IIT JEE , AIEEE Forum - Mathematics , Differential Calculus

waterdemon

If

,

are the roots of the equation ax²+bx+c=0 then_{[ x]}

_{[ ]}(ax²+bx+c+1)^2/x-

is =

a) 2a (

-

)

b)2 ln I a(

-

) I

c)e^2a(--

d)e^{a^2 I-I}

Answer is option (C)

titun

Let P = _{[ x][ ]}(ax²+bx+c+1)^2/x-

Taking log on both sides,

ln P = ln _{[ x][ ]}(ax²+bx+c+1)^2/x-= _{[ x][ ]}2 ln (ax² + bx + c + 1) / (x -

)

[ For simple functions, we can take log Lt f(x) = Lt log f(x) ]

Now, the limit is in 0/0 form. So, apply L.Hospital's Rule,

ln P = 2 _{[ x][ ]} (2ax + b)/(ax² + bx + c + 1) =2[ 2a

+ b ]

[ since, a

² + b

+ c = 0]

Therefore, ln P =2[2a

- a(

+

)] =2a(

-

) (since, -b/a = (

+

) )

So, the reqd. limit P = e^{2a( - )}

So, option (c) is correct.

Cheers !!

neeraj_agarwal_1990

u may also use ax^2 + bx +c= a(x-alpha)(x-beta)