Question

If $f"\left(x\right)>0\forall xϵR$ then for any two real numbers $x_{1}and{x}_2,(x_1\ne x_2)$

• A. $f\left(\frac{x_1+x_2}{2}\right)>\frac{f\left(x_1\right)+f\left(x_2\right)}{2}$
• B. $f\left(\frac{x_1+x_2}{2}\right)<\frac{f\left(x_1\right)+f\left(x_2\right)}{2}$
• C. $f^{\prime }\left(\frac{x_1+x_2}{2}\right)>\frac{f^{\prime }\left(x_1\right)+f^{\prime }\left(x_2\right)}{2}$
• D. $f^{\prime }\left(\frac{x_1+x_2}{2}\right)<\frac{f^{\prime }\left(x_1\right)+f^{\prime }\left(x_2\right)}{2}$

Answer 1

The correct option is B $f\left(\frac{x_1+x_2}{2}\right)<\frac{f\left(x_1\right)+f\left(x_2\right)}{2}$

Let A = $(x_1,f(x_1\left)\right)$ and B = $(x_2,f(x_2\left)\right)$ be any two points on the graph of y = f(x).
Since f"(x) $>$ 0, in the graph of the function tangent will always lie below the curve. Hence chord AB will lie completely above the graph of y = f(x).
Hence $\frac{f\left(x_1\right)+f\left(x_2\right)}{2}>f\left(\frac{x_1+x_2}{2}\right)$