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$f:[0,\infty)\rightarrow[0,\infty)$ be a continuous function such that $$\int_{0}^{\infty}f(t)dt<\infty$$ then, three statements a,b,c are given and we have to check their validity.

(a) The sequence ${f(n)}_{n\in \mathbb{N}}$ is bounded.

(b) $f(n)\rightarrow0\quad\textrm{as}\quad n\rightarrow\infty$.

(c) The series $\sum_{n=1}^{\infty} f(n)$ is convergent.

The question was asked in National Board for Higher Mathematics(India) in $2016$. I approached this way:

For(A) As the area is bounded then function must not grow on natural numbers because if grows then due to continuity argument the area much continue to increase. So must be unbounded.

For (B) As the function is unbounded then this option must not hold.

For (c) $$\Big|\sum_{n=1}^{\infty} f(n)\Big|<\int_{0}^{\infty}f(t)dt<\infty$$ So the sequence must be convergent.

But I'm wonder to see that none of the option is true. How ?

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    You can construct a counter-example by drawing a train of peaks. Although not the same question, [my previous answer has an image of such function](http://math.stackexchange.com/questions/1823185/find-a-continuous-function-f1-infty-to-bbb-r-such-that-fx-0-int/1823200#1823200). Of course, for your purpose you can devise a much simpler form of peaks than what appears in the link. – 2017-01-23
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    Let $t(x) = \max(0,1-|x|)$ and $f(x) = \sum_{k=2}^\infty k t(k^3(x-k))$. Then $f$ is continuous, $\int f = \sum_{k=2}^\infty {1 \over k^2}$ and $f(n) = n$. – 2017-01-23

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HINT. For each number $n$ take a triangle with the height $n$ and area $1/2^n$. In the remaing part of domain the function is equal to zero. Then the integral is finite, but none of your conditions is valid.

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    I constructed the function $f(x)=n.2^n(x-n+\frac{1}{2^n})\quad when \quad x\in[n-\frac{1}{2^n},n); \quad f(x)=n\quad when \quad x=n;\quad f(x)=-n.2^n(x-n-\frac{1}{2^n})\quad when \quad x\in(n,n+\frac{1}{2^n}] $ – 2017-01-24