Page 71 - ISC-12
P. 71

Matching-List & Matching-Matrix Type                     Jai Baba Ki13

          13. Match the following two lists:


                                                List-I                                      List-II



                (A) A line from the origin meets the lines                                  (p)  −4
                                                       8
                                                  x −
                      x − 2    y − 1    z + 1          3    y + 3    z − 1
                            =        =        and        =        =        at P
                        1       −2        1          2       −1        1
                                                                 2
                      and Q respectively. If length PQ = d, then d is


                (B)   The values of x satisfying                                            (q)  0

                                                             3
                                         −1
                         −1
                      tan (x + 3) − tan (x − 3) = sin   −1       are
                                                             5
                                         ~
                                                          ~
                (C)   Non-zero vectors ~a, b and ~c satisfy ~a · b = 0,                     (r)  4
                                                ~
                                                         ~
                               ~
                                                                        ~
                      ~
                      (b − ~a) · (b + ~c) = 0 and 2|b + ~c| = |b − ~a|. If ~a = µb + 4~c,
                      then the possible values of µ are
                (D)   Let f be the function on [−π, π] given by f(0) = 9 and                (s)  5
                                   9x

                              sin                                Z  π
                                   2                           2
                      f(x) =          for x 6= 0. The value of         f(x) dx is
                                   x
                              sin                              π    −π
                                   2

                                                                                            (t)  6







                                                       (p)  (q)   (r)  (s)   (t)
                                              (A)      p     q     r    s    t


                                              (B)      p     q     r    s    t

                                              (C)      p     q     r    s    t


                                              (D)      p     q     r    s    t



                                                                                                    [JEE 2010]
   66   67   68   69   70   71   72   73   74   75   76