Rd Sharma 2022 _mcqs for Class 10 Maths Chapter 2 - Polynomials

Answer:

For a quadratic polynomial, $a{x}^2+bx+c$, the zeros are precisely the x-coordinates of the points where the graph representing $y=a{x}^2+bx+c$ intersects the x-axis. 
The graph has one of the two shapes either open upwards like ∪ (parabolic shape) or open downwards like ∩ (parabolic shape) depending on whether a > 0 or a < 0.
Three cases are thus possible:
a) graph cuts x-axis at two distinct points (two zeroes)
b) graph cuts the x-axis at exactly one point (one zero)
c) the graph is either completely above the x-axis or completely below the x-axis (no zeroes)
In option (d), the graph is cutting the x-axis at three distinct points and it is not a parabola opening either upwards or downwards.
So, option (d) does not represent the graph of a quadratic polynomial.

Hence, the correct answer is option D.

Answer:

Let the quadratic polynomial be $f\left(x\right)=x^2+ax+b$.
Now, the zeroes are $\alpha \mathrm{and} -\alpha$.
So, the sum of the zeroes is zero.
$$\begin{gathered} \therefore \alpha +\left(-\alpha \right)=\frac{-a}{1}=-a \\ \Rightarrow a=0 \end{gathered}$$
So, the polynomial becomes $f\left(x\right)=x^2+b$, which is not linear
Also, the product of the zeros, 
$$\begin{gathered} \alpha \beta =\frac{b}{1}=b \\ \Rightarrow \alpha \left(-\alpha \right)=b \\ \Rightarrow -{\alpha }^2=b \end{gathered}$$
Thus, the constant term is negative. 

Hence, the correct answer is option A.

Answer:

Let the given polynomial be f(x) = x² + 3x + k.
Since 2 is one of the zero of the given plynomial, so (x − 2) will be a factor of the given polynomial.
Now, f(2) = 0 
$$\begin{gathered} \Rightarrow 2^2+3\times 2+k=0 \\ \Rightarrow 4+6+k=0 \\ \Rightarrow k=-10 \end{gathered}$$
Hence, the correct answer is option B.

Answer:

Since and are the zeros of the quadratic polynomials such that

If one of zero is 3 then

Substituting in we get

Let S and P denote the sum and product of the zeros of the polynomial respectively then

Hence, the required polynomials is

Hence, the correct choice is

Answer:

Quadratic Polynomial = $x^2-\left(\mathrm{sum} \mathrm{of} \mathrm{zeros}\right)x+\left(\mathrm{product} \mathrm{of} \mathrm{zeros}\right)$
                                    = $x^2-\left(-5\right)x+6$
                                    = $x^2+5x+6$

Hence, the correct answer is option(a).
 

Answer:

Given be the zeros of the polynomial

Product of the zeros =

The value of Product of the zeros is 6.

Hence, the correct choice is

Answer:

We are given then

One root of the polynomial is reciprocal of the other. Then, we have

1

Hence the correct choice is

Answer:

Since and are the zeros of the quadratic polynomial

We have

The value of is

Hence, the correct choice is _.

Answer:

Since and are the zeros of the quadratic polynomial

We have

The value of is.

Hence, the correct choice is

Answer:

Let be the zeros of polynomial such that

We have,

Putting in, we get

Therefore, the value of third zero is.

Hence, the correct alternative is.

Answer:

If one zero of the polynomial is reciprocal of the other. So

Now we have

Since

Therefore we have

Hence, the correct choice is

Answer:

Let and be the given zeros and be the third zero of x³ + x² − 5x − 5 = 0 then

By substitutingand in

Hence, the correct choice is

Answer:

Given be the zeros of the polynomial

Product of the zeros =

The value of Product of the zeros is 6.

Hence, the correct choice is

Answer:

Let and be the given zeros and be the third zero of the polynomial then

Substituting and in, we get

Hence, the correct choice is.

Answer:

Let andbe the given zeros and be the third zero of the polynomial . Then,

Substitutingandin

We get

Hence, the correct choice is

Answer:

Let, be the zeros of the polynomial and we are given that
$\alpha +\beta +\gamma$=.

