Question

If f(x) = $\left\{ \begin{matrix} \lbrack x\rbrack + \sqrt{\{ x\}}, & x < 1 \\ \frac{1}{\lbrack x\rbrack + \{ x\}^{2}}, & x \geq 1 \end{matrix} \right.\$

[·] denotes greatest integer function.

{} denotes fractional part function. Then

• A. f(x) is continuous at x = 1
• B. f(x) is not continuous at x = 1
• C. $\lim_{x \rightarrow 1}$f(x) does not exist
• D. f(x) is differentiable at x = 1

Answer 1

Answer: (A)

f(x) is continuous at x = 1

Answer 2

R.H.L. $\lim_{h \rightarrow 0}$ $\frac{1}{\lbrack 1 + h\rbrack + \{ 1 + h\}^{2}}$ = $\frac{1}{1 + h^{2}}$ = + 1

f(1) = $\frac{1}{1 + 0}$ = + 1

L.H.L. $\lim_{h \rightarrow 0}$ [1 – h] + $\sqrt{\{ 1 - h\}}$

= 0 + 1 = + 1

f(x) is continuous at x = 1