Can someone please do Q 8) a) alpha) i) and ii) from this trial paper:
http://www.angelfire.com/ab7/fourunit/marist2010.pdf
http://www.angelfire.com/ab7/fourunit/marist2010.pdf
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Harder 3U question (1 Viewer)
- Thread starter SonOfAGun
- Start date
i) If we pick any 2 points a and b on such a curve, for a < x < b, f(x) < corresponding value on the line segment ab. In particular the value of the mid-point of ab > f(mid-point of ab) = f([a+b]/2). i.e.
+f(b)}{2} > f(\frac {a+b}{2}) )
 + f(b) > 2f(\frac {a+b}{2}) )
ii) In the same way:
 + f(d) > 2f(\frac {c+d}{2}) )
 + f(\frac {c+d}{2})}{2} > f \left ( \frac {\frac {a+b}{2} + \frac {c+d}{2}}{2} \right ) )
 + f(\frac {c+d}{2}) > 2f(\frac {a+b+c+d}{4}) )
But
+f(b)}{2} + \frac {f(c)+f(d)}{2} > f(\frac {a+b}{2}) + f(\frac {c+d}{2}) )
+f(b)}{2} + \frac {f(c)+f(d)}{2} > 2 f(\frac {a+b+c+d}{4}) )
 + f(b) + f(c) + f(d) > 4f \left( \frac {a+b+c+d}{4} \right ))
Edit:
Turns out to be more or less equivalent to seanieg89's suggested approach.
ii) In the same way:
But
Edit:
Turns out to be more or less equivalent to seanieg89's suggested approach.
Last edited:
could you post the definite integrals one+f(b)}{2}\right)$ which bisects such a segment will lie above the corresponding point on the curve:$\left(\frac{a+b}{2},f\left(\frac{a+b}{2}\right)\right) $. \\This gives: $f\left(\frac{a+b}{2}\right)< \frac{f(a)+f(b)}{2}$ as required. \\The second part follows from applying i) to $\frac{a+b}{2} $ and $ \frac{c+d}{2}.$ \\If you don't like the geometric argument given above, I can think of an alternate approach which just uses properties of the definite integral.)