Math Question and Answers


Question and Answers Forum

All Questions Topic List
Differential Equation Questions
Previous in All Question Next in All Question
Previous in Differential Equation Next in Differential Equation
Question Number 200902 by Rupesh123 last updated on 26/Nov/23

	Answered by Rasheed.Sindhi last updated on 26/Nov/23
	$f(x)f(y)+1=2f(xy)+2(x+y) x=y=1: [f(1)]^2 +1=2f(1)+4 [f(1)]^2 −2f(1)−3=0 ( f(1)−3 )(f(1)+1 )=0 f(1)=3,−1 y=1: f(x)f(1)+1=2f(x)+2x+2 { ((f(1)=3: 3f(x)−2f(x)=2x+1)),((f(1)=−1: −f(x)−2f(x)=2x+1)) :} ⇒ { ((f(x)=2x+1)),((f(x)=−((2x+1)/3))) :}$
	$f (x) f (y) + 1 = 2 f (x y) + 2 (x + y)$ $x = y = 1 :$ ${[f (1)]}^{2} + 1 = 2 f (1) + 4$ ${[f (1)]}^{2} - 2 f (1) - 3 = 0$ $(f (1) - 3) (f (1) + 1) = 0$ $f (1) = 3, - 1$ $y = 1 :$ $f (x) f (1) + 1 = 2 f (x) + 2 x + 2$ ${\begin{cases} f (1) = 3 : 3 f (x) - 2 f (x) = 2 x + 1 \\ f (1) = - 1 : - f (x) - 2 f (x) = 2 x + 1 \end{cases}$ $\Rightarrow {\begin{cases} f (x) = 2 x + 1 \\ f (x) = - \frac{2 x + 1}{3} \end{cases}$
	Commented by Rupesh123 last updated on 26/Nov/23
	Perfect ��
	Commented by deleteduser1 last updated on 26/Nov/23

	$f (x) = \frac{- (2 x + 1)}{3} i s n o t a l w a y s t r u e . Y o u c a n c h e c k$ $t h i s b y s u b s t i t u t i n g b a c k i n t o t h e e q u a t i o n, y o u$ $g e t x = 1, y = 1;$ $f (x) = 2 x + 1 g i v e s 0 = 0 \Rightarrow {i t}^{'} s t r u e f o r a l l x$
	Commented by Rasheed.Sindhi last updated on 26/Nov/23

	$T rue sir! f (x) = \frac{- (2 x + 1)}{3} i s n o t t r u e f o r a l l$ $v a l u e s o f x . T h a n k s!$
	Answered by deleteduser1 last updated on 26/Nov/23

	${[f (0)]}^{2} - 2 f (0) + 1 = 0 \Rightarrow f (0) = 1$ $f (x) f (0) = 2 f (0) + 2 x - 1 \Rightarrow f (x) = 2 x + 1$
	Commented by Rupesh123 last updated on 26/Nov/23
	Nice!
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum