"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 :"

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Accepted Answer

"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"

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 :

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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 :

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