If $f(x)$ is a continuous and injective function, then which of the following may be correct. If $f(x)$ is a continuous and injective function for $x ≥
 0$ and $\int_0^x {f(t)dt < {x \over 2}\left( {f\left( 0 \right) + f\left( x \right)} \right)} $, then which of the following may be correct:
A)$f'(x) < 0$, $f''(x) < 0$
B)$f'(x) < 0, f''(x) > 0$
C)$f'(x) > 0, f''(x) < 0$
D)$f'(x) > 0, f''(x) > 0$
Since $f(x)$ is a continuous and injective function, it must be either strictly increasing or strictly decreasing.
Now, LHS of the given inequality is the algebraic area under the graph of $f(x)$ from $x=0$ to $x=x$. while the RHS is the algebraic area of the trapezium formed by the points, $(0,0), (0,f(0)), (x,0)$ and $(x,f(x))$ , so we may draw intuitively the possible graphs of the function as
as in this link..
So the answer seems to be option B and D, but is is their any method to solve this question, mathematically, and not by  intuition?
 A: First of all, the inequality from the hypothesis doesn't hold for $x=0$, since you get $0<0$. I imagine you'll want to change the $<$ to a $\le$.
Now, you're right, options B and D are the correct ones. And I'll say more, they aren't just possible, they are sufficient. And I'll add something more to that more, options A and C are sufficient for the contrary to the inequality in the same way. Let's see this.
B. Take any $x>0$ and define $g(t)=f(t)-f(x)$. Why do this? Well, we can do it because the inequality is equivalent for $g$ and $f$:
$\int_0^x {f(t)dt \le {x \over 2}( {f( 0 ) + f( x )} )}\Leftrightarrow$
$\int_0^x {f(t)dt \le {x \over 2}( {f( 0 ) - f( x )+2f(x)} )}\Leftrightarrow\int_0^x {f(t)dt \le {x \over 2}( {f( 0 ) - f( x )} )}+xf(x)$
$\Leftrightarrow\int_0^x {f(t)dt -xf(x)\le  {x \over 2}( {g( 0 )} )}\Leftrightarrow\int_0^x {f(t)dt -\int_0^xf(x)dt\le  {x \over 2}( {g( 0 )} )}$
$\Leftrightarrow\int_0^x {g(t)dt\le  {x \over 2}( {g( 0)} )}$ (and this is the inequality for $g$ because $g(x)=f(x)-f(x)=0$).
This last thing is why we want to do it. It's a nicer expression to handle, but it is also has a little bit nicer interpretation, because in the RHS now it's the area of a triangle what we are comparing the integral to. Note that $f'<0\Leftrightarrow g'<0$ and $f''>0\Leftrightarrow g''>0$.
Now consider the line $y=\frac{-g(0)}{x}t+g(0)$ with parameter $t$. I assert that $g(t)$ intersects the line in just two points, mainly $t=0$ and $t=x$. Suppose there was a third point of intersection and consider $h(t)=g(t)-\frac{-g(0)}{x}t-g(0)$. We have that $h$ has three roots, so by Rolle's theorem $h'$ has two roots between them, and by Rolle's theorem applied to $h'$ we conclude $h''$ has at least one root. But $h''(t)=g''(t)=f''(t)>0$, so there can't be a third point of intersection. Therefore our function $g$ in $(0,x)$ is either above the line or below it at every time.
$g$ takes two values in that line (at $t=0$ and $t=x$), so by the Mean Value theorem there has to be some $\theta\in(0,x)$ such that $g'(\theta)=\frac{g(x)-g(0)}{x-0}=\frac{g(0)}{x}$. If it were above the line in $(0,x)$ then for $t=0$ we would have $g'\ge\frac{-g(0)}{x}$, so for $g'$ to reach $\frac{-g(0)}{x}$ in $(0,x)$ we would need $g''\le0$, which is not the case. Then $g$ is always above the line and $g(t)\le \frac{-g(0)}{x}t+g(0)\quad\forall t\in[0,x]\Rightarrow\int_0^xg(t)dt\le\frac{x}{2}g(0)$.
D. Define $g(t)=f(t)-f(0)$. As before, we can because the inequality is equivalent for both:
$\int_0^x {f(t)dt \le {x \over 2}( {f( 0 ) + f( x )} )}\Leftrightarrow$
$\int_0^x {f(t)dt \le {x \over 2}( {2f( 0 ) + f( x )-f(0)} )}\Leftrightarrow\int_0^x {f(t)dt \le xf(0)+ {x \over 2}( {f( x ) - f( 0 )} )}$
$\Leftrightarrow\int_0^x {f(t)dt -xf(0)\le  {x \over 2}( {g( x )} )}\Leftrightarrow\int_0^x {f(t)dt -\int_0^xf(0)\le  {x \over 2}( {g( x )} )}\Leftrightarrow\int_0^x {g(t)dt\le  {x \over 2}( {g( x)} )}$ (and this is the inequality for $g$ because $g(0)=f(0)-f(0)=0$).
And as before, we are left with a triangle on the RHS. Note that $f'>0\Leftrightarrow g'>0$ and $f''>0\Leftrightarrow g''>0$.
Now consider the line $h(t)=\frac{g(x)}{x}t$ with parameter $t$. Similarly as before, $g(t)$ has only two intersection points; $t=0$ and $t=x$. By the Mean Value theorem $g$ there has to be some $\theta\in(0,x)$ such that $g'(\theta)=\frac{g(x)-g(0)}{x-0}=\frac{g(x)}{x}$. As before, if it were above the line we would have $g'(0)\ge\frac{g(x)}{x}$, but then we would need $g''\le0$ to reach $\frac{g(x)}{x}$ in $(0,x)$. Then $g(t)\le\frac{g(x)}{x}$ in $[0,x]$, so $\int_0^xg(t)dt\le\frac{x}{2}g(x)$.
A. Take $h(t)=-f(t)$. Then $h'>0$ and $h''>0$ as in option D, so $\int_0^xh(t)dt\le\frac{x}{2}(h(0)+h(x))\Rightarrow-\int_0^xf(t)dt\le\frac{x}{2}(-f(0)-f(x))\Rightarrow\int_0^xf(t)dt\ge\frac{x}{2}(f(0)+f(x))$.
C. Take $h(t)=-f(t)$. Then $h'<0$ and $h''>0$ as in option B. As before we end up with $\int_0^xf(t)dt\ge\frac{x}{2}(f(0)+f(x))$.
Note: B and D are sufficient conditions, but they aren't necessary. Indeed, we could consider getting one of the derivatives $0$, for example the second one (since it's more general). Then $f''=0$, which means that $f'$ is constant, so $f$ is a line. Indeed, this satisfies the inequality, because it gives us an equality (the area from the integral is in this case the are of the trapezoid).
This couldn't be happening if we would have kept the $<$ in the inequality, instead of $\le$. We had to change it for consistency, but it is true that we could have changed another thing: instead of $x\ge0$ we could consider $x>0$ and then the inequality can remain strict. In that case we couldn't let $f''=0$, but this doesn't mean B and D are necessary now; we could construct a function that starts below those lines we considered earlier for each $x$, but with a "brief moment" (a small interval for $x$) exceeding those lines. If we don't exceed those lines too much, then the area that will remain below the line will compensate the little area there is above it. Of course, for this to happen we need the first derivative to change sign, so $f''$ would be zero at some point.
