Paragraph:
Read the following passage and answer the questions.
For every function \(f(x)\) which is twice differentiable, these will be good approximation of \(\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}\). Now, if we take \(c=\frac{a+b}{2}\), then using above again, we get \(\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}\) and so on.
We get approximation for value of \(\int_a^b f(x) d x\).Question:
If \(\lim _{t \rightarrow a} \frac{\int_a^t f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^3}=0\), then degree of polynomial function \(f(x)\) at most is

\[
\begin{aligned}
\lim _{t \rightarrow a} \frac{\int_a^t f(t) d t-\frac{(t-a)}{2}(f(t)+f(a))}{(t-a)^3} & =0 \\
\Rightarrow \quad \lim _{h \rightarrow 0} \frac{\int_a^{a+h} f(t) d t-\frac{h}{2}(f(a+h)+f(a))}{h^3} & =0 \\
\Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(a+h)-\frac{1}{2}(f(a+h)+f(a))-\frac{h}{2}\left(f^{\prime}(a+h)\right)}{3 h^2} & =0
\end{aligned}
\]
\(f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)\)
\(\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{1}{2} f^{\prime}(a+h)-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0\)
\(\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{h}{2} f^{\prime}(a+h)}{6 h}=0\)
\(\Rightarrow \quad f^{\prime \prime}(a)=0, \forall a \in R\)
\(\Rightarrow(x)\) must have maximum degree 1.