MCQ Questions for Class 9 Maths Polynomials Quiz 12

$f(x) = 8$ then

$f(x)$ is called

0%
Constant polynomials
0%
linear polynomials
0%
quadratic polynomials
0%
Cubic polynomials

Explanation

$f(x) = 8$ then

$f(x)$ is called Constant polynomials

The degree of polynomial is

$x + 2$ is:

0%
$2$
0%
$1$
0%
$3$
0%
$4$

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

$2x^{2} + 5x^{3} + 7$ ?

0%
$3$
0%
$2$
0%
$5$
0%
$7$

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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$2$
0%
$3$
0%
$5$
0%
$4$

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

0%
$0$
0%
$1$
0%
$2$
0%
$3$

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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$-\sqrt{3}$
0%
$\dfrac{-\sqrt 3}{3}$
0%
$\dfrac{3}{\sqrt 3}$
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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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$3$
0%
$\dfrac {-1}{2}$
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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
0%
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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$4\pi ,\frac{\pi }{4}$
0%
$-4\pi ,\frac{-\pi }{4}$

For

$p(x) = 5x^2 - \dfrac{2}{3} x + 8$ , then value of

$p (3) =$

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$50$
0%
$51$
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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:

0%
-4
0%
$\frac{4}{\alpha}$
0%
$-\frac{4}{\alpha}$
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None of these

$p(x) = 1 + 7x - 9x^2 + \dfrac{2}{3} x^3$ , then

$p(-3) =$

0%
$-81$
0%
$-137$
0%
$-119$
0%
$-100$

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

$f(x)={ x }^{ 4 }-3{ x }^{ 3 }-9{ x }^{ 2 }+4$ .

0%
None
0%
One
0%
Two
0%
Three
0%
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.

$f(x)=x^{6}-10x^{5}-10x^{4}-10x^{3}-10x^{2}-10x+10$ , the value of

$f(11)$ is

0%
$1$
0%
$10$
0%
$11$
0%
$21$

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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$\frac{11}{27}$
0%
$\frac{11}{13}$
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11
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$\frac{11}{15}$
0%
$-\frac{22}{27}$

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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$-4$
0%
$\dfrac {17}{2}$
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$9$
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$\dfrac {-687}{2}$
0%
$\dfrac {427}{2}$

$Let\,x = \sqrt {3 - \sqrt 5 } \,and\,y = \sqrt {3 + \sqrt 5 } .$ 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

0%
$448$
0%
$610$
0%
$510$
0%
$540$

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

0%
0
0%
2
0%
4
0%
7

The number of polynomials having zeroes as

$-2$ and

$5$ is?

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$0$
0%
$2$
0%
$3$
0%
More than $3$

If a + b + 2 c = 0,

$c\neq 0$ , then equation $$

0%
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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$-1,1-2\omega ,1-{ 2\omega }^{ 2 }$
0%
$2,2\omega ,{ 2\omega }^{ 2 }$
0%
$2,1+2\omega ,1+{ 2\omega }^{ 2 }$

Number of common roots of the equation

$x^4+x^2+1=0$ and

$x^8-1=0$ is

0%
zero
0%
1
0%
2
0%
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

0%
(-4, 0)
0%
(1, 3)
0%
(0, 4)
0%
None of these

If the sum of two roots of the equation

${ x }^{ 3 }-p{ x }^{ 2 }+qx-r=0$ is zero, then

0%
pq=r
0%
qr=p
0%
pr=q
0%
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:

0%
$15$
0%
$-15$
0%
$21$
0%
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:

0%
$120$
0%
$-120$
0%
$114$
0%
$-114$

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

0%
$18$
0%
$3$
0%
$9$
0%
$6$

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:

0%
$120$
0%
$-120$
0%
$0$
0%
$6$

The product of the zero of the polynomal

${ x }^{ 2 }-4x+3$ is ....

0%
4
0%
1
0%
-4
0%
3

CBSE Questions for Class 9 Maths Polynomials Quiz 12 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

CBSE Questions for Class 9 Maths Polynomials Quiz 12 - MCQExams.com

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Practice Class 9 Maths Quiz Questions and Answers

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