Question

Suppose we define the definite integral using the following formula $\underset{a}{\overset{b}{\int }}f(x)dx=\frac{b-a}{2}(f(a)+f(b))$, 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 $c=\frac{a+b}{2},\underset{a}{\overset{b}{\int }}f(x)dx=\frac{b-a}{4}(f(a)+f(b)+2f(c))$.
If $f(x)$ is a polynomial and if $\underset{t\to a}{\lim }\frac{\underset{a}{\overset{t}{\int }}f(x)dx-\frac{(t-a)}{2}(f(t)+f(a))}{(t-a{)}^3}=0$ 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
$\underset{t\to a}{\lim }\frac{f(t)-\frac{1}{2}(f(t)+f(a))-\frac{(t-a)}{2}{f}^{\prime }(t)}{3(t-a{)}^2}=0$
$\underset{t\to a}{\lim }\frac{f(t)-f(a)-(t-a)f^{\prime }(t)}{6(t-a{)}^2}=0$
Applying L'Hospital rule again
$\underset{t\to a}{\lim }\frac{f^{\prime }(t)-f^{\prime }(t)-(t-a)f^{\prime \prime }(t)}{12(t-a)}=0$
$\Rightarrow \underset{t\to a}{\lim }-\frac{f^{\prime \prime }(t)}{12}=0$
$\Rightarrow f^{\prime \prime }(a)=0,$ for any a.
$\Rightarrow f(x)$ is atmost of degree 1.