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

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

Answers (1)

Answer:

 \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}

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Gurleen Kaur

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