Polynomials Test

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Polynomials Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

The repeated root of $$x^3+4x^2+5x+2=0$$ is

A
-5/3

B
-2

C
-1

D
1

Solution

$$x^{3}+4{x^{2}}+5{x}+2=(x+1)(x^{2}+3{x}+2)$$

$$=(x+1)(x+1)(x+2)$$

$$x^{3}+4{x^{2}}+5{x}+2=(x+1)^{2}(x+2)$$

The repeated root is $$-1$$
Question 2
1 / -0

The value of $$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$$ is:

A
$$2$$

B
$$3$$

C
$$5$$

D
$$1$$

Solution

Given, $$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$$.

We know, $$(a+b)(a-b)=a^2-b^2$$.

Then,
$$\displaystyle\left (5^{\tfrac {1}{2}} + 3^{\tfrac {1}{2}}\right ) \left (5^{\tfrac {1}{2}} - 3^{\tfrac {1}{2}}\right )$$
$$=\displaystyle 5^{\tfrac {1}{2}\times2} - 3^{\tfrac {1}{2}\times2}$$
$$=\displaystyle 5^{1} - 3^{1}$$
$$=\displaystyle 5- 3$$
$$=\displaystyle 2$$.

Hence, option $$A$$ is correct.
Question 3
1 / -0

Find the value of $$f(x) = x^2+\dfrac{1}{x^2}$$ at $$x= 1, 2$$.

A
$$f(1)=2, f(2)=\dfrac{17}{4}$$

B
$$f(2)=2, f(1)=\dfrac{17}{4}$$

C
$$f(1)=-2, f(2)=\dfrac{17}{4}$$

D
$$f(2)=2, f(1)=-\dfrac{17}{4}$$

Solution

Given: $$f(x)=x^2 + \dfrac{1}{x^2}$$
At $$x=1 ;$$

$$f(1)=1^{2}+\dfrac {1}{1^{2}}$$

$$\Rightarrow f(1)=2$$

At $$x=2 ;$$

$$f(2)=2^{2}+\dfrac {1}{2^{2}}$$

$$\Rightarrow f(2)=\dfrac {17}{4}$$
Question 4
1 / -0

The polynomial having atmost 3 zero

A
constant polynomial

B
linear polynomial

C
quadratic polynomial

D
cubic polynomial

Solution

A polynomial of degree $$3$$ is called a cubic polynomial.

For example, $$x^3 - 1,$$ and $$ 4a^3 - 100a^2 + a - 6$$ are cubic polynomials with atmost $$3$$ zeroes (having degree $$3$$).
Question 5
1 / -0

If one zero of the polynomial $$p(x)={x}^{3}-6{x}^{2}+11x-6$$ is $$3$$, find the other two zeroes.

A
$$0$$ and $$2$$

B
$$2$$ and $$-2$$

C
$$1$$ and $$2$$

D
$$2$$ and $$-3$$
Question 6
1 / -0

The number of rational roots of $$(2x+3)(2x+5)(x-1)(x-2)=30$$ is

A
$$4$$

B
$$3$$

C
$$1$$

D
$$5$$
Question 7
1 / -0

Using identities, evaluate:
$$71^2$$.

A
$$5041$$

B
$$3041$$

C
$$6041$$

D
$$2041$$

Solution

Given, $$71^2$$,
i.e. $$71^2$$ $$=(70+1)^2$$.

We know,
$$(x+y)^{ 2 }=x^{ 2 }+y^{ 2 }+2xy$$.

Then,
$$71^{ 2 }=(70+1)^{ 2 }\\ =70^{ 2 }+(2\times 70\times 1) + 1^2\\ =4900+140+1\\ =5040+1 \\=5041.$$

Hence, $$71^2=5041$$.

Therefore, option $$A$$ is correct.
Question 8
1 / -0

Find the zeros of the polynomial $$3\sqrt2 x^2+ 13x + 6 \sqrt2 $$ and verify the relationship between the zeros.

