Question

# If, is continuous at x=2, find the value of k.f(x)=⎧⎪⎨⎪⎩x3+x2−16x+20(x−2)2,x≠2k,x=2

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Solution

## Since the function f(x) is continuous x=2 then limx→2f(x)=f(2)......(1).Now,limx→2f(x)=limx→2x3+x2−16x+20(x−2)2 00 formUsing L'Hospital's rule we get,=limx→23x2+2x−162(x−2)00 formAgain using L'Hospital's rule we get,=limx→26x+22=7.From (1) we get, k=7.

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