Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be defined by \(f(x)=x^2+1\), Then pre-image of \(17\) is
Answer & explanation
Correct answer: option 1
$f(x)=x^2+1⇒f(x)=17=x^2+1$
$⇒x^2=16⇒x±4$
$⇒x∈\{4,-4\}$
Let \(f:\mathbb{R}\rightarrow \mathbb{R}\) be defined by \(f(x)=x^2+1\), Then pre-image of \(17\) is
Correct answer: option 1
$f(x)=x^2+1⇒f(x)=17=x^2+1$
$⇒x^2=16⇒x±4$
$⇒x∈\{4,-4\}$