Question:
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 :
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 :
\(\frac{f(c)-f(a)}{f(b)-f(c)}\)
is greater than :
Show Hint
Apply the Mean Value Theorem on [a, c] and [c, b] and use that \(f'\) is decreasing because \(f''<0\).
Updated On: Nov 5, 2026
- \(1\)
- \(\frac{b-c}{c-a}\)
- \(\frac{b+a}{b-a}\)
- \(\frac{c-a}{b-c}\)
Show Solution
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The Correct Option is D
Solution and Explanation
By the MVT there exist \(\xi_1\in(a,c)\) and \(\xi_2\in(c,b)\) with \(f(c)-f(a)=f'(\xi_1)(c-a)\) and \(f(b)-f(c)=f'(\xi_2)(b-c)\).
Since \(f''<0\), \(f'\) is decreasing, so \(f'(\xi_1)>f'(\xi_2)>0\).
Hence \(\frac{f(c)-f(a)}{f(b)-f(c)}=\frac{f'(\xi_1)}{f'(\xi_2)}\cdot\frac{c-a}{b-c}>\frac{c-a}{b-c}\).
Answer: (4) \(\frac{c-a}{b-c}\)
Since \(f''<0\), \(f'\) is decreasing, so \(f'(\xi_1)>f'(\xi_2)>0\).
Hence \(\frac{f(c)-f(a)}{f(b)-f(c)}=\frac{f'(\xi_1)}{f'(\xi_2)}\cdot\frac{c-a}{b-c}>\frac{c-a}{b-c}\).
Answer: (4) \(\frac{c-a}{b-c}\)
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