Question

Let \(f:\mathbb{R}\to\mathbb{R}\) be defined by \[ f(x)=5^{x^4+2} \] Then:

• A. \(f\) is one-one but not onto
• B. \(f\) is onto but not one-one
• C. \(f\) is both one-one and onto
• D. \(f\) is neither one-one nor onto

Hint:

If a function contains even powers like \(x^2,x^4,x^6\), always test \(f(x)\) and \(f(-x)\) to check whether the function is one-one.

Answer 1

The Correct Option is D

Step 1: Understand the function.
We are given $ f(x) = 5^{x^4+2} $, where $ f : \mathbb{R} \to \mathbb{R} $. We need to decide whether this function is one-one (injective) and onto (surjective).
Step 2: Check if the function is one-one.
A function is one-one if two different inputs always give two different outputs. Notice that the exponent is $ x^4 + 2 $. Since $ (-x)^4 = x^4 $, we have $ f(-x) = 5^{(-x)^4+2} = 5^{x^4+2} = f(x) $. For example, $ f(1) = 5^3 = f(-1) $, but $ 1 \neq -1 $. So the same output is produced by two different inputs. Therefore, $ f $ is NOT one-one.
Step 3: Find the range of the function.
For all real $ x $, we know $ x^4 \geq 0 $. So $ x^4 + 2 \geq 2 $. This means the exponent is always at least 2. Therefore $ f(x) = 5^{x^4+2} \geq 5^2 = 25 $. The range of $ f $ is $ [25, \infty) $.
Step 4: Check if the function is onto.
The codomain is all of $ \mathbb{R} $, which includes numbers like $ 0, 1, 10, -5 $, etc. But $ f(x) \geq 25 $ for every real $ x $. So values less than 25 (for example, $ y = 1 $) are never achieved by $ f $. Therefore, $ f $ is NOT onto.
Step 5: Confirm with a specific value.
Try $ y = 5 $. We would need $ 5^{x^4+2} = 5 $, which means $ x^4 + 2 = 1 $, so $ x^4 = -1 $. But $ x^4 \geq 0 $ for real $ x $, so no real solution exists. This confirms $ f $ is not onto.
Step 6: State the final conclusion.
Since $ f $ is neither one-one nor onto, the correct answer is option 4.
\[ \boxed{f \text{ is neither one-one nor onto}} \]