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Grade Select GradeDifferential Calculus

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-
  1. p
  2. -p
  3. 2p
  4. -2p

Profile image of Bharat Makkar
11 Years agoGrade Select Grade
Answers icon

2 Answers

Profile image of Jitender Singh
ApprovedApproved Tutor Answer11 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.
Profile image of Bharat Makkar
11 Years ago
THANK YOU SIR.....

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