JEE Main 2024 — Functions Question with Solution

To find the domain of the function $$f(x)=\frac{\sqrt{x^2-25}}{(4-x^2)}+{\log }_{10}(x^2+2x-15),$$ we need to consider the domain conditions for both the square root function and the logarithmic function.

The square root function $\sqrt{x^2-25}$ requires that the argument of the square root be non-negative, so $$x^2-25\ge 0.$$ This inequality is satisfied when $$x\le -5\text{or}x\ge 5.$$

The denominator of the rational part of $f(x)$, $(4-x^2)$, cannot be zero, otherwise, the function will become undefined due to division by zero. Thus, we must have $$4-x^2\ne 0.$$ This inequality is violated when $$x=\pm 2.$$

Combining these conditions gives us the domain for the rational part of the function: $$x\in (-\infty ,-5]\cup (5,\infty )\text{and}x\ne 2,-2.$$

Moving on to the logarithmic function, ${\log }_{10}(x^2+2x-15)$, the argument must be positive: $$x^2+2x-15>0.$$ This is a quadratic inequality, which we can factor to find the solution: $$(x+5)(x-3)>0.$$ From this, we see that the inequality is satisfied for $$x<-5\text{or}x>3.$$

The overall domain of $f(x)$ is the intersection of the domains for each piece. Taking the intersection of the two sets gives us: $$x\in (-\infty ,-5)\cup (5,\infty ),$$

Since the question states that the domain is of the form $(-\infty ,\alpha )\cup [\beta ,\infty )$, we can infer that $$\alpha =-5\text{and}\beta =5.$$

We calculate ${\alpha }^2+{\beta }^3$ as follows: $${\alpha }^2+{\beta }^3=(-5{)}^2+5^3=25+125=150.$$

So the correct answer, representing the sum of ${\alpha }^2$ and ${\beta }^3$, is: Option D $150$.