Let [t] denotes the greatest integer less than or equal to t and limx→0x[4x]=A. If function f(x)=[x2]sinπx is discontinuous, then possible value of x is
A
√A+21
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B
√A+5
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C
√A
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D
√A+1
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Solution
The correct option is D√A+1 f(x)=[x2]sinπx
It is continuous ∀x∈Z. f(x) is discontinuous at points where [x2] is discontinuous i.e. x2∈Z with an exception that f(x) is continuous as x is an integer. ∴ Points of discontinuity for f(x) would be at x=±√2,±√3,±√5,.......
Also, it is given that limx→0x[4x]=A (indeterminate form (0×∞)) ⇒limx→0x(4x−{4x}) ⇒4−limx→0x{4x}=A ⇒A=4 √A+5=3 √A+1=√5 √A+21=5 √A=2 ∴ Point of discontinuity for f(x) is x=√5