KCET · Maths · Functions
Let \(f: R \rightarrow R\) be defined by \(f(x)=x^2+1\). Then, the pre images of 17 and -3 . respectively are
- A \(\phi,\{4,-4\}\)
- B \(\{3,-3\}, \phi\)
- C \(\{4,-4\}, \phi\)
- D \(\{4,-4\},\{2,-2\}\)
Answer & Solution
Correct Answer
(C) \(\{4,-4\}, \phi\)
Step-by-step Solution
Detailed explanation
\(\because\) For pre image of 17 , we have
\(f(x)=x^2+1=17\)
\(\Rightarrow \quad x^2=16 \Rightarrow x= \pm 4\)
So, \(x \in\{-4,4\}\)
For pre image of -3 , we have
\(f(x)=x^2+1=-3\)
\(\Rightarrow \quad x^2=-4(\) not possible \()\)
Hence, no possible pre image.
\(f(x)=x^2+1=17\)
\(\Rightarrow \quad x^2=16 \Rightarrow x= \pm 4\)
So, \(x \in\{-4,4\}\)
For pre image of -3 , we have
\(f(x)=x^2+1=-3\)
\(\Rightarrow \quad x^2=-4(\) not possible \()\)
Hence, no possible pre image.
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