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-- Mathematics - Section B1
Relations and Functions
Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be defined by \(f(x)=x^2+1\), Then pre-image of \(17\) is
\(\{4,-4\}\)
\(\{3,-3\}\)
\(\{2,-2\}\)
None of these
$f(x)=x^2+1⇒f(x)=17=x^2+1$
$⇒x^2=16⇒x±4$
$⇒x∈\{4,-4\}$