Question

For every function $f(x)$ which is twice differentiable, these will be good approximation of
$\underset{a}{\overset{b}{\int }}f(x)dx=\left(\frac{b-a}{2}\right)\{f(a)+f(b)\},$
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 $c=\frac{a+b}{2}$
$\underset{a}{\overset{b}{\int }}f(x)dx=\frac{b-a}{4}\left[f(a)+f(b)+2\int (c)\right\}dx$
If $\underset{t\to a}{\lim }\frac{\underset{a}{\overset{t}{\int }}f(x)dx-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a{)}^3}=0$ 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, $\underset{t\to a}{\lim }\frac{\underset{a}{\overset{t}{\int }}f(x)dx-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a{)}^3}=0$
Using L'Hospital's rule, put $t-a=h$
$\Rightarrow \underset{h\to 0}{\lim }\frac{\underset{a}{\overset{a+h}{\int }}f(x)dx-\frac{h}{2}\{f(a+h)+f(a)\}}{h^3}=0$
$\Rightarrow \underset{h\to 0}{\lim }\frac{f(a+h)-\frac{1}{2}\{f(a+h)+f(a)\}-\frac{h}{2}\left\{f^{\prime }(a+h)\right\}}{3h^2}=0$
Again, using L' Hospital's rule,
$\underset{h\to 0}{\lim }\frac{f^{\prime }(a+h)-\frac{1}{2}{f}^{\prime }(a+h)-\frac{1}{2}{f}^{\prime }(a+h)-\frac{h}{2}{f}^{\prime \prime }(a+h)}{6h}=0$
$\Rightarrow \underset{h\to 0}{\lim }\frac{-\frac{h}{2}{f}^{\prime \prime }(a+h)}{6h}=0$
$\Rightarrow f^{\prime \prime }(a)=0,\forall a\in R$
$\Rightarrow f(x)$ must have maximum degree $1$.