Then,

$$\begin{gathered} \alpha +\beta +\gamma =\frac{-\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2} \\ = -\frac{(-3k)}{2} = \frac{3k}{2} \end{gathered}$$

It is given that

$\alpha +\beta +\gamma$=

Substituting $\alpha +\beta +\gamma =\frac{3k}{2}$, we get

The value of k is.

Hence, the correct alternative is

Answer:

Since and are the zeros of quadratic polynomial

So we have 

The value of a is

Hence, the correct alternative is.

Answer:

Polynomials having zeros −2 and 5 will be of the form
$p\left(x\right)=a{\left(x+2\right)}^n{\left(x-5\right)}^m$
Here, n and m can take any value from 1, 2, 3,...
Thus, the number of polynomials will be more than 3. 
Hence, the correct answer is option D.

Answer:

The given polynomial is f(x) = $\left(k-1\right)x^2+kx+1$.
Since $-3$ is one of the zeroes of the given polynomial, so $f\left(-3\right)=0$.
$$\begin{gathered} \left(k-1\right){\left(-3\right)}^2+k\left(-3\right)+1=0 \\ \Rightarrow 9\left(k-1\right)-3k+1=0 \\ \Rightarrow 9k-9-3k+1=0 \\ \Rightarrow 6k-8=0 \\ \Rightarrow k=\frac{4}{3} \end{gathered}$$
Hence, the correct answer is option A. 

Answer:

Let f(x) = x² + 99x + 127.

Product of the zeroes of f(x) = $127\times 1=127$      [Product of zeroes = $\frac{c}{a}$ when f(x) = ax² + bx + c]

Since the product of zeroes is positive, we can say that it is only possible when both zeroes are positive or both zeroes are negative.

Also, sum of the zeroes =  –99              [Sum of zeroes = $-\frac{b}{a}$ when f(x) = ax² + bx + c]

The sum being negative implies that both zeroes are positive is not correct.

So, we conclude that both zeroes are negative.

Hence, the correct answer is option B.

Answer:

We have to find the value of

Given and are the zeros of the quadratic polynomial f(x)

We have,

Hence, the correct choice is

Answer:

Let and be the zeros of the polynomial.Then,

And

Let S and R denote respectively the sum and product of the zeros of a polynomial

Whose zeros are and .then

Hence, the required polynomial whose sum and product of zeros are S and R is given by

So

Hence, the correct choice is

Answer:

Since and are the zeros of quadratic polynomial

We have

The value of is.

Hence, the correct choice is

Answer:

Given that, α and β are zeros of x² – 6x + k and 3α + 2β = 20
We know,
$\alpha +\beta =\frac{-\left(-6\right)}{1}=6$                    .....(1)

Solving both equations, we get
$\alpha =8 \mathrm{and} \beta =-2$

Also, the product of roots = $\alpha \beta =\frac{k}{1}=k$
$\Rightarrow k=-16$

Hence, the correct answer is option (c).
 

Answer:

If, is a zero of a polynomial thenis a factor of

Since 3 is the zero of the polynomial,

Thereforeis a factor of

Now, we divideby we get

Therefore we should add 2 to the given polynomial

Hence, the correct choice is

Answer:

We know that, if, is zero of a polynomial then is a factor of

Since 15 is zero of the polynomial f (x) = x² − 16x + 30, therefore (x − 15) is a factor of f (x)

Now, we divide by we get

Thus we should subtract the remainder from,

Hence, the correct choice is.

Answer:

Given that is a factor of and a + b=4

By solving and a + b = 4 by elimination method we get 

Multiply by we get,

. So 

By substituting b = 1 in a + b = 4 we get

Then a = 3, b = 1

Hence, the correct choice is

Answer:

We know that

Therefore,

The polynomial which when divided by gives a quotient and remainder 3, is

Hence, the correct choice is.

