Question

Let $f^{\prime }\left(x\right)>0$ and $f^{\prime \prime }\left(x\right)>0$ for all x. If $x_1<x_2$ then

• A. $f\left(\frac{x_1+x_2}{2}\right)<\frac{f\left(x_1\right)+f\left(x_2\right)}{2}$
• B. $f\left(\frac{x_1+2x_2}{3}\right)<\frac{f\left(x_1\right)+2f\left(x_2\right)}{3}$
• C. $f\left(\frac{x_1+3x_2}{4}\right)<\frac{f\left(x_1\right)+3f\left(x_2\right)}{4}$
• D. $f\left(\frac{x_1+3x_2}{4}\right)\ge \frac{f\left(x_1\right)+3f\left(x_2\right)}{4}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A $f\left(\frac{x_1+x_2}{2}\right)<\frac{f\left(x_1\right)+f\left(x_2\right)}{2}$
B $f\left(\frac{x_1+2x_2}{3}\right)<\frac{f\left(x_1\right)+2f\left(x_2\right)}{3}$
C $f\left(\frac{x_1+3x_2}{4}\right)<\frac{f\left(x_1\right)+3f\left(x_2\right)}{4}$

Given that $f^{\prime }\left(x\right)>0$
This implies that $f\left(x\right)$ is an increasing function.
Now $f^{\prime \prime }\left(x\right)>0$
This implies that graph of $f\left(x\right)$ is concave up.
The left hand side of the inequality in option A is y coordinate of a point on the curve with x-coordinate $(x_1+x_2)/2$, and the right hand side is y coordinate of a point with same x coordinate on the line joining $(x_1,f(x_1\left)\right)$ and $(x_2,f(x_2\left)\right)$. (as shown in figure)
Since the graph of the function is concave up, $y_Q<y_P$. Hence option A is true.
Similarly it can be shown that option B and C are also true.