Question

Let f be a continuous non-algebraic twice derivable increasing function in (0,1) & f"(x)>0. If f(0) = 0, f(1) = 1 and g(x) is inverse of f(x), then which of the following holds good?

• A. $\frac{f(x)}{x} \leq f'(x) \forall x \in (0,1)$
• B. $\frac{f(x)}{x} \geq f'(x) \forall x \in (0,1)$
• C. $\frac{f(x)}{x} < \frac{x}{g(x)} \forall x \in (0,1)$
• D. $\frac{f(x)}{x} > \frac{x}{g(x)} \forall x \in (0,1)$

Answer 1

Answer: (A), (C)

A, C

Answer 2

The problem asks us to identify the correct statement regarding a function $f(x)$ and its inverse $g(x)$.

Given properties of $f(x)$:

1. $f(x)$ is continuous, twice derivable, and increasing in $(0,1)$.
2. $f''(x) > 0$ in $(0,1)$, which means $f(x)$ is strictly convex.
3. $f(0) = 0$ and $f(1) = 1$.
4. $g(x)$ is the inverse of $f(x)$, so $f(g(x)) = x$ and $g(f(x)) = x$.

Analyzing Option (A) and (B): Comparing $\frac{f(x)}{x}$ and $f'(x)$

Consider the function $f(x)$ on the interval $[0, x]$ for any $x \in (0,1)$. Since $f(x)$ is continuous on $[0,x]$ and differentiable on $(0,x)$, by the Mean Value Theorem (MVT), there exists a $c \in (0,x)$ such that:

$f'(c) = \frac{f(x) - f(0)}{x - 0}$

Given $f(0) = 0$, this simplifies to:

$f'(c) = \frac{f(x)}{x}$

We are given that $f''(x) > 0$ for $x \in (0,1)$. This implies that $f'(x)$ is a strictly increasing function in $(0,1)$. Since $c \in (0,x)$, we have $c < x$. Because $f'(x)$ is strictly increasing, it follows that $f'(c) < f'(x)$. Substituting $f'(c) = \frac{f(x)}{x}$, we get:

$\frac{f(x)}{x} < f'(x)$ for all $x \in (0,1)$.

Therefore, option (A) $\frac{f(x)}{x} \leq f'(x)$ holds good (since it includes the strict inequality), and option (B) $\frac{f(x)}{x} \geq f'(x)$ is incorrect.

Analyzing Option (C) and (D): Comparing $\frac{f(x)}{x}$ and $\frac{x}{g(x)}$

Let's first establish a relationship between $f(x)$ and $x$. Consider the function $h(x) = f(x) - x$. We know $h(0) = f(0) - 0 = 0$. We know $h(1) = f(1) - 1 = 0$. Since $f''(x) > 0$, $f(x)$ is strictly convex. A strictly convex function whose graph passes through $(0,0)$ and $(1,1)$ must lie below the line segment connecting these two points. The line segment connecting $(0,0)$ and $(1,1)$ is $y=x$. Therefore, for $x \in (0,1)$, $f(x) < x$. This implies $h(x) < 0$ for $x \in (0,1)$.

Now, let's use the property $f(x) < x$ to analyze options (C) and (D). Option (C) is $\frac{f(x)}{x} < \frac{x}{g(x)}$. Since $x \in (0,1)$, $x > 0$. Also, since $f(x)$ is increasing and $f(0)=0$, $f(x) > 0$ for $x \in (0,1)$. Similarly, $g(x)$ is also increasing and $g(0)=0$, so $g(x) > 0$ for $x \in (0,1)$. We can multiply the inequality by $x \cdot g(x)$ (which is positive) without changing the inequality direction:

$f(x) \cdot g(x) < x^2$

Let $y = f(x)$. Since $g(x)$ is the inverse of $f(x)$, we have $x = g(y)$. Substitute $y = f(x)$ and $x = g(y)$ into the inequality $f(x) \cdot g(x) < x^2$:

$y \cdot g(y) < (g(y))^2$

Since $y \in (0,1)$, $g(y) \in (0,1)$, so $g(y) > 0$. We can divide by $g(y)$:

$y < g(y)$

Now, substitute back $y = f(x)$:

$f(x) < g(f(x))$

Since $g(f(x)) = x$ (by definition of inverse function):

$f(x) < x$

This is exactly the condition we derived from the strict convexity of $f(x)$ and its boundary conditions $f(0)=0, f(1)=1$. Since $f(x) < x$ for $x \in (0,1)$ is true, the equivalent inequality $\frac{f(x)}{x} < \frac{x}{g(x)}$ is also true. Therefore, option (C) holds good. Option (D) $\frac{f(x)}{x} > \frac{x}{g(x)}$ is incorrect.

Both (A) and (C) are correct statements.

Explanation:

1. For Option (A): Since $f''(x) > 0$, $f'(x)$ is strictly increasing. By MVT on $[0,x]$, $\frac{f(x)-f(0)}{x-0} = f'(c)$ for some $c \in (0,x)$. Given $f(0)=0$, we have $\frac{f(x)}{x} = f'(c)$. Since $c < x$ and $f'(x)$ is increasing, $f'(c) < f'(x)$. Thus, $\frac{f(x)}{x} < f'(x)$.
2. For Option (C): Since $f(x)$ is strictly convex and $f(0)=0, f(1)=1$, its graph lies below the line $y=x$ for $x \in (0,1)$. Therefore, $f(x) < x$. Let's check the given inequality $\frac{f(x)}{x} < \frac{x}{g(x)}$. Since $x, f(x), g(x)$ are all positive in $(0,1)$, we can multiply by $x g(x)$ to get $f(x)g(x) < x^2$. Let $y = f(x)$. Then $x = g(y)$. Substituting these into the inequality gives $y \cdot g(y) < (g(y))^2$. Since $g(y) > 0$, we can divide by $g(y)$ to get $y < g(y)$. Substituting back $y=f(x)$, we get $f(x) < g(f(x))$. Since $g(f(x)) = x$, the inequality simplifies to $f(x) < x$. As we established, $f(x) < x$ is true for $x \in (0,1)$ due to the convexity and boundary conditions. Therefore, option (C) is also correct.