Question:

The domain of the real valued function \[ f(x)=\frac{\log_2(x+3)}{\sqrt{x^2+3x+2}} \] is

Show Hint

For functions involving logarithms and square roots in denominator: \[ \text{(i) logarithm argument } \gt 0 \] and \[ \text{(ii) quantity under square root in denominator } \gt 0 \] must both hold simultaneously.
Updated On: Jun 26, 2026
  • \((-3,\infty)\)
  • \((-3,-1)\cup(-1,\infty)\)
  • \((-3,-2)\cup(-2,-1)\cup(-1,\infty)\)
  • \((-3,-2)\cup(-1,\infty)\)
Show Solution
collegedunia
Verified By Collegedunia

The Correct Option is D

Solution and Explanation

Step 1: Condition for logarithm to exist.
For \[ \log_2(x+3) \] to be defined, \[ x+3\gt 0 \] Thus, \[ x\gt -3 \]

Step 2: Condition for square root in denominator.
Since the square root is in the denominator, \[ x^2+3x+2\gt 0 \] Factorizing: \[ x^2+3x+2=(x+1)(x+2) \] Therefore, \[ (x+1)(x+2)\gt 0 \] This inequality is satisfied when: \[ x\lt -2 \] or \[ x\gt -1 \] Thus, \[ x\in(-\infty,-2)\cup(-1,\infty) \]

Step 3: Find the common interval.
From logarithm condition: \[ x\gt -3 \] From denominator condition: \[ x\in(-\infty,-2)\cup(-1,\infty) \] Taking intersection: \[ (-3,\infty)\cap\left[(-\infty,-2)\cup(-1,\infty)\right] \] we get: \[ (-3,-2)\cup(-1,\infty) \]

Step 4: Final conclusion.
Hence, the domain of the function is \[ \boxed{(-3,-2)\cup(-1,\infty)} \]
Was this answer helpful?
0
0