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Question: The function f(x) = [x]<sup>2</sup>− [x<sup>2</sup>] (where [y] is the greatest integer less than or...

The function f(x) = [x]2− [x2] (where [y] is the greatest integer less than or equal to y), is discontinuous at

A

All integers

B

All integers

C

All integers except 0

D

All integers except

Answer

All integers except

Explanation

Solution

We have f(x) = [x]2 – [x]2

At x = 0, L.H.L. = lim⁡h→0f(−h)lim⁡h→0[−h]2−[(−h)2]\lim_{h \rightarrow 0}f( - h)\lim_{h \rightarrow 0}\lbrack - h\rbrack^{2} - \left\lbrack ( - h)^{2} \right\rbrack

lim⁡h→0f(−1)2[h]2=lim⁡h→01−0=1\lim_{h \rightarrow 0}f( - 1)^{2}\lbrack h\rbrack^{2} = \lim_{h \rightarrow 0}1 - 0 = 1

R.H.L. = lim⁡h→0f(h)=lim⁡h→0[h]2−[h]2=lim⁡h→00−0=0\lim_{h \rightarrow 0}f(h) = \lim_{h \rightarrow 0}\lbrack h\rbrack^{2} - \lbrack h\rbrack^{2} = \lim_{h \rightarrow 0}0 - 0 = 0

∴ L.H.L ≠ R.H.L. ∴ f(x) is not continuous at x = 0.

At x = 1, L.H.L., lim⁡h→0f(1−h)=lim⁡h→0[1−h]2−[(1−h)2]\lim_{h \rightarrow 0}f(1 - h) = \lim_{h \rightarrow 0}\lbrack 1 - h\rbrack^{2} - \left\lbrack (1 - h)^{2} \right\rbrack

= lim⁡h→00−0=0\lim_{h \rightarrow 0}0 - 0 = 0

R.H.L. = lim⁡h→0f(1+h)=lim⁡h→0[1+h]2−[(1+h)2]\lim_{h \rightarrow 0}f(1 + h) = \lim_{h \rightarrow 0}\lbrack 1 + h\rbrack^{2} - \left\lbrack (1 + h)^{2} \right\rbrack

= lim⁡h→00−0=0\lim_{h \rightarrow 0}0 - 0 = 0

f(1) = [1]2 − [1]2 = 1 − 1 = 0

∴ L.H.L. = R.H.L. = f(1) ∴ f(x) is continuous at x = 1.