FOR WHAT VALUE OF K IS THE FUNCTION F(x)={(1+kx)

Question

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

Vikas TU · Accepted Answer

We have,f(x) = 1+kx√ − 1−kx√x; −1≤xLHL = 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

Differential Calculus> FOR WHAT VALUE OF K IS THE FUNCTION F(x)=...

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

NITIN KUMAR , 7 Years ago

Vikas TU

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Differential Calculus> FOR WHAT VALUE OF K IS THE FUNCTION F(x)=...

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

NITIN KUMAR , 7 Years ago

Vikas TU

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