IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

neeraj_agarwal_1990

Question:

If

are the roots of

and

is an even function, then

is equal to :

Options:

A.

B.

C.

D.	None of these

Answer

Answer:

B

goiit_user

has roots

&

Thus,
g(x) = (x-

)(x-

) - eq.1

We have,

P =

Replacing g(x ) as in eq .1, we get

P = _[

_]

^[^] {e^f(x-⁾} / { e^f(x-⁾+ ef^(x-⁾}.dx - eq.2

We know,
_[a]

^[b]f(x)dx = _[a]

^[b]f(a+b - x)dx
Applying the same in eq. 2

P = _[

_]

^[^] {e^f(^-x)} / { e^f(^-x)+ e^f(^-x)}.dx
As f(x) is even,
f(-x) = f(x),

P = _[

_]

^[^] {e^f(x-⁾} / { e^f(x-⁾+ e^f(x-⁾}.dx eq.3

Adding eq.2 and 3,
the numerator n denominator get cancelled and we get,

2P = _[

_]

^[^]1.dx

2P =

-

P = (

-

)/2

Thus, the solution is B
as

-

= ( -b+D)/2a - ( -b - D)/2a
= 2D/2a
= D/a

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ayshwarya

actually it is integral lowr alpa to higher lt beta

f[x-beta]
e / e^f[x-alpa] +e^f[x-beta]

=let it b I
hence 2I = S dx
I =[ alpa-beta] /2
=rt[b^2 -4ac] /2a

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