Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET20239 May 2023Evening ShiftMathematicsApplication of DerivativesActual

Let f^ (0)=-3 and f^ (x) 5 for all real values of x . The f (2) can have possible maximum value as

Options

  1. A10
  2. B5
  3. C7
  4. D13

Correct answer

C. 7

Step-by-step solution

Applying Lagrange's mean value theorem on interval [0,2] , we get there exist atleast one ' c ' (0,2) such that array ll & f (2)- f (0) 2-0 = f ^ ( c ) & f (2)- f (0)=2 f ^ ( c ) & f (2)= f (0)+2 f ^ ( c ) & f (2)=-3+2 f ^ ( c ) array Given that f ^ (x) 5 for all x array ll & f (2) -3+10 & f (2) 7 array Largest possible value of f (2) is 7 .

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