Step 1: Check one-one property.
A function is one-one if $f(a) = f(b) \implies a = b$.
Here, $f(x) = x^{2} + 5$.
Take $a=2, b=-2$:
\[
f(2) = 2^{2} + 5 = 9, f(-2) = (-2)^{2} + 5 = 9
\]
So, $f(2) = f(-2)$ but $2 \neq -2$.
Hence, the function is **not one-one (many-one)**.
Step 2: Check onto property.
The codomain is $\mathbb{R}$, but the range is $[5, \infty)$.
For example, there is no $x \in \mathbb{R}$ such that $f(x) = 3$.
Therefore, the function is **not onto**.
Step 3: Conclusion.
Since the function is neither one-one nor onto, the correct answer is (D).
Let $R$ be a relation defined on the set $\{1,2,3,4\times\{1,2,3,4\}$ by \[ R=\{((a,b),(c,d)) : 2a+3b=3c+4d\} \] Then the number of elements in $R$ is
Let \(M = \{1, 2, 3, ....., 16\}\), if a relation R defined on set M such that R = \((x, y) : 4y = 5x – 3, x, y (\in) M\). How many elements should be added to R to make it symmetric.