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Question

If the function f(x)=x2(k+2)x+2kx2forx22forx=2 is continuous at x=2, then k is equal to

A
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B
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C
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D
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E
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Solution

The correct option is C 0
Given, f(x)=x2(k+2)x+2kx2forx22forx=3
Since, f(x) is continuous at x=2.
Therefore, limx2f(x)=f(2)
limx2x2(k+2)x+2kx2=2ApplyLhospitalrule,weget\displaystyle \lim_{x \rightarrow 2} \dfrac {2x - (k + 2)}{1} = 2\Rightarrow 2(2) - (k + 2) = 2\Rightarrow k + 2 = 2\Rightarrow k = 0$

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