Q92 / 180 · COMEDK 2025 (25 May Shift) · Mathematics
If a function $f: R \rightarrow R$ is defined as $f(x)=x^2+1$, then the pre-image of -3 is
- A$\{-4,4\}$
- B$\{-\sqrt{2}, \sqrt{2}\}$
- C$\{-2,2\}$
- Correct answerD$\emptyset$
Solution
The pre-image of a value $y$ under a function $f(x)$ is the set of all $x$ such that $f(x) = y$.
Given $f(x) = x^2 + 1$, we set $f(x) = -3$.
$x^2 + 1 = -3$
$x^2 = -3 - 1$
$x^2 = -4$
Since the square of any real number $x$ must be non-negative ($x^2 \geq 0$ for all $x \in R$), there is no real number $x$ such that $x^2 = -4$.
Therefore, the set of pre-images is the empty set, denoted by $\emptyset$.
Answer: $\emptyset$
Given $f(x) = x^2 + 1$, we set $f(x) = -3$.
$x^2 + 1 = -3$
$x^2 = -3 - 1$
$x^2 = -4$
Since the square of any real number $x$ must be non-negative ($x^2 \geq 0$ for all $x \in R$), there is no real number $x$ such that $x^2 = -4$.
Therefore, the set of pre-images is the empty set, denoted by $\emptyset$.
Answer: $\emptyset$
Hence, the correct answer is option: 'D'.