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B
18
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C
13
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D
−8
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Solution
The correct option is D8 Multiplynumeratoranddenominatorby(√x+2+√3x−2)wewillget⇒ltx→2(x2−4)(√x+2+√3x−2)(√x+2−√3x−2)×(√x+2+√3x−2)⇒ltx→2(x2−4)(√x+2+√3x−2)x+2−(3x−2)⇒ltx→2(x−2)(x+2)(√x+2+√3x−2)−2x+4⇒ltx→2(x+2)(√x+2+√3x−2)−2⇒(2+2)(√2+2+√6−2)−2=4×4−2=−8