Choose the correct options in the following questions: (i) If y = elogx then dy/dx is: (ii) Radius of a circle is increasing at the rate of 2 m/s. Rate of change of its circumference is:

i) Step 1: Simplify the expression for $y$. We know that $e^{\log x} = x$. So, $y = x$.

Step 2: Differentiate $y$ with respect to $x$. $\frac{dy}{dx} = \frac{d}{dx}(x) = 1$ The correct option is (c). The final answer is $\boxed{1}$.

ii) Step 1: Identify the given rate and the formula for circumference. Let $r$ be the radius of the circle and $C$ be its circumference. Given rate of change of radius: $\frac{dr}{dt} = 2 \text{ m/s}$. The formula for the circumference of a circle is $C = 2\pi r$.

Step 2: Differentiate the circumference formula with respect to time $t$. $\frac{dC}{dt} = \frac{d}{dt}(2\pi r) = 2\pi \frac{dr}{dt}$

Step 3: Substitute the given value of $\frac{dr}{dt}$. $\frac{dC}{dt} = 2\pi (2 m/s) = 4\pi m/s$ The correct option is (a). The final answer is $\boxed{4\pi m/s}$.

iii) Step 1: Let the integral be $I$. $I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$

Step 2: Apply the property of definite integrals $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$. Here, $a = \frac{\pi}{6}$ and $b = \frac{\pi}{3}$. So, $a+b = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2}$. $I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos(\frac{\pi}{2} - x)}}{\sqrt{\sin(\frac{\pi}{2} - x)} + \sqrt{\cos(\frac{\pi}{2} - x)}} dx$ Using the identities $\cos(\frac{\pi}{2} - x) = \sin x$ and $\sin(\frac{\pi}{2} - x) = \cos x$: $I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$

Step 3: Add the original integral and the transformed integral. $2I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx + \int_{\pi/6}^{\pi/3} \frac{\sqrt{\sin x}}{\sqrt{\cos x} + \sqrt{\sin x}} dx$ $2I = \int_{\pi/6}^{\pi/3} \frac{\sqrt{\cos x} + \sqrt{\sin x}}{\sqrt{\sin x} + \sqrt{\cos x}} dx$ $2I = \int_{\pi/6}^{\pi/3} 1 \, dx$

Step 4: Evaluate the integral. $2I = [x]_{\pi/6}^{\pi/3}$ $2I = \frac{\pi}{3} - \frac{\pi}{6}$ $2I = \frac{2\pi}{6} - \frac{\pi}{6}$ $2I = \frac{\pi}{6}$ $I = \frac{\pi}{12}$ The correct option is (c). The final answer is $\boxed{\frac{\pi}{12}}$.

iv) Step 1: Recognize the standard integral form. The integral is $\int_0^1 \frac{dx}{1+x^2}$. We know that $\int \frac{1}{1+x^2} dx = \arctan x + C$.

Step 2: Evaluate the definite integral using the limits. $\int_0^1 \frac{dx}{1+x^2} = [\arctan x]_0^1$ $= \arctan(1) - \arctan(0)$

Step 3: Substitute the known values of $\arctan$. We know that $\arctan(1) = \frac{\pi}{4}$ and $\arctan(0) = 0$. $= \frac{\pi}{4} - 0$ $= \frac{\pi}{4}$ The correct option is (b). The final answer is $\boxed{\frac{\pi}{4}}$.