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Let f(x)=⎪ ⎪ ⎪ ⎪ ⎪⎪ ⎪ ⎪ ⎪ ⎪x3ax2+2x23x+2,0<x<1b2x2+bx+1,x=1(1+(lnc)tan2(x1))1/(lnx)2,1<x<2

be continuous at x=1. Then which of the following relations can hold good?

A
b+c=4
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B
ac=9
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C
cb=5
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D
ab=2
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Solution

The correct option is D ab=2
Given that f(x) is continuous at x=1
So, L.H.L. =limx1(x3ax2+2x23x+2) exists and is finite.
For being 00 form, a=3
Now apply L-Hospital rule
L.H.L. =limx1(3x23×2x2x3)=3

Also, f(1)=3
b2+b+1=3
(b1)(b+2)=0
b=2 or b=1

and limx1+(1+(lnc)tan2(x1))1/(lnx)2 (1 form)=3
exp(limx1+tan2(x1)lnc(lnx)2)=3
Let x=1+t
Then, exp(limt0+tan2tlnc(ln(1+t))2)=3
exp⎜ ⎜ ⎜ ⎜limt0+tan2tt2lnc(ln(1+t))2t2⎟ ⎟ ⎟ ⎟=3elnc=3c=3

We have a=3, b=2 or b=1, c=3

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