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 \[\Rightarrow \,\,\,\,\,\,\frac{f(c)-f(a)}{c-a}>\frac{f(b)-f(c)}{b-c}\] \[\Rightarrow \,\,\,\,\,\,\frac{f(c)-f(a)}{f(b)-f(c)}>\frac{c-a}{b-c}\]You need to login to perform this action.
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