#### Please solve RD Sharma class 12 chapter Increasing and Decreasing Functions exercise 16.2 question 38 maths textbook solution

$\left|f^{\prime}(x)<1\right| \text { for all } x \in[0,1]$

Given:

$\text { Let } f \text { defined on } [0,1] \text { be twice differentiable such that } \left|f^{\prime \prime}(x) \leq 1\right| \text { for all } x \in[0,1] \text { and } f(0)=f(1)$

To prove:

$\text { We have to show that }\left|f^{\prime}(x)<1\right| \text { for all } x \in[0,1]$

Hint:

Using mean value theorem

$\frac{\left|f^{\prime}(x)-f^{\prime}(c)\right|}{x-c}=f^{\prime \prime}(d)$

Solution:

As f(0) = f(1) and f is differentiable

Hence by rolls theorem

$f^{\prime}(c)=0 \text { for some } c \in[0,1]$

Let us now apply mean value theorem for point 0 and x $\in$ [0,1]

Hence

\begin{aligned} &\frac{\left|f^{\prime}(x)-f^{\prime}(c)\right|}{x-c}=f^{\prime \prime}(d) \\ &\Rightarrow \frac{\left|f^{\prime}(x)-0\right|}{x-c}=f^{\prime \prime}(d), \quad[\therefore f(c)=0] \\ &\Rightarrow \frac{\left|f^{\prime}(c)\right|}{x-c}=f^{\prime \prime}(d) \end{aligned}

\begin{aligned} &\text { As given that } f^{\prime \prime}(d) \leq 1 \text { for } x \in[0,1]\\ &\Rightarrow \frac{\left|f^{\prime}(c)\right|}{x-c} \leq 1\\ &\Longrightarrow\left|f^{\prime}(c)\right| \leq x-c \end{aligned}

Now, as both x and c lies in [0,1]

\begin{aligned} &\text { Hence } x-c \in[0,1] \\ &\Rightarrow\left|f^{\prime}(x)\right|<1 \text { for all } x \in[0,1] \end{aligned}