QUESTION PAPER CODE 65/1/1 EXPECTED ANSWERS/VALUE POINTS SECTION - A Q. No. 1-10. 1. x = 25 2. x = 6. 2x3/2 + 2 10.
1 5

Marks 3. 10 π 12

4. x = 2

5.

x = + 6
2π 3

x +c

7.

8. 5

9.

ˆ ˆ { r – (aˆi + bˆj + ck ) }⋅ (ˆi + ˆj + k ) = 0 or r⋅ ˆ+ˆ+k =a+b+c i j ˆ

(

)

1×10 =10 m

SECTION - B 11.
∀ (a, b) ∈ A × A

a + b = b + a ∴ (a, b) R (a, b) ∴ R is reflexive For (a, b), (c, d) ∈ A × A If (a, b) R (c, d) i.e. a + d = b + c ⇒ c + b = d + a then (c, d) R (a, b) ∴ R is symmetric For (a, b), (c, d), (e, f) ∈ A × A If (a, b) R (c, d) & (c, d) R (e, f) i.e. a + d = b + c & c + f = d + e Adding, a + d + c + f = b + c + d + e then (a, b) R (e, f) ∴ R is transitive
∴ R is reflexive, symmetric and transitive

1m

1m

⇒

a+f = b+e 1m

hence R is an equivalance relation [(2, 5)] = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}

½m ½m

2

12.

   1 + sin x + 1 – sin x   cot–1   1 + sin x – 1 – sin x          
2 2 x  x x x   cos + sin  +  cos − sin    2  2 2 2    2 2 x  x x x    cos + sin  −  cos − sin   2  2 2 2  

= cot–1

2½ m

x   2 cos 2  x x −1  = cot −1   = cot  cot  = 2 2   2 sin x   2 OR
5 2  1 1   LHS = 2  tan −1 + tan −1  + sec −1   7  5 8   

1½ m

1 1  +  −1 5 8  + tan −1 1 = 2 tan  7 1– 1     40     2⋅1  1  3  + tan −1 1 = tan −1  2  7 7 1–  1        3 

1½+½ m

= 2 tan −1

1 + tan −1 3

1m

= tan −1

3 1 25 π + tan −1 = tan −1 = tan −1 (1) = = RHS 4 7 25 4

1m

2y 13. LHS = 2z x – y–z

y–z–x 2z 2x

2y z–x–y 2x

3

x +y+z = 2z x –y–z x +y+z

x +y+z x +y+z 2z 2x 0 z – x – y R1 → R1 + R 2 + R 3 2x 0 1m

=

C 2 → C 2 – C1 2z 0 – (x + y + z ) ; C3 → C3 – C1 x – y –z x+y+z x+y+z
2 3

2m

= (x + y + z) . {0 . (x + y + z) + (x + y + z) } = (x + y + z)

1m