Problem Analysis:
- We are given a twice differentiable function $f(x)$ on the interval $[1, 3]$.
- Conditions provided: $f(1) = f(3)$ and $|f''(x)| \le 2$ for all $x \in [1, 3]$.
- The objective is to determine the constraint on the magnitude of the first derivative, $|f'(x)|$, for all $x$ within the interval $[1, 3]$.
Step 1: Apply Rolle's Theorem
The function $f(x)$ satisfies the conditions for Rolle's Theorem: it is continuous on $[1, 3]$, differentiable on $(1, 3)$, and the function values at the endpoints are equal ($f(1) = f(3)$). Therefore, Rolle's Theorem guarantees that there exists at least one point $c_1$ in the open interval $(1, 3)$ such that the first derivative is zero at that point, i.e., $f'(c_1) = 0$.
Step 2: Apply Mean Value Theorem to $f'(x)$
Since $f(x)$ is twice differentiable, its first derivative, $f'(x)$, is continuous and differentiable on $(1, 3)$. We can apply the Mean Value Theorem (MVT) to $f'(x)$ over an interval defined by $x$ and $c_1$. Let $x$ be any point in $[1, 3]$.
According to the MVT applied to $f'(t)$ over the interval between $x$ and $c_1$, there exists a point $c$ strictly between $x$ and $c_1$ such that: $f''(c) = \frac{f'(x) - f'(c_1)}{x - c_1}$
Substitute the result from Step 1 ($f'(c_1) = 0$) into the MVT equation: $f''(c) = \frac{f'(x) - 0}{x - c_1}$ $f''(c) = \frac{f'(x)}{x - c_1}$
Rearranging this equation to solve for $f'(x)$, we get: $f'(x) = f''(c) (x - c_1)$.
Step 3: Utilize the Bound on $f''(x)$
Now, consider the absolute value of $f'(x)$: $|f'(x)| = |f''(c)| |x - c_1|$
From the problem statement, we know that $|f''(x)| \le 2$ for all $x$ in $[1, 3]$. This implies $|f''(c)| \le 2$. Furthermore, $c$ lies strictly between $x$ and $c_1$. Since $x \in [1, 3]$ and $c_1 \in (1, 3)$, the distance $|x - c_1|$ is strictly less than the total length of the interval $[1, 3]$. Therefore, $|x - c_1| < (3 - 1) = 2$. Substitute these inequalities back into the expression for $|f'(x)|$:
$ |f'(x)| \le 2 \times |x - c_1| < 2 \times 2 $This calculation leads to the inequality:
$ |f'(x)| < 4 $This demonstrates that for any $x$ in the interval $[1, 3]$, the absolute value of the first derivative $f'(x)$ must be strictly less than 4.
Conclusion: Based on the derivation, the correct statement is $|f'(x)| < 4$, which corresponds to Option D.
Similar Questions
- Given $P(x) = x^4 + ax^3 + bx^2 + cx + d$ such that $x=0$ is the only real root of $P'(x) = 0$. If $P(-1) < P(1)$, then in the interval $[-1, 1]$
- Consider a function $f(x)$ which has exactly two roots at $x=a$. If $\lim_{x \to a}\left(\frac{\lambda f'(x)}{f(x)} - \frac{1}{x-a}\right) = m \, (\neq 0)$, then the value of $\lambda$ is
- Which of the following statements is always true?
- Let $A = [a, \infty)$ denotes the domain, then $f: [a, \infty) \to B$, which is defined by $f(x) = 2x^3 - 3x^2 + 6$ will have an inverse for the smallest real value of '$a$' if
- Number of elements in the range set of $f(x) = \left[ \frac{x}{15} \right] \left[ -\frac{15}{x} \right]$, for all $x \in (0, 90)$; (where $[\cdot]$ denotes the greatest integer function) is
- Let domain and range of $f(x)$ and $g(x)$ is $[0, \infty)$. If $f(x)$ is an increasing function, $g(x)$ is a decreasing function, $h(x) = f\{g(x)\}, h(0) = 0$ and $p(x) = h(x^3 - 2x^2 + 2x) - h(4)$, then for all $x \in (0, 2)$
For a real number $y$, consider $[y]$ denotes the greatest integer less than or equal to $y$.
If $f(x) = \frac{\tan(\pi[x-\pi])}{1+[x]^2}$, then- If the domain of $f(x)$ is $(0, 1)$, then the domain of $y = f(e^x) + f(\ln|x|)$ is
- A figure is bounded by the curves $y = x^2 + 1, y = 0, x = 0$ and $x = 1$. The point at which a tangent should be drawn to the curve $y = x^2 + 1$ for it to cut off trapezium of the greatest area from the figure is
- If $f$ be a real valued function defined for all real numbers $x$ such that for some fixed $a > 0$, it satisfies $f(x+a) = \frac{1}{2} + \sqrt{f(x) - (f(x))^2} \, \forall x$, then $f(x)$ is periodic with period
Important Questions from Differential Calculus
- Let the domain of the function $f(x) = \log_3 \log_5 \left(7 - \log_2 \left(x^2 - 10x + 85\right)\right) + \sin^{-1} \left(\left|\frac{3x - 7}{17 - x}\right|\right)$ be $(\alpha, \beta]$. Then $\alpha + \beta$ is equal to :
- Let $[\cdot]$ denote the greatest integer function, and let $f(x) = \min \{\sqrt{2}x, x^2\}$.
Let $S = \{x \in (-2, 2) : \text{the function } g(x) = |x|[x^2] \text{ is discontinuous at } x\}$.
Then $\sum_{x \in S} f(x)$ equals - If the function $f(x) = \frac{e^x\left(e^{\tan x - x} - 1\right) + \log_e(\sec x + \tan x) - x}{\tan x - x}$ is continuous at $x = 0$, then the value of $f(0)$ is equal to
- Let $(2\alpha, \alpha)$ be the largest interval in which the function $f(t) = \frac{|t+1|}{t^2}, t < 0$, is strictly decreasing. Then the local maximum value of the function $g(x) = 2\log_e(x - 2) + \alpha x^2 + 4x - \alpha$, $x > 2$, is ______
- Consider the following three statements for the function $f : (0, \infty) \rightarrow \mathbb{R}$ defined by $f(x) = |\log_e x| - |x - 1|$:
(I) $f$ is differentiable at all $x > 0$.
(II) $f$ is increasing in $(0, 1)$.
(III) $f$ is decreasing in $(1, \infty)$.
Then.
