If f " ( x ) > 0 ∀ x ϵ R then for any two real numbers x 1 a n d x 2 , ( x 1 ≠ x 2 ) Options: A . &nbsp f ( x 1 + x 2 2 ) > f ( x 1 ) + f ( x 2 ) 2 B . &nbsp f ( x 1 + x 2 2 ) < f ( x 1 ) + f ( x 2 ) 2 C . &nbsp f ′ ( x 1 + x 2 2 ) > f ′ ( x 1 ) + f ′ ( x 2 ) 2 D . &nbsp f ′ ( x 1 + x 2 2 ) < f ′ ( x 1 ) + f ′ ( x 2 ) 2

If f"(x) > 0 ∀ x ϵ R then f

Question -

If $f " (x) > 0 \forall x ϵ R$ then for any two real numbers $x_{1} a n d x_{2}, (x_{1} \neq x_{2})$

Options:

A .

f (\frac{x_{1} + x_{2}}{2}) > \frac{f (x_{1}) + f (x_{2})}{2}

B .

f (\frac{x_{1} + x_{2}}{2}) < \frac{f (x_{1}) + f (x_{2})}{2}

C .

f^{'} (\frac{x_{1} + x_{2}}{2}) > \frac{f^{'} (x_{1}) + f^{'} (x_{2})}{2}

D .

f^{'} (\frac{x_{1} + x_{2}}{2}) < \frac{f^{'} (x_{1}) + f^{'} (x_{2})}{2}

Answer: Option B
:
B
Let A =

(x_{1}, f (x_{1}))

and B =

(x_{2}, f (x_{2}))

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 (x_{1}) + f (x_{2})}{2} > f (\frac{x_{1} + x_{2}}{2})

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