11Use the substitution sinx=t, then expression for cos3x....
g(x) = ∫ cos4t dt , then g(x+π ) = ? { lower limit =0 upper limit= x}
a. g(x) + g(Ï€)
b. g(x) - g(Ï€)
c.g(x) *g(Ï€)
d. g(x)/g(Ï€)
-
UP 0 DOWN 0 0 4
4 Answers
1∫ cos4t dt= ∫ (1+cos2t / 2)2 dt
now,by expanding integrating and putting the limits,u would get:
g(x)= 1/4∫(3/2 + 2cos 2t + 1/2 cos4t)dt
= 3/8 x + 1/4 sin2x + 1/32 sin4x.
g(x+∩) = 3/8 (x+∩) + 1/4 sin2x + 1/32 sin4x
now,by checking the options,u would get, a) g(x+∩)=g(x) + g(∩)
1Objective method useful for JEE
Put x=0 : g(0) = 0
g(Ï€) = ?
option a : g(0) + g(Ï€) = g(Ï€)
option b : g(0) - g(Ï€) = -g(Ï€)
option c : g(0) *g(Ï€) = 0
option d : g(0)/g(Ï€) = 0
since g(Ï€) is non-zero only a is the correct option
