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```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)

kmin = 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}

fmax = f(5) = 7.

16.    Let ABC and ABC' be two non-congruent triangles with sides AB = 4, AC = AC' = 2√2 and angle B = 30o. 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 30o × 4 = 4.

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

Sol.   2

f(0) = 1, f'(x) = 3x2 + 1/2 ex/2

=> f'(g(x)) g'(x) = 1

Put x = 0 => g'(1) = 1/f'(0) = 2.

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