Question

If $f(x)=\left\\{\begin{array}{ll}{[x]+\sqrt{\\{x\\}}} & x<1 \\\ \frac{1}{[x]+\\{x\\}^{2}} & x \geq 1\end{array}\right.$, then [where [. ] and \\{ - \(\\}\) represent greatest integer and fractional part functions respectively] (A) \(\mathrm{f}(\mathrm{x})=\) is continuous at \(\mathrm{x}=1\) but not differentiable (B) \(\mathrm{f}(\mathrm{x})\) is not continuous at \(\mathrm{x}=1\)

Short answer

Answer: The function f(x) is not continuous at x = 1.

Step-by-step solution

Analyze the function f(x) for x < 1

For x < 1, the function f(x) is defined as [x] + sqrt{ x }. Since it is a combination of the greatest integer function and the square root function, we will analyze the behavior of both functions separately for x < 1.

Determine if f(x) is continuous at x = 1

In order to check the continuity, we need to check if the limits of both functions (x < 1 and x >= 1) are equal to f(1). For x < 1: \[ \lim_{x \to 1^-} ([x] + \sqrt{x}) = [1] + \sqrt{1} = 1 + 1 = 2\] For x >= 1: \[ \lim_{x \to 1^+} \frac{1}{[x] + x^2} = \frac{1}{[1] + 1^2} = \frac{1}{1+1} =\frac{1}{2} \] Since the limits at x = 1 for both functions are not equal, f(x) is not continuous at x = 1.

Determine if f(x) is differentiable at x = 1 (not necessary)

Since f(x) is not continuous at x = 1, there is no need to check for differentiability because a function must be continuous at a point in order to be differentiable. Based on the above analysis, we can conclude that: (A) f(x) is not continuous nor differentiable at x = 1 (B) f(x) is not continuous at x = 1 (Correct)