...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 ...
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