Shortcut!

Algebra Level 5

There is a really fast way of finding the ranges of functions of the form f ( x ) = a x + b c x + d f(x)=\dfrac{ax+b}{cx+d} where the f ( x ) f(x) is defined from R { d c } \mathbb{R}-\left\{\frac{-d}{c}\right\} to R \mathbb{R} .

The range is simply R { a c } \mathbb{R}-\left\{\frac{a}{c}\right\} . In other words, the range is R { the coefficient of x in the numerator the coefficient of x in the denominator } \mathbb{R}-\left\{\dfrac{\text{the coefficient of } x\ \text{in the numerator}}{\text{the coefficient of } x \ \text{in the denominator}}\right\} [verify this!].

For example, the range of f ( x ) = 3 x + 2 5 x + 9 f(x)=\dfrac{3x+2}{5x+9} is R { 3 5 } \mathbb{R}- \left\{\frac{3}{5}\right\} .

Now consider the following statements.

[ 1 ] [1] . The range of f ( x ) = 9 x 5 4 x + 3 f(x)=\dfrac{9x-5}{4x+3} is R { 9 4 } \mathbb{R}- \left\{\frac{9}{4}\right\} .

[ 2 ] [2] . The range of f ( x ) = 26 x 16 91 x 56 f(x)=\dfrac{26x-16}{91x-56} is R { 26 91 } \mathbb{R}- \left\{\frac{26}{91}\right\} .

[ 3 ] [3] . The range of f ( x ) = 2 x 6 17 x 41 f(x)=\dfrac{-2x-6}{17x-41} is R { 2 17 } \mathbb{R}- \left\{\frac{-2}{17}\right\} .

Which of them are correct?


This problem is from the set "MCQ Is Not As Easy As 1-2-3". You can see the rest of the problems here .

[ 1 ] [1] and [ 3 ] [3] . Only [ 1 ] [1] None of them are correct. In fact the method doesn't even work. All of them are correct.

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4 solutions

Connor Kenway
May 14, 2014

We can prove the shortcut given as follows- a x + b c x + d = y S o , a x + b = c x y + d y b d y = c x y a x b y d y = x ( c y a ) x = ( b d y ) / ( c y a ) S o w e w i l l b e a b l e t o f i n d r e a l x f o r a l l y e x c e p t y = a / c b u t i f a / c = b / d t h e n x = 0 / 0 w h i c h i s i n d e t e r m i n a t e i n s u c h a c a s e t h e f u n c t i o n b e c o m e s a c o n s t a n t e x c e p t f o r x = d / c . \frac { ax+b }{ cx+d } =y\\ So,\\ ax+b=cxy+dy\\ b-dy=cxy-ax\\ by-dy=x(cy-a)\\ x=(b-dy)/(cy-a)\\ So\quad we\quad will\quad be\quad able\quad to\quad find\quad real\quad x\quad for\quad all\quad y\quad except\quad y=a/c\quad \\ but\quad if\quad a/c=b/d\quad then\quad x=0/0\quad which\quad is\quad indeterminate\\ in\quad such\quad a\quad case\quad the\quad function\quad becomes\quad a\quad constant\quad except\quad for\quad x=-d/c.

The range of a linear rational function is defined for all real numbers except for a value that makes an equation false. This is mainly the value of the horizontal asymptote in which a/c. But plugging the value may hold the equation true or not. For (1) and (3), the equations don't hold. But for (2), -(26)(56) = -(16)(91). Hence, (1) and (3) are correct.

i didn't understand , pls clarify

Asif Mohammed L - 7 years, 1 month ago
Lee Isaac
Jun 4, 2015

26/91 is not in the simplest form. It should be 2/7. Thus, answer is [1] and [3].

This is not the reason why [ 2 ] [2] isn't correct. Even if you changed 26 91 \frac{26}{91} to 2 7 \frac{2}{7} in [ 2 ] [2] , it would still be false.

Can you figure out why?

Try plugging in some values for f ( x ) f(x) in [ 2 ] [2] .

Mursalin Habib - 6 years ago
Mursalin Habib
May 12, 2014

This is not a complete solution but in case you're wondering why [ 2 ] [2] isn't true, evaluate f ( anything other than 56 91 ) f(\text{anything other than}\ \dfrac{56}{91}) . Is it equal to 26 91 \dfrac{26}{91} ? Can you explain why?

It becomes a constant function of y= 2/7

Adeetya Tantia - 7 years, 1 month ago

in case if a=b=c=d then ur range doesnt hold gud .... does a,b,c,d has to b distinct .. ?? if so mention that ...

Ramesh Goenka - 7 years, 1 month ago

Log in to reply

The short-cut technique doesn't work whenever a c = b d \dfrac{a}{c}=\dfrac{b}{d} . I didn't mention that in the question on purpose. My goal was to encourage people to verify if the formula always works and not to take short-cuts for granted.

Mursalin Habib - 7 years, 1 month ago

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