CBSE[OUTSIDE DELHI]_X_Mathematics_2019_Set_I

The given quadratic equation is:

              kx (x–2) + 6 = 0

             kx² – 2kx + 6 = 0

        Comparing with ax² + bx + c = 0, we get

            a = k, b = – 2k and c = 6

        Since, roots are equal. So,

         D = 0 i.e., b² – 4ac = 0

                (–2k)² – 4k(6) = 0

                       4k² – 24k = 0

                        4k(k – 6) = 0

                         Either k = 0 (Neglect)

                              Or k = 6

          Therefore, the value of k is 6.

Since, 2² × 5³ × 3² × 17 = (2×5)×(2×5)×2×3²×17

                                             = 10 × 10 × 2 × 9 × 17
As there are two 10’s. So, number of zeroes in the end of number obtained by given product is two.

The multiples between 10 and 205 are:

         12, 16, 20, …, 204

         Here,          first term (a) = 12
       and common difference (d) = 4

        Let                   a_n = 204

               a + (n – 1) d = 204

              12 + (n – 1)3 = 204

                      (n – 1)3 = 204 – 12

                        (n – 1) = 192/3

                                n = 64 + 1

                                   = 65

  Therefore, there are 65 multiples of 4 between 10 and 205.

Given: a₃ = 16, a₇ – a₅ = 12

       Let first term of AP be ‘a’ and common 
       difference be ‘d’.

       Since, a_n = a + (n – 1)d

       So,     a₃ = a + (3 – 1)d

            Or 16 = a + 2d          …(i)

           a₇ – a₅ = 12

               (a + 6d) – (a + 4d) = 12

                        2d = 12

                          d = 6

                a + 2(6) = 16

                          a = 16 – 12
                             = 4

       The required AP is:

         a, a + d, a + 2d, a + 3d, …

    or  4, 4 + 6, 4 + 2(6), 4 + 3(6), …

    i.e., 4, 10, 16, 22,…