Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
NTA Abhyas JEE Main2020MathematicsApplication of DerivativesPractice

The slope of the tangent to the curve y = ∫ x x 2 c o s - 1 t 2 d t at x = 1 2 is equal to

Options

  1. Ac o s - 1 1 4 - π 3
  2. Bc o s - 1 1 4 + π 3
  3. C2 c o s - 1 1 4 - π 3
  4. D2 c o s - 1 1 4 + π 3

Correct answer

C. 2 c o s - 1 1 4 - π 3

Step-by-step solution

d y d x = c o s - 1 x 4 . 2 x - c o s - 1 x 2 . 1 dy dx x = 1 2 = cos - 1 1 4 . 2 - cos - 1 1 2 = 2 c o s - 1 1 4 - π 3

Practice Application of Derivatives on Quantrex Academy →

More from Application of Derivatives

Consider the function f: (0, ) (- , ) given by f(x) = x , _e(x) - x + 1 . Then which one of the following statements is TRUE? 2026Let P be the point on the parabola y = x^2 such that the slope of the tangent to the parabola at the point P is 4 . Let Q be the point in the first quadrant lying on the circle x^2 2026Let f: R R be a differentiable function such that f ( x+y 3 ) = f(x) + f(y) 3 for all x, y R , and f'(0) = 3 . Then the minimum value of the function g(x) = 3 + e^x f(x) , is: 2026_ 0 x (16 ( x 2 ) ^3 ( x 2 ) ) is equal to: 2026Let f(x) be a polynomial of degree 5 , and have extrema at x = 1 and x = -1 . If _ x 0 ( f(x) x^3 ) = -5 , then f(2) - f(-2) is equal to: 2026The number of critical points of the function f(x) = cases | x x |, & x 0 1, & x = 0 cases in the interval (-2 , 2 ) is equal to : 2026Let f be a differentiable function satisfying f(x)=1-2 x+ ₀^ x e ^ (x-t) f(t) dt , x R and let g (x)= ₀^ x (f( t )+2)¹⁵( t -4)⁶( t +12)¹⁷ dt , x R . If p and q are respectively the 2026Consider the following three statements for the function f:(0, ) R defined by f(x)= | _ e x |-|x-1| : (I) f is differentiable at all x>0 . (II) f is increasing in (0,1) . (III) f i 2026 Full Application of Derivatives list All NTA Abhyas JEE Main PYQs