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KVPY2012MathematicsApplication of Derivatives

Let n be a natural number and let 'a' be a real number. The number of zeros of x ^ 2 n +1 -(2 n +1) x + a =0 in the interval [-1,1] is -

Options

  1. A2 if a >0
  2. B2 if a < 0
  3. CAt most one for every value of a
  4. DAt least three for every value of a

Correct answer

C. At most one for every value of a

Step-by-step solution

array l f(x)=x^ 2 n+1 -(2 n+1) x+a f^ (x)=(2 n+1) x^ 2 n -(2 n+1) array =(2 n+1) (x^ 2 n -1 ) < 0 when x [-1,1] f ( x ) is strictly decreasing in [-1,1] f ( x ) cut x axis at most one point in given interval

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