### Question 1

Topic: SL 2.8

Given

\(f(x)=\frac{7 x+7}{2 x-4}\) for \(x \in \mathbb{R}, x \neq 2\).

Find

(a) zero of \(f(x)\).[2]

(b) For the graph of \(y=f(x)\)

(i) what is the vertical asymptote;

(ii) what is the horizontal asymptote.[2]

(c) Find \(f^{-1}(x)\), the inverse function of \(f(x)\).[3]

**▶️Answer/Explanation**

Ans:

(a) recognizing \(f(x)=0\)

\(f(x)=\frac{7 x+7}{2 x-4}\) =0

$

x=-1

$

(b) (i) \(\quad x=2\) (must be an equation with \(x\) )

x=2.

(ii) \(y=\frac{7}{2}\) (must be an equation with \(y\) )

(c) EITHER

interchanging \(x\) and \(y\)

$

2 x y-4 x=7 y+7

$

correct working with \(\mathrm{y}\) terms on the same side: \(2 x y-7 y=4 x+7\)

OR

$

2 y x-4 y=7 x+7

$

correct working with \(x\) terms on the same side: \(2 y x-7 x=4 y+7\) interchanging \(x\) and \(y\) OR making \(x\) the subject \(x=\frac{4 y+7}{2 y-7}\)

THEN

$

f^{-1}(x)=\frac{4 x+7}{2 x-7} \text { (or equivalent) }\left(x \neq \frac{7}{2}\right)

$

Detailed Explanation:

a) f(x)=0

\(f(x)=\frac{7 x+7}{2 x-4}\) =0

7x+7=0

x=-1.

b) i) For vertical asymptote, set the denominator equal to zero and solve for x.

2x-4=0

x=2.

b) ii) For horizontal asymptote,

Degree of numerator, N =1, Degree of denominator, D =1,

Here, N=D so,

The equation of horizontal asymptote is y= ratio of leading coefficients i.e. \(y=\frac{7}{2}\)

c) To get \(f^{-1}(x)\) ,

interchange x and y,

i.e. 2 x y-4 x=7 y+7

2 x y-7 y=4 x+7

y= \(\frac{4 x+7}{2 x-7} \)

$

f^{-1}(x)=\frac{4 x+7}{2 x-7} \text { (or equivalent) }\left(x \neq \frac{7}{2}\right)

$

### Question 2

Topic: SL 4.5

Given :

On a Monday at an amusement park, a sample of 40 visitors was randomly selected as they were leaving the park. They were asked how many times that day they had been on a ride called The Dragon. This information is summarized in the following frequency table.

It can be assumed that this sample is representative of all visitors to the park for the following day.

Find:

(a) For the following day, Tuesday, estimate

(i) the probability that a randomly selected visitor will ride The Dragon;

(ii) the expected number of times a visitor will ride The Dragon.

It is known that 1000 visitors will attend the amusement park on Tuesday. The Dragon can carry a maximum of 10 people each time it runs.

(b) Estimate the minimum number of times The Dragon must run to satisfy demand.[2]

**▶️Answer/Explanation**

Ans:(a) (i) summing frequencies of riders or finding complement ,

From the table, 6 visitors had the dragon ride 0 times, which means remaining 40-6 = 34 visitors had dragon ride at least once.

Therefore, the probability that a randomly selected visitor will ride The Dragon =

$

\text { probability }=\frac{34}{40}

$

(ii) attempt to find expected value:

To find the expected value, simply multiply each value of number of times on dragon by its probability and add the products.

$

\frac{16}{40}+\left(2 \times \frac{13}{40}\right)+\left(3 \times \frac{2}{40}\right)+\left(4 \times \frac{3}{40}\right)

$

$

\frac{60}{40}(=1.5)

$

Hence, The expected number of times a visitor will ride The Dragon =1.5

(b) The expected number of times a visitor will ride The Dragon =1.5

Total number of visitors = 1000

Expected number of total rides = 1.5*1000=1500

The Dragon can carry a maximum of 10 people each time it runs.

So, number of rides = 1500/10 =150.