Question:

Step 1: To determine if the function $f:\mathbb{R}\to\mathbb{R}$, $f(x)=\dfrac{x^2+2x-15}{x^2-4x+9}$ is one-one or onto, we first need to understand what it means for a function to be one-one and onto. A function is one-one if it assigns distinct outputs to distinct inputs, and it is onto if every possible output in its codomain is assigned to at least one input. Step 2: Let's start by checking if the function is one-one. For $f(x)$ to be one-one, we must have $f(x_1) = f(x_2)$ implying $x_1 = x_2$ for all $x_1, x_2$ in the domain of $f$. We can begin by setting $f(x_1) = f(x_2)$ and see if it leads to $x_1 = x_2$. So, we have $\dfrac{x_1^2+2x_1-15}{x_1^2-4x_1+9} = \dfrac{x_2^2+2x_2-15}{x_2^2-4x_2+9}$. Step 3: To simplify the equation from Step 2 and understand the relationship between $x_1$ and $x_2$, let's cross multiply to get rid of the fractions: $(x_1^2+2x_1-15)(x_2^2-4x_2+9) = (x_2^2+2x_2-15)(x_1^2-4x_1+9)$. Expanding both sides gives us a complicated equation, but the key insight is to determine if this equation forces $x_1 = x_2$ or if there are cases where $x_1 \neq x_2$ but the equation still holds. Step 4: Instead of expanding the equation fully, which can be very complex, let's consider the nature of the function. The function $f(x)$ is a rational function, and its behavior is determined by the numerator and denominator. For $f(x)$ to be onto, it must be able to achieve every real value. However, rational functions typically have restrictions on their range due to the nature of their numerator and denominator. Step 5: To check if $f(x)$ is onto, we need to see if for every real number $y$, there exists a real number $x$ such that $f(x) = y$. This means solving the equation $\dfrac{x^2+2x-15}{x^2-4x+9} = y$ for $x$. Rearranging gives us $x^2+2x-15 = y(x^2-4x+9)$, which simplifies to $x^2+2x-15 = yx^2 - 4yx + 9y$. Bringing all terms to one side to set the equation to 0 gives us $x^2(1-y) + x(2+4y) + (-15-9y) = 0$. For $f(x)$ to be onto, this equation must have a real solution for $x$ for every real $y$. Step 6: For the equation $x^2(1-y) + x(2+4y) + (-15-9y) = 0$ to have real solutions, its discriminant must be non-negative. The discriminant is given by $(2+4y)^2 - 4(1-y)(-15-9y) \geq 0$. Simplifying this inequality will tell us if there are any restrictions on $y$, which would indicate if $f(x)$ is onto. Step 7: Let's simplify the inequality from Step 6: $(2+4y)^2 - 4(1-y)(-15-9y) \geq 0$. Expanding gives $4 + 16y + 16y^2 + 4(15 + 9y + 15y + 9y^2) \geq 0$, which simplifies to $16y^2 + 16y + 4 + 60 + 36y + 60y + 36y^2 \geq 0$. Combining like terms yields $52y^2 + 112y + 64 \geq 0$. This is a quadratic inequality, and for the function to be onto, this inequality must be true for all real $y$. Step 8: To determine if $52y^2 + 112y + 64 \geq 0$ is true for all real $y$, we can examine the discriminant of the corresponding quadratic equation $52y^2 + 112y + 64 = 0$. The discriminant is $b^2 - 4ac = 112^2 - 4*52*64$. If the discriminant is negative, the quadratic equation has no real roots, meaning the inequality $52y^2 + 112y + 64 \geq 0$ holds for all real $y$, indicating $f(x)$ could be onto. However, if the discriminant is non-negative, the inequality may not hold for all $y$, suggesting $f(x)$ is not onto. Step 9: Calculating the discriminant from Step 8: $112^2 - 4*52*64 = 12544 - 13312 = -768$. Since the discriminant is negative, the quadratic $52y^2 + 112y + 64 = 0$ has no real roots, meaning $52y^2 + 112y + 64 > 0$ for all real $y$. However, this step alone does not conclusively determine the onto nature without considering the original function's behavior and possible restrictions on its range. Step 10: Given the complexity of directly proving onto or one-one nature through algebraic manipulations, let's reconsider the function $f(x) = \dfrac{x^2+2x-15}{x^2-4x+9}$. Notice that the denominator $x^2 - 4x + 9 = (x-2)^2 + 5$ is always positive, meaning it never equals zero, and thus $f(x)$ is defined for all real $x$. However, to be onto, $f(x)$ must be able to achieve every real value, which may not be possible given the structure of the function. Step 11: To conclude whether $f(x)$ is one-one or onto, let's analyze its behavior as $x$ approaches infinity and minus infinity. As $x$ becomes very large (positive or negative), the leading terms in the numerator and denominator dominate, suggesting that $f(x)$ approaches 1 as $x$ approaches infinity or minus infinity. This behavior indicates that $f(x)$ may not be able to achieve all real values, particularly values less than a certain minimum or greater than a certain maximum, hinting that it might not be onto. Step 12: Considering the function's behavior and the results from previous steps, we need to determine if $f(x)$ is one-one. If $f(x)$ is not one-one, there must exist distinct $x_1$ and $x_2$ such that $f(x_1) = f(x_2)$. Given the function's form and the previous analysis, it seems plausible that such $x_1$ and $x_2$ could exist, indicating $f(x)$ might not be one-one. Step 13: Given the analysis above, to conclusively determine if $f(x)$ is one-one or onto, let's examine if there are specific values of $x$ that could lead to the same $f(x)$ value, and if there are real values that $f(x)$ cannot achieve. The function $f(x) = \dfrac{x^2+2x-15}{x^2-4x+9}$ can be simplified or analyzed further to check for these conditions. Step 14: The final step is to conclude based on the analysis. Given that $f(x)$ has a complex structure that may prevent it from achieving all real values (onto) and considering the possibility of distinct $x$ values mapping to the same $f(x)$ value (not one-one), we need to match our findings with the provided options. The function's behavior and the mathematical analysis suggest that $f(x)$ is neither one-one nor onto, which corresponds to a specific option. The final answer is: $\boxed{4}$