If the domain of the function is , then the value of a + b + c + d + e is ________. [2026]

Question

If the domain of the function $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$ is $(- \infty, a] \cup {b} \cup [c, d) \cup (e, \infty)$ , then the value of a + b + c + d + e is ________. [2026]

Smart Achievers · Accepted Answer

(4)

We have, $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$

$\log_{0.6} | \frac{2 x - 5}{x^{2} - 4} | \geq 0$

$\Rightarrow | \frac{2 x - 5}{x^{2} - 4} | \leq 1$

$\Rightarrow | 2 x - 5 | \leq | x^{2} - 4 |; x \neq \frac{5}{2}, 2, - 2$

Also, $| 2 x - 5 | \leq | x^{2} - 4 |$

$\Rightarrow {(2 x - 5)}^{2} \leq {(x^{2} - 4)}^{2}$

$\Rightarrow (x^{2} - 4 + 2 x - 5) (x^{2} - 4 - 2 x + 5) \geq 0$

$\Rightarrow (x^{2} + 2 x - 9) (x^{2} - 2 x + 1) \geq 0$

$\Rightarrow (x^{2} + 2 x - 9) {(x - 1)}^{2} \geq 0$

Since, ${(x - 1)}^{2} \geq 0$

$\Rightarrow x^{2} + 2 x - 9 \geq 0$

$\Rightarrow x \in (- \infty, - 1 - \sqrt{10}] \cup [1} \cup (- 1 + \sqrt{10}, \infty)$

exclude x = 2, $\frac{5}{2}$ ( $∵$ –2 is already excluded)

Hence, $x \in (- \infty, - 1 - \sqrt{10}] \cup [1} \cup (- 1 + \sqrt{10}, 5 / 2) \cup (5 / 2, \infty)$

$x \in (- \infty, a] \cup [b} \cup [c, d) \cup (e, \infty)$ (Given)

$∴ a = - 1 - \sqrt{10}, b = 1, c = - 1 + \sqrt{10}, d = 5 / 2, e = 5 / 2$

$∴$ a + b + c + d + c = 4.

Solution

Q.
If the domain of the function $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$ is $(- \infty, a] \cup {b} \cup [c, d) \cup (e, \infty)$ , then the value of a + b + c + d + e is ________. [2026]

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Solution

Q. If the domain of the function f(x)=log(0.6)(|2x–5x2–4|) is (–∞,a]∪{b}∪[c,d)∪(e,∞), then the value of a + b + c + d + e is ________. [2026]

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Q.
If the domain of the function $f (x) = \sqrt{\log_{(0.6)} (| \frac{2 x - 5}{x^{2} - 4} |)}$ is $(- \infty, a] \cup {b} \cup [c, d) \cup (e, \infty)$ , then the value of a + b + c + d + e is ________. [2026]