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Suppose, that f is differentiable for all \mathrm{x} and that f^{\prime}(x) \leq 2 for all \mathrm{x}. If f(1)=2 and f(4)=8, then ff(2) has the value equal to
 

 

Option: 1

3


Option: 2

4


Option: 3

6


Option: 4

8


Answers (1)

Using LMVT for f in [1,2]
\begin{aligned} \forall c \in(1,2) \quad & \frac{f(2)-f(1)}{2-1}=f^{\prime}(c) \leq 2 \\ & f(2)-f(1) \leq 2 \Rightarrow f(2) \leq 4 \end{aligned}

Again, using LMVT in [2, 4]
\forall d \in(2,4) \frac{f(4)-f(2)}{4-2}=f^{\prime}(d) \leq 2 \\ \begin{aligned} \begin{array}{ccc} \therefore & & f(4)-f(2) \leq 4 \\ &&\;\;\;\;\; 8-f(2) \leq 4 \\ && \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;4 \leq f(2) \\ \Rightarrow && \;\;\;\;\;\;\; \;\;\;f(2) \geq 4 \end{array} \end{aligned}

\Rightarrow  From Eqs. (i) and (ii), we get
f(2)=4

Posted by

Ramraj Saini

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