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JEE Advanced2009MathematicsApplication of DerivativesActual

For the function f(x)=x 1 x , x 1 ,

Options

  1. Afor atleast one x in the interval [1, ) , f(x+2)-f(x) < 2
  2. B_ x f^ (x)=1
  3. Cfor all x in the interval [1, ) , f(x+2)-f(x)>2
  4. Df^ (x) is strictly decreasing in the interval [1, )

Correct answer

B. _ x f^ (x)=1

Step-by-step solution

Given, f(x)=x 1 x , x 1 aligned & f^ (x)= 1 x 1 x + 1 x & f^ (x)=- 1 x^3 ( 1 x ) aligned Now, _ x f^ (x)=0+1=1 Option (b) is correct. Now, x [1, ) 1 x (0,1] f^ (x) 1f^ (x) is strictly decreasing and _ x f^ (x)=1 So, graph of f^ (x) is shown as below. Now, in [x, x+2], x [1, ), f(x) is continuous and differentiable so by LMVT, f^ (x)= f(x+2)-f(x) 2 As f^ (x)>1 for all x [1, ) aligned & f(x+2)-f(x) 2 >1 & f(x+2)-f(x)>2 for all x [1, ) aligned

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