Answer:

The given quadratic equation is $x^2+\left(a+1\right)x+b=0$.
Since the zeroes of the given equation are 2 and –3.
So,
$\alpha =2$ and $\beta =-3$
Now,
$$\begin{gathered} \mathrm{Sum} \mathrm{of} \mathrm{zeroes}=-\frac{\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2} \\ \Rightarrow 2+\left(-3\right)=-\frac{\left(a+1\right)}{1} \\ \Rightarrow -1=-a-1 \\ \Rightarrow a=0 \end{gathered}$$
$$\begin{gathered} \mathrm{Product} \mathrm{of} \mathrm{zeroes}=\frac{\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} x^2} \\ \Rightarrow 2\times \left(-3\right)=\frac{b}{1} \\ \Rightarrow b=-6 \end{gathered}$$
So, a = 0 and b = −6

Hence, the correct answer is option D.

Answer:

Let and be the zeros of the polynomial

Therefore

The value of

Hence, the correct choice is

Answer:

Given that, α and β are zeros of ax² – 5x + c and $\alpha +\beta =\alpha \beta =10$          .....(1)
We know,
$\alpha +\beta =\frac{-\left(-5\right)}{a}=\frac{5}{a}$                    .....(2)

Solving both the equations, we get
$a=\frac{1}{2}$

Also, the product of roots = $\alpha \beta =\frac{c}{a}=10$
$\Rightarrow c=5$

Hence, the correct answer is option (d).

Answer:

Given that, α, β and γ are the zeroes of the polynomial x³ – x² – 10x – 8.
Now,
$$\begin{gathered} \mathrm{\alpha \beta }+\mathrm{\beta \gamma }+\mathrm{\gamma \alpha }=\frac{c}{a} \\ \mathrm{\alpha \beta \gamma }=\frac{-d}{a} \end{gathered}$$

So,
αβ + βγ + γα = −10              .....(1)
αβγ = 8                                .....(2)

From (1) and (2), we get 
αβ + βγ + γα + αβγ = −10 + 8
                               = −2

Hence, the correct answer is option (a).

Answer:

Let $g\left(x\right)=x^3+7x^2-2x-14$
Given that, $\sqrt{2}$ and $-\sqrt{2},$ are the zeros of g(x).
$\Rightarrow \left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=\left(x^2-2\right)$ is a factor of g(x).

graphic rqd

$\therefore g\left(x\right)=\left(x^2-2\right)\left(x+7\right)$
Thus, $x+7=0$ is a factor of g(x).
$\Rightarrow x=-7$ is a zero of g(x).

Hence, the correct answer is option (b).

Answer:

Let be the zeros of the polynomial then

Since a is a zero of the polynomial

Therefore,

Substituting .we get

Hence, the correct choice is

Answer:

Given that, α and β are the zeroes of the polynomial x² – (k + 6) x + 2 (2k – 1)
And $\mathrm{\alpha }+\mathrm{\beta }=\frac{\mathrm{\alpha \beta }}{2},$ 
Now,
$$\begin{gathered} \alpha +\beta =\frac{k+6}{1} .....\left(1\right) \\ \alpha \beta =\frac{2\left(2k-1\right)}{1} .....\left(2\right) \end{gathered}$$
From (1) and (2), we have
$$\begin{gathered} \Rightarrow k+6=\frac{2\left(2k-1\right)}{2} \\ \Rightarrow k+6=2k-1 \\ \Rightarrow k=7 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

Let  a – d, a, and a + d are zeroes of the polynomial x³ – 12x² + 44x + c
Sum of the roots = $\frac{-\mathrm{coefficient} \mathrm{of} x^2}{\mathrm{coefficient} \mathrm{of} x^3}$
$$\begin{gathered} \Rightarrow a-d+a+a+d=\frac{-\left(-12\right)}{1} \\ \Rightarrow 3a=12 \\ \Rightarrow a=4 \end{gathered}$$

Product of the roots, taken two at a time = $\frac{\mathrm{coefficient} \mathrm{of} x}{\mathrm{coefficient} \mathrm{of} x^3}$
$$\begin{gathered} \Rightarrow \left(a-d\right)a+a\left(a+d\right)+\left(a-d\right)\left(a+d\right)=44 \\ \Rightarrow 4\left(4-d\right)+4\left(4+d\right)+\left(4-d\right)\left(4+d\right)=44 \\ \Rightarrow 16-4d+16+4d+16-d^2=44 \\ \Rightarrow 48-d^2=44 \\ \Rightarrow 4=d^2 \\ \Rightarrow d=2 \end{gathered}$$

