11Q1.
Ans : B
bahut simple tha.
Put x = 0
f(x)f(-x) = [f(0)]2 = 9
let the fn f satisfy f(x).f'(-x)=f(-x)f'(x) for all x n f(o)=3
q1the value f(x)f(-x) for all x is
a]4
b]9
c]12
d]16
q2 ∫dx/(3+f(x))[limit -51 to 51]
a)17
b)34
c)102
d)0
q3
no of roots of f(x)=0 in[-2,2]
a1
b0
c2
d4
-
UP 0 DOWN 0 0 10
10 Answers
11In Q3. ,
" q3no of roots of f(x)=0 in[-2] " [7][7] something's missing.
11I = -51∫51 dx/(3 + f(x))
Let x = -t
dx = -dt
I = - 51∫-51 dt/(3 + f(-t))
= -51∫51 dt/(3 + f(-t))
f(y)f(-y) = 9
f(-y) = 9/f(y)
So,
I = -51∫51 dy/(3 + f(y)) = -51∫51 dy/{3 + ( 9/f(y) )}
2I = -51∫51 dy/(3 + f(y)) + {f(y)/3}dy/(3 + f(y))
2I = -51∫51 [ 1 + f(y)/3 ] dy/(3 + f(y))
2I = (1/3) -51∫51 dy
2I = (1/3)*[51 - (-51)]
So, I = 17
