IIT JEE 2009 Mathematics Paper2 Code 1 Solutions

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)    lim_x-->∞ 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, lim_x-->∞ (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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