If f(x)={√1+kx−√1−kxx,for−1≤x<02x2+3x−2,for0≤x≤1 is continuous at x = 0 , then k=
A
-4
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B
-3
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C
-2
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D
-1
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Solution
The correct option is C -2 Since f(x) is continuous at x = 0 the limits about x = 0 should be equal, i.e. L.H.L = R.H.L L.H.L=limx→0−√1+kx−√1−kxx=k R.H.L=limx→0+(2x2+3x−2)=−2 L.H.L = R.H.L ⇒ k = - 2.