question_answer

Let f be any function continuous on \[[a,b]\]and twice differentiable on \[(a,b)\]. If for all \[x\in (a,b),\] \[f'(x)>0\] and \[f''(x)<0,\] then for any \[c\in (a,b),\frac{f(c)-f(a)}{f(b)-f(c)}\]is greater than                   [JEE MAIN Held on 09-01-2020 Morning]

A) \[1\]          

B) \[\frac{b+a}{b-a}\]

C) \[\frac{c-a}{b-c}\]

D) \[\frac{b-c}{c-a}\]

Correct Answer: C

Solution :

[c]  \[\because \,\,\,\,\,\,\,\,\,\,\,\,f'(x)>0\] and \[f''(x)<0\] So graph of function \[f(x)\] is increasing and concave up \[\because \,\,\,\,\,\,\,\,\,a<c<b\] so \[f(a)<f(c)<f(b)\] Also slope of \[AC>\]slope of BC because \[f'(x)\] is decreasing function \[\Rightarrow \,\,\,\,\,\,\frac{f(c)-f(a)}{c-a}>\frac{f(b)-f(c)}{b-c}\] \[\Rightarrow \,\,\,\,\,\,\frac{f(c)-f(a)}{f(b)-f(c)}>\frac{c-a}{b-c}\]