Differential Calculus> Q. IF TWO ROOTS OF THE EQUATION- (p-1)(x...

Q. IF TWO ROOTS OF THE EQUATION- (p-1)(x2 + x + 1)2 – (p+1)( x4 + x2 +1) = 0 ARE REAL AND DISTINCT AND f(x) = 1-x/1+x , THEN f(f(x)) + f(f(1/x)) IS EQUAL TO-p -p 2p -2p

Bharat Makkar , 10 Years ago

Grade Select Grade

2 Answers

Jitender Singh

Last Activity: 10 Years ago

Ans:
Hello Student,
Please find answer to your question below

$f(x) = \frac{1-x}{1+x}$
$f(f(x)) = \frac{1-f(x)}{1+f(x)}$
$f(f(x)) = \frac{1-\frac{1-x}{1+x}}{1+\frac{1-x}{1+x}}$
$f(f(x)) = \frac{2x}{2}$
$f(f(x)) = x$
$f(f(\frac{1}{x})) = \frac{1-f(\frac{1}{x})}{1+f(\frac{1}{x})}$
$f(f(\frac{1}{x})) = \frac{1-\frac{x-1}{x+1}}{1+\frac{x-1}{x+1}}$
$f(f(\frac{1}{x})) = \frac{2}{2x}$
$f(f(\frac{1}{x})) = \frac{1}{x}$
$f(f(x)) + f(f(\frac{1}{x})) = x + \frac{1}{x}$
$(p-1)(x^2+x+1)^2 = (p+1)(x^4+x^2+1)$
$(p-1)(x^4+x^2+1+2x^3+2x+2x^2) = (p+1)(x^4+x^2+1)$
$2(p-1)(x^3+x+x^2) = 2(x^4+x^2+1)$
$(p-1)x(x^2+x+1) = (x^4+x^2+1)$
$p-1 = \frac{(x^4+x^2+1)}{(x^3+x^2+x)}$
$p = \frac{(x^4+x^2+1)}{(x^3+x^2+x)}+1$
$p = \frac{(x^4+x^3+2x^2+x+1)}{(x^3+x^2+x)}$
$p = \frac{(x^4+x^3+x^2+x^2+x+1)}{(x^3+x^2+x)}$
$p = \frac{x^2(x^2+x+1)+x^2+x+1)}{x(x^2+x+1)}$
$p = x+\frac{1}{x}$
Option(1) is correct.