If \( f(x) \) is defined as follows:
$$ f(x) = \begin{cases} 4, & \text{if } -\infty < x < -\sqrt{5}, \\ x^2 - 1, & \text{if } -\sqrt{5} \leq x \leq \sqrt{5}, \\ 4, & \text{if } \sqrt{5} \leq x < \infty. \end{cases} $$ If \( k \) is the number of points where \( f(x) \) is not differentiable, then \( k - 2 = \)
Step 1: {Check the points of non-differentiability}
We know that for a function to be differentiable at a point, both the left-hand derivative (LHD) and right-hand derivative (RHD) must be equal at that point. At \( x = -\sqrt{5} \), the left-hand derivative is 0, but the right-hand derivative is \( 2x = -2\sqrt{5} \). Therefore, \( f(x) \) is not differentiable at \( x = -\sqrt{5} \). Similarly, at \( x = \sqrt{5} \), the left-hand derivative is \( 2x = 2\sqrt{5} \), and the right-hand derivative is 0, meaning \( f(x) \) is not differentiable at \( x = \sqrt{5} \). Thus, \( k = 2 \).
Step 2: {Calculate \( k - 2 \)}
Since \( k = 2 \), we find: \[ k - 2 = 2 - 2 = 0 \]
Let $\left\lfloor t \right\rfloor$ be the greatest integer less than or equal to $t$. Then the least value of $p \in \mathbb{N}$ for which
\[ \lim_{x \to 0^+} \left( x \left\lfloor \frac{1}{x} \right\rfloor + \left\lfloor \frac{2}{x} \right\rfloor + \dots + \left\lfloor \frac{p}{x} \right\rfloor \right) - x^2 \left( \left\lfloor \frac{1}{x^2} \right\rfloor + \left\lfloor \frac{2}{x^2} \right\rfloor + \dots + \left\lfloor \frac{9^2}{x^2} \right\rfloor \right) \geq 1 \]
is equal to __________.
Evaluate the following limit: $ \lim_{n \to \infty} \prod_{r=3}^n \frac{r^3 - 8}{r^3 + 8} $.
In the given cycle ABCDA, the heat required for an ideal monoatomic gas will be:
A particle is moving in a straight line. The variation of position $ x $ as a function of time $ t $ is given as:
$ x = t^3 - 6t^2 + 20t + 15 $.
The velocity of the body when its acceleration becomes zero is: