Photo AI

Suppose $f(x)$ is a differentiable function such that $$\frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right)$$ for all $a$ and $b$ - HSC - SSCE Mathematics Extension 2 - Question 10 - 2017 - Paper 1

Question icon

Question 10

Suppose-$f(x)$-is-a-differentiable-function-such-that-$$\frac{f(a)+f(b)}{2}-\geq-f\left(-\frac{a+b}{2}-\right)$$-for-all-$a$-and-$b$-HSC-SSCE Mathematics Extension 2-Question 10-2017-Paper 1.png

Suppose $f(x)$ is a differentiable function such that $$\frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right)$$ for all $a$ and $b$. Which statement is always true... show full transcript

Worked Solution & Example Answer:Suppose $f(x)$ is a differentiable function such that $$\frac{f(a)+f(b)}{2} \geq f\left( \frac{a+b}{2} \right)$$ for all $a$ and $b$ - HSC - SSCE Mathematics Extension 2 - Question 10 - 2017 - Paper 1

Step 1

A. $\int_0^1 f(x) dx \geq \frac{f(0)+f(1)}{2}$

96%

114 rated

Answer

This statement is not always true. The given inequality suggests that the average value of f(a)f(a) and f(b)f(b) is greater than or equal to the average value at the midpoint. Thus, it does not guarantee that the integral will be greater than or equal to the average of the endpoints.

Step 2

B. $\int_0^1 f(x) dx \leq \frac{f(0)+f(1)}{2}$

99%

104 rated

Answer

This statement is true. The initial condition indicates that the function f(x)f(x) behaves in such a way that it can approach or remain below the average value at the endpoints, making this the correct answer.

Step 3

C. $f'\left( \frac{1}{2} \right) \geq 0$

96%

101 rated

Answer

This statement cannot be concluded from the given inequality, as it does not provide any information on the behavior of the derivative of f(x)f(x).

Step 4

D. $f'\left( \frac{1}{2} \right) \leq 0$

98%

120 rated

Answer

Like statement C, this cannot be concluded from the provided inequality, as there’s no implication regarding the sign of the derivative.

Join the SSCE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;