MCQ AND CASE STUDY OF POLYNOMIALS

Click here for The New pattern Examination (2021-22) Chapter 2 Polynomials

A polynomial is a finite expression constructed from variables and constants. using the operations of addition. subtraction. multiplication and taking non-negative integer powers it can be written as the sum of a finite number of terms.

POLYNOMIALS :-a mathematical expression of one or more algebraic terms each of which consists of a constant multiplied by one or more variables raised to a nonnegative integral power (such as a + bx + cx²)

Degree of a Polynomial : The highest power of x in a polynomial p(x) is called the degree of the polynomial.

Example 1. Write the degree of the polynomial:

x²+2x+4. The degree of the equation is 2 . i.e. the highest power of variable in the equation.

Value of a Polynomial : If p(x) is a polynomial in x and ‘a’ is a real number, then the value obtained by putting x = a in p(x), is the value of p(x) at

x = a and is denoted by p(a).

Example 2. Find the value of the polynomial

p(x) = 5x – 4x² + 3

On putting x = 0 ,

p(0) = 5(0) – 4(0)² + 3

= 0 – 0 + 3

p(0) = 3

constant polynomial : A polynomial having its highest degree zero is called a constant polynomial. It has no variables, only constants.

For example: f(x) = 12, g(x) = -51 , h(y) = 3/2 etc are constant polynomials.

linear polynomial : A polynomial having its highest degree one is called a linear polynomial. For example, f(x) = x- 2, g(x) = 2 x , h(x) = -9x + 8 are linear polynomials. In general g(x) = ax + b , a ≠ 0 is a linear polynomial.

Quadratic polynomial :- A quadratic polynomial is a polynomial of degree 2. A univariate quadratic polynomial has the form. . An equation involving a quadratic polynomial is called a quadratic equation.

Important questions for class 10 maths chapter 2 is prepared by CBSE after thorough research. These questions have the highest probability of coming in the examinations as per the New pattern 2021. class 10 polynomials important questions are prepared as per the examination guidelines to help you score well in your examinations.

Class 10 maths ch 2 important questions will give you a better understanding of the type of questions asked from this chapter. If you prepare all of these questions well, you will be able to solve any type of question that comes in the examination from this chapter. It will also help in testing your level of knowledge of this chapter and if you lack the understanding, you can put some extra effort.

The graph of a quadratic equation in two variables (y = ax²+ bx + c) is called a parabola. The following graphs are two typical parabolas their x-intercepts are marked by red dots, their y-intercepts are marked by a pink dot, and the vertex of each parabola is marked by a green dot:

We say that the first parabola opens upwards (is a U shape) and the second parabola opens downwards (is an upside down U shape). In order to graph a parabola we need to find its intercepts, vertex, and which way it opens.

Given y = ax² + bx + c , we have to go through the following steps to find the points and shape of any parabola:

Label a, b, and c.
Decide the direction of the parabola:

If a > 0 (positive) then the parabola opens upward.
If a < 0 (negative) then the parabola opens downward.

Find the x-intercepts :

Notice that the x-intercepts of any graph are points on the x-axis and therefore have y-coordinate 0. We can find these points by plugging 0 in for y and solving the resulting quadratic equation (0 = ax² + bx + c). If the equation factors we can find the points easily, but we may have to use the quadratic formula in some cases. If the solutions are imaginary, that means that the parabola has no x-intercepts (is strictly above or below the x-axis and never crosses it). If the solutions are real, but irrational (radicals) then we need to approximate their values and plot them.

Find the y-intercept:

The y-intercept of any graph is a point on the y-axis and therefore has x-coordinate 0. We can use this fact to find the y-intercepts by simply plugging 0 for x in the original equation and simplifying. Notice that if we plug in 0 for x we get: y = a(0)² + b(0) + c or y = c. So the y-intercept of any parabola is always at (0,c).

