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प्रश्न
The value of `1/"x"^2+1/"y"^2`, where x = `2 +sqrt3` and y `=2-sqrt3`, is
पर्याय
12
16
14
10
MCQ
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उत्तर
14
Explanation:
x = `2+sqrt3` , y =`2-sqrt3`
`1/"x"=1/(2+sqrt3)` `1/"y"=1/(2-sqrt3)`
By Rationalizing
`1/"x"=1/((2+sqrt3))((2-sqrt3))/((2-sqrt3))1/"y"=((2+sqrt3))/((2-sqrt3)(2+sqrt3))`
`1/"x"=2-sqrt3` `1/"y"=2sqrt3`
`1/"x"^2=(2-sqrt3)^2`
`1/"y"^2=(2+sqrt3)^2`1/"x"^2+1/"y"^2=2^2+(sqrt3)^2-4sqrt3+2^2+(sqrt3)^2+4sqrt3`
`1/"x"^2+1/"y"^2=4+4+3+3+=14`
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Number System (Entrance Exam)
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