Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET201921 Apr 2019Evening ShiftMathematicsApplication of DerivativesActual

If p and q are respectively the global maximum and global minimum of the function f ( x ) = x 2 e 2 x on the interval [ - 2 , 2 ] , then p e - 4 + q e 4 =

Options

  1. A0
  2. B4 e 8
  3. C4
  4. D4 e 8 + 1

Correct answer

C. 4

Step-by-step solution

It is given that, f ( x ) = x 2 e 2 x Differentiating the above equation we get, f ' ( x ) = 2 e 2 x x 2 + 2 x e 2 x ⇒ 0 = 2 e 2 x x 2 + x ⇒ x 2 + x = 0 ⇒ x = 0 ,   - 1 The maxima and minima is obtained at - 2 ,   - 1 ,   0 ,   2 as the function bound. Thus, f ( - 2 ) = ( - 2 ) 2 e - 4 = 4 e - 4 f ( - 1 ) = ( 1 ) 2 e - 2 f ( 0 ) = 0 f ( 2 ) = ( 2 ) 2 e 4 = 4 e 4 Thus, p = 4 e 4 and q = 0 . p e - 4 + q e 4 = 4 e 4 e - 4 + 0 = 4 e 0 = 4

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 AP EAMCET PYQs