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

If the function (f:[-1,1] R ) defined by (f(x)= array cc 2^x+1, & for x [-1,0) 1, & for x=0 2^x-1, & for x (0,1] array . ) then, in ([-1,1], f(x) ) has

Options

  1. Aa maximum
  2. Ba minimum
  3. Cboth maximum and minimum
  4. Dneither maximum nor minimum

Correct answer

D. neither maximum nor minimum

Step-by-step solution

The given function (f:[-1,1] R ) defined by (f(x)= [ array cc 2^x+1, & for x [-1,0) 1, & for x=0 2^x-1, & for x (0,1] array . ) The given function (f(x) ) in (x (-1,0) ) strictly increasing and (f (0⁻ ) 2 ) but (f(0)=1 ) and in interval (x (0,1) ), again it is strictly increasing, but (f (0⁺ )=0 ) So, function has neither maximum nor minimum. Hence, option (4) 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