 2 } }$$
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$$\displaystyle { { b }^{ 2 } }/{ { a }^{ 2 } },{ { b }^{ 2 } }/{ { a }^{ 2 } }$$
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$$\displaystyle { { a }^{ 2 } }/{ { { b }^{ 2 } } },\frac { { a }^{ 2 } }{ { { b }^{ 2 } } } $$
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$$\displaystyle { { a }^{ 2 } }/{ { { b }^{ 2 } } },-\frac { { a }^{ 2 } }{ { { b }^{ 2 } } } $$
Explanation
$$\displaystyle { a }^{ 4 }{ x }^{ 2 }-{ b }^{ 4 }=0$$
Since,$$(a^2-b^2)=(a-b)(a+b)$$
$$\displaystyle \left( { a }^{ 2 }x+{ b }^{ 2 } \right) \left( { a }^{ 2 }x-{ b }^{ 2 } \right) =0,{ a }^{ 2 }x={ b }^{ 2 }and \ ,{ a }^{ 2 }x=-{ b }^{ 2 }$$
$$\displaystyle x={ { b }^{ 2 } }/{ { a }^{ 2 } },x=-{ { b }^{ 2 } }/{ { a }^{ 2 } }$$
$$p(x) = \dfrac{37}{15}$$ is a _______ polynomial.
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linear
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quadratic
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cubic
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none of these
Explanation
The general form of a constant polynomial is $$p(x)=c$$ with constant $$c$$.
Since
$$p(x)=\frac { 37 }{ 15 }$$
is a polynomial with constant term
$$\frac { 37 }{ 15 }$$
and there is no variable in it.
Hence,
$$p(x)=\frac { 37 }{ 15 }$$
is a constant polynomial.
Is $$\displaystyle x=-10$$ is a solution of
$$\displaystyle { x }^{ 2 }+5x-9=0$$?
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Yes
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No
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None
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Can't say
Explanation
Put $$\displaystyle x=-10$$ in equation $$\displaystyle { x }^{ 2 }+5x-9=0$$.
$$\displaystyle { \left( -10 \right) }^{ 2 }+5\left( -10 \right) -9=100-50-9$$
$$\displaystyle =\quad 100-59=41\neq 0$$.
$$\displaystyle \therefore $$ The equation does not have $$x=-10$$, solution.
Therefore, option $$B$$ is correct.
Is $$\displaystyle x=\dfrac { 5 }{ 2 } $$ a solution of $$\displaystyle \left( 6x+16 \right) \left( 4x+10 \right) =0$$?
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Yes
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No
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Can't say
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None
Explanation
Put $$\displaystyle x=\dfrac { 5 }{ 2 } $$ in equation
$$\displaystyle \left( 6x+16 \right) \left( 4x+10 \right) =0$$
$$\displaystyle \left( 6\times \frac { 5 }{ 2 } +16 \right) \left( 4\times \frac { 5 }{ 2 } +10 \right) =\left( \frac { 30 }{ 2 } +16 \right) \left( \frac { 20 }{ 2 } +10 \right) $$
$$\displaystyle =\left( 15+16 \right) \left( 10+10 \right) $$
$$\displaystyle =\left( 31 \right) \left( 20 \right) \neq 0$$
$$\displaystyle \therefore \quad x=\frac { 5 }{ 2 } $$ is not a solution of equation
$$\displaystyle \left( 6x+16 \right) \left( 4x+10 \right)$$
Therefore, option $$B$$ is correct.
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