Polynomials Test

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Polynomials Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

The zeroes of an expression represented by the

following graph are:

A
1 and 1

B
lie between 1 and 1

C
does not lie between 1 and 1

D
None of these
Question 2
1 / -0

The zero polynomial has :

A
One zero

B
two zeroes

C
infinite zeroes

D
No zero

Solution

Zeroes of the polynomial mean to find those values of $$x$$ for which the value of the polynomial becomes zero.
i.e. $$p(x)=0$$
Now, we have a zero polynomial $$p(x)=0$$.
This is the zero polynomial and will be true for all values of x belonging to complex numbers.
Hence, infinite no. of solutions.
Question 3
1 / -0

If $$\alpha, \beta$$ are real and $$\alpha^2, -\beta^2$$ are the roots of $$a^2 x^2 + x+1-a^2=0;\ (a > 1)$$, then $$\beta^2=$$

A
$$a^2$$

B
$$1$$

C
$$1-a^2$$

D
$$1+a^2$$

Solution

$$\alpha^2, -\beta^2$$ are the roots of $$a^2x^2+x+1-a^2=0$$
$$\Rightarrow \alpha^2 - \beta^2=\displaystyle -\frac{1}{a^2},$$
And
$$\Rightarrow\alpha^2 (-\beta^2)=\dfrac{1-a^2}{a^2}$$

$$(\alpha^2 + \beta^2)^2 = (\alpha^2 - \beta^2)^2 + 4\alpha^2 \beta^2$$

$$=\displaystyle \frac{1}{a^4}+4 \left ( \frac{a^2-1}{a^2} \right ) = \frac{1}{a^4}+ 4 - \frac{4}{a^2}$$

$$=\displaystyle \left ( 2- \frac{1}{a^2} \right )^2 \Rightarrow \alpha^2 + \beta^2 =2 - \frac{1}{a^2}$$

$$\beta^2 = \displaystyle \frac{1}{2} \left [ (\alpha^2 + \beta^2) - (\alpha^2 - \beta^2) \right ]$$

$$=\displaystyle \frac{1}{2}\left [ 2 - \frac{1}{a^2} + \frac{1}{a^2} \right ] =1$$
Question 4
1 / -0

If the zeros of the polynomial $$f(x)=k^2x^2-17x+k+2, (k > 0)$$ are reciprocal of each other, then the value of $$k$$ is

A
$$2$$

B
$$-1$$

C
$$-2$$

D
$$1$$

Solution

We know that the product of roots of equation $$ax^{2}+bx+c=0$$ is equal to $$\dfrac{c}{a}$$
Let $$\alpha, \dfrac {1}{\alpha}$$ are the roots of $$k^2x^2-17x+(k+2)$$

$$\alpha \times \dfrac {1}{\alpha}=\dfrac {k+2}{k^2}$$

$$\Rightarrow k^2=k+2$$

$$\Rightarrow k^2-k-2=0$$

$$\Rightarrow (k-2)(k+1)=0$$

$$\Rightarrow k=2$$ and $$k=-1$$

But $$k > 0$$ $$\therefore k=2$$
Question 5
1 / -0

If the roots of $$x^2 - bx + c = 0$$ are each decreased by $$2$$, then resulting equation is $$x^2 - 2x + 1 = 0$$, if and only if

A
$$b=6,c=9$$

B
$$b=3,c=5$$

C
$$b=2,c=-1$$

D
$$b=-4,c=3$$

Solution

We know that equation $$ax^{2}+bx+c=0$$
Then sum of roots $$=\dfrac{-b}{a}$$ and product of roots$$=\dfrac{c}{a}$$
For the 1st equation,
Let the roots be $$\alpha$$ and $$\beta$$
$$\alpha +\beta =b$$ and $$\alpha \beta =c$$ ...(1)

As per the given condition,
$$ (\alpha -2 +\beta -2 )=2$$ ...(2)
$$\therefore \alpha + \beta = 6$$
$$\therefore b= 6$$ ...from (1) & (2)

Also, $$ (\alpha-2)(\beta-2) = 1$$ ...(3)
$$\therefore \alpha \beta-2(\alpha +\beta) +4=1$$
$$\therefore c -2(6) = -3$$
$$\therefore c = 9$$
Question 6
1 / -0

If the roots of the given equation $$2x^2+3(\lambda -2)x+\lambda +4=0$$ be equal in magnitude but opposite in sign, then value of $$\lambda$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$\dfrac 23$$

Solution

Given equation is $$2x^2+3(\lambda -2)x+\lambda +4=0$$
We know that equation $$ax^{2}+bx+c=0$$
Then sum of roots $$=\dfrac{-b}{a}$$
Let $$\alpha$$, $$\beta$$ be the roots of the given equation
$$\therefore \alpha+\beta=-\dfrac{3(\lambda-2)}{2}$$
But it is given that
$$\alpha=-\beta$$
Therefore, $$ -\beta+\beta=-\dfrac{3(\lambda-2)}{2}$$

$$ 0=-\dfrac{3(\lambda-2)}{2}$$

$$ 0\times 2=-{3(\lambda-2)}$$

$$ \dfrac{0}{-3}=(\lambda-2)$$

$$ 0=(\lambda-2)$$
$$\lambda-2=0$$
Hence, $$\therefore \lambda=2$$
Question 7
1 / -0

If $$\alpha, \beta, \gamma $$ are the roots of the equation $$2x^3-3x^2+6x+1=0$$, then $$\alpha^2 +\beta^2+ \gamma^2$$ is equal to

