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JEE Main201911 Jan 2019Evening ShiftMathematicsApplication of DerivativesActual

Let f(x)= x a²+x² - d-x b²+(d-x)² , x R wherea, b and d are non-zero real constants. Then :

Options

  1. Af is an increasing function of x
  2. Bf is a decreasing function of x
  3. Cf^ is not a continuous function of x
  4. Df is neither increasing nor decreasing function of x

Correct answer

A. f is an increasing function of x

Step-by-step solution

f(x)= x a²+x² - (d-x) b²+(d-x)² = x a²+x² + (x-d) b²+(x-d)² array l f^ (x)= a²+x² - x(2 x) 2 a²+x² (a²+x² ) = a²+x²-x² (a²+x² )^ 3 / 2 + b²+(x-d)²-(x-d)² (b²+(x-d)² )^ 3 / 2 + b²+(x-d)² - (x-d) 2(x-d) 2 b²+(x-d)² (a²+(x-d)² ) f^ (x)>0, x R array f(x) is increasing function. Hence, f(x) is increasing function

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