Let f(x)= ([a] 2 – 5[a] + 4)x 3 – (6{a} 2 – 5 {a} + 1)x – (tan x) sgn (x) be an even function ∀ x ∈ R, then the sum of all possible values of '3a' is
(where [ ⋅ ] denotes G.I. F and { ⋅ } fractional part functional part function)
Text Solution
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Sol. f(x) = ([a] 2 –5[a] + 4) x 3 – (6{a} 2 – 5{a} + 1) x – tan x(sgn x)
x > 0
f(x) = ([a] 2 –5[a] + 4) x 3 – (6{a} 2 – 5{a} + 1) x – tan x
x < 0
f(x) = ([a] 2 –5[a] + 4) x 3 – (6{a} 2 – 5{a} + 1) x + tan x
Given that function is even function ∀ x ∈ R
So f(x) – f(–x) = 0 ∀ n ∈ R
2x 3 ([a] 2 – 5[a] + 4) – 2x (6{a} 2 – 5{a} + 1) = 0
So this equation should be independent from x ∴ coff. of x 3 & x will be zero.
[a] 2 – 5[a] + 4 = 0 ; 6{a} 2 –5{a} + 1 = 0
[a] = 1, 4 ; {a} = 1/2,1/3
a = 1+ 1/2, 4 + 1/2, 1 + 1/3, 4 + 1/3 = 3/2, 9/2, 4/3, 13/3
Sum = 3/2 + 9/2 + 4/3 + 13/3 = 6 + 17/3 = 35/3
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