Q.The domain of the function $f$ defined by $f(x) = \sqrt{4 - x} + \dfrac{1}{\sqrt{x^2 - 1}}$ is equal to
(A) $(-\infty,\ -1) \cup (1,\ 4]$
(B) $(-\infty,\ -1] \cup (1,\ 4]$
(C) $(-\infty,\ -1) \cup [1,\ 4]$
(D) $(-\infty,\ -1) \cup [1,\ 4)$

The key idea is to find the domain of each component of the function and then take their intersection. For a function to be defined, all its constituent parts must be defined simultaneously.

1. For the term $\sqrt{4 - x}$, the expression under the square root must be non-negative:

   $4 - x \ge 0 \implies x \le 4$.

   This gives the domain $(-\infty, 4]$.

2. For the term $\dfrac{1}{\sqrt{x^2 - 1}}$, two conditions must be met:

   • The expression under the square root must be non-negative: $x^2 - 1 \ge 0$.
   • The denominator cannot be zero: $\sqrt{x^2 - 1} \ne 0$. Combining these, we need $x^2 - 1 > 0$. $x^2 > 1 \implies |x| > 1$. This means $x < -1$ or $x > 1$. This gives the domain $(-\infty, -1) \cup (1, \infty)$. …

The domain of a function is the set of all possible input values ($x$) for which the function produces a real output. When a function involves square roots or fractions, we must impose conditions to ensure that these operations are mathematically valid in the real number system.

For a term like $\sqrt{A}$, the expression $A$ inside the square root must be non-negative, i.e., $A \ge 0$. If $A$ were negative, $\sqrt{A}$ would be an imaginary number, which is outside the scope of real-valued functions.

For a term like $\frac{1}{B}$, the denominator $B$ must not be zero, i.e., $B \ne 0$. Division by zero is undefined.

When we have a term like $\frac{1}{\sqrt{B}}$, both conditions apply: $B$ must be non-negative (for the square root) AND $B$ must be non-zero (for the denominator). Combining these, $B$ must be strictly positive, i.e., $B > 0$.

The given function $f(x)$ is a sum of two terms: $f(x) = \sqrt{4 - x} + \dfrac{1}{\sqrt{x^2 - 1}}$. For $f(x)$ to be defined, both terms must be defined simultaneously. This means we need to find the domain for each term separately and then take the intersection of these individual domains.

1. Determine the domain for the first term: $\sqrt{4 - x}$

   For this term to be defined in the real numbers, the expression under the square root must be non-negative.

   $4 - x \ge 0$

   Adding $x$ to both sides gives:

   $4 \ge x$

   This can also be written as $x \le 4$.

   In interval notation, the domain for the first term, let's call it $D_1$, is $(-\infty, 4]$.

2. Determine the domain for the second term: $\dfrac{1}{\sqrt{x^2 - 1}}$

   For this term, we have a square root in the denominator. This means two conditions must be met:

   • The expression under the square root must be non-negative: $x^2 - 1 \ge 0$.
   • The denominator cannot be zero: $\sqrt{x^2 - 1} \ne 0$, which implies $x^2 - 1 \ne 0$. Combining these, the expression under the square root must be strictly positive: $x^2 - 1 > 0$ We can factor the left side as a difference of squares: $(x - 1)(x + 1) > 0$ To solve this inequality, we find the critical points where the expression equals zero, which are $x = 1$ and $x = -1$. These points divide the number line into three intervals: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. We test a value from each interval:
   • For $x < -1$ (e.g., $x = -2$): $(-2 - 1)(-2 + 1) = (-3)(-1) = 3$. Since $3 > 0$, this interval is part of the domain.
   • For $-1 < x < 1$ (e.g., $x = 0$): $(0 - 1)(0 + 1) = (-1)(1) = -1$. Since $-1 \not> 0$, this interval is not part of the domain.
   • For $x > 1$ (e.g., $x = 2$): $(2 - 1)(2 + 1) = (1)(3) = 3$. Since $3 > 0$, this interval is part of the domain. So, the domain for the second term, $D_2$, is $(-\infty, -1) \cup (1, \infty)$. …