which of the following statement is TRue ?..... let $f : R \rightarrow R$ be a continious and nonnegative function. 
which of the following statement  is TRue ?
a) if there exist  $ c \in (0,1)$ such that  $f(c) = 100$ then $\int_{0}^{1} f(x) dx \ge  \frac {1}{2}.$
b)$\int_{0}^{1} f(x) dx > \frac {1}{2}$.then $f(c) > \frac{1}{2}$ for  some  $c \in (0,1).$
c)$\int_{0}^{1} f(x) dx = \frac {1}{2}$ then there exist  $c\in (0,1)$ such that $f(c) = \frac {1}{2}$
d) None of these
My answer  : option b) and C)   is true.. by intermediate theorem
For option a)  if  i take $f(x) = 200x$ now  put $x = \frac{1}{2}$..then  $\int_{0}^{1} f(x) dx = \frac {200 x^2}{2} |_0^1$...we will not get $\int_{0}^{1} f(x) dx =  \frac {1}{2}.$..so option a) is false
Is its right or  wrong ??  Pliz tell me
Any hints/ solution
 A: You're in the right direction by invoking the MVT to prove (b) and (c).
It remains to show that (a) is false by providing a counterexample: consider a "peak" function which is piecewise linear by joining the points $(0,0)$, $(c-\delta,0),(c,100),(c+\delta,0)$ and $(1,0)$.  This traces a triangle with height and base width $100$ and $2\delta$ respectively.  The area under the graph is $100\delta$.  To make it smaller than $0.5$, set $\delta < 0.05$.
A: In the same vein as GNU Supporter's answer, but not as elementary (and neat :D)

Consider another "peak function", namely 
$$f(x) = 100\ e^{-a(x-\frac{1}{2})^2}\ .$$
Clearly $f(\frac{1}{2}) = 100.$ We can compute
\begin{align*}
\int_0^1 100\ e^{-a(x-1/2)^2} \ dx 
&= 100 \int_{-\frac{1}{2}}^{\frac{1}{2}} e^{-a x^2} \ dx \\
&= \frac{100}{\sqrt{a}} \int_{-\frac{\sqrt{a}}{2}}^{\frac{\sqrt{a}}{2}} e^{-x^2} \ dx \\
&= \frac{100 }{\sqrt{a}} \sqrt{\pi}\ \mathsf{erf}\left(\frac{\sqrt{a}}{2}\right)
\end{align*}
where $\mathsf{erf}$ is the error function. Now, all we have to do is find for what values of $a$
\begin{align*}
\frac{100 }{\sqrt{a}} \sqrt{\pi}\ \mathsf{erf}\left(\frac{\sqrt{a}}{2}\right) < \frac{1}{2}\ .
\end{align*}
This can be solved with WolframAlpha, and the answer is roughly $a\geq 125\ 664$. 
So an example would be with $a = 100^3$, and we get
\begin{align*}
\int_0^1 100\ e^{-100^3 (x-1/2)^2} \ dx \approx 0.177 < \frac{1}{2}
\end{align*}