Now, 
$$\begin{gathered} a-d=4-2=2 \\ a+d=4+2=6 \end{gathered}$$

$$\begin{gathered} \mathrm{Product} \mathrm{of} \mathrm{roots}=\frac{\mathrm{constant} \mathrm{term}}{\mathrm{coefficient} \mathrm{of} x^3}=\frac{-c}{1} \\ \Rightarrow 2\times 4\times 6=-c \\ \Rightarrow c=-48 \end{gathered}$$

Hence, the correct answer is option (d).

Answer:

We have to find the value of

Given be the zeros of the polynomial

We know that

So

Hence, the correct choice is

Answer:

Let $\alpha ,\beta \mathrm{and} \gamma$ are the zeros of the polynomial x³ – 2x² + qx – r.
$\mathrm{Sum} \mathrm{of} \mathrm{zeros}=\frac{-\mathrm{coefficient} \mathrm{of} x^2}{\mathrm{coefficient} \mathrm{of} x^3}$
$$\begin{gathered} \Rightarrow \alpha +\beta +\gamma =\frac{-\left(-2\right)}{1} \\ \Rightarrow \alpha +\beta +\gamma =2 \\ \Rightarrow 0+\gamma =2 \left(\because \alpha +\beta =0\right) \\ \Rightarrow \gamma =2 \end{gathered}$$

Now,
$$\begin{gathered} \Rightarrow {\left(2\right)}^3-2\times {\left(2\right)}^2+q\times 2-r=0 \\ \Rightarrow 8-8+2q-r=0 \\ \Rightarrow 2q-r=0 \\ \Rightarrow 2q=r \end{gathered}$$

Hence, the correct answer is option (a).
 

Answer:

Given that,  f(x) = 2x³ – kx² + 5x + 9 is a factor of x + 2.
$\Rightarrow x=-2$ is a zero of 2x³ – kx² + 5x + 9

$$\begin{gathered} \Rightarrow 2\times {\left(-2\right)}^3-k{\left(-2\right)}^2+5\times \left(-2\right)+9=0 \\ \Rightarrow -16-4k-1=0 \\ \Rightarrow -4k-17=0 \\ \Rightarrow k=\frac{-17}{4} \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

Given that, a – b, a, and a + b are zeroes of the polynomial x³ – 3x² + x + 1.

Sum of the roots= $\frac{-\mathrm{coefficient} \mathrm{of} x^2}{\mathrm{coefficient} \mathrm{of} x^3}$
$$\begin{gathered} \Rightarrow a-b+a+a+b=\frac{-\left(-3\right)}{1} \\ \Rightarrow 3a=3 \\ \Rightarrow a=1 \end{gathered}$$

Product of the roots, taken two at a time = $\frac{-\mathrm{constant} \mathrm{term}}{\mathrm{coefficient} \mathrm{of} x^3}$
$$\begin{gathered} \Rightarrow \left(1-b\right)\left(1+b\right)=-1 \\ \Rightarrow 1-b^2=-1 \\ \Rightarrow b^2=2 \\ \Rightarrow b=\pm \sqrt{2} \end{gathered}$$

Thus, $a+b=1\pm \sqrt{2}$.

Hence, the correct answer is option (d).


 

Answer:

Let f(x) = x² + ax + a.
Product of the zeroes of f(x) = a        [Product of zeroes = $\frac{c}{a}$ when f(x) = ax² + bx + c]
Since the product of zeroes is positive, so the zeroes must be either both positive or both negative.
Also, sum of the zeroes = –a           [Sum of zeroes = $-\frac{b}{a}$ when f(x) = ax² + bx + c]
So, the sum of the zeroes is negative, so the zeroes cannot be both positive.

Hence, the correct answer is option B.

Answer:

Let the given quadratic polynomial be f(x) = $a{x}^2+bx+c$.
Suppose $\alpha$ and $\beta$ be the zeroes of the given polynomial.
Since $\alpha$ and $\beta$ are equal so they will have the same sign i.e. either both are positive or both are negative. 
So, $\alpha \beta >0$
But, $\alpha \beta =\frac{c}{a}$
∴ $\frac{c}{a}>0$, which is possible only when both have same sign
Hence, the correct answer is option C.

