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Image is attached . The function fx is discontinous at ?

Drake , 10 Years ago
Grade 12
anser 1 Answers
Jitender Singh

Last Activity: 10 Years ago

Ans:
f(x) = [x^{2}]-[x]^{2}
Lets check at x = 0
LHL =\lim_{h\rightarrow 0} ([(-h)^{2}]-[-h]^{2}) = -1
RHL =\lim_{h\rightarrow 0} ([(h)^{2}]-[h]^{2}) = 0
f(0) = 0
So, f(x) is discontinuous at x = 0
Lets check at x = 1
LHL =\lim_{h\rightarrow 0} ([(1-h)^{2}]-[1-h]^{2}) = 0
RHL =\lim_{h\rightarrow 0} ([(1+h)^{2}]-[1+h]^{2}) = 0
f(1) = 0
So, f(x) is continuous at x = 1
Lets check at x = 2
LHL =\lim_{h\rightarrow 0} ([(2-h)^{2}]-[2-h]^{2}) = 3-1 = 2
RHL =\lim_{h\rightarrow 0} ([(2+h)^{2}]-[2+h]^{2}) = 4-4 = 0
f(2) = 0
You should have given the options. Otherwise, you can check like this.
Thanks & Regards
Jitender Singh
IIT Delhi
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