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Differential Calculus

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Joined:	11 Jun 2007
Post:	1048

12 Jun 2007 15:46:52 IST

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736

doubt in continuity problem

None

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)

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Comments (2)

Titun

Forum Expert

Joined: 23 Dec 2006

Posts: 374

12 Jun 2007 16:02:04 IST

Like 2 people liked this

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

Blazing goIITian

Joined: 22 Jan 2007

Posts: 2039

12 Jun 2007 21:46:56 IST

Like 0 people liked this

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

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