SATHEE: Indefinite Integration 1 Question 43

Indefinite Integration 1 Question 43

44. If $lim_{t \to a} \frac{\int_{a}^{t} f (x) d x - \frac{(t - a)}{2} f (t) + f (a)}{(t - a)^{3}} = 0$ ,

then degree of polynomial function $f (x)$ atmost is

(a) 0

(b) 1

(d) 2

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Solution:

Given, $lim_{t \to a} \frac{\int_{a}^{t} f (x) d x - \frac{(t - a)}{2} f (t) + f (a)}{(t - a)^{3}} = 0$

Using L’Hospital’s rule, put $t - a = h$

$\begin{array}{ll} \Rightarrow lim_{h \to 0} \frac{\int_{a}^{a + h} f (x) d x - \frac{h}{2} f (a + h) + f (a)}{h^{3}} & = 0 \\ \Rightarrow lim_{h \to 0} \frac{f (a + h) - \frac{1}{2} f (a + h) + f (a) - \frac{h}{2} f^{'} (a + h)}{3 h^{2}} & = 0 \end{array}$

Again, using L’ Hospital’s rule,

$lim_{h \to 0} \frac{f^{'} (a + h) - \frac{1}{2} f^{'} (a + h) - \frac{1}{2} f^{'} (a + h) - \frac{h}{2} f^{''} (a + h)}{6 h} = 0$

$\begin{aligned} \Rightarrow & lim_{h \to 0} \frac{- \frac{h}{2} f^{''} (a + h)}{6 h} & = 0 \\ \Rightarrow & f^{''} (a) & = 0, \forall a \in R \end{aligned}$

$\Rightarrow f (x)$ must have maximum degree 1 .

Indefinite Integration 1 Question 43

44. If $lim_{t \to a} \frac{\int_{a}^{t} f (x) d x - \frac{(t - a)}{2} f (t) + f (a)}{(t - a)^{3}} = 0$ ,

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Indefinite Integration 1 Question 43

44. If limt→a∫atf(x)dx−(t−a)2f(t)+f(a)(t−a)3=0,

Solution:

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44. If $lim_{t \to a} \frac{\int_{a}^{t} f (x) d x - \frac{(t - a)}{2} f (t) + f (a)}{(t - a)^{3}} = 0$ ,