Definite Integration Question 43
Question 43
- 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 $f(x)$ atmost is (a) 0 (b) 1 (c) 3 (d) 2
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Solution:
- Given, $\lim {t \rightarrow a} \frac{\int{a}^{t} f(x) d x-\frac{(t-a)}{2}{f(t)+f(a)}}{(t-a)^{3}}=0$
Using L’Hospital’s rule, put $t-a=h$
$$ \begin{array}{ll} \Rightarrow \quad \lim {h \rightarrow 0} \frac{\int{a}^{a+h} f(x) d x-\frac{h}{2}{f(a+h)+f(a)}}{h^{3}} & =0 \ \Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(a+h)-\frac{1}{2}{f(a+h)+f(a)}-\frac{h}{2}\left{f^{\prime}(a+h)\right}}{3 h^{2}} & =0 \end{array} $$
Again, using L’ Hospital’s rule,
$\lim _{h \rightarrow 0} \frac{f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0$
$$ \begin{aligned} \Rightarrow & & \lim _{h \rightarrow 0} \frac{-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h} & =0 \ \Rightarrow & & f^{\prime \prime}(a) & =0, \forall a \in R \end{aligned} $$
$\Rightarrow f(x)$ must have maximum degree 1 .
