The number of integral values of x satisfying √−x2+10x−16<x−2 is
A
3
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B
2
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C
4
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D
5
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Solution
The correct option is A 3 Thequantityinsquarerootmustbenon−negative−x2+10x−16≥0or,x2−10x+16≤0or,(x−2)(x−8)≤0 ⇒2≤x≤8......[1]Alsoonsquaringthegivenequation,weget−x2+10x−16<x2−4x+4⇒2x2−14x+20>0⇒x2−7x+10>0⇒x>5orx<2.......[2]From[1]and[2]weget,5<x≤8⇒x=6,7,8