There is a really fast way of finding the ranges of functions of the form f ( x ) = c x + d a x + b where the f ( x ) is defined from R − { c − d } to R .
The range is simply R − { c a } . In other words, the range is R − { the coefficient of x in the denominator the coefficient of x in the numerator } [verify this!].
For example, the range of f ( x ) = 5 x + 9 3 x + 2 is R − { 5 3 } .
Now consider the following statements.
[ 1 ] . The range of f ( x ) = 4 x + 3 9 x − 5 is R − { 4 9 } .
[ 2 ] . The range of f ( x ) = 9 1 x − 5 6 2 6 x − 1 6 is R − { 9 1 2 6 } .
[ 3 ] . The range of f ( x ) = 1 7 x − 4 1 − 2 x − 6 is R − { 1 7 − 2 } .
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 .
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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
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 ] isn't correct. Even if you changed 9 1 2 6 to 7 2 in [ 2 ] , it would still be false.
Can you figure out why?
Try plugging in some values for f ( x ) in [ 2 ] .
This is not a complete solution but in case you're wondering why [ 2 ] isn't true, evaluate f ( anything other than 9 1 5 6 ) . Is it equal to 9 1 2 6 ? Can you explain why?
It becomes a constant function of y= 2/7
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 ...
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The short-cut technique doesn't work whenever c a = d b . 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.
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We can prove the shortcut given as follows- c x + d a x + b = 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 .