let f(x) = ([a]2-5[a]+4)x3 - (6{a}2-5{a} + 1)x - (tanx)sgnx, be an even function for all x belongs to R, then sum of all possible values of "a" is
(where [] and {} denotes greatest integer function and fractional part functions respectively)
a. 17/6
b. 53/6
c. 31/3
d. 35/3
1Hint-(tanx)(sgnx) is even function ...(product of two odd functions is even fucntion)
now x is an odd function and x3 is also odd fuction
so for f(x) to be even coefficent of x and x^3 must be 0
so u got to solve thes
2 eq
[a]2-5[a]+4=0
and
6{a}2-5{a}+1=0
1is it answer is d- 35/3.please post me if i m right.then i will give u solution.
