Statement 1: Consider the function \[f(x)\] \[f\left( \frac{{{x}_{1}}+{{x}_{2}}}{2} \right)<\frac{f({{x}_{1}})+f({{x}_{2}})}{2}\] |
Statement 2: \[f'(x)>0,\,f''(x)>0\] where \[{{x}_{1}}<{{x}_{2}}\]. |
A) Both statements are true, and Statement-2 explains Statement-1.
B) Both statements are true, but Statement-2 does not explain Statement-1.
C) Statement-1 is True, Statement-2 is False.
D) Statement-1 is False, Statement-2 is true.
Correct Answer: A
Solution :
\[\because \,\,f'(x)>0\] and \[f''(x)>0\] \[\therefore \] Coordinates of \[A({{x}_{1}},\,f({{x}_{1}}))\] and \[B({{x}_{2}},\,f({{x}_{2}}))\]. Let M is midpoint of AB. \[\therefore \,\,M\equiv \,\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\,\frac{f({{x}_{1}})+f({{x}_{2}})}{2} \right)\] and \[N\equiv \,\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\,f\left( \frac{{{x}_{1}}+{{x}_{2}}}{2} \right) \right)\] \[\Rightarrow \,\,f\left( \frac{{{x}_{1}}+{{x}_{2}}}{2} \right)<\frac{f({{x}_{1}})+f({{x}_{2}})}{2}\].You need to login to perform this action.
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