Question

For every function $f(x)$ which is twice differentiable, these will be good approximation of
$\int\limits_{a}^{b} f(x) d x=\left(\frac{b-a}{2}\right)\{f(a)+f(b)\},$
for more acurate results 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} $
$\int\limits_{a}^{b} f(x) d x =\frac{b-a}{4}\left[f(a)+f(b)+2 \int(c)\right\} d x$
If $\displaystyle\lim _{t \rightarrow a} \frac{\int\limits_{a}^{t} f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^{3}}=0$ then degree of polynomial function $f(x)$ atmost is
(1) 0
(2) 1
(3) 3
(4) 2

• A. 0
• B. 1
• C. 3
• D. 2

Answer 1

Answer: (B)

Given, $\displaystyle\lim _{t \rightarrow a} \frac{\int\limits_{a}^{t} f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^{3}}=0$
Using L'Hospital's rule, put $t-a=h$
$\Rightarrow \displaystyle\lim _{h \rightarrow 0} \frac{\int\limits_{a}^{a+h} f(x) d x-\frac{h}{2}\{f(a+h)+f(a)\}}{h^{3}}=0$
$\Rightarrow \displaystyle\lim _{h \rightarrow 0} \frac{f(a+h)-\frac{1}{2}\{f(a+h)+f(a)\}-\frac{h}{2}\left\{f^{\prime}(a+h)\right\}}{3 h^{2}}=0$
Again, using L' Hospital's rule,
$\displaystyle\lim _{h \rightarrow 0} \frac{f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0 $
$\Rightarrow \displaystyle\lim _{h \rightarrow 0} \frac{-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0$
$\Rightarrow f^{\prime \prime}(a)=0, \forall a \in R$
$\Rightarrow f(x) $ must have maximum degree $ 1 $.