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FOR WHAT VALUE OF K IS THE FUNCTION F(x)={(1+kx) ½ -(1-kx) 1/2 /x ,-1 2x+1/x-1,0 continous at x=0.

FOR WHAT VALUE OF K IS THE FUNCTION  F(x)={(1+kx)½ -(1-kx)1/2 /x ,-1
2x+1/x-1,0
continous at x=0.

Grade:12

1 Answers

Vikas TU
14149 Points
6 years ago
We have,f(x) = 1+kx√ − 1−kx√x; −1≤x
LHL = limx→0− f(x) = limx→0−(1+kx√ − 1−kx√x) 
put x = 0−h, as x→0−, at that point h→0 
LHL = limh→0 [1+k(0−h)√ − 1−k(0−h)√(0−h)] 
LHL = limh→0[1−kh√ − 1+kh√−h] 
LHL = limh→0[1+kh√ − 1−kh√h] 
LHL =limh→0[1+kh√ − 1−kh√h×1+kh√ + 1−kh√1+kh√ + 1−kh√] 
LHL = limh→0[1+kh−1+khh(1+kh√ + 1−kh√)] 
LHL = limh→0[2khh(1+kh√ + 1−kh√)] 
LHL = limh→0[2k(1+kh√ + 1−kh√)] = 2k1√ + 1√ = 2k2 = k 
RHL =limx→0+ f(x) = limx→0+(2x+1x−1) 
put x = 0+h, as x→0+, at that point h→0 
RHL = limh→0[2(0+h)+1(0+h)−1] 
RHL = limh→0[2h+1h−1] = 0+10−1 = −1 
f(0) = 2×0+10−1 = −1 
Since f(x) is persistent at x = 0, 
thenLHL = RHL = f(0) 
⇒k = −1 = −1 
Thus, k = −1 

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