Question

P(x) is polynomial of degree 3 which have maximum value 8 at x = 1, min 6 at x = 2 find P(0) ​
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Answer 1

Answer:

Let the polynomial be P(x) = ax³ + bx² + cx + d

According to given conditions:-

P(1)=a+bc+d=10

P(1)=a+b+c+d=6

P(1)=3a2b+c=0

P(1)=6a+2b=0

3a + b = 0

Solving for a,b,c,d we get P(x) = x³ - 3x² - 9x + 5

$\to{ \sf{p(0) = ({0)}^{3} - 3 ({0)}^{2} - 9(0) + 5 }}$

$\to{ \sf{0 - 0 - 0 + 5}}$

${ \to{ \sf{5}}}$

Therefore,

• p(0) = 5

Hope it's Help you dear ❤

If Not It's Ok ☺

Answer 2

Answer:

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Step-by-step explanation:

Given :

P(x) is polynomial of degree 3 which have maximum value 8 at x = 1, and min 6 at x = 2

To Find :

The value of p(0)

Solution :

Given : 2,1 are the critical points of p(x)

Thus,

\sf\:p(x)'=\lambda(x-1)(x-2)p(x)

′

=λ(x−1)(x−2)

Where ,λ is a constant

\sf\:p(x)'=\lambda(x^2-2x-x+2)p(x)

′

=λ(x

2

−2x−x+2)

\sf\:p(x)'=\lambda(x^2-3x+2)p(x)

′

=λ(x

2

−3x+2)

Now integrate

\sf\:p(x)=\lambda\int(x^2-3x+2)p(x)=λ∫(x

2

−3x+2)

\sf\:p(x)=\lambda(\frac{x^3}{3}-\frac{-3x^2}{2}+2x)+cp(x)=λ(

3

x

3

−

2

−3x

2

+2x)+c

It is given that p(1)=8 then ,

\sf\:p(1)=\lambda(\frac{1^3}{3}-\frac{-3\times1^2}{2}+2)+cp(1)=λ(

3

1

3

−

2

−3×1

2

+2)+c

\sf\:p(1)=\lambda(\frac{1}{3}-\frac{-3}{2}+2)+cp(1)=λ(

3

1

−

2

−3

+2)+c

\sf\:p(1)=\lambda(\frac{14-9}{6})+cp(1)=λ(

6

14−9

)+c

\sf\:p(1)=\frac{5\lambda}{6}+cp(1)=

6

5λ

+c

\sf\:8=\frac{5\lambda}{6}+c....(1)8=

6

5λ

+c....(1)

It is also given that p(2)=6 ,then

\sf\:p(2)=\lambda(\frac{2^3}{3}-\frac{-3\times2^2}{2}+2\times2)+cp(2)=λ(

3

2

3

−

2

−3×2

2

+2×2)+c

\sf\:p(2)=\lambda(\frac{8}{3}-\frac{-3\times4}{2}+4)+cp(2)=λ(

3

8

−

2

−3×4

+4)+c

\sf\:p(2)=\lambda(\frac{8-6}{3})+cp(2)=λ(

3

8−6

)+c

\sf\:p(2)=\frac{2\lambda}{3}+cp(2)=

3

2λ

+c

\sf\:6=\frac{2\lambda}{3}+c....(2)6=

3

2λ

+c....(2)

Subtract Equation (2) from (1)

Then ,

\sf\:2=\dfrac{5\lambda}{6}-\dfrac{2\lambda}{3}2=

6

5λ

−

3

2λ

\sf\:2=\dfrac{(5-4)\lambda}{6}2=

6

(5−4)λ

\sf\:2=\dfrac{\lambda}{6}2=

6

λ

\sf\lambda=12λ=12

Put ,λ=12 in equation (1)

\sf\:8=\dfrac{5}{6}\times12+c8=

6

5

×12+c

\sf\:8=10+c8=10+c

\sf\:c=-2c=−2

Thus,

\sf\:p(x)=12(\frac{x^3}{3}-\frac{-3x^2}{2}+2x)-12p(x)=12(

3

x

3

−

2

−3x

2

+2x)−12

We have to find p(0)

Then ,

\sf\:p(0)=12(\frac{0^3}{3}-\frac{-3\times0^2}{2}+2\times0)-2p(0)=12(

3

0

3

−

2

−3×0

2

+2×0)−2

\sf\:p(0)=-2p(0)=−2

Therefore ,p(0)= -2