Find the vertex (h,k):

To find the x-coordinate for the vertex we use the following formula:

To find the y-coordinate for the vertex we plug in h in the original equation:
k = a(h)² + b(h) + c

Plot the points and graph the parabola

Example (1)

Graph y = x² + 2x – 8

In this problem: a = 1, b = 2 , and c = -8.

Since “a” is positive we’ll have a parabola that opens upward (is U shaped).
To find the x-intercepts we plug in 0 for y:
0 = x² + 2x – 8 (which factors)
0 = (x + 4)(x – 2)
x = -4 or x = 2
So this parabola has two x-intercepts: (-4,0) and (2,0).

To find the y-intercept we plug in 0 for x:
y = (0)² + 2(0) – 8 = -8
So the y-intercept of the parabola is (0,-8).

To find the vertex we use:

and to find k, we plug in -1 in for x:
k = (-1)² + 2(-1) – 8
k = 1 – 2 – 8 = -9
The vertex of this parabola is at (-1, -9)

Example (2)

Graph y = -3x² + 3

In this problem a = -3, b = 0 and c = 3.

Since “a” is negative this parabola is going to open downward (upside down U shape).

To find the x-intercepts we plug in 0 for y:
0 = -3x² + 3 (this equation factors)
0 = -3(x² – 1)
0 = -3(x – 1)(x + 1) and since -3 can not equal zero:
x = 1 or x = -1
The x-intercepts are: (1,0) and (-1,0)

The y-intercept is found by plugging 0 for x:
y = -3(0)² + 3 = 3
So, the y-intercept is at (0,3).

And to find the vertex:

k = -3(0)² + 3 = 3
So the vertex is at (0,3).

Notice that in this problem the vertex and the y-intercept are the same point.

Example (3)

Graph y = x² + 4x + 7
a = 1, b = 4, and c = 7

Since a 0 the parabola opens up (is U shaped).

To find the x -intercept we plug in 0 for y:
0 = x² + 4x + 7 (this expression does not factor so we have to use the quadratic formula)

Since the roots are imaginary the parabola has no x-intercepts.

We find the y-intercepts by plugging in 0 for x:
y = 0² + 4(0) + 7 = 7
The y-intercept is (0,7).

The vertex:

So the vertex is at (-2, 3).

CASE STUDY 1:

The below picture are few natural examples of parabolic shape which is represented by a quadratic polynomial. A parabolic arch is an arch in the shape of a parabola. In structures, their curve represents an efficient method of load, and so can be found in bridges and in architecture in a variety of forms.

1. In the standard form of quadratic polynomial, ax² + bx + c, a, b and c are

a) All are Polynomials.

b) All are rational numbers.

c) ‘a’ is a non zero real number and b and c are any Polynomials.

d) All are integers.

Answers: c) ‘a’ is a non zero real number and b and c are any Polynomials.

2. If the roots of the quadratic polynomial are equal, where the discriminant D = b² – 4ac, then

a) D > 0

b) D < 0

c) D ≥ 0

d) D = 0

Answers: d) D = 0

3. If α and 1/α are the zeroes of the quadratic polynomial 2x² – x + 8k, then k is

a) 4

b) 1/4

c) –1/4

d) 2

Answers: b) 1/4

4. The graph of x²+1 = 0

a) Intersects x‐axis at two distinct points.

b)Touches x‐axis at a point.

c) Neither touches nor intersects x‐axis.

d)Either touches or intersects x‐ axis.

Answers: c) Neither touches nor intersects x‐axis.

5. If the sum of the roots is –p and product of the roots is –1/p, then the quadratic polynomial is

a) k(–px² + x/p + 1)

b) k(px² – x/p – 1)

c) k(x² + px – 1/p)

d) k(x² – px + 1/p)

Answers: c) k(x² + px – 1/p)

CASE STUDY 2:

An asana is a body posture, originally and still a general term for a sitting meditation pose, and later extended in hatha yoga and modern yoga as exercise, to any type of pose or position, adding reclining, standing, inverted, twisting, and balancing poses. In the figure, one can observe that poses can be related to representation of quadratic polynomial.

