Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
AP EAMCET201920 Apr 2019Morning ShiftMathematicsApplication of DerivativesActual

Let (M ) and (m ) respectively denote the maximum and the minimum values of ([f( )]^2 ), where (f( )= a^2 ^2 +b^2 ^2 ) (+ a^2 ^2 +b^2 ^2 ). Then (M-m= )

Options

  1. A(a^2+b^2 )
  2. B((a-b)^2 )
  3. C(a^2 b^2 )
  4. D((a+b)^2 )

Correct answer

B. ((a-b)^2 )

Step-by-step solution

If ( aligned & f( )= a^2 ^2 +b^2 ^2 + a^2 ^2 +b^2 ^2 & [f( )]^2=a^2 ^2 +b^2 ^2 +a^2 ^2 +b^2 ^2 & +2 (a^2 ^2 +b^2 ^2 ) (a^2 ^2 +b^2 ^2 ) aligned ) ( [f( )]^2 ) will be maximum, if ( ^2 = ^2 = 1 2 ) and will be minimum, if either ( ^2 =0 ) or ( ^2 =0 ). ( aligned & M= (a^2+b^2 )+2 ( a^2 2 + b^2 2 )^2 =2 (a^2+b^2 ) & and m= (a^2+b^2 )+2 a^2 b^2 =a^2+b^2+2 a b & M-m=2 (a^2+b^2 )- [a^2+b^2+2 a b ] & =a^2+b^2-2 a b=(a-b)^2 aligned ) Hence, option (2) is correct.

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