Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202314 May 2023Morning ShiftMathematicsApplication of DerivativesActual

The maximum value of the function f(x)=3 x^3-18 x^2+27 x-40 on the set S = x R / x^2+30 11 x is

Options

  1. A122
  2. B-122
  3. C-222
  4. D222

Correct answer

A. 122

Step-by-step solution

aligned S & = x R / x^2+30 11 x & = x R / x^2-11 x+30 0 & = x R /(x-5)(x-6) 0 & = x R / x [5,6] f (x) & =3 x^3-18 x^2+27 x-40 f ^ (x) & =9 x^2-36 x+27 & =9(x-1)(x-3)>0 x [5,6] aligned f (x) is increasing in [5,6] . aligned Maximum value & = f (6) & =3(6)^3-18(6)^2+27(6)-40 & =122 aligned

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 MHT CET PYQs