Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
COMEDK2021MathematicsApplication of Derivatives

Find the maximum value of f(x)= 1 4 x²+2 x+1 .

Options

  1. A3 4
  2. B4 3
  3. C1 3
  4. DNone of these

Correct answer

B. 4 3

Step-by-step solution

To find the maximum value of f(x) = 1 4x² + 2x + 1 , we need to find the minimum value of the denominator g(x) = 4x² + 2x + 1 . The expression g(x) = 4x² + 2x + 1 is a quadratic in the form ax² + bx + c where a = 4 , b = 2 , and c = 1 . Since a > 0 , the quadratic has a minimum value at x = - b 2a . x = - 2 2 4 = - 2 8 = - 1 4 . Substituting x = - 1 4 into g(x) to find the minimum value: g (- 1 4 ) = 4 (- 1 4 )² + 2 (- 1 4 ) + 1 = 4 ( 1 16 ) - 1 2 + 1 = 1 4 - 1 2 + 1 = 1 - 2 + 4 4 = 3 4 . Since the minimum value of

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 COMEDK PYQs