Question

Suppose we define the definite integral using the following formula $\int\limits_a^b f(x) d x=\frac{b-a}{2} (f(a)+f(b))$, for more accurate result for $c \in(a, b), F(c)=\frac{c-a}{2}(f(a)+f(c))+\frac{b-c}{2}(f(b)+f(c))$. When $c=\frac{a+b}{2}, \int\limits_a^b f(x) d x=\frac{b-a}{4}(f(a)+f(b)+2 f(c))$.
If $f(x)$ is a polynomial and if $\displaystyle\lim _{t \rightarrow a} \frac{\int\limits_a^t f(x) d x-\frac{(t-a)}{2}(f(t)+f(a))}{(t-a)^3}=0$ for all a. then the degree of $f(x)$ can atmost be

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

Applying L'Hospital rule
$\displaystyle \lim _{t \rightarrow a} \frac{f(t)-\frac{1}{2}(f(t)+f(a))-\frac{(t-a)}{2} f^{\prime}(t)}{3(t-a)^2}=0$
$ \displaystyle\lim _{t \rightarrow a} \frac{f(t)-f(a)-(t-a) f^{\prime}(t)}{6(t-a)^2}=0$
Applying L'Hospital rule again
$\displaystyle\lim _{t \rightarrow a} \frac{f^{\prime}(t)-f^{\prime}(t)-(t-a) f^{\prime \prime}(t)}{12(t-a)}=0$
$\Rightarrow \displaystyle \lim _{t \rightarrow a}-\frac{f^{\prime \prime}(t)}{12}=0 $
$\Rightarrow f^{\prime \prime}(a)=0, $ for any a.
$\Rightarrow f(x) $ is atmost of degree 1.