Form a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. (2020)

 

1.    Form a quadratic polynomial, the sum and product of whose zeroes are -3 and 2 respectively. (2020)

Soln:

Given that:

Sum of zeros a + b       =  -3

Product of zeros ab     =  2

We know that,

          The quadratic polynomial will be,

          x² – (a + b) x + ab

          i.e., x² – (-3)x + 2

          Þ x² + 3x + 2

Hence the quadratic polynomial is x² + 3x + 2

If one zero of the quadratic polynomial x2 + 3x + k is 2, then find the value of k

 

1.    If one zero of the quadratic polynomial x² + 3x + k is 2, then find the value of k

Soln:

          Given that,

                   P(x)   =        x² + 3x + k

          Also given that 2 is one of the zero of p(x)

          i.e., p(2)     =        0

                   2² + 3x2 + k         = 0

                   4 + 6 + k              = 0

                   10 + k                  = 0

                   k                           = -10

Hence the value of ‘k’ is  -10 .

The sums of first n terms of three arithmetic progression are S1, S2 and S3 respectively. The first term of each AP is 1 and their common differences are 1, 2 and 3 respectively. Prove than S1 + S3 = 2S2 (CBSE 2016 OD)

 

This Question appeared in the 2016 CBSE Board Examinations.  This type of questions are in repeating model of questions.  Please work out more questions

Find the values of frequencies 𝑥 and 𝑦 in the following frequency distribution table, if 𝑁 = 100 and median is 32. (2019)

 

Find the values of frequencies 𝑥 and 𝑦 in the following frequency distribution table, if 𝑁 = 100 and median is 32. (2019)

Marks	No. of students
0 – 10	10
10 – 20	𝑥
20 – 30	25
30 – 40	30
40 – 50	𝑦
50 – 60	10
Total	100

 

Soln:

First find out the cumulative frequency (cf)

Marks	No. of students (f)	Cumulative frequency (f)
0 – 10	10	10
10 – 20	𝑥	10+x
20 – 30	25	35+x
30 – 40	30	65+x
40 – 50	𝑦	65+x+y
50 – 60	10	75+x+y
Total	100

Now, Given that there are 100 students

          f = 100

          Þ 75+x+y = 100

          Þ x + y = 100 – 75 = 25            …. (1)

Also given that

          Median = 32

          That is, median lies in the range 30-40.

Therefore, 30-40 is the median class.

        

          Here,  L = Lower class containing the median

                   N = Total number of students

                   f  = Frequency of the class containing median

                    cf  = Cumulative frequency before the median class

                   h = class interval = upper limit – lower limit

In our problem

          L = 30

          N = 100

          f = 30

          cf = 35 + x

          h = 10 – 0 = 10

Substitute these values, we get

         

          2 x 3 = 15 – x

          x = 15 - 6

          x = 9

Substituting the value of x in eqn (1) we get

          y = 25 – x

          y = 25 – 9

        y = 16

Hence, the value of x = 9 and value of y = 16.