Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Manipal MET2012MathematicsApplication of Derivatives

Local maximum value of the function _e x x is

Options

  1. A1
  2. Be
  3. C1 e
  4. DNone of these

Correct answer

C. 1 e

Step-by-step solution

Let f(x)= _e x x On differentiating, w.r.t. x , we get f^ (x)= 1 x^2 - _e x x^2 For maximum or minimum value of f(x) , array rlr Put & f^ (x) & =0 & 1- _e x x^2 & =0 & _e x & =1 array x=e , which lies in (0, ) . For x=e, f^ (x)=-v e Hence, y is maximum at x=e and its maximum value = _e e e = 1 e

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