$f(x)=e^{2x}-(c+1)e^x+2x+\cos 2+\sin 1,$ is monotonically increasing for all $x\in R,$ If the exhaustive set of all possible values of $c$ such that $f(x)=e^{2x}-(c+1)e^x+2x+\cos 2+\sin 1,$ is monotonically increasing for all $x\in R,$ is $(-\infty,\lambda],$then find the value of $\lambda.$

$f(x)=e^{2x}-(c+1)e^x+2x+\cos 2+\sin 1$
$f'(x)=2e^{2x}-(c+1)e^x+2>0$.....as $f(x)$ is monotonically increasing function.
$2e^{2x}-(c+1)e^x+2$ is quadratic in $e^x$,so its discriminant has to be negative.
$(c+1)^2-16<0$ gives me $-5<c<3$
But i cannot find $\lambda$ from this,Am i wrong somewhere?
 A: You are fine up until $2e^{2x}-(c+1)e^x+2>0$. You need to do the next step carefully as $e^x>0$.
There are three cases to consider, either the discriminant is negative or both the root(s) are negative (and the discriminant is non-negative) or there is one root.
You have done the former correctly so lets consider the latter two. Calculate the roots then for both to be negative consider the larger of the two roots, it must be negative:
$$\frac{c+1+\sqrt{(c+1)^2-16}}{2}<0$$
For the larger root to exist we need $c\leq-5$ or $c\geq3$. Clearly if $c\geq3$ then the root is positive. This gives the other set of solutions $c\leq-5$.
Lastly, for one there to be one exactly root $c=3$ or $c=-5$.
So combining these sets of solutions gives $c\leq3$, which leads to your answer of $\lambda=3$.
A: So it is required that
$$
e^{\,2\,x}  - {{\left( {c + 1} \right)} \over 2}e^{\,x}  + 1 > 0\quad \left| {\; - \infty  < x < \infty } \right.
$$
the point to take care is that .."for all $x \in R$".
So this translates into:
$$
y = z^{\,2}  - {{\left( {c + 1} \right)} \over 2}z + 1 > 0\quad \left| {\;0 < \forall z} \right.
$$
which is a parabola in $z$, with vertical axis, passing through $(0,1)$, which is a fix point (under varying $c$), vertex in $$
\left( {{{\left( {c + 1} \right)} \over 4},1 - {{\left( {c + 1} \right)^{\,2} } \over {16}}} \right)
$$
and crossing the $z$ axis in
$$
z_{\,0}  = \left( {{{c + 1} \over 4}} \right) \pm \sqrt {\left( {{{c + 1} \over 4}} \right)^{\,2}  - 1}  = {1 \over 4}\left( {c + 1 \pm \sqrt {\left( {c - 3} \right)\left( {c + 5} \right)} } \right)
$$
Now it is easy to see that until $c<3$ then $y>0$ for all $0<z$, while for $3<=c$ the parabola will become non-positive for a segment included in $0<z$, which breaks the condition "for all $0<z$" $=>$ "for all $x \in R$".
