Polynomials Test

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Polynomials Test - 35

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Question 1
1 / -0

STATEMENT -1 : From the given graph of y $$=$$ p(x), number of zeroes of p(x) is 1.
STATEMENT -2 : The number of points where the graph cuts the x-axis gives the zeroes of any polynomial p(x).

A
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

B
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

C
Statement - 1 is True, Statement - 2 is False

D
Statement - 1 is False, Statement - 2 is True

Solution

We know that the number of points where the graph cuts the x-axis gives the zeroes of any polynomial p(x). Since the polynomial is a linear line parallel to x-axis and is not cutting the x-axis at any point, therefore this polynomial has no zeroes/roots.
Question 2
1 / -0

Find the number of zeroes for the graphs shown respectively.

A
$$1, 3, 3$$

B
$$1, 2, 1$$

C
$$3, 1, 3$$

D
$$3, 3, 1$$

Solution

We know that the zeros of a polynomial can be found by finding where the graph of the polynomial crosses or touches the $$x-axis$$.

Now, consider the given graphs:

(i) The number of times the graph touches the $$x-axis$$ is one.

Hence, the number of zeroes of the polynomial is $$1$$.

(ii) The number of times the graph touches the $$x-axis$$ is three.

Hence, the number of zeroes of the polynomial is $$3$$.

(iii) The number of times the graph touches the $$x-axis$$ is three.

Hence, the number of zeroes of the polynomial is $$3$$.
Question 3
1 / -0

If $$\alpha$$ and $$\beta$$ are the zeros of the polynomial $$f(x)=6x^2-3-7x$$, then $$(\alpha+1)(\beta+1)$$ is equal to

A
$$\dfrac {5}{2}$$

B
$$\dfrac {5}{3}$$

C
$$\dfrac {2}{5}$$

D
$$\dfrac {3}{5}$$

Solution

We know that equation $$ax^{2}+bx+c=0$$
Then sum of roots $$=\dfrac{-b}{a}$$ and product of roots$$=\dfrac{c}{a}$$
If $$\alpha $$, $$\beta $$ are the zeros of the quadratic polynomial $$f(x)=6x^2-3-7x$$, then
$$\left( \alpha +\beta \right) =\dfrac { -b }{ a } =-\dfrac { (-7) }{ 6 }=\dfrac { 7 }{ 6 }, \quad \alpha \beta =\dfrac { c }{ a }=\dfrac { -3 }{ 6 }=\dfrac { -1 }{ 2 }$$
Now, $$(\alpha+1)(\beta +1)$$
$$=\alpha\beta + \alpha + \beta + 1$$
$$=\dfrac { -1 }{ 2 }+\dfrac { 7 }{ 6 }+1$$
$$=\dfrac { -3+7+6 }{ 6 } $$
$$=\dfrac { 10 }{ 6 }$$
$$=\dfrac { 5 }{ 3 }$$
Question 4
1 / -0

If $$\alpha, \beta$$ be the zeros of the quadratic polynomial $$2-3x-x^2$$, then $$\alpha+\beta=$$

A
$$2$$

B
$$9$$

C
$$1$$

D
none of these

Solution

If $$\alpha$$ and $$ \beta $$ are the zeros of the polynomial then
$$\left( x-\alpha \right) \left( x-\beta \right)$$ are the factors of the polynomial
Thus, $$\left( x-\alpha \right) \left( x-\beta \right) $$ is the polynomial.
So, the polynomial $$={ x }^{ 2 }-\alpha x-\beta x+\alpha \beta $$
$$ ={ x }^{ 2 }-\left( \alpha +\beta \right) x+\alpha \beta ....(i)$$
Now,the quadratic polynomial is
$$2-3x-{ x }^{ 2 } = $$ $${ x }^{ 2 }+3x-2 ....(ii)$$
Now, comparing equation (i) and (ii),we get,
$$-\left( \alpha +\beta \right)=3$$
$$\alpha +\beta =-3$$
Question 5
1 / -0

The sum and product of zeros of the quadratic polynomial are - 5 and 3 respectively the quadratic polynomial is equal to

A
$$x^2+2x+3$$

B
$$x^2-5x+3$$

C
$$x^2+5x+3$$

D
$$x^2+3x-5$$

Solution

$${\textbf{Step - 1: Stating the framing of quadratic equation}}{\text{.}}$$

$${\text{We can form a quadratic equation if we know that sum and product of the roots,}}$$

  $${\text{So a quadratic equation can be written as,}}$$

  $${x^2} - {\text{(sum of roots)}}x + {\text{(product of roots)}} = 0$$

$${\textbf{Step - 2: Framing the equation}}{\text{.}}$$

  $${\text{We know the sum of roots is - 5 and its product is 3}}{\text{.}}$$

  $${\text{So we can write the equation as,}}$$

  $${x^2} - ( - 5)x + 3 = 0$$

  $${x^2} + 5x + 3 = 0$$

$$\textbf{Hence option C is correct}$$
Question 6
1 / -0

What is the type of polynomial $$11\, =\, -4x^{2}\, -\, x^{3}$$?

A
Cubic

B
Quadratic

C
Linear

D
None of these

Solution

A cubic polynomial is a polynomial of degree $$3$$. A cubic polynomial has the form $$f(x)=a_3x^3+a_2x^2+a_1x+a_0$$.

