Q 1 :

The zeroes of the polynomial x23xm(m+3) are

  • m,m+3

     

  • -m,m+3

     

  • m,-(m+3)

     

  • -m,-(m+3)

     

(2)

Let p(x)=x2-3x-m(m+3)

p(x)=x2-(m+3)x+mx-m(m+3)

=x{x-(m+3)}+m{x-(m+3)}

For zeros of p(x)

p(x)=(x+m){x-(m+3)}=0x=-m,m+3

 Its zeros are -m,m+3



Q 2 :

If α and β are the zeros of a polynomial f(x)=px22x+3p and α + β = αβ, then p is

  • -2/3

     

  • 2/3

     

  • 1/3

     

  • -1/3

     

(2)

Given, f(x)=px2-2x+3p

Since α and β are the zeroes of the given polynomial.

 α+β=-(-2)p=2p, and αβ=3pp=3

 α+β=αβ      (given)

 2p=3p=23



Q 3 :

The zeroes of a polynomial x2+px+q are twice the zeroes of the polynomial 4x25x6. The value of p is :

  • -5/2

     

  • 5/2

     

  • -5

     

  • 10

     

(1)

Given polynomials : x2+px+q   ...(i)

and 4x2-5x-6       ...(ii)

Zero of polynomial 4x2-5x-6 are: x=2 and x=-3/4

Now, zero of polynomial x2+px+q are 4 and -3/2

 Sum of zeroes =-p1

4-32=-p1-p=52p=-52



Q 4 :

If the sum of zeroes of the polynomial p(x)=2x2-k2x+1 is 2, then value of k is :

  • 2

     

  • 2

     

  • 22

     

  • 1/2

     

(2)

Sum of zeroes =2=-(-k2)2k=2

 



Q 5 :

If α and β are zeroes of the polynomial 5x2+3x7, the value of 1α+1β is

  • −3/7

     

  • 3/5

     

  • 3/7

     

  • −5/7

     

(3)

α+β=-ba=-35,  αβ=ca=-75

Now, 1α+1β=α+βαβ=-35-75=37

 



Q 6 :

Consider the polynomial p(x) = x³  3x² + 2x. Which of the following statements are true about its graph and zeros?

(i) The polynomial has a zero at x = 0.
(ii) The polynomial has a zero at x = 1.
(iii) The graph touches the x-axis at x = 1.
(iv) The graph intersects the x-axis at x = 2.

 

Choose the correct option from the following:

  • both (i) and (ii)

     

  • both (ii) and (iii)

     

  • (i), (ii) and (iv)   

     

  • both (i) and (iv)

     

(3)

p(x) = x³  3x² + 2x p(x) = x(x²  3x + 2) p(x) = x(x  2)(x  1)

Factoring p(x) gives x(x – 1)(x – 2), indicating zeros at x = 0, 1, and 2. The graph intersects the x-axis at these points.



Q 7 :

For the polynomial r(x) = x² – 4x + 4, which of the following statements are correct about its graph and zeros?

 

 

(i) The graph touches the x-axis at x = 2.
(ii) The graph intersects the x-axis at x = 2.
(iii) The polynomial is equivalent to (x – 2)².
(iv) The vertex of the graph is at the point (2, 0).

 

Choose the correct option from the following:

  • (i), (iii) and (iv)

     

  • (ii) and (iii)

     

  • (i), (ii) and (iv)

     

  • (i) and (iv)

     

(1)

Given polynomial r(x) can be rewritten as (x – 2)², indicating a repeated zero at x = 2. The graph touches the x-axis at this point and does not cross it. The vertex of the graph is at point (2, 0).



Q 8 :

For the polynomial t(x) = x³ – 6x² + 11x – 6, which statements are true about its graph and zeros?

(i) The polynomial has a zero at x = 1.
(ii) The polynomial has a zero at x = 2.
(iii) The graph of the polynomial intersects the x-axis at x = 3.
(iv) The polynomial can be factorised as (x – 1)(x – 2)(x – 3).

Choose the correct option from the following:

  • (i), (iii) and (iv)

     

  • (i), (ii), (iii) and (iv)      

     

  • (i), (ii) and (iv)

     

  • (i) and (iv)

     

(2)

At x = 1, t(1) = 1 – 6 + 11 – 6 = 0

So (x – 1) is a factor of t(x).

∴ t(x) = (x – 1)(x² – 5x + 6)

⇒ t(x) = (x – 1)(x – 2)(x – 3)

Factoring t(x) as (x – 1)(x – 2)(x – 3) reveals zeros at x = 1, 2, 3. The graph intersects the x-axis at these points.



Q 9 :

Shown below is a part of the graph of a polynomial h(x).

On dividing h(x) by which of the following will the remainder be zero?

(i) (x – 2)
(ii) (x + 2)
(iii) (x – 4)
(iv) (x + 4)

Choose the correct option from the following:

  • only (ii)

     

  • only (i) and (iii)

     

  • only (ii) and (iv)

     

  • cannot be determined without knowing the polynomial h(x).

     

(1)

Since the given graph intersects the x-axis at x = –2, therefore (x + 2) is a factor of h(x).

So, on dividing h(x) by (x + 2), we get remainder zero

 



Q 10 :

If the sum of zeros of the polynomial p(x) = 2x²  k2x + 1 is 2, then value of k is:

  • √2         

     

  • 2

     

  • 2√2      

     

  • 1/2

     

(2)

Given polynomial is p(x) = 2x² – k√2x + 1.

Let α, β be its zeros then

α + β = –b/a = –(–k√2)/2 = k√2/2 = √2

[Here a = 2, b = –k√2, c = 1]

⇒ k√2 = 2√2 ⇒ k = 2