The value of \(\mathrm{k}(\mathrm{k}>0)\) for which the function \(\mathrm{f}(\mathrm{x})=\left[\left(\mathrm{e}^{\mathrm{x}}-1\right)^{4} /\left\\{\operatorname{Sin}\left(\mathrm{x}^{2} / \mathrm{k}^{2}\right) \log \left[1+\left(\mathrm{x}^{2} / 2\right)\right]\right\\}\right]\) \(\mathrm{x} \neq 0, \mathrm{f}(0)=8\) may be continuous at \(\mathrm{x}=0\) is: (a) 1 (b) 2 (c) 4 (d) 3

The function f(x) is continuous at x=0 if the following conditions are met: 1. The function f(x) is defined at x=0. 2. The limit of f(x) exists as x approaches 0. 3. The limit of f(x) as x approaches 0 is equal to f(0). Step 2: Evaluate the limit of the function as x approaches 0

First, let's rewrite the given function f(x) as: f(x) = \(\frac{(e^x - 1)^4}{\sin(\frac{x^2}{k^2}) \log(1 + \frac{x^2}{2})}\) Now to find the value of k that makes the function continuous at x=0, we need to find \(\lim_{x\to 0} f(x)\). We will use L'Hopital's rule to evaluate the limit. Step 3: Applying L'Hopital's rule

Differentiating both the numerator and denominator of f(x) with respect to x, we get: \[\frac{d}{dx}((e^x - 1)^4) = 4(e^x - 1)^3e^x\] \[\frac{d}{dx}\left(\sin\left(\frac{x^2}{k^2}\right)\log\left(1 + \frac{x^2}{2}\right)\right) = \frac{2x\cos\left(\frac{x^2}{k^2}\right)}{k^2}\cdot\log\left(1 + \frac{x^2}{2}\right) + \frac{x\sin\left(\frac{x^2}{k^2}\right)}{1 + \frac{x^2}{2}}\] Now apply L'Hopital's rule: \[\lim_{x\to 0} f(x) = \lim_{x\to 0}\frac{4(e^x - 1)^3e^x}{\frac{2x\cos(\frac{x^2}{k^2})}{k^2}.\log(1 + \frac{x^2}{2}) + \frac{x\sin(\frac{x^2}{k^2})}{1 + \frac{x^2}{2}}}\] Denoting the limit as L and replace x=0 in the above equation, we get: L = \(\frac{4(1 - 1)^3(1)}{\frac{2(0)\cos(\frac{0}{k^2})}{k^2}\cdot\log(1 + \frac{0}{2}) + \frac{0\sin(\frac{0}{k^2})}{1 + \frac{0}{2}}}\) = 0 Since f(0) = 8, this means the limit when x approaches 0 should be equal to f(0) which is 8. However, our limit calculation has resulted in L = 0, which means there is no value of k that will make f(x) continuous at x = 0. Therefore, none of the given options (a) 1, (b) 2, (c) 4 or (d) 3 is correct.