Polynomials Test

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Polynomials Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

Zero of the polynomial $$p(x) = \sqrt{3}x + 3$$ is

A
$$-\sqrt{3}$$

B
$$\dfrac{-\sqrt 3}{3}$$

C
$$\dfrac{3}{\sqrt 3}$$

D
$$3\sqrt 3$$

Solution
Question 2
1 / -0

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

A
$$50$$

B
$$51$$

C
$$55$$

D
$$35$$

Solution

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, $$p(3)=51$$.
Question 3
1 / -0

Identify the zeroes of the given polynomial.

$$p(z)=4z^2-15z\pi -4\pi ^3$$

A
$$4\pi ,\frac{-\pi }{4}$$

B
$$-4\pi ,\frac{\pi }{4}$$

C
$$4\pi ,\frac{\pi }{4}$$

D
$$-4\pi ,\frac{-\pi }{4}$$
Question 4
1 / -0

$$ \alpha, \beta, \gamma$$ be the zeroes of the expression
$$ax^{3} +

bx^{2} + 4x + 7$$, then the value of
$$ \alpha\beta + \beta\gamma + \gamma\alpha$$ is:

A
-4

B
$$\frac{4}{\alpha}$$

C
$$-\frac{4}{\alpha}$$

D
None of these
Question 5
1 / -0

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

A
$$-81$$

B
$$-137$$

C
$$-119$$

D
$$-100$$

Solution

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, $$p(-3)=-119$$.
Question 6
1 / -0

If $$f(x)=x^{6}-10x^{5}-10x^{4}-10x^{3}-10x^{2}-10x+10$$, the value of $$f(11)$$ is

A
$$1$$

B
$$10$$

C
$$11$$

D
$$21$$

Solution

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, $$f(11)=21$$.
Question 7
1 / -0

Two roots of the equation $$4x^3 - px^2+ qx - 2p = 0$$ are 4 and 7. What is the third root?

A
$$\frac{11}{27}$$

B
$$\frac{11}{13}$$

C
11

D
$$\frac{11}{15}$$

E
$$-\frac{22}{27}$$

Solution
Question 8
1 / -0

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

A
$$-4$$

B
$$\dfrac {17}{2}$$

C
$$9$$

D
$$\dfrac {-687}{2}$$

E
$$\dfrac {427}{2}$$
Question 9
1 / -0

Calculate the number of real numbers $$k$$ such that $$f(k)=2$$ if $$f(x)={ x }^{ 4 }-3{ x }^{ 3 }-9{ x }^{ 2 }+4$$.

A
None

B
One

C
Two

D
Three

E
Four

Solution

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.
Question 10
1 / -0

The number of polynomials having zeroes as $$-2$$ and $$5$$ is?

A
$$0$$

B
$$2$$

C
$$3$$

D
More than $$3$$

Solution

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Re-Attempt Weekly Quiz Competition

Polynomials Test - 44

Result

Polynomials Test - 44

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SHARING IS CARING

Question 1

$$-\sqrt{3}$$

$$\dfrac{-\sqrt 3}{3}$$

$$\dfrac{3}{\sqrt 3}$$

$$3\sqrt 3$$

Solution

Question 2

$$50$$

$$51$$

$$55$$

$$35$$

Solution

Question 3

$$4\pi ,\frac{-\pi }{4}$$

$$-4\pi ,\frac{\pi }{4}$$

$$4\pi ,\frac{\pi }{4}$$

$$-4\pi ,\frac{-\pi }{4}$$

Question 4

-4

$$\frac{4}{\alpha}$$

$$-\frac{4}{\alpha}$$

None of these

Question 5

$$-81$$

$$-137$$

$$-119$$

$$-100$$

Solution

Question 6

$$1$$

$$10$$

$$11$$

$$21$$

Solution

Question 7

$$\frac{11}{27}$$

$$\frac{11}{13}$$

11

$$\frac{11}{15}$$

$$-\frac{22}{27}$$

Solution

Question 8

$$-4$$

$$\dfrac {17}{2}$$

$$9$$

$$\dfrac {-687}{2}$$

$$\dfrac {427}{2}$$

Question 9

None

One

Two

Three

Four

Solution

Question 10

$$0$$

$$2$$

$$3$$

More than $$3$$

Solution

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