My approach as follows :
As f(0) = 0 = f(1) let f(x) = kx(x-1)
f'(x) = k(x-1) + kx
f''(x) = 2k
Therefore, 2k - 2kx + 2k - 2kx + kx(x-1) >= e^x
kx^2 - 5kx + 4k >= e^x
Since, e^x is always positive ;
Therefore, k > 0 and D < 0
25k^2 - 16k^2 < 0
9k^2 < 0 OR k < 0
From above, no value of k is possible.
Please help me to solve the question in the image. I cannot figure out further steps.
Here's a big hint for $1$. Let's not try to guess the form of $f$, but rather use properties of the derivative at critical points.
Observe that $e^x$ has minimum of $1$ on $[0,1]$.
So $f''(x) - 2f'(x) + f(x) \geq 1$.
Now suppose f(x) takes a maximum at $x = x_0$. Then $f'(x_0) = 0$.
So we have $f''(x_0) + f(x_0) \geq 1$.
But we know even more, because we know something about the value of $f''(x_0)$ when $f(x_0)$ is a local maximum (the graph is concave down!). Use this to refine the last equality and then see which of the bounds on $f$ still make sense.