\(\mathrm{f}(\mathrm{x})=\mathrm{x} \cdot \sin (1 / \mathrm{x})\) and \(\mathrm{x} \in[-1,1]\). Also \(\mathrm{f}(0)=0\) then. (a) \(\mathrm{f}(\mathrm{x})\) is continuous in \([-1,1]\) (b) Roll's theorem is applicable in \([-1,1]\) (c) First mean value theorem is applicable in \([-1,1]\) (d) none of these

A function is continuous at a point if the limit exists at that point, and the limit is equal to the function's value at that point. We need to check the continuity of the function \(f(x) = x \cdot \sin(1/x)\) over the interval \([-1, 1]\). If \(\mathrm{x}\neq 0\), the function is continuous because the product of two continuous functions (x and \(\sin(1/x)\)) is continuous. To check the continuity at x=0, we need to find the limit as x approaches 0. Let's calculate the limit: \(\lim_{x\to 0} x \cdot \sin(1/x)\) divide by x, the limit becomes: \(\lim_{x\to 0} \sin(1/x)\) Since sine function has output value in interval \([-1, 1]\), the limit as x approaches 0 for \(\sin(1/x)\) lies between -1 and 1. Hence, \(\lim_{x\to 0} x \cdot \sin(1/x) = 0\), and since f(0) = 0, the function is continuous at x = 0. Therefore, f(x) is continuous in the interval [-1, 1].

To check differentiability of the function, we need to find its derivative, and see if it exists in the given interval. The derivative of f(x) when x ≠ 0: \(f'(x) = \sin(1/x) + x \cdot (-\frac{1}{x^2})\cos(1/x)\) We cannot be sure about differentiability at x = 0, so let's check it separately. As x approaches 0, we find the limit: \(\lim_{x\to 0} f'(x)\) Since it is difficult to find the limit directly, we will use the Squeeze Theorem. Firstly, we have: -1 ≤ \(\sin(1/x)\) ≤ 1 Now, we add x\(\cdot\)(-1/x²)\(\cos(1/x)\) to each term: -1 + x\(\cdot\)(-1/x²)\(\cos(1/x)\) ≤ \(\sin(1/x) + x\cdot(-1/x²)\cos(1/x)\) ≤ 1 + x\(\cdot\)(-1/x²)\(\cos(1/x)\) Since the left part approaches 0 as x approaches 0, and the right part also approaches 0 as x approaches 0, we can conclude that the middle part approaches 0 as well, according to the Squeeze theorem. Therefore, the function is differentiable at x = 0. Now, we can conclude that f(x) is differentiable in the interval [-1, 1].

a) The function f(x) is continuous in [-1, 1]. b) Roll's theorem states that if a function is continuous on [a, b] and differentiable on (a, b), and its values at a and b are equal ( f(a) = f(b) ), then there exists at least one point c in the interval (a, b) such that its derivative at c is 0 ( f'(c) = 0 ). Since f is continuous, differentiable but not equal in the given interval [-1, 1] i.e. f(-1) ≠ f(1), Roll's theorem is not applicable in [-1, 1]. c) First mean value theorem states that if a function is continuous on [a, b] and differentiable on (a, b), then there exists a point c in the interval (a, b) such that f'(c) = (f(b) - f(a)) / (b - a). Since f is continuous and differentiable in the given interval [-1, 1], the First Mean Value Theorem is applicable in [-1, 1]. d) Since the function f(x) is continuous, differentiable, and the First Mean Value Theorem is applicable in [-1, 1], "none of these" is incorrect. So, the correct answers are: - f(x) is continuous in [-1, 1]. - First Mean Value Theorem is applicable in [-1, 1].