Answer:

Given that, 2 and $\frac{1}{2}$ are zeros of px² + 5x + r.
Now,
Sum of the zeros = $\frac{-\mathrm{coefficient} \mathrm{of} x}{\mathrm{coefficient} \mathrm{of} x^2}$
$$\begin{gathered} \Rightarrow 2+\frac{1}{2}=\frac{-5}{p} \\ \Rightarrow \frac{5}{2}=\frac{-5}{p} \\ \Rightarrow p=-2 \end{gathered}$$

And
Product of the zeros = $\frac{\mathrm{constant} \mathrm{term}}{\mathrm{coefficient} \mathrm{of} x^2}$
$$\begin{gathered} 2\times \frac{1}{2}=\frac{r}{p} \\ \Rightarrow \frac{r}{-2}=1 \\ \Rightarrow r=-2 \end{gathered}$$

Hence, the correct answer is option (b).

Answer:

(i) The wire is bent in the shape of a parabola.

Hence, the correct answer is option (c).

(ii) Since the graph cuts the x-axis at two points.
∴ Number of zeros = 2
Hence, the correct answer is option (a).

(iii) The zeros of a polynomial would be where the graph touches the x-axis.
The above graph is touching the x-axis at −1 and 3. Thus, the zeros of the polynomial are −1 and 3.

Hence, the correct answer is option (b).

(iv) We know, the zeros of the polynomial are −1 and 3.
$\begin{array}{rcl}\therefore \mathrm{Polynomial}& =& \left(x+1\right)\left(x-3\right)\\ & =& x^2-3x+x-3\\ & =& x^2-2x-3\end{array}$

Thus, the expression of the polynomial is $x^2-2x-3$.

Hence, the correct answer is option (c).

(v) At x = – 1,
Value of polynomial = ${\left(-1\right)}^2-2\times \left(-1\right)-3$
                                  = 1 + 2 − 3
                                  = 0

Hence, the correct answer is option (d).

Answer:

(i) We have, the highway overpass is represented by x² – 2x – 8.
Solving by splitting the middle term,
$$\begin{gathered} \Rightarrow x^2-2x-8=0 \\ \Rightarrow x^2-4x+2x-8=0 \\ \Rightarrow x\left(x-4\right)+2\left(x-4\right)=0 \\ \Rightarrow \left(x+2\right)\left(x-4\right)=0 \end{gathered}$$
$\therefore x=-2,4$

Hence, the correct answer is option (b).

(ii) Number of zeros of the polynomial representing the number of points where the graph of a polynomial intersects the x-axis.

Hence, the correct answer is option (a).

(iii) The graph of a quadratic polynomial is a parabola.

Hence, the correct answer is option (c).

(iv) Given that,
Sum of the zeros = 0
One zero = 6
Now,  Another zero = $-$6

∴ Required polynomial = $\left(x+6\right)\left(x-6\right)$
                                      = $x^2-36$

Hence, the correct answer is option (b).

(v) We have,
f(x) = (x – 2)² + 4
       = $x^2+4-4x+4$
       = $x^2-4x+8$

 As (x − 2)² and 4 both are positive terms, their sum must be positive and we know that sum of two positive terms can never be zero.
Thus, given expression has no zero.

Hence, the correct answer is option (c).

Answer:

(i) Given polynomial = x² – 8x – 20
By splitting the middle terms, we have
$\begin{array}{rcl}{x}^2-8x-20& =& x^2-10x+2x-20\\ & =& x\left(x-10\right)+2\left(x-10\right)\\ & =& \left(x+2\right)\left(x-10\right)\end{array}$

∴ Zeros of the polynomial are –2, 10.

Hence, the correct answer is option (b).

(ii) Given that,
Sum of zeros = –4
Product of zeros = –12

The required polynomial is of the form $f\left(x\right)=x^2-\left(\mathrm{sum} \mathrm{of} \mathrm{zeros}\right)x+\left(\mathrm{product} \mathrm{of} \mathrm{zeros}\right)$.

