If f ( 2 x 1 + x 2 ) ≤ 2 f ( x 1 ) + f ( x 2 ) then the result of f ( n x 1 + x 2 + …… + x n ) ≤ n f ( x 1 ) + f ( x 2 ) + … . + f ( x n ) is true.

If f ( 2 x 1 + x 2 ) ≤ 2 f ( x 1 ) + f ( x 2 ) then the result of f | TestHub

Passage / Comprehension

Let $f$ be a function satisfying $f (\frac{x _{1} + x _{2}}{2}) \leq \frac{f ( x _{1} ) + f ( x _{2} )}{2}$ , for all $x_{1}, x_{2}$ in its domain. If we draw the graph of a continuous function satisfying this inequality, we will notice that the chord joining any two points on the curve will always be above the portion of the curve between the two points. We can also prove that if the given inequality is true, then a similar result follows for four values $x_{1}, x_{2}, x_{3}, x_{4}$ . Indeed:

$f (\frac{x _{1} + x _{2} + x _{3} + x _{4}}{4}) = f (\frac{\frac{x _{1} + x _{2}}{2} + \frac{x _{3} + x _{4}}{2}}{2}) \leq \frac{f ( \frac{x _{1} + x _{2}}{2} ) + f ( \frac{x _{3} + x _{4}}{2} )}{2}$

$\leq \frac{\frac{f ( x _{1} ) + f ( x _{2} )}{2} + \frac{f ( x _{3} ) + f ( x _{4} )}{2}}{2} \leq \frac{f ( x _{1} ) + f ( x _{2} ) + f ( x _{3} ) + f ( x _{4} )}{4}$

If $f (\frac{x _{1} + x _{2}}{2}) \leq \frac{f ( x _{1} ) + f ( x _{2} )}{2}$ then the result of $f (\frac{x _{1} + x _{2} + \dots\dots + x _{n}}{n}) \leq \frac{f ( x _{1} ) + f ( x _{2} ) + \dots . + f ( x _{n} )}{n}$ is true.

Options:

Answer:

Solution:

If $f (\frac{x _{1} + x _{2}}{2}) \leq \frac{f ( x _{1} ) + f ( x _{2} )}{2}$ , then the function will be convex and the convexity will remain true for any number of variables. Algebraically it can be shown in the following manner: (i) Whenever it is true for k it is also true for 2k. (ii) Whenever it is true for k then it is also true for k-1.

Mathematics - Application of Derivative Question with Solution | TestHub

Options:

Doubts & Discussion