Ncert – Class 10- Polynomials- Hots Questions
Question -1
In the CCA competition, the participants are asked to draw Mandala Art on a
sheet of length 7 cm more than the breadth. After drawing the art Linkan found
that 8 cm2 area of the sheet is left blank. Represent the area of the Mandala art
in polynomial form of polynomial and find the zeroes.
Solution
Let the breadth of the paper = x
Length = x + 7
Area of the sheet = x (x+7)
= x2 + 7x
Now area of the Mandala art = x2 + 7x – 8
To find the zeroes of the art
x2 + 7x – 8 = 0
=> x2 + 8x – x – 8 = 0
=>x (x + 8) – 1 (x+8) = 0
=> (x + 8) (x -1) = 0
=> x = 1, – 8
Question -2
The zeroes of the polynomial are -1 and 3. Find the expression for the polynomial in x as well as the value of the polynomial at x = -1
Solution
Here sum of the zeroes = 2
Product of the zeroes = -3
So, the polynomial will be
x2 – 2x – 3
at x= -1, the value will be
1 + 2 -3 = 0
Question -3
p and q are the zeroes of the polynomial 2x2 + 5x – 4. Without finding the actual values of p and q, evaluate (1-p) (1-q).
Solution
Here the polynomial is
F(x) = 2x2 + 5x – 4
P + q = -5/2
Pq = -2
Now (1-p) (1-q)= 1 – q – p + pq
= 1 –(p+q) + pq
= 1+ 5/2 -2
= 1.5
Question -4
If 𝛼 𝑎𝑛𝑑 𝛽 are zeroes of the polynomial
𝑓(𝑥) = 𝑥2 − 𝑥 − 2 ,then without finding 𝛼 𝑎𝑛𝑑 𝛽 find a polynomial whose
zeroes are 2𝛼 + 1 𝑎𝑛𝑑 2𝛽 + 1 .
Solution
𝑓(𝑥) = 𝑥2 − 𝑥 − 2, 𝛼 + 𝛽 = 1, αβ = −2. Let S and P denote
respectively the sum and the product of zeros of the required
polynomial.
𝑆 = (2𝛼 + 1) + (2𝛽 + 1) = 2(𝛼 + 𝛽) + 2 = 2 × 1 + 2 = 4
𝑃 = (2𝛼 + 1)(2𝛽 + 1) = 4𝛼𝛽 + 2(𝛼 + 𝛽) + 1 = 4(−2) + 2(1) + 1
= −5
Required polynomial =𝑘(𝑥2 − 𝑆𝑥 + 𝑃) = 𝑘(𝑥2 − 4𝑥 − 5)
Question -5
If 𝛼 and 𝛽 are the zeroes of the polynomial 2x2– 3x + 1 , form a quadratic polynomial whose
zeroes are 3𝛼 and 3𝛽.
Solution
𝛼 + 𝛽 = 3/2, 𝛼 𝛽 = 1/2
Sum of the zeroes of the required polynomial = 3 𝛼 + 3 𝛽 = 9/2
Product = 9 𝛼 𝛽 = 9/2
So the polynomial is x2– 9/2 x + 9/2
Question -6
Sunitha throws a ball upwards, from a rooftop, which is 20 m above from ground. It will reach a maximum height and then fall back to the ground. The height of the ball from the ground at time t is h , which is given by h = – 4t2 +16t + 20
What are the two possible times to reach the ball at the same height of 32 m?
Solution
h = – 4t2+16t + 20 .
32= – 4t2 +16t + 20
4t2– 16t + 12 = 0
For solving t = 1 and 3.
Question 7
A parabolic shape object used for receiving signals from a satellite, especially television signals. A satellite dish works in the same way as the reflector for a torch or car headlights. A signal is produced or reflected from a focal point. If the parabola represents the polynomial p(t) = t2– p(t+1) – c and 𝛼 and 𝛽 are its zeroes then prove that (𝛼 + 1) (𝛽 + 1) = 1 – c .
Solution
p(t) = t2– p(t+1) – c= t2– pt – p – c
(𝛼 + 1) (𝛽 + 1) = 𝛼 𝛽 + 𝛼 + 𝛽 + 1 = – p – c +p + 1 = 1 – c
Question 8
Find the value of k such that the polynomial x2-(k+6)x+2(2k-1) has sum of its zeroes equal to half of
their product.
Solution
K= 7
Question -9
Quadratic polynomial 2x2-3x+1 has zeroes as α and β. Now Form a quadratic polynomial whose zeroes are 3α and 3β Respectively.
Solution
Required quadratic polynomial is ½ (2x2-9x+9)
Question 10
Find the zeroes of the quadratic polynomial 7y2-11/3y-2/3
And verify the relationship between the zeroes and the
Solution
Y = 2/3 and y= -1/7