Therefore,
Required polynomial = $x^2-\left(-4\right)x+\left(-12\right)$
                                   = $x^2+4x-12$

Hence, the correct answer is option (c).

(iii) Given that,
Product of zeros = 8
⇒ (−2) × other zero = 8
⇒ other zero = −4
Therefore, the sum of the zeros = −6

Therefore,
Required polynomial = $x^2-\left(-6\right)x+\left(8\right)$
                                   = $x^2+6x+8$

Hence, the correct answer is option (a).

(iv) Given that, 6x² – 7x – 3
By splitting the middle term,
$\begin{array}{rcl}6x^2-7x-3& =& 6x^2-9x+2x-3\\ & =& 3x\left(2x-3\right)+1\left(2x-3\right)\\ & =& \left(3x+1\right)\left(2x-3\right)\end{array}$

Therefore, the zeros of required polynomial are $\frac{-1}{3} \mathrm{and} \frac{3}{2}$.
Since zeros of the quadratic polynomial are reciprocal of the above zeros, i.e.,  –3 and $\frac{2}{3}$.
Now, the sum of zeros = $\frac{-7}{3}$
Product of zeros = –2

The required polynomial is of the form $f\left(x\right)=x^2-\left(\mathrm{sum} \mathrm{of} \mathrm{zeros}\right)x+\left(\mathrm{product} \mathrm{of} \mathrm{zeros}\right)$.
 $\begin{array}{rcl}{x}^2+\frac{7}{3}x-2& =& \frac{1}{3}\left(3x^2+7x-6\right)\\ & =& k\left(3x^2+7x-6\right), \mathrm{where} k=\frac{1}{3}\end{array}$

Thus, the required polynomial is $3x^2+7x-6=0$.

Hence, the correct answer is option (d).

(v) If the parabola representing quadratic polynomial ax² + bx + c touches x-axis, then its roots are real and equal.

Hence, the correct answer is option (b).

Answer:

(i) Since the quadratic polynomial p(x) = ax² + bx + c is in the downward direction.
∴ a < 0

Hence, the correct answer is an option (b).

(ii) Since the above graph is intersecting the graph at –4 and 2.
Thus, zeros of the polynomial are –4 and 2.

Hence, the correct answer is an option (b).

(iii) Since the two zeros of the polynomial are –4 and 2, the polynomial formed by them is $\left(x+4\right)\left(x-2\right)$.
Now,
Required polynomial p(x) =  $\left(x+4\right)\left(x-2\right)$
                                          = $x^2+4x-2x-8$
                                          = $x^2+2x-8$

Hence, the correct answer is an option (b).

(iv) Required polynomial p(x) = $x^2+2x-8$
At x = 0,
$\begin{array}{rcl}p\left(0\right)& =& {\left(0\right)}^2+2\left(0\right)-8\\ & =& -8\end{array}$
 
Hence, the correct answer is an option (a).

(v) If the parabola shown is moved rightward through one unit, then the two zeros of the polynomial are 3 and −3.
Thus,
Sum of zeros = 0
Product of zeros = −9

Therefore,
$\begin{array}{rcl}\mathrm{Required} \mathrm{polynomial}& =& x^2-\left(\mathrm{sum} \mathrm{of} \mathrm{zeros}\right)x+\left(\mathrm{product} \mathrm{of} \mathrm{zeros}\right)\\ & =& x^2-\left(0\right)x+\left(-9\right)\\ & =& x^2-9\end{array}$

Hence, the correct answer is an option (a).

Answer:

(i)  Since polynomial y = f(x) intersects the x-axis at 3 points. Thus, the number of zeros of the polynomial is 3.

Hence, the correct answer is an option (b).

(ii)  Since polynomial y = f(x) intersects the x-axis at 3 points.
Thus, y = f(x) represents a cubic polynomial.

Hence, the correct answer is option (d).

(iii) The coordinates where the curve intersects the x-axis are (2, 0), (0, 0) and (–2, 0).

Hence, the correct answer is an option (c).

