try it

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

10 Answers

11
rkrish ·

Q1.

Ans : B

bahut simple tha.
Put x = 0
f(x)f(-x) = [f(0)]2 = 9

11
rkrish ·

In Q3. ,

" q3no of roots of f(x)=0 in[-2] " [7][7] something's missing.

1
vector ·

that s not ma doubt it s for ppl on d forum

1
vector ·

neways edited

11
rkrish ·

Q2.

Ans : A

11
rkrish ·

I = -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

39
Dr.House ·

hey 2nd one i am getting 0....

1
vector ·

2nd 1 ans is 17

39
Dr.House ·

shit came till last and at last step went wrong .

1
vector ·

ans 3rd b

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