Question

Let us define a function, $f:\left[ {0,1} \right] \to R$. Suppose $f$ is a twice differentiable function,$f\left( 0 \right) = 0 = f\left( 1 \right)$, and satisfies the inequation ${f'’}\left( x \right) - 2{f'}\left( x \right) + f\left( x \right) \geqslant {e^x}$for $x \in \left[ {0,1} \right]$. Choose the correct relation below.
1) $0 < f\left( x \right) < \infty $
2) $\dfrac{{ - 1}}{2} < f\left( x \right) < \dfrac{1}{2}$
3) $\dfrac{{ - 1}}{4} < f\left( x \right) < 1$
4) $ - \infty < f\left( x \right) < 0$

Answer 1

Hint: This is a good problem. First we should try to solve this problem by solving the inequation. After some calculations we can observe that it is nothing but the double differentiation of a function in the form of something multiplied to $f\left( x \right)$. Then by observing the inequation we can get the graph of that function. From that graph we can get a relation of that function and then find about the $f\left( x \right)$.

Complete step-by-step answer:
Let us take,
$g\left( x \right) = {e^{ - x}}.f\left( x \right)$,
Now let us find the value of.
Now observe the given inequation,
${f"}\left( x \right) - 2{f'}\left( x \right) + f\left( x \right) \geqslant {e^x}$,
Divide the inequation with ${e^x}$ , we get
$ \Rightarrow {e^{ - x}}{f’’}\left( x \right) - 2{e^{ - x}}{f'}\left( x \right) + {e^{ - x}}f\left( x \right) \geqslant 1$
Observe that the left hand side of the above inequation is nothing but .
That means
Nothing but,
This means $g\left( x \right)$ has a concave curve.
But we have $g\left( x \right) = {e^{ - x}}.f\left( x \right)$, and $f\left( 0 \right) = 0 = f\left( 1 \right)$
$ \Rightarrow g\left( 0 \right) = 0 = g\left( 1 \right)$
Now by the above information we can get the graph of $g\left( x \right)$.
Graph of $g\left( x \right)$ :
This means the graph of $g\left( x \right)$ is entirely below the X-axis.
$ \Rightarrow g\left( x \right) \leqslant 0$ in $\left[ {0,1} \right]$ ,
Now substitute the value of $g\left( x \right)$ in above inequation,
$ \Rightarrow g\left( x \right) = {e^{ - x}}.f\left( x \right) \leqslant 0$ in $\left[ {0,1} \right]$ ,
$ \Rightarrow {e^{ - x}}.f\left( x \right) \leqslant 0$ in $\left[ {0,1} \right]$ ,
$ \Rightarrow f\left( x \right) \leqslant 0$ in $\left[ {0,1} \right]$ ,
$ \Rightarrow - \infty < f\left( x \right) \leqslant 0$ in $\left[ {0,1} \right]$ ,
So the correct option is 4.
So, the correct answer is “Option 4”.

Note: : As I said this problem is a good problem it is tough to get the idea that the given inequation is a double differential of a function. And observing that we used the properties of the double differential in this problem, that it is concave when the double differential is negative and the convex double differential is positive, that helped us a lot.