IIT JEE 2009 Mathematics Paper2 Code 1 Solutions

14.    The smallest value of k, for which both the roots of the equation x² - 8kx + 16(k² - k + 1) = 0 are real, distinct and have values at least 4, is

Sol.   2

         

        x² - 8kx + 16 (k² - k + 1) = 0

        Since , D > 0 => k > 1                                  ...... (1)

        Also -b/2a > 4 => 8k/2 > 4

        =>    k > 1                                             ...... (2)

       Also f(4) > 0 Þ 16 - 32 k + 16 (k² - k + 1) > 0

       k² - 3k + 2 > 0

       k < 1 υ k > 2                                     ...... (3)

       From equations (1), (2) and (3)

       k_min = 2

15.    The maximum value of the function f(x) = 2x³ - 15x² + 36x - 48 on the set A = {x|x² + 20 < 9x|} is

Sol.   7

        We have ,

        f'(x) = 6(x - 2)(x - 3)

        so f(x) is increasing in (3, ∞)

        Also, A = {4 < x < 5}

        f_max = f(5) = 7.

16.    Let ABC and ABC' be two non-congruent triangles with sides AB = 4, AC = AC' = 2√2 and angle B = 30^o. The absolute value of the areas of these triangles is

Sol.   4

        cosβ = (a²+16-8)/(2×a×4)

                 

        => √3/2=(a²+8)/8a

        => a₂ - 4√3 a + 8 = 0

        => a₁ + a₂ = 4√3, a₁a₂ = 0

        => |a₁ - a₂| = 4

        => |Δ₁ - Δ₂| = 1/2 × 4 sin 30^o × 4 = 4.

 

17.    If the function f(x) = x³ + e^x/2 and g(x) = f^-1(x), then the value of f'(1) is

Sol.   2

        f(0) = 1, f'(x) = 3x² + 1/2 e^x/2

        => f'(g(x)) g'(x) = 1

         Put x = 0 => g'(1) = 1/f'(0) = 2.

 

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