seupeljewj

2022-06-29

Find the biggest negative value of a , for which the maximum of

$f(x)=sin(24x+\frac{\pi a}{100})$ is at ${x}_{0}=\pi $

The answer is $a=-150$ , but I don't understand the solving way. I would appreciate if you'd help me please

$f(x)=sin(24x+\frac{\pi a}{100})$ is at ${x}_{0}=\pi $

The answer is $a=-150$ , but I don't understand the solving way. I would appreciate if you'd help me please

Colin Moran

Beginner2022-06-30Added 21 answers

Step 1

The problem involves determining the closest maximum of $\mathrm{sin}(24x)$ to ${x}_{0}=\pi $ and then shift the curve to the right amount to put it in ${x}_{0}=\pi $

Maximums are at the points $x=\frac{1}{48}(4\pi \cdot n+\pi )$ so you have that the closest maximum is at $n=11$ so ${x}_{closest}=\frac{45}{48}\pi $

This means that you have to shift the curve to the right by $\frac{3}{48}\pi $ to respect the condition that the maximum is in ${x}_{0}=\pi $ Can you finish from here? (you can use $x=y-\frac{3}{48}\pi $ to shift the curve )

The problem involves determining the closest maximum of $\mathrm{sin}(24x)$ to ${x}_{0}=\pi $ and then shift the curve to the right amount to put it in ${x}_{0}=\pi $

Maximums are at the points $x=\frac{1}{48}(4\pi \cdot n+\pi )$ so you have that the closest maximum is at $n=11$ so ${x}_{closest}=\frac{45}{48}\pi $

This means that you have to shift the curve to the right by $\frac{3}{48}\pi $ to respect the condition that the maximum is in ${x}_{0}=\pi $ Can you finish from here? (you can use $x=y-\frac{3}{48}\pi $ to shift the curve )

Layla Velazquez

Beginner2022-07-01Added 11 answers

Step 1

Recall that the general solution of $\mathrm{sin}\theta =1$ is:

$\theta =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}$

So the general solution for finding all maximum values of f(x) is:

$\begin{array}{rl}24x+\frac{\pi a}{100}& =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}\\ 24x& =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n,\text{where}n\in \mathbb{Z}\\ x& =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n),\text{where}n\in \mathbb{Z}\end{array}$

$\begin{array}{rl}24x+\frac{\pi a}{100}& =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}\\ 24x& =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n,\text{where}n\in \mathbb{Z}\\ x& =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n),\text{where}n\in \mathbb{Z}\end{array}$

$\begin{array}{rl}24x+\frac{\pi a}{100}& =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}\\ 24x& =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n,\text{where}n\in \mathbb{Z}\\ x& =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n),\text{where}n\in \mathbb{Z}\end{array}$

In particular, we know that there is some ${n}_{0}\in \mathbb{Z}$ such that ${x}_{0}=\pi $ so:

$\begin{array}{rl}\pi & =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi {n}_{0})\\ 24\pi & =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi {n}_{0}\\ \frac{\pi a}{100}& =\frac{\pi}{2}-24\pi +2\pi {n}_{0}\\ a& =-2350+200{n}_{0}\end{array}$

Since $a<0$ , we know that ${n}_{0}<\frac{2350}{200}=11.75$ . Rounding down to the nearest integer, we find that ${n}_{0}=11$ so that $a=-2350+200(11)=-150$ , as desired.

Recall that the general solution of $\mathrm{sin}\theta =1$ is:

$\theta =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}$

So the general solution for finding all maximum values of f(x) is:

$\begin{array}{rl}24x+\frac{\pi a}{100}& =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}\\ 24x& =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n,\text{where}n\in \mathbb{Z}\\ x& =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n),\text{where}n\in \mathbb{Z}\end{array}$

$\begin{array}{rl}24x+\frac{\pi a}{100}& =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}\\ 24x& =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n,\text{where}n\in \mathbb{Z}\\ x& =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n),\text{where}n\in \mathbb{Z}\end{array}$

$\begin{array}{rl}24x+\frac{\pi a}{100}& =\frac{\pi}{2}+2\pi n,\text{where}n\in \mathbb{Z}\\ 24x& =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n,\text{where}n\in \mathbb{Z}\\ x& =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi n),\text{where}n\in \mathbb{Z}\end{array}$

In particular, we know that there is some ${n}_{0}\in \mathbb{Z}$ such that ${x}_{0}=\pi $ so:

$\begin{array}{rl}\pi & =\frac{1}{24}(\frac{\pi}{2}-\frac{\pi a}{100}+2\pi {n}_{0})\\ 24\pi & =\frac{\pi}{2}-\frac{\pi a}{100}+2\pi {n}_{0}\\ \frac{\pi a}{100}& =\frac{\pi}{2}-24\pi +2\pi {n}_{0}\\ a& =-2350+200{n}_{0}\end{array}$

Since $a<0$ , we know that ${n}_{0}<\frac{2350}{200}=11.75$ . Rounding down to the nearest integer, we find that ${n}_{0}=11$ so that $a=-2350+200(11)=-150$ , as desired.

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