Question ID: #1175
If the domain of the function $f(x)=\log_{7}(1-\log_{4}(x^{2}-9x+18))$ is $(\alpha,\beta)\cup(\gamma,\delta)$, then $\alpha+\beta+\gamma+\delta$ is equal to
- (1) 18
- (2) 16
- (3) 15
- (4) 17
Solution:
$$f(x)=\log_{7}(1-\log_{4}(x^{2}-9x+18))$$
$$1 – \log_{4}(x^{2}-9x+18) > 0$$
$$x^{2}-9x+18 > 0$$
$$(x-3)(x-6) > 0$$
$$x \in (-\infty, 3) \cup (6, \infty) \quad \text{— (i)}$$
$$\log_{4}(x^{2}-9x+18) < 1$$
$$x^{2}-9x+18 < 4^1$$
$$x^{2}-9x+14 < 0$$
$$(x-2)(x-7) < 0$$
$$x \in (2, 7) \quad \text{— (ii)}$$
$$(\text{i}) \cap (\text{ii})$$
$$x \in (2, 3) \cup (6, 7)$$
$$(\alpha, \beta) \cup (\gamma, \delta) = (2, 3) \cup (6, 7)$$
$$\alpha = 2, \beta = 3, \gamma = 6, \delta = 7$$
$$\alpha + \beta + \gamma + \delta = 2 + 3 + 6 + 7 = 18$$
Ans. (1)
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