1. The shape of the poses shown is

a) Spiral

b) Ellipse

c) Linear

d) Parabola

Answer: d) Parabola

2. The graph of parabola opens downwards, if _______

a) a ≥ 0

b) a = 0

c) a < 0

d) a > 0

Answer: c) a < 0

3. In the graph, how many zeroes are there for the polynomial?

a) 0

b) 1

c) 2

d) 3

Answer: c) 2

4. The two zeroes in the above shown graph are

a) 2, 4

b) -2, 4

c) -8, 4

d) 2, -8

Answer: b) -2, 4

CASE STUDY 3:

Basketball and soccer are played with a spherical ball. Even though an athlete dribbles the ball in both sports, a basketball player uses his hands and a soccer player uses his feet. Usually, soccer is played outdoors on a large field and basketball is played indoor on a court made out of wood. The projectile (path traced) of soccer ball and basketball are in the form of parabola representing quadratic polynomial.

1. The shape of the path traced shown is

a) Spiral

b) Ellipse

c) Linear

d) Parabola

Answer: d) Parabola

2. The graph of parabola opens upwards, if _______

a) a = 0

b) a < 0

c) a > 0

d) a ≥ 0

Answer: c) a > 0

3. Observe the following graph and answer

In the above graph, how many zeroes are there for the polynomial?

a) 0

b) 1

c) 2

d) 3

Answer: d) 3

4. The three zeroes in the above shown graph are

a) 2, 3,-1

b) -2, 3, 1

c) -3, -1, 2

d) -2, -3, -1

Answer: c) -3, -1, 2

5. What will be the expression of the polynomial?

a) x³ + 2x² – 5x – 6

b) x³ + 2x² – 5x + 6

c) x³ + 2x² + 5x – 6

d) x³ + 2x² + 5x + 6

Answer: a) x³ + 2x² – 5x – 6

Case Study 4 Questions

While playing in garden, Sahiba saw a honeycomb and asked her mother what is that. She replied that it’s a honeycomb made by honey bees to store honey. Also, she told her that the shape of the honeycomb formed is parabolic. The mathematical representation of the honeycomb structure is shown in the graph.

Based on the above information, answer the following questions.
(i) Graph of a quadratic polynomial is in ___________ shape.

(a) straight line	(b) parabolic
(c) circular	(d) None of these

(ii) The expression of the polynomial represented by the graph is

(a) x²– 49

(b) x²– 64

(d) x²– 81

(iii) Find the value of the polynomial represented by the graph when x = 6.

(a) -2

(b) -1

(d) 1

(iv) The sum of zeroes of the polynomial x² + 2x – 3 is

(a) -1

(b) -2

(d) 1

(v) If the sum of zeroes of polynomial at² + 5t + 3a is equal to their product, then find the value of a.

(a) -5

(b) -3

$(c) \frac{5}{3}$

$(d) \frac{- 5}{3}$

Pankaj’s father gave him some money to buy avocado from the market at the rate of p(x) = x²– 24x + 128. Let α , $β$ are the zeroes of p(x).
Based on the above information, answer the following questions.

(i) Find the value of α and $β$ , where α < $β$ .

(a) -8 , -16 (b) 8, 16 (c) 8 , 15 (d) 4 , 9

(ii) Find the value of $α$ + $β$ + $α$ $β$ .

(a) 151 (b) 158 (c) 152 (d) 155

(iii) The value of p(2) is

(a) 80 (b) 81 (c) 83 (d) 84

(iv) If $α$ and $β$ are zeroes of $x^{2} + x - 2, then \frac{1}{α} + \frac{1}{β} =$

(a) 1/2 (b) 1/3 (c) 1/4 (d) 1/5

(v) If sum of zeroes of $q (x) = k x^{2} + 2 x + 3 k$ is equal to their product, then k =

(a) 2/3 (b) 1/3 (c) -2/3 (d) -1/3
In a soccer match, the path of the soccer ball in a kick is recorded as shown in the following graph.