A
$$x=\dfrac{\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$$

B
$$x={2\sqrt2}\text{ and } \dfrac{-3}{\sqrt{2}}$$

C
$$x=\dfrac{-2\sqrt2}{3}\text{ and } {-3}{\sqrt{2}}$$

D
$$x=\dfrac{2\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$$

Solution

Let us first factorize the given equation $$3\sqrt 2x^2+13x+6\sqrt2$$ as shown below:

$$\Rightarrow 3\sqrt { 2 } x^{ 2 }+13x+6\sqrt { 2 } \\ =3\sqrt { 2 } x^{ 2 }+9x+4x+6\sqrt { 2 } \\ =3x(\sqrt { 2 } x+3)+2\sqrt { 2 } (\sqrt { 2 } x+3)\\ =(3x+2\sqrt { 2 } )(\sqrt { 2 } x+3)$$

Therefore, the zeroes of the given polynomial are:

$$(3x+2\sqrt { 2 } )=0\\ \Rightarrow x=-\dfrac { 2\sqrt { 2 } }{ 3 }=\alpha\ \\ (\sqrt { 2 } x+3)=0\\ \Rightarrow x=-\dfrac { 3 }{ \sqrt { 2 } }=\beta$$

Now, the sum and product of the roots is as follows:

$$\alpha +\beta =-\dfrac { 2\sqrt { 2 } }{ 3 } -\dfrac { 3 }{ \sqrt { 2 } } =-\left( \dfrac { 2\sqrt { 2 } }{ 3 } +\dfrac { 3 }{ \sqrt { 2 } } \right) =\dfrac { -4-9 }{ 3\sqrt { 2 } } =-\dfrac { 13 }{ 3\sqrt { 2 } } =-\dfrac { b }{ a } \\ \alpha \beta =\left( -\dfrac { 2\sqrt { 2 } }{ 3 } \right) \left( -\dfrac { 3 }{ \sqrt { 2 } } \right) =2=\dfrac { 6\sqrt { 2 } }{ 3\sqrt { 2 } } =\dfrac { c }{ a }$$

Hence verified.
Question 9
1 / -0

The value of $${ \left( 1.02 \right) }^{ 2 }+{ \left( 0.98 \right) }^{ 2 }$$, corrected to three decimal places is:

A
$$2.001$$

B
$$2.004$$

C
$$2.006$$

D
$$1.995$$

Solution

Given, $$(1.02)^2+(0.98)^2$$.

We know, $$(a+b)^2= a^2+2ab+b^2$$.

Then,
$$(1.02)^2+(0.98)^2$$
$$=(1.02)^2+(0.98)^2+(2\times 1.02\times 0.98)-(2\times 1.02\times 0.98$$)
$$=(1.02+0.98)^2-1.9992$$
$$=(2)^2-1.9992$$
$$=4-1.9992$$
$$=2.0008$$.

When corrected to three decimal digit,
$$2.0008$$ $$=2.001$$.

Therefore, option $$A$$ is correct.
Question 10
1 / -0

Which of the following is the zeroes of the polynomial $$x^2 + 4x + 4 =$$?

A
$$2$$

B
$$-2$$

C
$$4$$

D
$$-4$$

Solution

Now,
$$x^2 + 4x + 4$$
$$=(x+2)^2$$.
If $$x^2+4x+4=0$$ then $$x=-2,-2$$.
So the the zeroes of $$x^2+4x+4 $$ is $$-2$$.

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Polynomials Test - 41

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Polynomials Test - 41

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SHARING IS CARING

Question 1

-5/3

-2

-1

1

Solution

Question 2

$$2$$

$$3$$

$$5$$

$$1$$

Solution

Question 3

$$f(1)=2, f(2)=\dfrac{17}{4}$$

$$f(2)=2, f(1)=\dfrac{17}{4}$$

$$f(1)=-2, f(2)=\dfrac{17}{4}$$

$$f(2)=2, f(1)=-\dfrac{17}{4}$$

Solution

Question 4

constant polynomial

linear polynomial

quadratic polynomial

cubic polynomial

Solution

Question 5

$$0$$ and $$2$$

$$2$$ and $$-2$$

$$1$$ and $$2$$

$$2$$ and $$-3$$

Question 6

$$4$$

$$3$$

$$1$$

$$5$$

Question 7

$$5041$$

$$3041$$

$$6041$$

$$2041$$

Solution

Question 8

$$x=\dfrac{\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$$

$$x={2\sqrt2}\text{ and } \dfrac{-3}{\sqrt{2}}$$

$$x=\dfrac{-2\sqrt2}{3}\text{ and } {-3}{\sqrt{2}}$$

$$x=\dfrac{2\sqrt2}{3}\text{ and } \dfrac{-3}{\sqrt{2}}$$

Solution

Question 9

$$2.001$$

$$2.004$$

$$2.006$$

$$1.995$$

Solution

Question 10

$$2$$

$$-2$$

$$4$$

$$-4$$

Solution

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