Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202616 April 2026Morning ShiftMathematicsApplication of DerivativesActual

The function f(x) = ⁻¹( x + x) is an increasing function in the interval.....

Options

  1. A(0, 2 )
  2. B( - 2 , 2 )
  3. C( 4 , 2 )
  4. D( - 2 , 4 )

Correct answer

D. ( - 2 , 4 )

Step-by-step solution

Given f(x) = ⁻¹( x + x) Differentiating with respect to x , we get: f'(x) = 1 1 + ( x + x)^2 d dx ( x + x) f'(x) = x - x 1 + ( x + x)^2 For f(x) to be an increasing function, f'(x) > 0 . Since the denominator 1 + ( x + x)^2 is always positive for all real x , the sign of f'(x) depends entirely on the numerator. x - x > 0 x > x From the standard trigonometric graphs, the inequality x > x holds true in the interval ( - 2 , 4 ) . Therefore, the function is increasing in the interval ( - 2 , 4 ) . Answer: ( - 2 , 4 )

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