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CBSE Questions for Class 9 Maths Polynomials Quiz 12 - MCQExams.com
CBSE
Class 9 Maths
Polynomials
Quiz 12
If $$f(x) = 8$$then$$ f(x)$$ is called
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Constant polynomials
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linear polynomials
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quadratic polynomials
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Cubic polynomials
Explanation
If $$f(x) = 8$$ then $$ f(x)$$ is called Constant polynomials
The degree of polynomial is $$x + 2$$ is:
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$$2$$
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$$1$$
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$$3$$
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$$4$$
Explanation
$$\textbf{Step 1: Apply the property of polynomial.}$$
$$\text{We have, x + 2}$$
$$\text{We know that the highest power of the variable in a polynomial in one variable is }$$
$$\text{called the degree of the polynomial.}$$
$$\text{Here, the highest power of variable x is 1.}$$
$$\textbf{Thus, degree of polynomial is 1, option - (B).}$$
Write the correct alternative answer for the following question:
Which is the degree of the polynomial $$ 2x^{2} + 5x^{3} + 7 $$ ?
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$$3$$
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$$2$$
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$$5$$
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$$7 $$
Explanation
The given polynomial is
$$ 2x^{2} + 5x^{3} + 7 $$
.
We know,
in a polynomial in one variable,
the highest power of the polynomial term with a non-zero coefficient, is called the
degree
of the polynomial.
Here, the highest power of the variable $$x$$ is $$3$$.
$$\therefore$$ The degree of the given polynomial is
$$3$$.
Therefore, option $$A$$ is correct.
The degree of the polynomial $$x^4 + x^3$$ is:
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$$2$$
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$$3$$
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$$5$$
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$$4$$
Explanation
The highest power of the variable in a polynomial in one variable is called the
degree
of the polynomial.
Here, in the given polynomial $$x^4 + x^3$$, the highest power of the variable $$x$$ is $$4$$.
$$\therefore$$ The degree of the given polynomial is
$$4$$.
Therefore, option $$D$$ is correct.
The number of zeroes of linear polynomial at most is
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$$0$$
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$$1$$
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$$2$$
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$$3$$
Explanation
$$\textbf{Step-1: Explanation for number of zeroes of linear polynomial}$$
$$\text{The number of zeroes of a polynomial is equal to the degree of the polynomial}$$
$$\text{As the degree of a linear polynomial is 1, the number of zeroes is 1}$$
$$\textbf{Therefore, the number of zeroes of the linear polynomial at most is 1}$$
Zero of the polynomial $$p(x) = \sqrt{3}x + 3$$ is
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$$-\sqrt{3}$$
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$$\dfrac{-\sqrt 3}{3}$$
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$$\dfrac{3}{\sqrt 3}$$
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$$3\sqrt 3$$
One zero of $$p(x)=2x+1$$ will be
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$$\dfrac {1}{2}$$
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$$3$$
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$$\dfrac {-1}{2}$$
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$$1$$
Explanation
$$p(x)=2x+1$$
For zeroes $$p(x)=0$$
$$0=2x+1$$
$$\Rightarrow x=\dfrac {-1}{2}$$
State True/False:-
One of the zeroes of the polynomial $$x^{2} + 2x - 15$$ is $$11$$
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True
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False
Explanation
To find the zeroes of the polynomial, we have to proceed in the following way:-
Let f(x)=$$x^2+2x-15$$
For zeroes
$$f(x)=0$$
$$\Rightarrow x^2+2x-15=0$$
$$\Rightarrow x^2+5x-3x-15=0$$
$$\Rightarrow x(x+5)-3(x+5)=0$$
$$\Rightarrow (x-3)(x+5)=0$$
$$x-3=0$$ or $$x+5=0$$
$$x=3$$ or $$x=-5$$
Hence the zeroes of the polynomial $$x^2+2x-15$$ are $$3,-5$$.
Hence, given statement is
false.