Based on the above i!;formation, answer the following questions.
(i) The shape of path of the soccer ball is a

(a) Circle (b) Parabola (c) Line (d) None of these

(ii) The axis of symmetry of the given parabola is

(a) y – axis (b) x – axis

(c) line parallel to y – axis (d) line parallel to x – axis

(iii) The zeroes of the polynomial, represented in the given graph, are

(a) – 1 , 7 (b) 5 ,- 2 (c) – 2 , 7 (d) – 3 , 8

(iv) Which of the following polynomial has -2 and -3 as its zeroes?

$(a) x^{2} - 5 x - 5$ $(b) x^{2} + 5 x - 6$ $(c) x^{2} + 6 x - 5$ $(d) x^{2} + 5 x + 6$

(v) For what value of ‘x’, the value of the polynomial $f (x) = (x - 3)^{2} + 9 is 9 ?$

(a) 1 (b) 2 (c) 3 (d) 4

5
Shweta and her husband Sunil who is an architect by profession, visited France. They went to see Mont Blanc Tunnel which is a highway tunnel between France and Italy, under the Mont Blanc Mountain in the Alps, and has a parabolic cross-section. The mathematical representation of the tunnel is shown in the graph.

Based on the above information, answer the following questions.
(i) The zeroes of the polynomial whose graph is given, are

(a) -2 , 8 (b) -2 , -8 (c) 2 , 8 (d) -2 , 0

(ii) What will be the expression of the polynomial given in diagram?

$(a) x^{2} - 6 x + 16$ $(b) - x^{2} + 6 x + 16$ $(c) x^{2} + 6 x + 16$ $(d) - x^{2} - 6 x - 16$

(iii) What is the value of the polynomial, represented by the graph, when x = 4?

(a) 22 (b) 23 (c) 24 (d) 25

(iv) If the tunnel is represented by x² + 3x – 2, then its zeroes are

(a) -1 , – 2 (b) 1 , – 2 (c) -1 , 2 (d) 1 , 2

(v) If one zero is 4 and sum of zeroes is -3, then representation of tunnel as a polynomial is

$(a) x^{2} - x + 24$ $(b) - x^{2} - 3 x + 28$ $(c) x^{2} + x + 28$ $(d) x^{2} - x + 28$

Shray, who is a social worker, wants to distribute masks, gloves, and hand sanitizer bottles in his block. Number of masks, gloves and sanitizer bottles distributed in 1 day can be represented by the zeroes $α, β, γ, (α > β > γ)$ of the polynomial $p (x) = x^{3} - 18 x^{2} + 95 x - 150 .$

Based on the above information, answer the following questions.
(i) Find the value of $α, β, γ .$

(a) – 10 , – 5 ,- 3 (b) 3 , 6 , 5

(c) 10 , 5 , 3 (d) 4 , 8 , 9

(ii) The sum of product of zeroes taken two at a time is

(a) 91 (b) 92 (c) 94 (d) 95

(iii) Product of zeroes of polynomial p(x) is

(a) 150 (b) 160 (c) 170 (d) 180

(iv) The value of the polynomial p(x), when x = 4 is

(a) 5 (b) 6 (c) 7 (d) 8

(v) If $α, β, γ$ are the zeroes of a polynomial g(x) such that $α + β + γ = 3, α β + β γ + γ α = - 16$ and $α β γ = - 48$ g(x) =

$(a) x^{3} - 2 x^{2} - 48 x + 6$ $(b) x^{3} + 3 x^{2} + + 16 x - 48$

$(c) x^{3} - 48 x^{2} - 16 x + 3$ $(d) x^{3} - 3 x^{2} - 16 x + 48$

ASSERTION REASONING QUESTIONS

DIRECTION : In the following questions, a statement of assertion (A) is followed by a statement of reason (R). Mark the correct choice as:
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation
of assertion (A). (b)Both assertion (A) and reason (R) are true but reason (R) is not the correct
explanation of assertion (A).

