JEE Mathematics 2024 Question with Solution

Given: $f(x) = \frac{x}{x^2 - 6x - 16}$ with domain $x \in \mathbb{R} \setminus \{-2, 8\}$.

Find: The intervals on which the function increases or decreases.

Use the quotient rule:

$$\frac{d}{dx}\left(\frac{u(x)}{v(x)}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2}$$

Let $u(x) = x$ and $v(x) = x^2 - 6x - 16$. Then $u'(x) = 1$ and $v'(x) = 2x - 6$.

So,

$$f'(x) = \frac{1 \cdot (x^2 - 6x - 16) - x(2x - 6)}{(x^2 - 6x - 16)^2}$$

Simplifying the numerator,

$$= \frac{x^2 - 6x - 16 - (2x^2 - 6x)}{(x^2 - 6x - 16)^2}$$ $$= \frac{x^2 - 6x - 16 - 2x^2 + 6x}{(x^2 - 6x - 16)^2}$$ $$= \frac{-x^2 - 16}{(x^2 - 6x - 16)^2}$$

Now, $-x^2 - 16 < 0$ for every real value of $x$, and $(x^2 - 6x - 16)^2 > 0$ wherever the function is defined.

Therefore, $f'(x) < 0$ on each interval of the domain. Hence, the function is decreasing in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$.

The correct option is B.

Given: $f(x) = \frac{x}{x^2 - 6x - 16}$.

Find: Its monotonicity intervals.

From the derivative obtained in the solution,

$$f'(x) = \frac{-(x^2 + 16)}{(x^2 - 6x - 16)^2}$$

Here, $x^2 + 16 > 0$ for all real $x$, so $-(x^2 + 16) < 0$ always. Also, the denominator is a square, so it is positive wherever defined.

Hence $f'(x)$ is negative throughout the domain except at the discontinuity points $x = -2$ and $x = 8$, where the function is not defined.

Therefore, the function decreases in $(-\infty, -2) \cup (-2, 8) \cup (8, \infty)$, so the correct option is B.