Question

If $f\left(x\right)$ is a polynomial and $\underset{t\to a}{lim}\frac{\underset{a}{\overset{t}{\int }}f\left(x\right)dx-\frac{(t-a)}{2}\left(f\right(t)+f(a\left)\right)}{(t-a{)}^3}=0$ for all $a$, then the degree of $f\left(x\right)$ can atmost be

Answer 1

We have,
$\underset{t\to a}{lim}\frac{\underset{a}{\overset{t}{\int }}f\left(x\right)dx-\frac{(t-a)}{2}\left(f\right(t)+f(a\left)\right)}{(t-a{)}^3}=0$
Applying L' Hospital's rule, $\underset{t\to a}{lim}\frac{f\left(t\right)-\frac{1}{2}\left(f\right(t)+f(a\left)\right)-\frac{(t-a)}{2}{f}^{\prime }\left(t\right)}{3(t-a{)}^2}=0\Rightarrow \underset{t\to a}{lim}\frac{f^{\prime }\left(t\right)-\frac{1}{2}{f}^{\prime }\left(t\right)-\frac{(t-a)}{2}{f}^{\prime \prime }\left(t\right)-\frac{1}{2}{f}^{\prime }\left(t\right)}{6(t-a)}=0\Rightarrow \underset{t\to a}{lim}\frac{f^{\prime \prime }\left(t\right)}{12}=0\Rightarrow f^{\prime \prime }\left(a\right)=0\text{for any}a\therefore f\left(a\right)\text{is atmost of degree}1$