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Question

If L=limx23x2+kx+k72x2+x6 exists and is finite, where k is a real number, then the values of k and L are

A
k=5,L=1
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B
k=5,L=1
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C
kR,L=32
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D
k=5,L=1
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Solution

The correct option is B k=5,L=1
L=limx23x2+kx+k72x2+x6
For 00 form and limit to exist, numerator should tend to 0 as x2
3(2)2+k(2)+k7=0k=5L=limx23x2+5x22x2+x6 =limx2(3x1)(x+2)(2x3)(x+2) =6143=1

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