JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 21)
Let P(x) be a real polynomial of degree 3 which vanishes at x = $$-$$3. Let P(x) have local minima at x = 1, local maxima at x = $$-$$1 and $$\int\limits_{ - 1}^1 {P(x)dx} $$ = 18, then the sum of all the coefficients of the polynomial P(x) is equal to _________.
Answer
8
Explanation
P'(x) = a(x + 1)(x $$-$$ 1)
$$ \therefore $$ P(x) = $${{a{x^3}} \over 3}$$ $$-$$ ax + C
P($$-$$3) = 0 (given)
$$ \Rightarrow $$ a($$-$$9 + 3) + C = 0
$$ \Rightarrow $$ 6a = C ..... (i)
Also, $$\int\limits_{ - 1}^1 {P(x)dx} = 18 $$
$$\Rightarrow \int\limits_{ - 1}^1 {\left( {a\left( {{{{x^3}} \over 3} - x} \right) + C} \right)} dx = 18$$
$$ \Rightarrow 0 + 2C = 18 \Rightarrow C = 9$$
from (i)
$$a = {3 \over 2}$$
$$ \therefore $$ $$P(x) = {{{x^3}} \over 2} - {3 \over 2}x + 9$$
Sum of co-efficient
= $${1 \over 2} - {3 \over 2} + 9$$ = $$-$$1 + 9 = 8
$$ \therefore $$ P(x) = $${{a{x^3}} \over 3}$$ $$-$$ ax + C
P($$-$$3) = 0 (given)
$$ \Rightarrow $$ a($$-$$9 + 3) + C = 0
$$ \Rightarrow $$ 6a = C ..... (i)
Also, $$\int\limits_{ - 1}^1 {P(x)dx} = 18 $$
$$\Rightarrow \int\limits_{ - 1}^1 {\left( {a\left( {{{{x^3}} \over 3} - x} \right) + C} \right)} dx = 18$$
$$ \Rightarrow 0 + 2C = 18 \Rightarrow C = 9$$
from (i)
$$a = {3 \over 2}$$
$$ \therefore $$ $$P(x) = {{{x^3}} \over 2} - {3 \over 2}x + 9$$
Sum of co-efficient
= $${1 \over 2} - {3 \over 2} + 9$$ = $$-$$1 + 9 = 8
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