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: The value of $\alpha + \beta + \gamma$ if the domain is $[-\alpha, \beta) - \{\gamma\}$.

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

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

From the left inequality,

$$-4 \leq 2 - |x| \Rightarrow |x| \leq 6$$

so

$$-6 \leq x \leq 6$$

For $\left(\log_e(3-x)\right)^{-1}$ to be defined, we need

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

and also

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

which implies

$$3-x \neq 1 \Rightarrow x \neq 2$$

Combining all conditions,

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

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

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

Therefore,

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

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

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

Find: The value of $\alpha + \beta + \gamma$.

First analyze the inverse cosine part:

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

From

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

we get

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

which is always true.

From

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

we get

$$-4 \leq 2 - |x| \Rightarrow -6 \leq -|x| \Rightarrow |x| \leq 6$$

Hence,

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

Now analyze $\left(\log_e(3-x)\right)^{-1}$. The logarithm must exist, so

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

Also, because it is in the denominator, it cannot be zero:

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

This happens when

$$3-x \neq 1 \Rightarrow x \neq 2$$

Thus the final domain is

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

So,

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

and therefore

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

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