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KVPY2015MathematicsQuadratic Equation

Let p ( x ) be a polynomial such that p(x)-p^ (x)=x^ n , where n is a positive integer. Then p(0) equals.

Options

  1. An !
  2. B(n-1) !
  3. C1 n !
  4. D1 (n-1) !

Correct answer

A. n !

Step-by-step solution

Let P ( x )= a ₀ x ^ n + a ₁ x ^ n -1 + a ₂ x ^ n -2 . a _ n = _ r =0 ^ n a _ r x ^ n - r array l P¹(x)= _ r=0 ^ n-1 a_ r (n-r) x^ n-r-1 P(x)-P¹(x)=a₀ x^ n + _ r=1 ^ n a_ r -a_ r+ (n-r+1) x^ n-r =x^ n array So, a₀=1 and a_ r =a_ r-1 (n-r+1) a _ r a _ r -1 = n - r +1 Now, P(0)=a_ n = a_ n a_ n-1 a_ n-1 a_ n-2 a₂ a₁ a₁ a₀ a₀=(1 2 . ., n )= n !

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