If f(x) = {x^2} + 1 , then {f^{ - 1}}(17) and {f^{ - 1}}( - 3) will be

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1.Relation and Function

easy

If $f(x) = {x^2} + 1$, then ${f^{ - 1}}(17)$ and ${f^{ - 1}}( - 3)$ will be

A

$4, 1$

B

$4, 0$

C

$3, 2$

D

None of these

Solution

(d) Let $y = {x^2} + 1$ ==> $x = \pm \sqrt {y – 1} $

==> ${f^{ – 1}}(y) = \pm \sqrt {y – 1} $

==> ${f^{ – 1}}(x) = \pm \sqrt {x – 1} $

==> ${f^{ – 1}}(17) = \pm \sqrt {17 – 1} = \pm 4$

and ${f^{ – 1}}( – 3) = \pm \sqrt { – 3 – 1} = \pm \sqrt { – 4} $, which is not possible.

Standard 12

Mathematics

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