(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Assertion: Degree of a zero polynomial is not defined.

Reason: Degree of a non-zero constant polynomial is 0

a) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
b) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
c) Assertion is correct but Reason is incorrect
d) Assertion is incorrect but Reason is correct

Correct answer is option ‘B’. Can you explain this answer?

1. Assertion : P(x) = $4x^{3}- x^{^2} + 5x^{4}+3x -2$ is a polynomial of degree 3.

Reason : The highest power of x in the polynomial P(x) is the degree of the polynomial.
(a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans: The highest power of x in the polynomial P(x) = 4x³ – x² + 5 $x^{_{4}}$ + 3x – 2 is 4.
Therefore, the degree of the polynomial P(x) is 4.
Correct option is (d) Assertion (A) is false but reason (R) is true.

2. Assertion : x³ + x has only one real zero.
Reason : A polynomial of nth degree must have n real zeroes.
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : Reason is false [a polynomial of nth degree has at most n zeroes.]
Again, x³ + x = x(x² + 1)
which has only one real zero i.e. x = 0
[x² + 1 ≠ 0 for all x ∈ R]
Assertion is true.
Correct option is (c) Assertion (A) is true but reason (R) is false

3. Assertion : If one zero of poly-nominal p(x) = (k² + 4)x² + 13x + 4k is reciprocal of
other, then k = 2.
Reason : If (x – a) is a factor of p(x), then p(a) = 0 i.e. a is a zero of p(x).
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : Let α, 1/α be the zeroes of p(x) then we have
Product of Zeroes = $\alpha \times \frac{1}{\alpha } = \frac{4k}{k^{2}+4}= 1$

⇒ k² – 4k + 4 = 0
⇒ (k – 2)² = 0 ⇒ k = 2
(b) Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).

4. Assertion : x² + 4x + 5 has two zeroes.
Reason : A quadratic polynomial can have at the most two zeroes.
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : p(x) = 0 ⇒ x² + 4x + 5 = 0
Discriminant, D = b² – 4ac
= 4² – 4 x 1 x 5
= 16 – 20 = – 4 < 0
Therefore, no real zeroes are there.

(d) Assertion (A) is false but reason (R) is true.

5. Assertion : The graph y = f(x) is shown in figure, for the polynomial f (x). The number

of zeros of f(x) is 3.
Reason : The number of zero of the polynomial f(x) is the number of point of
which f(x) cuts or touches the axes.

Ans : As the number of zeroes of polynomial f(x) is the number of points at
which f(x) cuts (intersects) the x –axis and number of zero in the given figure is 3.
So A is correct but R is not correct.
Correct Option (c) Assertion (A) is true but reason (R) is false.

6. Assertion: Degree of a zero polynomial is not defined.
Reason: Degree of a non-zero constant polynomial is 0
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct
explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : We know that, the constant polynomial 0 is called a zero polynomial.
The degree of a zero polynomial is not defined. So, Assertion is true.
Now, the degree of a non-zero constant polynomial is zero.
So, Reason is true.
Since both Assertion and Reason are true and Reason is not a correct explanation of Assertion.
Correct option is (b)

7. Assertion : x² + 7x + 12 has no real zeroes.
Reason : A quadratic polynomial can have at the most two zeroes.
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : x² + 7x + 12 = 0
⇒ x² + 4x + 3x + 12 = 0
⇒ x(x + 4) + 3(x + 4) = 0
⇒ (x + 4) (x + 3) = 0
⇒ (x + 4) = 0 or (x + 3) = 0
⇒ x = −4 or x = −3
Therefore, x² + 7x + 12 has two real zeroes.
Correction option is (d) Assertion (A) is false but reason (R) is true.

