\( P(x) \) is polynomial of degree 3 such that,

\( P(I)=1/(1+I) \),

Where \(I\) belongs to (1,2,3,4).

Then find the value of \(P(5)\).

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThe answer should be \(\frac{2}{15}\). Use Lagrange Interpolation

Log in to reply

Can u solve more easily? Or Please explain your solution!

Log in to reply

Actually, in such questions where a polynomial \(f(x)\) follows a pattern for certain values in its domain, we generally use a trick like defining a new polynomial \(g(x)\) whose roots are known, and then calculating \(f(x)\). For instance, here we could have done this by defining \(g(x) = f(x) - \frac{1}{x+1}\). Then the roots of \(g(x)\) are \(1,2,3,4\). Then, \(g(x) = k(x-1)(x-2)(x-3)(x-4)\), giving us \(f(x) = k(x-1)(x-2)(x-3)(x-4) + \frac{1}{x+1}\). But then, the degree of \(f(x)\) will be \(4\), but we are given its degree to be \(3\). And in such cases, I think we have only one option, i.e. to use Lagrange Interpolation.

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Let \(g(x)=(x+1)f(x)-1\) You have to express it in this form in order to make \(g(x)\) a polynomial.Anyways, \(g(x)=0\) for \(x=1,2,3,4\).So \((x+1)f(x)-1=g(x)=c(x-1)(x-2)(x-3)(x-4)\) for some constant \(c\).Putting \(x=-1\) we get: \[-1=c(-2)(-3)(-4)(-5)\implies c=\frac{-1}{120}\] Therefore \(g(x)=(x+1)f(x)-1=\frac{-1}{120}(x-1)(x-2)(x-3)(x-4)\) Now simply putting \(x=5\) gives: \[6\times f(5)-1=\frac{-1}{120} (4)(3)(2)(1)=\frac{-1}{5}\\ \implies \boxed{f(5)=\frac{2}{15}}\]

Log in to reply

Log in to reply

Log in to reply

Since you are evaluating for consecutive integers you could have used method of differences.

Log in to reply

Log in to reply