$(1):$ If $P(x)$ is a polynomial of Degree $4$ such that $P(-1) = P(1) = 5$ and

$P(-2)=P(0)=P(2)=2\;,$Then Max. value of $P(x).$

$(2):$ If $f(x)$ is a polynomial of degree $6$ with leading Coefficient $2009.$ Suppose

further that $f(1)=1\;,f(2)=3\;,f(3)=5\;,f(4)=7\;,f(5)=9$ and $f'(2)=2,$

Then sum of all digits of $f(6)$ is

$\bf{My\; Try\; For\; (1):}$ Given $x=-2\;,x=0\;,x=+2$ are the roots of $P(x)=0.$

So $(x+2)\;,(x-0)\;,(x-2)$ are factors of $P(x)=0$. So we can write $P(x)$ as

$P(x) = A\cdot x\cdot (x-2)\cdot (x+2)(x-r)\;,$ So we have calculate value of $A$ and $r$

Now put $x=-1\;,$ we get $P(-1)=-3A\cdot (1+r)\Rightarrow -3A\cdot (1+r)=5............................(1)$

Now put $x=1\;,$ we get $P(1)=-3A\cdot (1-r)\Rightarrow -3A\cdot (1-r)=5..................................(2)$

So from $(1)$ and $(2)\;,$ We get $r=0$ and $\displaystyle A=-\frac{5}{3}.$

So Polynomial $\boxed{\boxed{\displaystyle P(x)=-\frac{5}{3}\cdot x^2\cdot (x^2-4)}}$

$\bf{My\; Try\; For \; (2):}$Let $f(x)=2x-1\;\forall\; x=1\;,2\;,3\;,4\;,5.$

So we can say that $(x-1)\;,(x-2)\;,(x-3)\;,(x-4)\;,(x-5)$ are the roots of $f(x)-2x+1=0$

So $f(x)-2x+1=2009\cdot \underbrace{(x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)}\cdot\underbrace{(x-r)}$

Now How can i solve after that

Help me and plz explain me, is my $(1)$ Try is right or not


  • $\begingroup$ If you have $P(-2) = P(0) = P(2) = 2 \neq 0$, why can you say that $x(x+2)(x-2)$ is a factor of $P(x)$? $\endgroup$ – Platehead Nov 14 '14 at 2:57

Question 1 directly states that $P(-2)=P(0)=P(2)=2$. Why on earth are you then saying that they're roots of the polynomial?

What you can say is that $x=-2,0,2$ are roots of $P(x)=2$, or equivalently $P(x)-2=0$. So we can say that, if $P(x)$ is 4th degree, then $P(x)-2$ is as well, and hence $P(x)-2=A(x-\alpha)(x-2)x(x+2)$ for some $A,\alpha$. Then you can insert your values for $P(1)$ and $P(-1)$.

Also, you haven't answered the question. Read it carefully, see what it's asking, and make sure you answer it.

As for 2, you're on the right track. The next step would probably be to look at what the derivative of that expression will look like at $x=2$.

  • $\begingroup$ Oh Sorry I did not Noticed that, $\endgroup$ – juantheron Nov 14 '14 at 4:08

For part 2, here's one way to simplify the differentiation. We have

$$f(x)-2x+1=2009\cdot (x-1)\cdot(x-2)\cdot (x-3)\cdot (x-4)\cdot (x-5)\cdot(x-r)$$

Substituting $x + 3 \to x$,

$$f(x + 3) - 2(x+3)+1 = 2009(x+2)(x+1)(x)(x-1)(x-2)(x+3-r)$$ $$f(x + 3) -2x - 5 = 2009x(x^2-1)(x^2-4)(x + 3 -r)$$ $$f(x + 3) - 2x - 5 = 2009(x^5 - 5x^3 + 4x)(x + 3 -r)$$

Applying the chain rule to LHS and product rule to RHS,

$$f'(x+3) - 2 = 2009(5x^4 - 15x^2 + 4)(x + 3 - r) + 2009(x^5 - 5x^3 + 4x)$$

Substituting $x = -1$,

$$f'(2) - 2 = 2009(5 - 15 + 4)(2-r) + 2009(1-5+4)$$ $$0 = 2009\cdot(-6)\cdot(2-r)$$

to give $r=2$. Hence,

$$f(6) - 12 + 1 = 2009\cdot5!\cdot4$$ $$f(6) = 964331$$


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