8. Assertion : If the sum of the zeroes of the quadratic polynomial x² – 2kx + 8 are is 2 then value of k is 1.
Reason : Sum of zeroes of a quadratic polynomial ax² + bx + c is –b/a
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : Relation is true as we know that Sum of zeroes = $\frac{-b}{a}$
$= -\frac{-2k}{1}= 2 = k = 1$

So, Assertion is true.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

9. Assertion : If the product of the zeroes of the quadratic polynomial x² + 3x + 5k
is -10 then value of k is -2.
Reason : Sum of zeroes of a quadratic polynomial ax² + bx + c is –b/a
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : Reason is true as we know that Sum of zeroes = $\frac{-b}{a}$ , Also we know that Product of zeroes = $\frac{c}{a}$
= $\frac{5k}{1}=-10=k = -2$
So, Assertion is true. But Reason is not the correct explanation of assertion. Correct option is (b)

10. Assertion : 3 − $2\sqrt{5}$ is one zero of the quadratic polynomial then other zero will be 3 + $2\sqrt{5}$ .
Reason : Irrational zeros (roots) always occurs in pairs.
(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : As irrational roots/zeros always occurs in pairs therefore, when one zero is
3 − $2\sqrt{5}$ then other will be 3 + $2\sqrt{5}$ .
So, both A and R are correct and R explains A.
Correct option is (a) Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).

11. Assertion : A quadratic polynomial, sum of whose zeroes is 8 and their product is 12 is x² – 20x + 96.
Reason: If and be the zeroes of the polynomial f(x), then polynomial is given by f(x) = x² $-\left ( \alpha +\beta \right )x+ \alpha \beta$

(a)Both assertion (A) and reason (R) are true and reason (R) is the correct explanation of assertion (A).
(b)Both assertion (A) and reason (R) are true but reason (R) is not the correct explanation of assertion (A).
(c)Assertion (A) is true but reason (R) is false.
(d)Assertion (A) is false but reason (R) is true.

Ans : Reason is correct. If and be the zeroes of the required polynomial f(x),
then ( $\alpha$ + $\beta$ ) = 8 and $\alpha$ $\beta$ = 12
∴ f(x) = x² – ( $\alpha$ + $\beta$ )x +
⇒ f(x) = x² – 8x + 12
So, Assertion is not correct
Correct option is (d)

1.The value of x, for which the polynomials x ² – 1 and
x² – 2x +1 vanish simultaneously, is

(a) 2 (b) -2 (c) -1 (d) 1

2. If $\alpha$ and $\beta$ are zeroes and the quadratic polynomial
f(x) = x² – x – 4 , then the value of $\frac{1}{\alpha }+\frac{1}{\beta }-\alpha \beta$ is
(a) $\frac{15}{4}$ (b) – $\frac{15}{4}$ (c) 4 (d) 15

3. The value of the polynomial x⁸– x⁵+ x² – x+1 is
(a) positive for all the real numbers
(b) negative for all the real numbers
(c) 0
(d) depends on value of x

4. On dividing x³ – 3x ² + x + 2 by a polynomial g(x),
the quotient and remainder were x – 2 and – 2 x + 4
respectively, then g(x) is equal to
(a) x²+x + 1 (b) x²+1 (c) x²– x + 1 (d) x²– 1

5. If x = 0.7 then 2x is
(a) 1. $\bar{4}$ (b) 1. $\bar{5}$ (c) 1 . $\bar{54}$ (d) 1. $\bar{45}$

6. The difference between two numbers is 642. When the
greater is divided by the smaller, the quotient is 8
and the remainder is 19, then find the sum of cube of
numbers.
(a) 391322860 (b) 319322860 (c) 319322680 (d) 391223860

