JEE Mathematics 2024 Question with Solution

Given: $f(x) = \frac{x^2 + 2x - 15}{x^2 - 4x + 9}$

Find: Whether the function is one-one and/or onto for $f : \mathbb{R} \to \mathbb{R}$.

Let $g(x) = x^2 - 4x + 9$.

The discriminant of $g(x)$ is

$$D = (-4)^2 - 4(1)(9) = 16 - 36 = -20$$

Since $D < 0$, we have $g(x) > 0 \ \forall x \in \mathbb{R}$. So the denominator is always positive and the function is defined for all real $x$.

Now,

$$f(x) = \frac{(x + 5)(x - 3)}{x^2 - 4x + 9}$$

Evaluate at two different points:

$$f(-5) = 0, \quad f(3) = 0$$

Since the function takes the same value at two distinct points $-5$ and $3$, the function is not one-one.

To find whether it is onto, let

$$y = \frac{x^2 + 2x - 15}{x^2 - 4x + 9}$$

Then

$$y(x^2 - 4x + 9) = x^2 + 2x - 15$$

Rearranging,

$$x^2(y - 1) - 2x(2y + 1) + (9y + 15) = 0$$

For real $x$ to exist, the discriminant must satisfy

$$D = 4(2y + 1)^2 - 4(y - 1)(9y + 15) \geq 0$$

So,

$$D = 4 \left[(2y + 1)^2 - (y - 1)(9y + 15)\right]$$

Expanding,

$$D = 4 \left[4y^2 + 4y + 1 - (9y^2 + 6y - 15)\right]$$ $$D = 4 \left[-5y^2 - 2y + 16\right]$$

Factorizing,

$$D = 4(-5y + 8)(y + 2)$$

For $D \geq 0$,

$$y \in \left[-2, \frac{8}{5}\right]$$

Thus the range of $f(x)$ is $\left[-2, \frac{8}{5}\right]$, which is not all of $\mathbb{R}$. Therefore, the function is not onto.

Hence, $f(x)$ is neither one-one nor onto.

The correct option is C.

Given: $f(x) = \frac{x^2 + 2x - 15}{x^2 - 4x + 9}$

Find: Whether the function is injective and surjective.

First check injectivity using direct values. Since

$$x^2 + 2x - 15 = (x + 5)(x - 3)$$

we get

$$f(-5) = 0, \quad f(3) = 0$$

with $-5 \neq 3$. Therefore, equal outputs occur for different inputs, so the function is many-one.

Now check surjectivity by solving for $x$ in terms of $y$:

$$y(x^2 - 4x + 9) = x^2 + 2x - 15$$ $$(y - 1)x^2 - 2(2y + 1)x + (9y + 15) = 0$$

This is a quadratic equation in $x$. A real value of $x$ exists only when its discriminant is non-negative:

$$\Delta = [-2(2y+1)]^2 - 4(y-1)(9y+15)$$ $$\Delta = 4(2y+1)^2 - 4(y-1)(9y+15)$$ $$\Delta = 4\big[(2y+1)^2 - (y-1)(9y+15)\big]$$ $$\Delta = 4\big[4y^2 + 4y + 1 - (9y^2 + 6y - 15)\big]$$ $$\Delta = 4(-5y^2 - 2y + 16)$$ $$\Delta = 4(-5y + 8)(y + 2)$$

Hence,

$$(-5y + 8)(y + 2) \geq 0$$

which gives

$$y \in \left[-2, \frac{8}{5}\right]$$

So the range is not all real numbers. Therefore the function is not onto.

Thus the function is neither one-one nor onto, so the correct option is C.