Answer:

(i) Since, the polynomial y = f(x) intersect the x-axis at only one point. Thus, the number of zeros of the polynomial is 1.

Hence, the correct answer is option (a).

(ii) Since the polynomial y = g(x) intersect the x-axis at two points. Thus, the number of zeros of the polynomial is 2.

Hence, the correct answer is option (a).

(iii) The graph in Figure 1 has points (−2, −8), (−1, −1), (0, 0), (1, 1) and (2, 8).

For the point (−2, −8), the values of the given polynomials are as follows:

Answer:

(i) The shape of the hanging wire is a parabola.

Hence, the correct answer is option (c).

(ii) Since it is a parabola. Therefore the expression of the polynomial will be y = ax² + bx + c.

Hence, the correct answer is option (b).

(iii) The number of zeros of the polynomial is equal to the number of points where the graph of the polynomial intersects the x-axis.

Hence, the correct answer is option (a).

(iv) Given that,
Sum of zeros =  –3
Product of zeros = 5

Therefore,
Required polynomial of parabola = $x^2-\left(\mathrm{sum} \mathrm{of} \mathrm{zeros}\right)x+\left(\mathrm{product} \mathrm{of} \mathrm{zeros}\right)$
                                                      = $x^2-\left(-3\right)x+5$
                                                      = $x^2+3x+5$

Hence, the correct answer is option (d).

(v) The graph of a quadratic polynomial is a parabola.

Hence, the correct answer is option (c).

Answer:

Statement-1 (Assertion): The polynomial f(x) = x² – 2x + 2 has two real zeros.

Now, Discriminant (D) = $b^2-4ac$
                                      = ${\left(-2\right)}^2-4\times 2\times 1$
                                     = −4 < 0

Thus,  f(x) = x² – 2x + 2 does not have real zeros. So, Statement-1 is false.

Statement-2 (Reason): A quadratic polynomial can have at most two real zeroes.
A quadratic polynomial can have at most two real zeroes as its degree is 2.

Thus, Statement-2 is true but Statement-1 is false.

Hence, the correct answer is option (d).

Answer:

Statement-1 (Assertion): A quadratic polynomial having $\frac{1}{2}$ and $\frac{1}{3}$ as its zeroes is 6x² – 5x + 1.
Given that, $\frac{1}{2}$ and ​$\frac{1}{3}$ are zeros of polynomial.
$\therefore$ Required polynomial = $x^2-\left(\frac{1}{2}+\frac{1}{3}\right)x+\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)$
                                       = $\frac{6x^2-5x+1}{6}$
                                       = $k\left(6x^2-5x+1\right)$

Thus, Statement-1 is true.

Statement-2 (Reason): Quadratic polynomials having α and β as zeroes are given by f(x) = k{x² – (α + β) x + αb), where k is a non-zero constant.
A polynomial with α and β as zeroes is given by k{x² – (sum of zeros) x + (product of zeros)}. Therefore, quadratic polynomials having α and β as zeroes are given by f(x) = k{x² – (α + β) x + αb), where k is a non-zero constant.

Thus, Statement-2 is true and is a correct explanation for Statement-1.

Hence, the correct answer is option (a).

 

Answer:

Statement- 1 (Assertion): If one root of the quadratic polynomial f(x) = (k – 1) x² – 10x + 3, k ≠ 1 is reciprocal of the other, then k = 4.
Statement- 1:  f(x) = (k – 1) x² – 10x + 3, k ≠ 1
Let one root of the polynomial be $\alpha$, then the other root will be $\frac{1}{\alpha }$.
Now, product of roots = $\frac{\mathrm{Constant} \mathrm{Term}}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{3}{k-1}$
$$\begin{gathered} \Rightarrow \alpha \times \frac{1}{\alpha }=\frac{3}{k-1} \\ \Rightarrow 1=\frac{3}{k-1} \\ \Rightarrow k-1=3 \\ \Rightarrow k=4 \end{gathered}$$

Thus, Statement-1 is true.

Statement- 2 (Reason): The product of roots of the quadratic polynomial ax² + bx + c, a ≠ 0 is $\frac{a}{c}.$
Now, product of roots = $\frac{\mathrm{Constant} \mathrm{Term}}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{c}{a}$.
Thus, Statement-2 is false.

