IBDP Math AI:Topic:AHL 2.8:Transformations of graphs:IB style Questions HL Paper 1

Question

The graph of the function f (x) = lnx is translated by points \((_{b}^{a})\) so that it then passes through the and (e3 , 1 + ln2)

Find the value of a and the value of b .

▶️Answer/Explanation

The new function is \(f(x-a)+b (=ln (x-a)+ b)f(o)= ln (-a)+b=1 f(e^{3})= ln (e^{3}-a)+ b = 1+ ln2 ln (-a)=ln (e^{3}-a)-ln2 ln (-a)= ln (\frac{e^{3}-a}{2})- a = (\frac{e^{3}-a}{2}) -2a e -a a = -e^{3} (= -20.0855..)b=1 -lm 1-3=-2\)

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

(a)     Express the quadratic \(3{x^2} – 6x + 5\) in the form \(a{(x + b)^2} + c\), where a, b, c \( \in \mathbb{Z}\).

(b)     Describe a sequence of transformations that transforms the graph of \(y = {x^2}\) to the graph of \(y = 3{x^2} – 6x + 5\).

▶️Answer/Explanation

Markscheme

(a)     attempt at completing the square     (M1)

\(3{x^2} – 6x + 5 = 3({x^2} – 2x) + 5 = 3{(x – 1)^2} – 1 + 5\)     (A1)

\( = 3{(x – 1)^2} + 2\)     A1

\((a = 3,{\text{ }}b = – 1,{\text{ }}c = 2)\)

(b)     definition of suitable basic transformations:

\({{\text{T}}_1} = \) stretch in y direction scale factor 3     A1

\({{\text{T}}_2} = \) translation \(\left( {\begin{array}{*{20}{c}}
  1 \\
  0
\end{array}} \right)\)     A1

\({{\text{T}}_3} = \) translation \(\left( {\begin{array}{*{20}{c}}
  0 \\
  2
\end{array}} \right)\)     A1

[6 marks]

Question

The quadratic function \(f(x) = p + qx – {x^2}\) has a maximum value of 5 when x = 3.

a.Find the value of p and the value of q .[4]

 

b.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]

a.

\(g(x) = – 4 + 6(x – 3) – {(x – 3)^2}{\text{ }}( = – 31 + 12x – {x^2})\)     M1A1

Note: Accept any alternative form which is correct.

Award M1A0 for a substitution of (x + 3) .

[2 marks]

b.

Question

Let \(f(x) = {x^3} + a{x^2} + bx + c\) , where a , b , \(c \in \mathbb{Z}\) . The diagram shows the graph of y = f(x) .

a.Using the information shown in the diagram, find the values of a , b and c .[4]

 

b.If g(x) = 3f(x − 2) ,

(i)     state the coordinates of the points where the graph of g intercepts the x-axis.

(ii)     Find the y-intercept of the graph of .[3]

 
▶️Answer/Explanation

Markscheme

METHOD 1

f(x) = (+ 1)(− 1)(− 2)     M1

\( = {x^3} – 2{x^2} – x + 2\)     A1A1A1

a = −2 , b = −1 and c = 2

METHOD 2

from the graph or using f(0) = 2

c = 2     A1

setting up linear equations using f(1) = 0 and f(–1) = 0 (or f(2) = 0)     M1

obtain a = −2 , b = −1     A1A1

[4 marks]

a.

(i)     (1, 0) , (3, 0) and (4, 0)     A1

(ii)     g(0) occurs at 3f(−2)     (M1)

= −36     A1

[3 marks]

b.
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