First slide

Evaluation of definite integrals

Question

$Let f (x) be function on R - {1, 0} satisfying \int_{0}^{x} tf (t) dt = x^{2} \int_{0}^{x} f (t) dt . And also given that$ $f (2) = - 1 . Then which of the following is/are true.$

Difficult

A

$Number of the real solutions of f (x) = e is 1$
B

$Number of real solution of \frac{f^{'} (x)}{f (x)} = - e, is 1$
C

$Function y = f (x) has atleast one point of local extremum$
D

$Area of the region enclosed by y = f (x), x -axis, y -axis and the lines x = - 2, x = - 1$ $is \frac{5}{72}$

Solution

$\int_{0}^{x} tf (t) dt = x^{2} \int_{0}^{x} f (t) dt \Rightarrow d i f f e r e n t i a t e w . r . t . x b y l e i b n i t z r u l e x f (x) = x^{2} f (x) + 2 x^{} \int_{0}^{x} f (t) dt \Rightarrow (1 - x) f^{} (x) = 2 \int_{0}^{x} f (t) dt \Rightarrow (1 - x) f^{'} (x) - f (x) = 2 f (x)$
$\begin{array}{l} f^{'} (x) = \frac{3 f (x)}{1 - x} \\ \frac{f^{'} (x)}{f (x)} = \frac{3}{1 - x} \Rightarrow i n t e g t a t e o n b o t h s i d e s t h e n \log f^{} (x) = \log {(1 - x)}^{- 3} + \log c \\ \sin c e f (2) = - 1 \Rightarrow c = 1 \\ f^{} (x) = \frac{1}{(1 - x)^{3}} \\ (1) f (x) = e \Rightarrow \frac{1}{(1 - x)^{3}} = e t h e n n u m b e r o f v a l u e s o f x i s = 1 \\ (2) \frac{f^{'} (x)}{f (x)} = \frac{3}{1 - x} = - e \Rightarrow \frac{3}{x - 1} = e t h e n n u m b e r o f v a l u e s o f x = 1 \\ (3) f (x) i s a n i n c r e a \sin g f u n c t i o n \sin c e f^{'} (x) > 0 \\ (4) area = \int_{- 2}^{- 1} \frac{1}{(1 - x)^{3}} dx = \frac{5}{72} \end{array}$

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