Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Since, from the given graph it seems f(a) and f(b)are equal(or approx. equal since there is no scaling given). Then f(a)/a>f(b)/b as a

enter image description here

enter image description here

share|improve this question
    
Since, from the given graph it seems f(a) and f(b)are equal(or approx. equal since there is no scaling given). Then f(a)/a>f(b)/b as a<b.The only graph that satisfies this is Fig:4. But the answer is given to be (B). Where have I gone wrong? –  Rajath Krishna R Aug 24 '13 at 15:36
add comment

1 Answer 1

up vote 1 down vote accepted

Depending on the scale, $\frac{1}{x}$ could be close to constant on $[a,b]$, which would explain how Figure 2 could be correct.

If you calculate:

$$g'(x)=\frac{xf'(x)-f(x)}{x^2}$$

you can see that at $f$'s maximal x-value, $g'$ should be negative. This seems to rule out Fig 4. You've already ruled out Figs 1 and 3, but the reasoning does not apply to Figure 2 because there may be a difference between $g(a)$ and $g(b)$ that is too small to see.

You could also rule out Figures 1 and 3 by looking at $g'(b)$. The formula says it should be negative, but in Figures 1 and 3, it's positive.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.