Here, the polynomial $$11=-4x^2-x^3$$ or $$x^3+4x^2+11=0$$ is a polynomial of degree $$3$$ with coefficients $$a_1=0$$, $$a_2=4$$, $$a_3=1$$ and constant $$a_0=11$$.

Hence, $$11=-4x^2-x^3$$ is a cubic polynomial.
Question 7
1 / -0

Let $$p(x) = ax^2 + bx + c$$ be a quadratic polynomial. It can have at most

A
One zero

B
Two zeros

C
Three zeros

D
None of these

Solution

The polynomial $$ax^2+bx+c$$ has three terms. The first one is $$ax^2$$, the second is $$bx$$, and the third is $$c$$.

The exponent of the first term is $$2$$.

The exponent of the second term is $$1$$ because $$bx = bx^1$$.

The exponent of the third term is $$0$$ because $$c = cx^0$$.

Since the highest exponent is $$2$$, therefore, the degree of $$ax^2+bx+c$$ is $$2$$.

Since, the degree of the polynomial is $$2$$, hence, the polynomial $$ax^2+bx+c$$ can have zero, one or two zeroes.

Hence, the polynomial $$ax^2+bx+c$$ can have at most two zereos.
Question 8
1 / -0

If $$\alpha, \beta, \gamma$$ are zeros of the polynomial $$ax^3+bx^2+cx+d$$, then $$\alpha \beta \gamma = $$

A
$$-\dfrac {d}{a}$$

B
$$\dfrac {d}{a}$$

C
$$\dfrac {c}{a}$$

D
$$-\dfrac {c}{a}$$

Solution

If $$\alpha , \beta ,\gamma$$ are the zeros of the cubic polynomial
$$ax^{ 3 }+bx^{ 2 }+cx+d$$ then,

$$\alpha\beta\gamma= -\dfrac { \text{constant term} }{ \text{coefficient of}\ { x }^{ 3 } } $$

$$=-\dfrac{d}{a}$$
Question 9
1 / -0

If one of the zeros of the cubic polynomial $$x^3 + ax^2 + bx + c$$ is $$-1$$, then the product of the other two zeros is

A
$$b - a + 1$$

B
$$b - a - 1$$

C
$$a - b + 1$$

D
$$a - b - 1$$

Solution

Let $$\alpha , \beta$$ be the other zeros of the given polynomial $$x^3+ax^2+bx^2+c$$

Sum of the zeros $$= \dfrac { - \text{coefficient of } x^ 2 }{ \text{coefficient of } x^ 3 } $$

$$\Rightarrow -1+\alpha+\beta=\dfrac { -a }{ 1 }=-a $$

$$\Rightarrow \alpha+\beta= -a+1 $$ (i)

Again,
$$(-1)\alpha+\alpha\beta+(-1)\beta=\dfrac { - \text{coefficient of } x}{ \text{coefficient of } x^ 3 } $$

$$\Rightarrow -\alpha+\alpha\beta-\beta=\dfrac { b}{ 1 }$$

$$=\alpha\beta=b+\alpha+\beta$$
$$\alpha+\beta= -a+1 $$ , from (i))

$$=b-a+1$$
Question 10
1 / -0

If the zeros of the quadratic polynomial $$x^2 + (a + 1) x + b$$ are $$2$$ and $$-3$$, then

A
$$a=-7, b=-1$$

B
$$a=5, b=-1$$

C
$$a=2, b=-6$$

D
$$a=0, b=-6$$

Solution

$${ x }^{ 2 }+(a+1)x+b$$ is the quadratic polynomial.

$$2$$ and $$-3$$ are the zeros of the quadratic polynomial.

Thus, $$2+(-3)=\dfrac { -(a+1) }{ 1 } $$

$$=>\dfrac { (a+1) }{ 1 }=1$$

$$=>a+1=1$$

$$=>a=0$$

Also, $$2\times (-3) = b$$
$$=>b=-6$$

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Polynomials Test - 35

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Polynomials Test - 35

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Question 1

Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1

Statement - 1 is True, Statement - 2 is False

Statement - 1 is False, Statement - 2 is True

Solution

Question 2

$$1, 3, 3$$

$$1, 2, 1$$

$$3, 1, 3$$

$$3, 3, 1$$

Solution

Question 3

$$\dfrac {5}{2}$$

$$\dfrac {5}{3}$$

$$\dfrac {2}{5}$$

$$\dfrac {3}{5}$$

Solution

Question 4

$$2$$

$$9$$

$$1$$

none of these

Solution

Question 5

$$x^2+2x+3$$

$$x^2-5x+3$$

$$x^2+5x+3$$

$$x^2+3x-5$$

Solution

Question 6

Cubic

Quadratic

Linear

None of these

Solution

Question 7

One zero

Two zeros

Three zeros

None of these

Solution

Question 8

$$-\dfrac {d}{a}$$

$$\dfrac {d}{a}$$

$$\dfrac {c}{a}$$

$$-\dfrac {c}{a}$$

Solution

Question 9

$$b - a + 1$$

$$b - a - 1$$

$$a - b + 1$$

$$a - b - 1$$

Solution

Question 10

$$a=-7, b=-1$$

$$a=5, b=-1$$

$$a=2, b=-6$$

$$a=0, b=-6$$

Solution

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