Question
The graph of \( y=f(\left| x\right|)\) for \(-6\leq x\leq 6\) is shown in the following diagram.
(a) On the following axes, sketch the graph of \(y=\left| f(\left| x\right|)\right|\) for \(-6\leq x\leq 6\).
It is given that f is an odd function.
(b) On the following axes, sketch the graph of \(y=f(x)\) for \(-6\leq x\leq 6\).
It is also given that \(\int_{0}^{4}f(\left| x\right|)dx=1.6\).
(c) Write down the value of
(i) \(\int_{-4}^{0}f(x)dx\);
(ii) \(\int_{-4}^{4}(f(\left| x\right|)+f(x))dx\).
▶️Answer/Explanation
Detailed Solution
(a) To find the graph for \(y=\left| f(\left| x\right|)\right|\) , the part below the -axis is taken as mirror image above the X-axis (reflection of all negative sections in x-axis), therefore the graph is as follows
(a) 
(b) To find the graph for \(y=f(x)\) for \(-6\leq x\leq 6\) , since the mod is removed from x and it is given that y=f(x) is an odd function, therefore the graph will be same for x≥0 whereas for x<0 it will be symmetric about the origin that is just same behavior as the graph of y = \(x^{3}\) as shown below
(b) 
(c) (i) From the graph in part (b), we can conclude that the area \(\int_{-4}^{0}f(x)dx\) = \(-\int_{0}^{4}f(\left| x\right|)dx\) = \(-1.6\)
(ii) Since f(x) is an odd function therefore, \(\int_{-4}^{4} f(x))dx\) = 0, and \(\int_{-4}^{4}(f(\left| x\right|) = 2 \int_{0}^{4}(f(\left| x\right|) \)=\(2\times (-1.6)\) = \(3.2\)
Question
The following diagram shows the graph of y = arctan( \(2x+1+\frac{\pi}{4} \)) for \(x \in \mathbb{R}\) with asymptotes at \(y=-\frac{\pi}{4}\) and \(y=\frac{3\pi}{4}\)
(a) Describe a sequence of transformations that transforms the graph of y = arctan x to the graph of y = arctan( \(2x+1+\frac{\pi}{4} \)) for \(x \in \mathbb{R}\) [3]
(b) Show that arctan arctan p+ arctan q= arctan \(\frac{p+q}{1-pq}\) where p , q > 0 and pq < 1 [4]
(c) Verify that \(arctan(2x+1) = arctan(\frac{x}{x+1})+\frac{\pi}{4}\) for x ∈ \(\mathbb{R}\) , x > 0 . [3]
(d) Using mathematical induction and the result from part (b), prove that
\(\sum_{r=1}^{n}arctan\frac{1}{2r^2}=arctan\frac{n}{n+1}\) for n ∈ \(\mathbb{Z}^+\)
▶️Answer/Explanation
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 $n\in\mathbb{Z}^+$.
Question
Part of the graph of a function, f , is shown in the following diagram. The graph of y = f (x) has a y-intercept at (0, 3), an x-intercept at (a, 0) and a horizontal asymptote y = -2.
Consider the function g(x) = |f (| x |)|.
(a) On the following grid, sketch the graph of y = g(x), labelling any axis intercepts and giving the equation of the asymptote
b) Find the possible values of k such that \({(g(x))}^2 = k\) has exactly two solutions .
Reflect f in x or y axis.
▶️Answer/Explanation
(a)
(b) To determine the possible values of \( k \) such that the equation \( g(x) = f'(|x|) = k \) has exactly two solutions, we need to analyze the graph of \( g(x) \) based on the given graph of \( f(x) \).
The graph of \( f(x) \) has a \( y \)-intercept at \( (0, 3) \), an \( x \)-intercept at \( (\alpha, 0) \), and a horizontal asymptote at \( y = -2 \).
Since \( g(x) \) involves the absolute value of \( f(x) \) and the absolute value of \( x \), the graph of \( g(x) \) will be the top half of the graph of \( f(x) \) reflected about the \( x \)-axis for \( x < 0 \).
The horizontal asymptote at \( y = -2 \) for \( f(x) \) becomes \( y = 2 \) for \( g(x) \), and the \( y \)-intercept remains at \( (0, 3) \). Therefore, the graph of \( g(x) \) will have a minimum point at \( (0, 3) \), and it will approach the horizontal line \( y = 2 \) from above.
For the equation \( g(x) = k \) to have exactly two solutions, the horizontal line \( y = k \) must intersect the graph of \( g(x) \) at exactly two points.
This can only occur if:
\(
4 \leq k < 9
\)
since the \( y \)-value of \( g(x) \) at \( x = 0 \) is 3, and the graph approaches \( y = 2 \) without ever touching it again.
Hence, the line \( y = k \) must be above \( y = 3 \) but below \( y = 9 \), which is the maximum value of \( f(x) \) before taking the absolute value.
Thus, the correct range for \( k \) is:
\(
4 \leq k < 9
\)
Question
The quadratic function \(f(x) = p + qx – {x^2}\) has a maximum value of 5 when x = 3.
Find the value of p and the value of q .[4]
The graph of f(x) is translated 3 units in the positive direction parallel to the x-axis. Determine the equation of the new graph.[2]
▶️Answer/Explanation
Markscheme
METHOD 1
\(f'(x) = q – 2x = 0\) M1
\(f'(3) = q – 6 = 0\)
q = 6 A1
f(3) = p + 18 − 9 = 5 M1
p = −4 A1
METHOD 2
\(f(x) = – {(x – 3)^2} + 5\) M1A1
\( = – {x^2} + 6x – 4\)
q = 6, p = −4 A1A1
[4 marks]
\(g(x) = – 4 + 6(x – 3) – {(x – 3)^2}{\text{ }}( = – 31 + 12x – {x^2})\) M1A1
Note: Accept any alternative form which is correct.
Award M1A0 for a substitution of (x + 3) .
[2 marks]