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  • CBSE Math XII Board Paper – SET 1 (Delhi)

    MATHEMATICS

    Time allowed : 3 hours] [Maximum marks : 100

    i. All questions are compulsory.

    ii. The question paper consists of 29 questions divided into three sections A, B and C. Section A comprises of 10 questions of one mark each, Section B

    comprises of 12 questions of four marks each, and Section C comprises of 7

    questions of six marks each.

    iii. All questions in section A are to be answered in one word, one sentence or as per the exact requirements of the question.

    iv. There is no overall choice. However, internal choice has been provided in 4 questions of four marks each and 2 questions of six marks each. You have to

    attempt only one of the alternatives in all such questions.

    v. Use of calculators is not permitted.

  • 1. State the reason for the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)}

    not to be transitive.

    Solution:

    It is known that a relation R in a set A is transitive if (a1, a2) R and (a2, a3)  R 

    (a1, a3)  R  a1, a2, a3  A.

    It can be observed that (1, 2), (2, 1)  R, but (1,1)  R.

    Thus, the relation R in the set {1, 2, 3} is not transitive.

    2. Write the value of 1 π 1

    sin sin 3 2

            

    Solution:

    1

    1

    1

    π 1 sin sin

    3 2

    1 Let sin

    2

    1 sin

    2

    π π π sin sin sin sin 2π

    6 6 6

    π 2π

    6

    π 1 π π sin sin sin 2π

    3 2 3 6

    9π 9π sin sin

    6 6

    3π π sin sin π

    2 2

    x

    x

    x

    x

           

      

       

     

      

                 

       

      

                  

        

              

       

            

     

    π sin 1

    2

        

     

    Thus, the value of the given expression is 1.

    3. For a 2  2 matrix, A = [aij] whose elements are given by ij i

    a j

     , write the value of

    a12.

    Solution:

    Any element in the i th

    row and j th

    column is ij i

    a j

    For a12, the value of i = 1 and j = 2.

  • 12

    1

    2 a 

    Thus, the value of a12 is 1

    2 .

    4. For what value of x, the matrix 5 1

    2 4

    x x      

    is singular?

    Solution:

    The matrix 5 1

    2 4

    x x A

           

    is singular

       

    0

    5 1 0

    2 4

    4 5 2 1 0

    20 4 2 2 0

    6 18

    18 3

    6

    A

    x x

    x x

    x x

    x

    x

     

       

        

        

       

       

    Thus, the required value of x is 3.

    5. Write A –1

    for 2 5

    1 3 A

          

    Solution:

     

       

    1

    1

    2 5 A

    1 3

    3 51 A

    1 2det A

    3 51

    1 22 3 1 5

    3 51

    1 26 5

    3 5 A

    1 2

          

        

     

        

        

        

      

        

     

  • 6. Write the value of  sec sec tanx x x dx

    Solution:

     secx (secx + tanx) dx

    = (sec 2 x + secx tanx)dx

    = sec 2 x dx + secx tanx dx

    = tanx + secx + c , where c is an arbitrary constant

    7. Write the value of 2 16

    dx

    x 

    Solution:

    2

    2 2

    16

    4

    dx

    x

    dx

    x

     

    11 tan 4 4

    x C

        

      , where C is the arbitrary constant

    8. For what value of ‘a’ the vectors ˆˆ ˆ2 3 4i j k  and ˆˆ ˆ6 8ai j k  are collinear?

    Solution:

    Two vectors x and y are collinear if x y , where  is a constant.

    It is given that ˆ ˆˆ ˆ ˆ ˆ2 3 4 and 6 8i j k ai j k    are collinear.

     ˆ ˆˆ ˆ ˆ ˆ2 3 4 6 8i j k ai j k      for some constant  . 2 , 3 6 ,4 8

    1 3 6 or 4 8 gives

    2

    a  

      

         

         

    Now, 2 =  a

    1 2

    2

    4

    a

    a

       

      

    Thus, the value of a is –4.

    9. Write the direction cosines of the vector ˆˆ ˆ2 5i j k   .

    Solution 9:

    The given vector is ˆˆ ˆ2 5i j k   .

  • The direction cosines of the given vector is given by

                      2 2 2 2 2 2 2 2 2

    2 1 5 , ,

    2 1 5 2 1 5 2 1 5

        

                  

    2 1 5 , ,

    30 30 30

           

    10. Write the intercept cut off by the plane 2x + y – z = 5 on x-axis.

    Solution:

     2 5 ... 1x y z   Dividing both sides of the equation (1) by 5, we obtain

     

    2 1

    5 5 5

    1 ... 2 5 5 5

    2

    y z x

    x y z

      

        

    It is known that the equation of a plane in intercept form is 1 x y z

    a b c    , where a, b

    and c are the intercepts cut off by the plane at x, y, and z- axes respectively.

    Therefore, for the given equation, 5

    , 5, and 5 2

    a b c   

    Thus, the intercept cut off by the given plane on the x-axis is 5

    . 2

  • SECTION – B

    Question numbers 11 to 22 carry 4 marks each.

    11. Consider the binary operation * on the set {1, 2, 3, 4, 5} defined by a * b = min.

    {a, b}. Write the operation table of the operation *.

    Solution:

    The binary operation  on the set {1, 2, 3, 4, 5} is defined by a  b = min {a, b}

    1  1 = min {1, 1} = 1

    1  2 = min {1, 2} = 1

    2  3 = min {2, 3} = 2

    4  5 = min {4, 5} = 4

    The operation table for the given operation  on the given set can be written as:

     1 2 3 4 5

    1 1 1 1 1 1

    2 1 2 2 2 2

    3 1 2 3 3 3

    4 1 2 3 4 4

    5 1 2 3 4 5

    12. Prove the following:

    1 1 sin 1 sin π

    cot , 0, 2 41 sin 1 sin

    x x x x

    x x

          

            

    OR

    Find the value of 1 1tan tan x x y

    y x y

               

    Solution:

    2

    2 2

    2

    2 2

    1 sin cos sin 2sin cos cos sin 2 2 2 2 2 2

    1 sin cos sin 2sin cos cos sin 2 2 2 2 2 2

    x x x x x x x

    x x x x x x x

            

     

            

     

  • 1

    2 2

    1

    2 2

    1

    1 sin 1 sin cot

    1 sin 1 sin

    cos sin cos sin 2 2 2 2

    cot

    cos sin cos sin 2 2 2 2

    cos sin cos sin 2 2 2 2

    cot

    cos sin cos sin 2 2 2

    x x

    x x

    x x x x

    x x x x

    x x x x

    x x x

          

       

         

               

                     

             

         

         

    1

    1

    2

    2cos 2cot

    2sin 2

    cot cot 2

    2

    x

    x

    x

    x

    x

         

           

       

          

        

     

    Thus, the result is proved.

    OR

     

    1 1

    1 1

    1 1 1 1 1

    1 1 1

    1

    tan tan

    1

    tan tan

    1

    1

    tan tan tan tan tan 1

    1 1

    tan tan tan 1

    tan

    x x y

    y x y

    x

    x y

    xy

    y

    x

    x a by a b

    xy ab

    y

    x x

    y y

    x

     

     

        