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If p(x) be a polynomial of degree three that has a local maximum value 8 at x = 1 and a local minimum value 4 at x = 2; then p(0) is equal to :

JEE · Math · previous-year question

  1. A.6
  2. B.12
  3. C.-12correct
  4. D.-24

Answer

C. -12

Explanation

Since p(x) has relative extreme at x = 1 & 2 so p'(x) = 0 at x = 1 & 2 $$ \therefore $$ Let p'(x) = A(x – 1) (x – 2) $$ \Rightarrow $$ p(x) = $$\int {A\left( {{x^2} - 3x + 2} \right)dx} $$ $$ \Rightarrow $$ p(x) = $${A\left( {{{{x^2}} \over 3} - 3{x^2} + 2x} \right)}$$ + C ...(1) As P(1) = 8 From (1) $$8 = A\left( {{1 \over 3} - {3 \over 2} + 2} \right) + C$$ $$ \Rightarrow $$ 8 = $${{5A} \over 6} + C$$ $$ \Rightarrow $$ 48 = 5A + 5C ...(2) Also P(2) = 4 From (1) $$4 = A\left( {{8 \over 3} - 6 + 4} \right) + C$$ $$ \Rightarrow $$ 4 = $${{2A} \over 3} + C$$ $$ \Rightarrow $$ 12 = 2A + 3C ......(3) Form (3) & (4), C = –12, A = 24 Now p(0) = C = -12

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