If domain of \(f(x) = \sin^{-1}\left(\frac{5-x}{2x+3}\right) + \frac{1}{\log_{e}(10-x)}\) is \((-\infty, \alpha] \cup (\beta, \gamma) - \{\delta\}\) then value of \(6(\alpha + \beta + \gamma + \delta)\) is equal to :
Show Hint
- 60
- 70
- 80
- 90
The Correct Option is B
Solution and Explanation
Step 1: Understanding the Question:
We need to find the domain of the function \(f(x)\), which is the intersection of the domains of its two component parts. Then, we compare the resulting domain with the given format to find the values of \(\alpha, \beta, \gamma, \delta\). There seems to be a typo in the provided format, we assume it should be \((-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}\) based on standard results.
Step 2: Domain of the Inverse Sine Function:
For \(g(x) = \sin^{-1}\left(\frac{5-x}{2x+3}\right)\), the argument must be in the interval \([-1, 1]\). Also, the denominator cannot be zero, so \(2x+3 \neq 0 \Rightarrow x \neq -3/2\).
\[ -1 \leq \frac{5-x}{2x+3} \leq 1 \] This gives two inequalities:
Inequality 1: \(\frac{5-x}{2x+3} \leq 1 \Rightarrow \frac{5-x}{2x+3} - 1 \leq 0 \Rightarrow \frac{5-x-(2x+3)}{2x+3} \leq 0 \Rightarrow \frac{2-3x}{2x+3} \leq 0 \Rightarrow \frac{3x-2}{2x+3} \geq 0\).
This holds for \(x \in (-\infty, -3/2) \cup [2/3, \infty)\).
Inequality 2: \(\frac{5-x}{2x+3} \geq -1 \Rightarrow \frac{5-x}{2x+3} + 1 \geq 0 \Rightarrow \frac{5-x+2x+3}{2x+3} \geq 0 \Rightarrow \frac{x+8}{2x+3} \geq 0\).
This holds for \(x \in (-\infty, -8] \cup (-3/2, \infty)\).
The domain of the sine inverse part is the intersection of these two sets: \(x \in (-\infty, -8] \cup [2/3, \infty)\).
Step 3: Domain of the Logarithmic Function:
For \(h(x) = \frac{1}{\log_{e}(10-x)}\), we have two conditions:
The argument of the logarithm must be positive: \(10-x>0 \Rightarrow x<10\).
The denominator cannot be zero: \(\log_{e}(10-x) \neq 0 \Rightarrow 10-x \neq 1 \Rightarrow x \neq 9\). So, the domain of the logarithmic part is \(x \in (-\infty, 10)\) and \(x \neq 9\).
Step 4: Finding the Overall Domain and the Final Answer:
The domain of \(f(x)\) is the intersection of the domains from Step 2 and Step 3. We need to find the intersection of \((-\infty, -8] \cup [2/3, \infty)\) and \((-\infty, 10) - \{9\}\). \[ ( (-\infty, -8] \cup [2/3, \infty) ) \cap (-\infty, 10) - \{9\} \] \[ = ((-\infty, -8] \cap (-\infty, 10)) \cup ([2/3, \infty) \cap (-\infty, 10)) - \{9\} \] \[ = (-\infty, -8] \cup [2/3, 10) - \{9\} \] Comparing this with the format \((-\infty, \alpha] \cup [\beta, \gamma) - \{\delta\}\) (assuming the typo correction), we get: \(\alpha = -8\), \(\beta = 2/3\), \(\gamma = 10\), \(\delta = 9\).
The required value is \(6(\alpha + \beta + \gamma + \delta)\): \[ 6\left(-8 + \frac{2}{3} + 10 + 9\right) = 6\left(11 + \frac{2}{3}\right) = 6\left(\frac{33+2}{3}\right) = 6\left(\frac{35}{3}\right) = 2 \times 35 = 70 \]
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