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# FOR WHAT VALUE OF K IS THE FUNCTION  F(x)={(1+kx)½ -(1-kx)1/2 /x ,-12x+1/x-1,0continous at x=0.

Vikas TU
14149 Points
3 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