JEE Mathematics 2026 Question with Solution

Given:

$$f(x)=\sin^{-1}\!\left(\frac{5-x}{3+2x}\right)+\frac{1}{\ln(10-x)}$$

Find: The value of $6(\alpha+\beta+\gamma+\delta)$ when the domain is written as $(-\infty,\alpha]\cup[\beta,\gamma)-\{\delta\}$.

For $\sin^{-1}$ to be defined,

$$-1 \le \frac{5-x}{3+2x} \le 1$$

Solving the two inequalities shown in the solution:

$$\frac{5-x}{3+2x} \le 1 \Rightarrow x \ge \frac{2}{3}$$

and

$$\frac{5-x}{3+2x} \ge -1 \Rightarrow x \le -8$$

Hence,

$$x \in (-\infty,-8] \cup \left[\frac{2}{3},\infty\right)$$

For $\frac{1}{\ln(10-x)}$, the conditions are:

$$10-x>0 \Rightarrow x<10$$

and

$$\ln(10-x)\ne 0 \Rightarrow 10-x\ne 1 \Rightarrow x\ne 9$$

So,

$$x<10, \quad x\ne 9$$

Taking the intersection,

$$(-\infty,-8] \cup \left[\frac{2}{3},10\right) - \{9\}$$

Comparing with $(-\infty,\alpha] \cup [\beta,\gamma) - \{\delta\}$, we get

$$\alpha=-8, \quad \beta=\frac{2}{3}, \quad \gamma=10, \quad \delta=9$$

Now,

$$6(\alpha+\beta+\gamma+\delta)=6\left(-8+\frac{2}{3}+10+9\right)$$ $$=6\left(\frac{35}{3}\right)=70$$

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

Given: The function is a sum of two terms. Find: The common domain of both terms.

1. First find where

$$\sin^{-1}\!\left(\frac{5-x}{3+2x}\right)$$

is defined by ensuring the fraction lies in $[-1,1]$.

2. Then find where

$$\frac{1}{\ln(10-x)}$$

is defined by requiring the logarithm argument to be positive and the denominator to be non-zero.

3. Intersect the resulting sets exactly as shown:

$$(-\infty,-8] \cup \left[\frac{2}{3},\infty\right) \, \cap \, \left(( -\infty,10) - \{9\}\right)$$

which gives

$$(-\infty,-8] \cup \left[\frac{2}{3},10\right) - \{9\}$$

So the comparison values are

$$\alpha=-8, \quad \beta=\frac{2}{3}, \quad \gamma=10, \quad \delta=9$$

and hence the required value is $70$.](streamdown:incomplete-link)