1.Relation and Function
easy

यदि $f(x) = {x^2} + 1$, तब ${f^{ - 1}}(17)$ तथा ${f^{ - 1}}( - 3)$ का मान क्रमश: होगा

A

$4, 1$

B

$4, 0$

C

$3, 2$

D

इनमें से कोई नहीं

Solution

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

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

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

तथा ${f^{ – 1}}( – 3) = \pm \sqrt { – 3 – 1} = \pm \sqrt { – 4} $, जो संभव नहीं है।

Standard 12
Mathematics

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