If f (x) is twice differentiable and continuous function in x ∈ [a, b] also f' (x) 0 and f'' (x)

Q: Can't I just plug in a function like $f(x) = \sqrt{x}$ to solve this quickly?

While testing a specific function like $f(x) = \sqrt{x}$ is a brilliant exam-day strategy to find the answer, in JEE Advanced, we must master the rigorous proof. The LMVT approach ensures your logic holds for *any* function satisfying these conditions, not just the ones you can guess.

Q: Wait, why didn't the inequality sign flip when I cross-multiplied?

That is a fantastic catch! We only flip the inequality if we multiply by a negative number. Here, because $f(x)$ is strictly increasing and $b > c$, we know $f(b) > f(c)$, which means $f(b) - f(c)$ is strictly positive. We are safe!

Q: Why is $f'(\alpha) > f'(\beta)$?

Because $f''(x) < 0$ implies the slope $f'(x)$ is strictly decreasing. Since $\alpha$ is to the left of $\beta$ (because $\alpha \in (a, c)$ and $\beta \in (c, b)$), the slope at $\alpha$ must be greater than the slope at $\beta$.

Question

If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f' (x) > 0$ and $f'' (x) < 0$ and $c \in (a, b)$ then $\frac{f(c)-f(a)}{f(b)-f(c)}$ is greater than

Options:

(A) $\frac{b-c}{c-a}$

(B) 1

(C) $\frac{a+b}{b-c}$

(D) $\frac{b-c}{a-c}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine you are standing on a path defined by a function $f(x)$. You are given two critical constraints: the function is strictly increasing ($f'(x) > 0$) and it is concave down ($f''(x) < 0$). Think of this as a path that is always going uphill, but it is getting "tired"—the steepness is fading as you walk. This creates the characteristic "umbrella" shape. We consider three points on this path: $A$ at $x=a$, $C$ at $x=c$, and $B$ at $x=b$, where $a < c < b$. We aim to determine the relationship between the vertical rise between $A$ and $C$ and the vertical rise between $C$ and $B$. ### The Bridge of Lagrange Whenever you encounter a ratio of differences such as $\frac{f(c)-f(a)}{c-a}$, your mind should immediately jump to the **Mean Value Theorem (LMVT)**. LMVT serves as the bridge between the average slope of a secant line and the instantaneous slope of a tangent line. For the interval $[a, c]$, there exists some $\alpha \in (a, c)$ such that the slope of the secant $AC$ is exactly $f'(\alpha)$: $$\frac{f(c)-f(a)}{c-a} = f'(\alpha)$$ Similarly, for the interval $[c, b]$, there exists some $\beta \in (c, b)$ such that the slope of the secant $CB$ is $f'(\beta)$: $$\frac{f(b)-f(c)}{b-c} = f'(\beta)$$ ### The Inequality Dance Now, we incorporate the property of concavity. Since $f''(x) < 0$, the slope function $f'(x)$ is strictly decreasing. Because $\alpha$ lies in the interval $(a, c)$ and $\beta$ lies in the interval $(c, b)$, we know for a fact that $\alpha < \beta$. Therefore, the slope at $\alpha$ must be greater than the slope at $\beta$: $$f'(\alpha) > f'(\beta)$$ Substituting our LMVT results into this inequality, we obtain: $$\frac{f(c)-f(a)}{c-a} > \frac{f(b)-f(c)}{b-c}$$ ### The Final Victory We are now ready to isolate the ratio of the vertical rises. We want to evaluate the expression $\frac{f(c)-f(a)}{f(b)-f(c)}$. Since $f(x)$ is strictly increasing, the term $f(b) - f(c)$ is guaranteed to be positive. This allows us to cross-multiply the terms without reversing the inequality sign. Rearranging the terms, we arrive at the final result: $$\frac{f(c)-f(a)}{f(b)-f(c)} > \frac{c-a}{b-c}$$ You have just proven a fundamental property of concave functions using nothing but the elegance of calculus. **Well done!**

Animated Solution for Mathematics - Differentiation: If $f (x)$ is twice differentiable and continuous function in $x \in [a, b]$ also $f^{'} (x) > 0$ and $f^{''} (x) < 0$ and $c \in (a, b)$ then $\frac{f ( c ) - f ( a )}{f ( b ) - f ( c )}$ is greater than

Select Answer:

Visualized Solution

Visualizing the Function $f (x)$

Lagrange's Mean Value Theorem

LMVT on Interval $[a, c]$

LMVT on Interval $[c, b]$

Analyzing the Slopes

Substituting the LMVT Results

Final Rearrangement

The Sigma Insight: Mean Value Theorems

Analyzing the Setup

The Bridge of Lagrange

The Inequality Dance

The Final Victory

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