Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
KCET2024MathematicsApplication of Derivatives

If f(x)=x e^ x(1-x) , then f(x) is

Options

  1. Aincreasing in R
  2. Bdecreasing in R
  3. Cdecreasing in [- 1 2 , 1 ]
  4. Dincreasing in [- 1 2 , 1 ]

Correct answer

D. increasing in [- 1 2 , 1 ]

Step-by-step solution

Given, f(x)=x e^ x(1-x) f^ (x)=e^ x(1-x) +x e^ x(1-x) (1-2 x)=e^ x(1-x) (1+x(1-2 x))=-e^ x(1-x) (2 x^2-x-1 )=-e^ x(1-x) (2 x^2-2 x+x-1 )=-e^ x(1-x) (x-1)(2 x+1) So, f(x) is increasing when f^ (x) 0 and decreasing when f^ (x) 0 . Hence, f is increasing in [ -1 2 , 1 ] .

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 KCET PYQs