Identify the zeroes of the given polynomial.
$$p(z)=4z^2-15z\pi -4\pi ^3$$
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$$4\pi ,\frac{-\pi }{4}$$
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$$-4\pi ,\frac{\pi }{4}$$
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$$4\pi ,\frac{\pi }{4}$$
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$$-4\pi ,\frac{-\pi }{4}$$
For $$p(x) = 5x^2 - \dfrac{2}{3} x + 8$$, then value of $$p (3) = $$
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$$50$$
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$$51$$
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$$55$$
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$$35$$
Explanation
The polynomial given is
$$p(x)=5x^{ 2 }-\frac { 2 }{ 3 } x+8$$
after substituting the value $$x=3$$ in the polynomial we get:
$$p(3)=5x^{ 2 }-\frac { 2 }{ 3 } x+8=5(3)^{ 2 }-\left( \frac { 2 }{ 3 } \times 3 \right) +8=\left( 5\times 9 \right) -2+8=45-2+8=53-2=51$$
Hence, $$p(3)=51$$.
$$ \alpha, \beta, \gamma$$
be the zeroes of the expression
$$ax^{3} +
bx^{2} + 4x + 7$$, then the value of
$$ \alpha\beta + \beta\gamma + \gamma\alpha$$ is:
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-4
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$$\frac{4}{\alpha}$$
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$$-\frac{4}{\alpha}$$
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None of these
If $$p(x) = 1 + 7x - 9x^2 + \dfrac{2}{3} x^3$$, then $$p(-3) =$$
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$$-81$$
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$$-137$$
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$$-119$$
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$$-100$$
Explanation
Consider the polynomial
$$p(x)=1+7x-9x^{ 2 }+\frac { 2 }{ 3 } x^{ 3 }$$
and substitute $$x=-3$$ in the polynomial:
$$p(-3)=1+(7\times -3)-9(-3)^{ 2 }+\frac { 2 }{ 3 } (-3)^{ 3 }=1-21-(9\times 9)+\left( \frac { 2 }{ 3 } \times -27 \right) =1-21-81-18=-119$$
Hence, $$p(-3)=-119$$.
Calculate the number of real numbers
$$k$$ such that $$f(k)=2$$
if
$$f(x)={ x }^{ 4 }-3{ x }^{ 3 }-9{ x }^{ 2 }+4$$.
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None
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One
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Two
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Three
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Four
Explanation
Given, $$f(x)=x^4-3x^3-9x^2+4, f(k)=2$$
Therefore, $$y = 2= f(x)$$
$$\Rightarrow 2=x^4-3x^3-9x^2+4$$
$$\Rightarrow x^4-3x^3-9x^2+2=0$$
Then plug in consecutive numbers for $$x$$ and if the sign of $$f(x)$$ changes,
$$f(-2) = 6 $$
$$f(-1) = -3$$ so there's one
$$f(0) = 2$$ so that's two
$$f(1 ) = -9$$ so that's three
And eventually it has to become positive again so there are four.
If $$f(x)=x^{6}-10x^{5}-10x^{4}-10x^{3}-10x^{2}-10x+10$$, the value of $$f(11)$$ is
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$$1$$
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$$10$$
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$$11$$
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$$21$$
Explanation
The given expression $$f(x)=x^6-10x^5-10x^4-10x^3-10x^2-10x+10$$ can be rewritten as
$$f(x)=x^{ 6 }-10(x^{ 5 }+x^{ 4 }+x^{ 3 }+x^{ 2 }+x-1)$$.
Now, substitute $$x=11$$ in the above expression as shown below:
$$\Rightarrow f(x)=x^{ 6 }-10(x^{ 5 }+x^{ 4 }+x^{ 3 }+x^{ 2 }+x-1)$$
$$\Rightarrow f(11)=11^{ 6 }-10(11^{ 5 }+11^{ 4 }+11^{ 3 }+11^{ 2 }+11-1)$$
$$ \Rightarrow f(11)=1771561-10(161051+14641+1331+121+11-1)$$
$$ \Rightarrow f(11)=1771561-(10\times 177154)$$
$$ \Rightarrow f(11)=1771561-1771540$$
$$\Rightarrow f(11)=21$$
Hence, $$f(11)=21$$.
Two roots of the equation $$4x^3 - px^2+ qx - 2p = 0$$ are 4 andWhat is the third root?
