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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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