A
$$-\dfrac{15}{4}$$

B
$$\dfrac{15}{4}$$

C
$$\dfrac{9}{4}$$

D
$$4$$

Solution

Given equation $$2x^3 - 3x^2 + 6x + 1 = 0$$,
Sum of roots:
$$\alpha +\beta +\gamma =\dfrac{-b}{a}$$
So, $$\alpha +\beta +\gamma =\dfrac{3}{2}$$,
Product of roots:
$$\alpha \beta \gamma =\dfrac{-d}{a}$$
$$\alpha \beta \gamma =\dfrac{-1}{2}$$,
Sum of products taken $$2$$ at a time:
$$\Sigma \alpha \beta =\alpha\beta+\beta\gamma+\alpha\gamma=\dfrac{c}{a}$$
$$\alpha \beta +\beta\gamma+\alpha\gamma=\dfrac62 =3$$

Now, we know that
$$(\alpha+\beta+\gamma)^2 = \alpha^2+\beta^2+\gamma^2 + 2(\alpha\beta+\beta\gamma+\gamma\alpha)$$

Re-arranging, we get
$$\alpha^2 +\beta^2+ \gamma^2 =(\alpha +\beta+ \gamma)^2-2(\Sigma \alpha \beta )$$
$$\alpha^2+\beta^2+\gamma^2 = \left(\dfrac{3}{2} \right)^2-2\times3$$
$$=\dfrac{9}{4}-6$$
$$=-\dfrac{15}{4}$$

Hence, option A.
Question 8
1 / -0

If $$'k'$$ be the ratio of the roots of the equation $$x^{2} - px +q = 0$$, the value of $$\dfrac{k}{1 + k^{2}}$$ is

A
$$\dfrac{q^{2} - 2p}{p}$$

B
$$\dfrac{q}{p^{2} - 2q}$$

C
$$\dfrac{p}{q^{2} - p}$$

D
$$\dfrac{p}{p^{2} - 2q}$$

Solution

We know that equation $$ax^{2}+bx+c=0$$
Then sum of roots $$=\dfrac{-b}{a}$$
Let the roots of the equation be a and b.
Then, $$a + b = p$$
$$ab = q$$

$$\dfrac{a}{b} = k$$
Now, $$\dfrac{k}{1+k^2}$$
$$ = \dfrac{\dfrac{a}{b}}{1+ (\dfrac{a}{b})^2}$$
$$= \dfrac{ab}{a^2+b^2}$$
$$= \dfrac{ab}{(a+b)^2 - 2ab}$$
$$= \dfrac{q}{p^2 - 2q}$$
Question 9
1 / -0

If $$\alpha, \beta$$ be the zeros of the quadratic polynomial $$2x^2+5x+1$$, then the value of $$\alpha+\beta+\alpha \beta=$$

A
-2

B
-1

C
1

D
none of these

Solution

We know that equation $$ax^{2}+bx+c=0$$
Then sum of roots $$=\dfrac{-b}{a}$$ and product of roots$$=\dfrac{c}{a}$$
Let $$\alpha ,\beta $$ are the zeros of the quadratic equation.

Therefore,
$$\displaystyle \alpha +\beta =-\dfrac { 5 }{ 2 }$$ and $$\alpha \beta =\dfrac { 1 }{ 2 }$$
Now,
$$\displaystyle\alpha +\beta +\alpha \beta = -\dfrac { 5 }{ 2 } +\dfrac { 1 }{ 2 } $$

$$ = \dfrac { -4 }{ 2 } $$

$$ =-2$$
Question 10
1 / -0

Quadratic polynomial having sum of it's zeros is 5 and product of it's zeros is - 14 is

A
$$x^2-5x-14$$

B
$$x^2-10x-14$$

C
$$x^2-5x+14$$

D
none of these

Solution

If $$\alpha$$ $$\beta$$ be the zeros of the quadratic polynomial,
as given sum of it's zeros $$5$$ and product of it's zeros $$- 14$$
then $$\alpha + \beta =5$$ and $$\alpha \beta =-14$$

$$\left( x-\alpha \right) \left( x-\beta \right)$$ is the quadratic polynomial.
$$=>{ x }^{ 2 }-2x-\beta x+\alpha \beta $$
$$=>{ x }^{ 2 }-\left( \alpha +\beta \right) x+\alpha \beta $$
$$=>{ x }^{ 2 }-5x+\left( -14 \right) $$
$$=>{ x }^{ 2 }-5x-14$$
Therefore quadratic polynomial is
$$ { x }^{ 2 }-5x-14 $$

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Polynomials Test - 34

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Polynomials Test - 34

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Question 1

1 and 1

lie between 1 and 1

does not lie between 1 and 1

None of these

Question 2

One zero

two zeroes

infinite zeroes

No zero

Solution

Question 3

$$a^2$$

$$1$$

$$1-a^2$$

$$1+a^2$$

Solution

Question 4

$$2$$

$$-1$$

$$-2$$

$$1$$

Solution

Question 5

$$b=6,c=9$$

$$b=3,c=5$$

$$b=2,c=-1$$

$$b=-4,c=3$$

Solution

Question 6

$$1$$

$$2$$

$$3$$

$$\dfrac 23$$

Solution

Question 7

$$-\dfrac{15}{4}$$

$$\dfrac{15}{4}$$

$$\dfrac{9}{4}$$

$$4$$

Solution

Question 8

$$\dfrac{q^{2} - 2p}{p}$$

$$\dfrac{q}{p^{2} - 2q}$$

$$\dfrac{p}{q^{2} - p}$$

$$\dfrac{p}{p^{2} - 2q}$$

Solution

Question 9

-2

-1

1

none of these

Solution

Question 10

$$x^2-5x-14$$

$$x^2-10x-14$$

$$x^2-5x+14$$

none of these

Solution

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