7. Lowest value of x² + 4x + 2 is
(a) 0 (b) 2 (c) 2 (d) 4

8. If a³ – 3 a² b + 3ab² – b³is divided by ( a -b) , then the
remainder is
(a) a² – ab + b² (b) a²+ ab + b²(c) 1 (d) 0

9. A quadratic polynomial when divided by x+2 leaves
a remainder of 1 and when divided by x-1, leaves a
remainder of 4. What will be the remainder if it is
divided by (x+2)(x-1)?
(a) 1 (b) 4 (c) x + 3 (d) x -3

10. If a quadratic polynomial curve in the shape of semicircle is shown below.
Then, the equation of this curve.
(a) -x²+ 2 (b) x²+ 2 (c) $\frac{1}{2}$ x²+2 (d) – $\frac{1}{2}$ x ²+2

11. If the sum of the zeroes of the polynomial
f(x) = 2x²– 3kx²+ 4x – 5 is 6, then the value of k is
(a) 2 (b) -2 (c) 4 (d) -4

12. If a cubic polynomial with the sum of its zeroes, sum
of the products and its zeroes taken two at a time and
product of its zeroes as 2, -5 and -11 respectively,
then the cubic polynomial is
(a) x³+ 7x – 6 (b) x³+ 7x + 6 (c) x³– 7x – 6 (d) x³– 7x + 6

13. If $\alpha$ and $\beta$ are the zeroes of the quadratic polynomial
f(x) = ax²+ bx + c , then the value of a⁴+ $\beta$ ⁴is
(a) $\frac{(b^{2}-2ac)^{2}+a^{2}c^{2}}{a^{4}}$ (b) $\frac{(b^{2}+2ac)^{2}-a^{2}c^{2}}{a^{4}}$

(c) $\frac{(b^{2}-2ac)^{2}-2a^{2}c^{2}}{a^{4}}$ (d) $\frac{(b^{2}+2ac)^{2}+2a^{2}c^{2}}{a^{4}}$
The polynomial f (x) ax³ + bx – c is divisible by the
polynomial g (x) = x² +bx+ c , c ≠ 0, if
(a) ab = 2 (b) ab = 1
(c) ac = 2 (d) c b

15. If one of the zeroes of a quadratic polynomial of the
form x² +ax+ b is the negative of the other, then
which of the following is correct?
(a) Polynomial has linear factors
(b) Constant term of polynomial is negative
(c) Both (a) and (b) are correct
(d) Neither (a) nor (b) is correct

16. If $\alpha$ $\beta$ , and $\gamma$ are the zeroes of the polynomial
p(x)= ax³ +3bx² + 3cx +d and having relation
2 $\beta$ =a + $\gamma$ , then 2b³– 3abc + a² d is
(a) -1 (b) 1 (c) 0 (d) None of the above

17. If the square of difference of the zeroes of the quadratic
polynomial x²+ px + 45 is equal to 144, then the value
of p is
(a) ±9 (b)±12 (c)±15 (d) ±18

18. If $\alpha$ and $\beta$ are zeroes and the quadratic polynomial p(S)=3S²+6S+4, then the value of
$\frac{\alpha }{\beta }$ + $\frac{\beta }{\alpha }$ + $2\left ( \frac{1}{\alpha }+\frac{1}{\beta } \right )$ +3α β is
(a) 7 (b) 6 (c) 8 (d) 10

19. Find the zeroes of the quadratic polynomial y²– 3y + 2
with the help of the graph.
(a) 1,- 2 (b) $\frac{-1}{4},\frac{3}{2}$ (c) 6,-1 (d) 1 , 2

20. If the sum of the zeroes of the equation
$\frac{1}{x+a}+\frac{1}{x+b}=\frac{1}{c}$ is zero, then the product of zeroes
of the equation is?
(a) $\frac{a^{2}+b^{2}}{2}$ (b) $-\left ( \frac{a^{2}+b^{2}}{2} \right )$ (c) $\frac{ab}{2}$ (d) $\left ( \frac{a+b^{}}{2} \right )^{2}$