Question

Hey ♥.

A new question from class 11th
Maths.
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Ques:- Find the inverse of ( x^2 + x + 1 )
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Give relevant answer.



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Answer 1

Given : y = x^2 + x + 1

= > x = y^2 + y + 1

= > y^2 + y + 1 - x = 0

It is in the form of ax^2 + bx + c= 0, we get a = 1, b = 1, c = 1 - x. the solutions are:

(1)

$= \ \textgreater \ x = \frac{-b + \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-1 + \sqrt{1^2 - 4 * 1 * (1 - x)} }{2(1)}$

$= \ \textgreater \ \frac{-1 + \sqrt{1 - 4(1 - x)} }{2}$

$= \ \textgreater \ \frac{-1 + \sqrt{1 - 4 + 4x} }{2}$

$= \ \textgreater \ \frac{-1 + \sqrt{4x - 3} }{2}$


(2)

$x = \frac{-b- \sqrt{b^2 - 4ac} }{2a}$

$= \ \textgreater \ \frac{-1 - \sqrt{(1)^2 - 4(1)(1 - x)} }{2(1)}$

$= \ \textgreater \ \frac{-1 - \sqrt{1 - 4 + 4x} }{2}$

$= \ \textgreater \ \frac{-1 - \sqrt{4x - 3} }{2}$

$= \ \textgreater \ \frac{-1 - \sqrt{4x - 3} }{2}$


Therefore:

$f^{-1} (x) = \frac{-1 + \sqrt{4x - 3} }{2} , \frac{-1 - \sqrt{4x - 3} }{2}$




Hope this helps!

Answer 2

Hi,

Please see the attached file!


Thanks