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$2f(\frac{x}{2})+3f(\frac{2-x}{3})=g(x)$, $0 \leq x < 3$, $f''(x)>0$; if $g(x)$ is strictly increasing in $(a,b)$ and $g(x)$ is strictly decreasing $(c,d)$ then $25ad-bc$ is ?

My Attempt:

On taking derivative on both sides:

$$f'(\frac{x}{2}) - f'(\frac{2-x}{3}) =g'(x)$$

$g(x)$ is strictly increasing in $(a,b)$ implies for some domain $g'(x)>0$ which implies that:

$$\frac{x}{2}>\frac{2-x}{3}$$

$$\implies x>\frac{4}{5}$$

as $f'(x)$ is an increasing function ( $\because f''(x)>0$).

So

$a=\frac{4}{5}$

$b=3$

$c=0$

$d=\frac{4}{5}$

Is my method correct?

The answer comes out as $25(\frac{4}{5})(\frac{4}{5})=16$

  • 1
    $f''(x)>0$ and so $f'(x)$ is an increasing function @MichaelBurr – 2017-02-25
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    $g(x)$ strictly increasing in $(a,b)$ $\Rightarrow g(a+ c) \geq g(a)$ for $0\leq c \leq b$ – 2017-02-25
  • 0
    Aren't you assuming $g^{'}(a) > 0$? – 2017-02-25
  • 0
    @mrnovice In $(a,b)$ $g$ is increasing. I had earlier made some mistakes while typing which I corrected now. – 2017-02-25

0 Answers 0