JEE MAIN - Mathematics (2020 - 2nd September Morning Slot - No. 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 :
6
12
-12
-24
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
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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