The function \(\mathrm{f}(\mathrm{x})=2 \log (\mathrm{x}-2)-\mathrm{x}^{2}+4 \mathrm{x}+1\) increasing on the interval (a) \((2,3)\) (b) \((1,2)\) (c) \((2,4)\) (d) \((1,3)\)

To find the derivative of the function \(f(x) = 2\log(x-2) - x^2 + 4x + 1\), we'll use the rules of differentiation: Derivative of \(2\log(x-2)\) is \(\frac{d}{dx}(2\log(x-2)) = 2\cdot \frac{1}{x-2}\) Derivative of \(-x^2\) is \(\frac{d}{dx}(-x^2) = -2x\) Derivative of \(4x\) is \(\frac{d}{dx}(4x) = 4\) Derivative of \(1\) is \(\frac{d}{dx}(1) = 0\) Now add the derivatives together: \(f'(x) = \frac{2}{x-2} - 2x + 4\)

Now we will check in which of the given intervals, \(f'(x)>0\). (a) Check the interval (2,3): Select a number within this interval, for example, \(x = 2.5\), and check if the derivative is positive: \(f'(2.5) = \frac{2}{0.5} - 2(2.5) + 4 = 4 - 5 + 4 = 3\) Since the derivative is positive, the function is increasing in the interval (2,3). (b) Check the interval (1,2): Select a number within this interval, for example, \(x = 1.5\), and check if the derivative is positive: \(f'(1.5) = \frac{2}{-0.5} - 2(1.5) + 4 = -4 - 3 + 4 = -3\) Since the derivative is negative, the function is not increasing in the interval (1,2). (c) Check the interval (2,4): Select a number within this interval, for example, \(x = 3\), and check if the derivative is positive: \(f'(3) = \frac{2}{1} - 2(3) + 4 = 2 - 6 + 4 = 0\) Since the derivative is not positive, the function is not increasing in the interval (2,4). (d) Check the interval (1,3): This interval includes both interval (1,2) and (2,3), and as we found before, the function is not increasing in the interval (1,2). Therefore, the function isn't increasing in the entire interval (1,3). So, the function is increasing in the interval \( (2,3) \), which corresponds to option (a).