I was solving this question in my assignment:
I didn't have any idea to approach this , so i assumed that the first two are correct and came to the conclusion that the graph would be one of these:
From this , i got [C] and [D] correct too . But i cannot seem to understand why [A] and [B] should be correct always.
Consider a contradiction:
let f(x) be a increasing function
i.e. $f'(x)>0$ now we have 2 cases to satisfy the given condition
CASE 1 $f(x)>0$ and $f''(x)<0$ now assume this holds true thus $f(c)=k$ where k is positive number
now consider $f(c-w)$ such that $c-w<c$ and as $f'(x)>0$ therefore $f(c-w)<f(w)$ or $f(c-w)<k$ or $f(c-w)=k-p$
we can say there is a '$w$'[actually infinately many] such that $ p>k$ due to which $f(c-w)<0$ [which condraticts the assumption of f(x)>0 for all x] as the derivative is always positive and the function keeps decreasing towards the negative side and in fact increasingly decreasing towards negative side
CASE 2 $f(x)<0$ and $f''(x)>0$ now assume this holds true thus $f(c)=k$ where k is negative number
now consider $f(c+w)$ such that $c+w>c$ and as $f'(x)>0$ therefore $f(c+w)>f(w)$ or $f(c+w)>k$ or $f(c+w)=k+p$
we can say there is a '$w$'[actually infinately many] such that $ p>|k|$ due to which $f(c+w)>0$ [which condraticts the assumption of f(x)<0 for all x] as the derivative is always positive and the function keeps increasing and in fact increasingly increasing [`increases at a higher rate as value of x increases] as the double derivative is always assumed positive
similarly considering a decreasing function will NOT lead to a contradiction