    Question

# Let f(x)=⎧⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪⎩x3−ax2+2x2−3x+2,0<x<1b2x2+bx+1,x=1(1+(lnc)⋅tan2(x−1))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 a−b=2Given that f(x) is continuous at x=1 So, L.H.L. =limx→1−(x3−ax2+2x2−3x+2) exists and is finite. For being 00 form, a=3 Now apply L-Hospital rule L.H.L. =limx→1−(3x2−3×2x2x−3)=3 Also, f(1)=3 ⇒b2+b+1=3 ⇒(b−1)(b+2)=0 ⇒b=−2 or b=1 and limx→1+(1+(lnc)⋅tan2(x−1))1/(lnx)2 (1∞ form)=3 ⇒exp(limx→1+tan2(x−1)⋅lnc(lnx)2)=3 Let x=1+t Then, exp(limt→0+tan2t⋅lnc(ln(1+t))2)=3 ⇒exp⎛⎜ ⎜ ⎜ ⎜⎝limt→0+tan2tt2⋅lnc(ln(1+t))2t2⎞⎟ ⎟ ⎟ ⎟⎠=3⇒elnc=3⇒c=3 We have a=3, b=−2 or b=1, c=3  Suggest Corrections  0      Similar questions
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