Solveeit Logo
Login

Question

Question: If f”(x) \> 0 ∀ x ∈ R then for any two real num bers x<sub>1</sub> and x<sub>2</sub>, (x<sub>1</sub>...

If f”(x) > 0 ∀ x ∈ R then for any two real num bers x1 and x2, (x1≠x2)

A

f(x1+x22)>f(x1)+f(x2)2f \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } \right) > \frac { f \left( x _ { 1 } \right) + f \left( x _ { 2 } \right) } { 2 }

B

f(x1+x22)<f(x1)+f(x2′)2f \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } \right) < \frac { f \left( x _ { 1 } \right) + f \left( x _ { 2 } ^ { \prime } \right) } { 2 }

C

f′(x1+x22)>f′(x1)+f′(x2)2f ^ { \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′(x1+x22)<f′(x1)+f′(x2)2f ^ { \prime } \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } \right) < \frac { f ^ { \prime } \left( x _ { 1 } \right) + f ^ { \prime } \left( x _ { 2 } \right) } { 2 }

Answer

f(x1+x22)<f(x1)+f(x2′)2f \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } \right) < \frac { f \left( x _ { 1 } \right) + f \left( x _ { 2 } ^ { \prime } \right) } { 2 }

Explanation

Solution

Let A = (x1, f(x1)) and B = (x3, f(x2)) be any two points on the graph of y = f(x). Chord AB will lie completely above the graph of y = f(x).

Hence f(x1)+f(x2)2>f(x1+x22)\frac { f \left( x _ { 1 } \right) + f \left( x _ { 2 } \right) } { 2 } > f \left( \frac { x _ { 1 } + x _ { 2 } } { 2 } \right)