Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Manipal MET2016MathematicsApplication of Derivatives

The points on the curve x y^2=1 which are nearest to the origin, are

Options

  1. A[ ( 1 2 )^ 1 / 3 , ( 1 2 )^ -1 / 6 ]
  2. B[ ( 1 2 )^ 1 / 3 , 2^ -1 / 6 ]
  3. C[2^ 1 / 3 , ( 1 2 )^ -1 / 6 ]
  4. DNone of these

Correct answer

A. [ ( 1 2 )^ 1 / 3 , ( 1 2 )^ -1 / 6 ]

Step-by-step solution

Let P(x, y) be such a point, then O P^2=x^2+y^2 Let s=O P^2=x^2+ 1 x d s d x =2 x- 1 x^2 For maximum or minimum, put d s d x =0 2 x- 1 x^2 =0 x^3= 1 2 x= ( 1 2 )^ 1 3 Also, d^2 s d x^2 =2+ 2 x^3 Now, ( d^2 s d x^2 )_ x= ( 1 2 )^ 1 / 3 =2+ 2 1 / 2 0 s is minimum at x= ( 1 2 )^ 1 / 3 . Thus, for nearest point x= ( 1 2 )^ 1 / 3 , y= ( 1 2 )^ -1 / 6

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the quadratic equation a x^2+b x+c=0 , where 2 a+3 b+6 c=0 and let g(x)= a x^3 3 + b x^2 2 +c x . Statement-I : The given quadratic equation ax ^2+ bx + c =0 has at least 2025The difference between the absolute maximum and absolute minimum values of the function f(x)=2 x^3-15 x^2+36 x-30 on [-1,4] is 2025If f(x)=x e^ x(1-x) , x R , then f(x) is 2025The angle between the curves y ^2= x and x ^2= y at the point (1,1) is 2025If the tangent of the curve 4 y^3=3 a x^2+x^3 drawn at the point (a, a) forms a triangle of area 25 24 sq.units with the coordinate axes then a = 2025If the function f(x)= x- ^2 x is defined on the interval [- , ] , then f is strictly increasing in the interval 2025If the Lagrange's mean value theorem is applied to the function f(x)=e^x defined on the interval [1,2] and the value of c (1,2) is k , then e^ k-1 = 2025If the tangent to the curve x y+a x+b y=0 at (1,1) makes an angle Tan ⁻¹ 2 with X -axis, then ab a + b = 2025 Full Application of Derivatives list All Manipal MET PYQs