IBDP Maths analysis and approaches Topic :SL 2.11 :Transformations of graphs, translations and stretches HL Paper 1

Ans:

(a) By replacing the $x$ in $y=\arctan x$ with $x+1$, we obtain $y=\arctan \left(x+1\right)$, which is a translation of $-1$ units parallel to the $x$-axis.
Next, by replacing the $x$ in $y=\arctan \left(x+1\right)$ with $2x$, we obtain $y=\arctan \left(2x+1\right)$, which is a stretch parallel to the $x$-axis by scale factor $\frac{1}{2}$.
Finally, by replacing the $y$ in $y=\arctan \left(2x+1\right)$ with $y-\frac{\pi}{4}$, we obtain $y-\frac{\pi}{4}=\arctan \left(2x+1\right)$, i.e., $y=\arctan \left(2x+1\right)+\frac{\pi}{4}$, which is a translation of $\frac{\pi}{4}$ units parallel to the $y$-axis.<br>
(b) Taking tangent on both sides in $\arctan p+\arctan q=\arctan\left(\frac{p+q}{1-pq}\right)$, we have $\tan\left(\arctan p+\arctan q\right)=\frac{p+q}{1-pq}$
$$\begin{eqnarray}
\text{LHS} &=& \tan\left(\arctan p+\arctan q\right) \nonumber \\
&=& \frac{p+q}{1-pq} \nonumber \\
&=& \text{RHS}.
\end{eqnarray}$$
(c) Observe that $\arctan 1=\frac{\pi}{4}$. Then, using (b), we have
$$\begin{eqnarray}
\text{RHS} &=& \arctan \left(\frac{x}{x+1}\right)+\frac{\pi}{4} \nonumber \\
&=& \arctan \left(\frac{x}{x+1}\right)+\arctan 1 \nonumber \\
&=& \arctan \left(\frac{1+\frac{x}{x+1}}{1-\frac{x}{x+1}}\right) \nonumber \\
&=& \arctan \left(\frac{2x+1}{x+1-x}\right) \nonumber \\
&=& \arctan \left(2x+1\right) \nonumber \\
&=& \text{LHS}.
\end{eqnarray}$$
(d) Let $\text{P}_n$ be the proposition “$\sum_{r=1}^{n} \arctan\left(\frac{1}{2r^2}\right)=\arctan\left(\frac{n}{n+1}\right)$” for $n\in\mathbb{Z}^+$.<br>
When $n=1$,<br>
$$\begin{eqnarray}
\text{LHS}=\sum_{r=1}^{1} \arctan\left(\frac{1}{2r^2}\right)=\arctan\left(\frac{1}{2}\right) \nonumber \\
\text{RHS}=\arctan\left(\frac{1}{1+1}\right)=\arctan\left(\frac{1}{2}\right).
\end{eqnarray}$$
Since $\text{LHS}=\text{RHS}$, $\text{P}_1$ is true.<br>
Assume that $\text{P}_k$ is true for some $k\in\mathbb{Z}^+$, i.e.,
$$\begin{eqnarray}
\sum_{r=1}^{k} \arctan\left(\frac{1}{2r^2}\right)=\arctan\left(\frac{k}{k+1}\right).
\end{eqnarray}$$
To show that $\text{P}_{k+1}$ is also true, i.e.,
$$\begin{eqnarray}
\sum_{r=1}^{k+1} \arctan\left(\frac{1}{2r^2}\right)=\arctan\left(\frac{k+1}{k+2}\right).
\end{eqnarray}$$
$$\begin{eqnarray}
\text{LHS} &=& \sum_{r=1}^{k+1} \arctan\left(\frac{1}{2r^2}\right) \nonumber \\
&=& \arctan\left(\frac{k}{k+1}\right)+\arctan\left(\frac{1}{2\left(k+1\right)^2}\right) \nonumber \\
&=& \arctan \left[\frac{\frac{k}{k+1}+\frac{1}{2\left(k+1\right)^2}}{1-\frac{k}{2\left(k+1\right)^3}}\right] \nonumber \\
&=& \arctan \left[\frac{2k\left(k+1\right)^2+\left(k+1\right)}{2\left(k+1\right)^3-k}\right] \nonumber \\
&=& \arctan \left[\frac{\left(k+1\right)\left(2k^2+2k+1\right)}{\left(k+2\right)\left(2k^2+2k+1\right)}\right] \nonumber \\
&=& \arctan \left(\frac{k+1}{k+2}\right) \nonumber \\
&=& \text{RHS}.
\end{eqnarray}$$
Since $\text{P}_1$ is true and $\text{P}_k$ is true implies $\text{P}_{k+1}$ is also true, by the principle of mathematical induction, $\text{P}_n$ is true for $n\in\mathbb{Z}^+$.