Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
COMEDK2021MathematicsApplication of Derivatives

The equation of normal to the curve y=(1+x)^ y + ⁻¹ ( ² x ) at x=0 is

Options

  1. Ax+y=1
  2. Bx-y=1
  3. Cx+y=-1
  4. Dx-y=-1

Correct answer

A. x+y=1

Step-by-step solution

Given the curve y = (1+x)^ y + ⁻¹( ²x) . At x = 0 , y = (1+0)^ y + ⁻¹( ²0) = 1^ y + 0 = 1 . Thus, the point of contact is (0, 1) . To find the slope of the tangent, differentiate the equation with respect to x : Let u = (1+x)^ y . Then u = y (1+x) . Differentiating with respect to x : 1 u du dx = dy dx (1+x) + y 1+x . So, du dx = (1+x)^ y [ dy dx (1+x) + y 1+x ] . At (0, 1) , du dx = 1¹ [ dy dx (1) + 1 1 ] = 1 . Now differentiate ⁻¹( ²x) with respect to x : d dx [ ⁻¹( ²x)] = 1 1 - ( ²x)² 2 x x = 2x 1 - ⁴x . At x =

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