If the function f(x) = x^2 - ax - 2 is strictly decreasing on (2, 3) then a lies in the interval.

The derivative of x³ + 2x² - 5x + 1 is:

Match **List-I** with **List-II** | List-I | List-II | |---|---| | **Function** | **Increasing on the interval** | | (A) $f(x) = -x^2 - 2x + 1$ | (I) $(-\infty, -1)$ | | (B) $f(x) = x^2 + 1$ | (II) $(1, \infty)$ | | (C) $f(x) = x^2 - 2x + 3$ | (III) $(-\infty, 0)$ | | (D) $f(x) = -x^2$ | (IV) $(0, \infty)$ | Choose the correct answer from the options given below:

For what value of $\alpha$, the function $f$ defined by $f(x) = \begin{cases} \alpha(x^2 - 2x + 1), & \text{if } x \leq 0 \\ 2x + 1, & \text{if } x > 0 \end{cases}$ is continuous at $x = 0$?

$\int_0^{\pi/2} \sqrt{1 - \sin 2x}\,dx$ is equal to:

The equation of the tangent to the curve $y = \frac{(x - 3)}{(x - 1)(x - 2)}$ at the point, where it cuts x-axis is: