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by Pharo » Mon Apr 02, 2012 2:54 am
Are you sure this is a valid question? I can see no way of drawing a linear function with these requirements. (This also does not fit in the "Data Sufficiency" type of questions)

The first condition states that f is an increasing function in the {1,2} range (thus the function will look like : y = mx + b)
The second condition states that f is a decreasing function in the {3,4} range. (thus the function will look like : y = -mx + b)

The two conditions are mutually exclusive for a linear function. (i.e. a linear function can only have one behaviour; either increasing or decreasing).
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by sanju09 » Wed Apr 04, 2012 4:14 am
Pharo wrote:Are you sure this is a valid question? I can see no way of drawing a linear function with these requirements. (This also does not fit in the "Data Sufficiency" type of questions)

The first condition states that f is an increasing function in the {1,2} range (thus the function will look like : y = mx + b)
The second condition states that f is a decreasing function in the {3,4} range. (thus the function will look like : y = -mx + b)

The two conditions are mutually exclusive for a linear function. (i.e. a linear function can only have one behaviour; either increasing or decreasing).
Reasoning is a bit less than perfect, but yes, the question sounds dubious to me.
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by Shalabh's Quants » Sun Apr 08, 2012 5:58 am
Pharo wrote:Are you sure this is a valid question? I can see no way of drawing a linear function with these requirements. (This also does not fit in the "Data Sufficiency" type of questions)

The first condition states that f is an increasing function in the {1,2} range (thus the function will look like : y = mx + b)
The second condition states that f is a decreasing function in the {3,4} range. (thus the function will look like : y = -mx + b)

The two conditions are mutually exclusive for a linear function. (i.e. a linear function can only have one behaviour; either increasing or decreasing).
There looks to be mathematical fallacy in question...

Say linear fn is f(x)=ax+b;

Then, f(1)=a.1+b=a+b;

f(2)=a.2+b=2a+b;

As f(1)<f(2)...that means a+b < 2a+b => a>0;-----(1)

Again, f(3)=a.3+b=3a+b;

f(4)=a.4+b=4a+b;

As f(3)>f(4)...that means 3a+b > 4a+b => a<0;------(2)

Results from (1) & (2) are contradicting.
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by shubham_k » Sun Apr 08, 2012 6:17 am
Hi,
The only way the given three conditions can be true is by assuming that the given function is piece wise linear.
Image

Image
following attachments show the examples of piecewise linear functions. But even after assuming that it would be difficult to comment on f(0) due to insufficient information.
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by Pharo » Sun Apr 08, 2012 6:33 am
sanju09 wrote:
Pharo wrote:Are you sure this is a valid question? I can see no way of drawing a linear function with these requirements. (This also does not fit in the "Data Sufficiency" type of questions)

The first condition states that f is an increasing function in the {1,2} range (thus the function will look like : y = mx + b)
The second condition states that f is a decreasing function in the {3,4} range. (thus the function will look like : y = -mx + b)

The two conditions are mutually exclusive for a linear function. (i.e. a linear function can only have one behaviour; either increasing or decreasing).
Reasoning is a bit less than perfect, but yes, the question sounds dubious to me.
How would you make it perfect?
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by Shalabh's Quants » Sun Apr 08, 2012 8:23 am
Md.Nazrul Islam wrote:Let f be a linear function with properties that f(1)< f(2) ,f (3)> f(4) and f(5)=5 ,which of the statement is true .

a) f(0)>8 b)f(0) =0 c) f(0)=5 d) f(0) =2 .

There looks to be mathematical fallacy in question...

Say linear fn is f(x)=ax+b;

Then, f(1)=a.1+b=a+b;

f(2)=a.2+b=2a+b;

As f(1)<f(2)...that means a+b < 2a+b => a>0;-----(1)

Again, f(3)=a.3+b=3a+b;

f(4)=a.4+b=4a+b;

As f(3)>f(4)...that means 3a+b > 4a+b => a<0;------(2)

Results from (1) & (2) are contradicting.
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by pemdas » Wed Feb 27, 2013 4:38 pm
This one got my attention

Firstly, the linear function can either increase or decrease, BUT it cannot increase and decrease at the same time. If the latter is a case, then we don't have a linear function.
The question explicitly states that f is a linear function, i.e. f(x)=ax+b where a is the slope and b is y-intercept. With the condition f(1)<f(2) the slope is positive (a>0). For f(3)>f(4), the slope is negative (a<0). Finally, with f(5)=5 the slope is unknown and we learn only about one point (x,y) as (5,5).

The answer choices given propose only one input of "x" for f(x), this is x=0. So effectively, we may ignore our just recently "over-verbolized" theory of a function slope. We need only y-intercept to answet this question, because x=0. Is there any opportunity we can find the y-intercept from the data given in our question?
Because we are given not one but several functional relationships and our functions' y-intercepts are unknown, we may not answer this question. This is an example of the bad question-answer combination.
Md.Nazrul Islam wrote:Let f be a linear function with properties that f(1)< f(2) ,f (3)> f(4) and f(5)=5 ,which of the statement is true .

a) f(0)>8 b)f(0) =0 c) f(0)=5 d) f(0) =2 .
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by Tommy Wallach » Wed Feb 27, 2013 6:47 pm
Hey Guys,

Yep. Ignore questions like this. (It only has four answer choices!).

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