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Question

Find the value of k, for which f(x)=⎪ ⎪ ⎪⎪ ⎪ ⎪1+kx1kxxif1x<02x+1x1if0x<1⎪ ⎪ ⎪⎪ ⎪ ⎪ is continuous at x=0.

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Solution

The given function f(x) is continuous at x = 0.

limx0f(x)

0+101=limx0(1+kx1kxx)

1=limx0(1+kx1kxx)(1+kx+1kx1+kx+1kx)

1=limx0(1+kx1+kxx[1+kx+1kx])

1=limx0(2k1+kx+1kx)

When k=1

1=2k2

k=1

Therefore the value of k is 1.


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