Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
COMEDK2013MathematicsApplication of Derivatives

If a tangent to the curve y=6 x-x² is parallel to the line 4 x-2 y-1=0 , then the point of tangency on the curve is

Options

  1. A(2,8)
  2. B(8,2)
  3. C(6,1)
  4. D(4,2)

Correct answer

A. (2,8)

Step-by-step solution

Let P (x₁, y₁ ) be the required point. The given curve is aligned y &=6 x-x² d i j d x &=6-2 x ( d y d x )_ (x₁, h ) &=6-2 x₁ aligned Since, the tangent at (x₁, y₁ ) is parallel to the line 4 x-2 y-1=0 . Slope of the tangent at (x₁, y₁ )= Slope of the line 4 x-2 y-1=0 ( d y d x )_ (x, b) =2 6-2 x₁=2 x₁=2 y₁=6 x₁-x₁²=6 2-2²=8 Thus, required point is (2,8) .

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 COMEDK PYQs