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6. For function f(x) = s cos 1/x x > 1,
(A) for atleast one x in interval [1, ∞), f(x + 2) - f(x) < 2
(B) limx-->∞ f'(x) = 1
(C) for all x in the interval [1, ∞), f(x + 2) - f(x) > 2
(D) f'(x) is strictly decreasing in the interval [1, ∞)
Sol. (B, C, D)
For f(x) = x cos1/x x > 1
f'(x) = cos(1/x) + 1/x sin 1/x --> 1 for x --> ∞
also f"(x) = 1/x2 sin 1/x - 1/x2 sin 1/x - 1/x3 cos 1/x
= - 1/x3 cos 1/x < 0 for x > 1
=> f'(x) is decreasing for [1, ∞)
=> f'(x + 2) < f'(x). Also, limx-->∞ (x + 2) - f(x)
= -------------------------- = 2
\ f(x+2) - f(x) > 2 for all x > 1
Hence , the answer.
7. For 0 < q < Π/2, the solution(s) of
∑m=16 cosec (θ +(m-1)π /4)cosec(θ +mπ/4) = 4√2 is(are)
(A) π/4 (B) π/6 (C) π/12 (D) 5π/12
Sol. (C, D)
Given solutions
=> √2[cot θ - cot (θ + π/4) + cot(θ + π/4) - cot(θ + π/2) +...
...+ cot(θ + 5π/4) - cot(θ + 3π/2)] = 4√2
=> tan θ + cot θ = 4 => tan θ = 2 + √3
=> Hence , θ = Π/12 or 5Π/12
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IIT JEE 2009 Mathematics Paper2 Code 1 Solutions...