JEE Mathematics 2024 Question with Solution

Given: $f(x) = \cos^{-1}\left(\frac{2 - |x|}{4}\right) + \left(\log_e(3 - x)\right)^{-1}$

Find: $\alpha + \beta + \gamma$ when the domain is of the form $[-\alpha, \beta) - \{\gamma\}$.

For $\cos^{-1}\left(\frac{2 - |x|}{4}\right)$ to be defined, its argument must satisfy

$$-1 \leq \frac{2 - |x|}{4} \leq 1$$

This gives

$$-4 \leq 2 - |x| \leq 4$$

So,

$$-6 \leq -|x| \leq 2$$

which leads to

$$|x| \leq 6$$

Hence,

$$x \in [-6, 6]$$

Now consider $\left(\log_e(3 - x)\right)^{-1}$. For this to be defined:

1. $\log_e(3 - x)$ must exist, so

$$3 - x > 0$$

that is,

$$x < 3$$

2. Since it is in the denominator, $\log_e(3 - x) \neq 0$. Therefore,

$$\log_e(3 - x) \neq 0 \Rightarrow 3 - x \neq 1 \Rightarrow x \neq 2$$

Combining all conditions,

$$x \in [-6, 6] \cap (-\infty, 3) \setminus \{2\} = [-6, 3) \setminus \{2\}$$

Thus,

$$\alpha = 6, \quad \beta = 3, \quad \gamma = 2$$

Therefore,

$$\alpha + \beta + \gamma = 6 + 3 + 2 = 11$$

So, the correct option is C.](streamdown:incomplete-link)

Analyze the two parts of the function separately.

For the inverse cosine part,

$$-1 \leq \frac{2 - |x|}{4} \leq 1$$

From the right inequality,

$$\frac{2 - |x|}{4} \leq 1 \Rightarrow 2 - |x| \leq 4 \Rightarrow -|x| \leq 2$$

which is always true.

From the left inequality,

$$-1 \leq \frac{2 - |x|}{4} \Rightarrow -4 \leq 2 - |x| \Rightarrow -6 \leq -|x| \Rightarrow |x| \leq 6$$

Hence,

$$x \in [-6, 6]$$

For the logarithmic reciprocal part,

$$\left(\log_e(3 - x)\right)^{-1}$$

we need both:

$$3 - x > 0 \Rightarrow x < 3$$

and

$$\log_e(3 - x) \neq 0$$

Now,

$$\log_e(3 - x) = 0 \Rightarrow 3 - x = 1 \Rightarrow x = 2$$

So,

$$x \neq 2$$

Therefore the overall domain is

$$[-6, 3) - \{2\}$$

Comparing with $[-\alpha, \beta) - \{\gamma\}$, we get

$$\alpha = 6, \quad \beta = 3, \quad \gamma = 2$$

Hence,

$$\alpha + \beta + \gamma = 11$$

Therefore, the correct option is C.](streamdown:incomplete-link)