Hence, the correct answer is option (c).

Answer:

Statement-1 (Assertion): If α and b are zeroes of the quadratic polynomial x² + 7x + 12, then $\frac{12}{\mathrm{\alpha }}+\frac{12}{\mathrm{\beta }}-24\mathrm{\alpha \beta }=395.$
Given that, α and b are zeroes of the quadratic polynomial x² + 7x + 12.
By splitting the middle terms, we have
$\begin{array}{rcl}{x}^2+7x+12& =& x^2+4x+3x+12\\ & =& x\left(x+4\right)+3\left(x+4\right)\\ & =& \left(x+4\right)\left(x+3\right)\end{array}$

So, let $\alpha =-4 \mathrm{and} \beta =-3$. Then,
$\begin{array}{rcl}\frac{12}{\mathrm{\alpha }}+\frac{12}{\mathrm{\beta }}-24\mathrm{\alpha \beta }& =& \frac{12}{\left(-4\right)}+\frac{12}{\left(-3\right)}-24\times \left(-4\right)\left(-3\right)\\ & =& -3-4-288\\ & =& -295\ne 395\end{array}$
Thus, Statement-1 is false.

Statement-2 (Reason): If α and b are zeroes of the quadratic polynomial ax² + bx + c, then $\mathrm{\alpha }+\mathrm{\beta }=-\frac{b}{a}$ and $\mathrm{\alpha \beta }=\frac{c}{a}.$
Given that, α and b are zeroes of the quadratic polynomial ax² + bx + c.

Now,
Sum of roots = $\frac{-\mathrm{Coefficient} \mathrm{of} x}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{-b}{a}$
Product of roots = $\frac{\mathrm{Constant} \mathrm{Term}}{\mathrm{Coefficient} \mathrm{of} x^2}=\frac{c}{a}$

Therefore, if α and b are zeroes of the quadratic polynomial ax² + bx + c, then $\mathrm{\alpha }+\mathrm{\beta }=-\frac{b}{a}$ and $\mathrm{\alpha \beta }=\frac{c}{a}.$
Thus, Statement-2 is true.

Hence, the correct answer is option (d).


 

Answer:

Statement-1 (Assertion): If α, β and γ are the zeroes of the polynomial 6x³ + 3x² – 5x + 1, then α^–1 + β^–1 + γ^–1 = 5.

Given that, α, β and γ are the zeroes of the polynomial 6x³ + 3x² – 5x + 1.
Now,
$\begin{array}{rcl}\alpha +\beta +\gamma & =& \frac{-\mathrm{Coefficient} \mathrm{of} x^2}{\mathrm{Coefficient} \mathrm{of} x^3}\\ & =& \frac{-3}{6}\\ & =& \frac{-1}{2}\end{array}$

And
$\begin{array}{rcl}\alpha \beta \gamma & =& \frac{-\mathrm{Constant} \mathrm{term}}{\mathrm{Coefficient} \mathrm{of} x^3}\\ & =& \frac{-1}{6}\end{array}$

$\begin{array}{rcl}\therefore {\alpha }^{-1}+{\beta }^{-1}+{\gamma }^{-1}& =& \frac{1}{\alpha }+\frac{1}{\beta }+\frac{1}{\gamma }\\ & =& \frac{\alpha +\beta +\gamma }{\alpha \beta \gamma }\\ & =& \frac{{\displaystyle \frac{-1}{2}}}{{\displaystyle \frac{-1}{6}}}\\ & =& 3\ne 5\end{array}$

Thus, Statement-1 is false.

Statement-2 (Reason): If α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then a + β + γ = $–\frac{b}{a}.$
Given that, α, β, γ are the zeroes of the cubic polynomial ax³ + bx² + cx + d. Now,

$\begin{array}{rcl}\alpha +\beta +\gamma & =& \frac{-\mathrm{Coefficient} \mathrm{of} x^2}{\mathrm{Coefficient} \mathrm{of} x^3}\\ & =& \frac{-b}{a}\end{array}$

Thus, Statement-2 is true.

Hence, the correct answer is option (d).