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Let $\left [ p \right ]$ indicates the floor function $\leq p$ and $\lim_{p\rightarrow 0}p^{2}\left [ \frac{49}{p^{2}} \right ]= L$ . Then the function defined as $f\left ( p \right )= \left [ p^{4} \right ]\sin \left ( \pi p \right )+\left [ p^{4} \right ]\cos \left ( \pi p \right )$ becomes discontinuous, for which of the following values of $p$ ?
Option: 1
$\sqrt{L+15}$

Option: 2
$\sqrt{L-15}$

Option: 3
$\sqrt{L-13}$

Option: 4
$\sqrt[4]{L-33}$

Answers (1)

The following points are noteworthy.

The Floor Function indicates the greatest integer function denoted mathematically as $\left [ p\right]$ , for only real values. This function $\left [ p\right]$ rounds downs the real number having any fractional or decimal part (if any) to the nearest integral value less than the indicated number.
For the real number $p$ that can be an integer or a fraction or a decimal, $\left [ p\right]$ is used to indicate the greatest integer function and $\left [ p\right]$ is used to indicate the fractional or the decimal part of $p$ . Thus, $p= \left [ p\right]+\left [ p\right]$ .
The functions $\sin \left ( \pi p \right )$ and $\cos \left ( \pi p \right )$ are both continuous for $\forall p \in R$ .

Now, the provided function is:

$f\left ( p \right )= \left [ p^{4} \right ]\sin \left ( \pi p \right )+\left [ p^{4} \right ]\cos \left ( \pi p \right )$
$f\left ( p \right )= \left [ p^{4} \right ]\left ( \sin \left ( \pi p \right )+\cos \left ( \pi p \right ) \right )$

It is evident that $f\left ( p \right )$ is continuous for $\forall p \in Z$ , and is discontinuous at those points where $\left [ p^{4} \right ]$ is discontinuous i.e. only for non-integral values of $p$ .

Hence, it follows that the points of discontinuity for $f\left ( p \right )$ are at
$p=\pm \sqrt[4]{2},\pm \sqrt[4]{3},\pm \sqrt[4]{5},\cdots \cdots ,\pm \sqrt{2},\pm \sqrt{3},\pm \sqrt{5, }\cdots$

Again, the provided limit function isas follows:
$\lim_{p\rightarrow 0}p^{2}\left [ \frac{49}{p^{2}} \right ]= L$
$\lim_{p\rightarrow 0}p^{2}\left ( \frac{49}{p^{2}}-\left \{ \frac{49}{p^{2}} \right \} \right )= L$
$\lim_{p\rightarrow 0}\left ( p^{2}\frac{49}{p^{2}} \right )-\lim_{p\rightarrow 0}\left ( p^{2}\left \{ \frac{49}{p^{2}} \right \} \right )= L$
$49-0= L$
$L= 49$

Now, evaluate the following:

$\sqrt{L+15}$
$= \sqrt{49+15}$
$= \sqrt{64}$
$= 8$

$\sqrt{L-15}$
$= \sqrt{49-15}$
$= \sqrt{34}$

$\sqrt{L-13}$
$= \sqrt{49-13}$
$= \sqrt{36}$
$= 6$

$\sqrt[4]{L-33}$
$= \sqrt[4]{49-33}$
$= \sqrt[4]{16}$
$= 2$

Therefore, $f\left ( p \right )$ would be discontinuous at $\sqrt{34}$ or $\sqrt{L-15}$