2023 May Mathematics analysis and approaches paper 1 TZ1 HL Concept Questions and Answers

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)
$