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Question : 10 of 160
Marks:
+1,
-0
Solution:
A={x∈R∕√x2−8x+15∈R} ⇒‌‌x2−8x+15≥0 ⇒‌‌(x−5)(x−3)≥0 ⇒‌‌x∈(−∞,3]∪[5,∞] ∴‌‌A=(−∞,3]∪[5,∞] B={x∈R∕‌<‌} ⇒‌−‌<0 ‌| (x−3)(2x−11) |
| (2x−5)(2x−11) |
−‌| (x−6)(2x−5) |
| (2x−11)(2x−5) |
<0‌| (x−3)(2x−11)−(x−6)(2x−5) |
| (2x−5)(2x−11) |
<0 | (x⋅2x−x⋅11−3⋅2x+3⋅11)−(x⋅2x−x⋅5−6⋅2x+6⋅5) |
| (2x−5)(2x−11) |
<0| (2xx−11x−6x+33)−(2xx−5x−12x+30) |
| (2x−5)(2x−11) |
<0 ⇒‌‌‌<0 ⇒‌‌‌>0
⇒‌‌x∈(‌,‌)∪(‌,∞) ∴‌‌B=(‌,‌)∪(‌,∞) A∩B=(‌,3]∪[5,‌)
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