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$$\frac{11}{27}$$
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$$\frac{11}{13}$$
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11
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$$\frac{11}{15}$$
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$$-\frac{22}{27}$$
The equation $$8x^{6} + 72x^{5} + bx^{4} + cx^{3} - 687x^{2} - 2160x - 1700 = 0$$, as shown in the figure, has two complex roots. The product of these complex roots is
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$$-4$$
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$$\dfrac {17}{2}$$
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$$9$$
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$$\dfrac {-687}{2}$$
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$$\dfrac {427}{2}$$
$$Let\,x = \sqrt {3 - \sqrt 5 } \,and\,y = \sqrt {3 + \sqrt 5 } .$$ If the value of expression $$x - y + 2{x^2}y + 2x{y^2} - {x^4}y + x{y^4}$$ can be expressed in the form $$\sqrt p + \sqrt q $$ $$where\,p,\,q \in N,$$ then $$(p+q)$$ has the value equal
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$$ 448 $$
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$$610$$
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$$510$$
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$$540$$
Explanation
The zeros of the polynomial $${ x }^{ 2 }-9$$ are
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3,3
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$$\sqrt { 3 } ,-\sqrt { 3 } $$
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3,-3
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$$-\sqrt { 3 } ,-\sqrt { 3 } $$
Number of root of equation
3|x|-|2-x|=1
is
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0
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2
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4
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7
The number of polynomials having zeroes as $$-2$$ and $$5$$ is?
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$$0$$
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$$2$$
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$$3$$
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More than $$3$$
If a + b + 2 c = 0, $$c\neq 0$$, then equation $$
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At least one root in (0, 1)
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At least one root in (0, 2)
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At least one root in (-1, 1)
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None of these
The roots of the equation $$\left( x-1 \right) ^{ 3 }+8=0$$ are
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$$-1,1+2\omega ,1+{ 2\omega }^{ 2 }$$
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$$-1,1-2\omega ,1-{ 2\omega }^{ 2 }$$
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$$2,2\omega ,{ 2\omega }^{ 2 }$$
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$$2,1+2\omega ,1+{ 2\omega }^{ 2 }$$
Number of common roots of the equation $$x^4+x^2+1=0$$ and $$x^8-1=0$$ is
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zero
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1
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2
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4
For equation $${ x }^{ 3 }-{ 6x }^{ 2 }+9x+k=0$$ to have exactly one root in (1, 3), the set of values of k is
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(-4, 0)
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(1, 3)
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(0, 4)
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None of these
If the sum of two roots of the equation $${ x }^{ 3 }-p{ x }^{ 2 }+qx-r=0$$is zero, then
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pq=r
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qr=p
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pr=q
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pqr=1
Let $$f ( x )$$ be a polynomial of degree 5 with leading coefficient unity, such that $$f ( 1 ) = 5 , f ( 2 ) = 4 , f ( 3 ) = 3 , f ( 4 ) = 2$$ and $$f ( 5 ) = 1 ,$$ then
Sum of the roots of $$f ( x )$$ is equal to:
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$$15$$
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$$-15$$
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$$21$$
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Can't be determine
Let $$f ( x )$$ be a polynomial of degree 5 with leading coefficient unity, such that $$f ( 1 ) = 5 , f ( 2 ) = 4 , f ( 3 ) = 3 , f ( 4 ) = 2$$ and $$f ( 5 ) = 1 ,$$ then
Product of the roots of $$f ( x )$$ is equal to:
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$$120$$
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$$-120$$
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$$114$$
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$$-114$$
Let for $$a \neq a_1 \neq 0, f(x) = ax^2 +bx +c, g(x) = a_1x^2 +b_1x+c_1$$ and $$p(x) =f(x) -g(x)$$. If $$p(x) =0$$ only for $$x= -1$$ and $$p(-2) =2$$ then the value of $$p(2)$$ is
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$$18$$
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$$3$$
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$$9$$
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$$6$$
Let $$f ( x )$$ be a polynomial of degree 5 with leading coefficient unity, such that $$f ( 1 ) = 5 , f ( 2 ) = 4 , f ( 3 ) = 3 , f ( 4 ) = 2$$ and $$f ( 5 ) = 1 ,$$ then
$$f ( 6 )$$ is equal to:
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$$120$$
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$$-120$$
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$$0$$
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$$6$$
The product of the zero of the polynomal $${ x }^{ 2 }-4x+3$$ is ....
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4
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1
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-4
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3
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