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Question

Match List I with the List II and select the correct answer using the code given below the lists :

List IList II(A)If limx(x2+1x+1axb)=0, then (P)a=32,bR(B)If limx0(1+ax+bx2)2/x=e3, then(Q)a=1,b=12(C)If limx0(aexbx)=2, then(R)a=1,b=1(D)If limx{(x2x+1)axb}=0, then(S)a=2,b=2

Which of the following is the only CORRECT combination?

A
(A)(Q), (D)(Q)
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B
(A)(R), (D)(S)
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C
(A)(R), (D)(Q)
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D
(A)(S), (D)(R)
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Solution

The correct option is C (A)(R), (D)(Q)
limx(x2+1x+1axb)=0limx(1+x2ax2bxaxbx+1)=0limx((1a)x2(a+b)x+1bx+1)=0
For existence of limit,
1a=0a=1(a+b)=0b=1

limx{(x2x+1)axb}=0limx{(x2x+1)axb}(x2x+1)+ax+b(x2x+1)+ax+b=0limxx2x+1(ax+b)2(x2x+1)+ax+b=0limxx2(1a2)x(2ab+1)+(1b2)x(11x+1x2+a+bx)=0limxx(1a2)(2ab+1)+1b2x11x+1x2+a+bx=0

For existence of limit,
1a2=0a=±1
But at a=1, denominator becomes 0
So, a=1
2b12=0b=12

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