The shortest distance between the point ( 3 2 , 0 ) and the curve y= x ,(x>0) , is-
Mathematics · JEE Main · NTA Exams — Limit, Continuity and Differentiability
The shortest distance between the point \(\left(\frac{3}{2}, 0\right)\) and the curve \(y=\sqrt{x},(x>0)\), is-
- \(\frac{\sqrt{3}}{2}\)
- \(\frac{5}{4}\)
- \(\frac{3}{2}\)
- \(\frac{\sqrt{5}}{2}\)
Answer
(D) 5 2
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