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A polynomial $p(x)$ satisfies the following:

• $p(1) = p(3) = p(5) = 1$
• $p(2) = p(4) = -1$

The minimum degree of such a polynomial is

1. $1$
2. $2$
3. $3$
4. $4$

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yes, option $D$ is the correct answer.

Here is how $p(x$) should look like:

Value of $P(x$) will be zero at circled $(O)$ points, so they will be the roots of the polynomial $P(x)$.
Hence, the minimum degree of $P(x)$ will be $4$.

@ anurag https://gateoverflow.in/2245/gate1997_4-4#viewbutton

in this graph will not touch at y=0 then how did we calculate degree ... using graph (other than arjun sir solution )

The value  of x is a root of polynomial ,p(x), if at that value p(x) = 0

p(1) = 1 , p(2) = -1 , x is going from 1 to 2 , p(x) is going from 1 to -1 , it is clear somewhere it goes though 0 in between, for x in  [1,2 ] ,  minimum one root lies there

similarly minimum one root , in between [2,3],[3,4], [4,5].

[Note : here  https://gateoverflow.in/2245/gate1997_4-4 , here we don't have no such given value of x for which p(x) crosses though 0. ]

yes ,sir this is clear ... i was talking about link which i gave there is no p(x)=0 then how we will get point using graph ...i am not getting , minimum one root , in between [2,3],[3,4], [4,5]. :(

p(2) = -1 p(3) = 1 , -1 to 1 , go though 0 . so root lies in [2,3] .

In that link, it is guess , polynomial is degree 1, p(x) = ax+b , and should satisfy p(0) = 5, p(1) = 4, p(2) = 9 and p(3) = 20 if it does not then guess it is of degree2 , p(x) = ax2+bx+c and should satisfy p(0) = 5, p(1) = 4, p(2) = 9 and p(3) = 20. . and so on .

p(2) = -1 p(3) = 1 , -1 to 1 , go though 0 . so root lies in [2,3] .   am getting this no issue ...my problem is this   if i want to solve  this  p(0) = 5, p(1) = 4, p(2) = 9 and p(3) = 20  by  graph then i all are going in increasing order rt ... that we cant find degree using graph ??

I believe graphs can give intuition for https://gateoverflow.in/2245/gate1997_4-4#viewbutton too.

On plotting all those points it can be observed that we will get a bowl shaped plot and any equation of the form "y = ax + b" can not have a bowl shaped plot since y = ax + b represents a straight line whose slope is 'a' & y intercept is 'b'..

You can also check it by differentiating y = ax + b with respect to x, it will give 'a' which is a constant, but in bowl shaped plot our slope is varying

So definitely the polynomial should have the degree strictly greater than 1.

& from the graph of y = x^2(which also is a polynomial of degree 2) we can observe that it has a bowl shape.

This implies that there exists a polynomial(x^2) who is of degree 2 and whose plot is bowl shaped.

So our polynomial could have 2 as its minimum degree since it has a bowl shaped plot.

We can do by plotting graph and there is pattern of graphs of different degree, as polynomial of degree 2 will be parabolic. But for exact plotting large no of data should be given . Else it is not good to use it.
ok yes ... you mean type of parabola ... not straight line means not linear rt ... ok clear , praveen sir also right ...  thank you ... praveen sir, anurag :

i would like to add some important point here,

if a polynomial has minimum degree n then it will have atmost 'n' , x-intercepts and atmost 'n-1' turns. here in this graph we have 3 turns, we can say that it has minimum degree of 3+1 = 4.

sometimes we may get less than 'n' , x-intercepts, so we can check using number of turns in the graph.

example in this question https://gateoverflow.in/2245/gate1997-4-4

if we draw graph like above we will not get any x intercept but it has 1 turn, so we can say that it has minimum degree 1+1=2

Do by this method  ,  this is more easy .

by

By above method not able get the same difference

1    -1     1    -1      1

-2     2     -2    2

4      -4     4

-8       8

@Dileep kumar M 6

Do one more time

-8   8

16

4th difference same.So, Degree = 4

Easiest!

Does this method fail in any special case or order?
This should be the best answer...
@tuhin datta this method never failed
easiest

it means when we get all values same it will provide degree

@Rajesh Panwar

yes
Minimum Degree is 4 as there are atleast 4 roots possible for this polynomial

### 1 comment

how can we say that at least 4 roots exist?please xplain

For a linear polynomial $p$, you'll always have $p(n+1)−p(n)$ the same. If you write down a table

1     2    3     4   5
p(1) p(2) p(3) p(4) p(5)

which in our case would be this:

1    2    3    4   5
1   -1    1   -1   1


and then write the differences $p(2)−p(1)$, $p(3)−p(2)$, etc in a row beneath, you'd get (again in our case)

1    2    3    4   5
1   -1    1   -1   1
-2    2   -2   2


That new row is called the "first differences". For a linear function, the entries would all be the same. You can also write down second, third and fourth differences:

1    2    3    4   5
1   -1    1   -1   1
-2    2   -2   2
4   -4   4
-8   8
16

For a function with degree 2, the second differences will all be the same.
In our case fourt difference is same. So degree is 4


Answer is $D$

by

Any reference/reason for why this behavior is as it is ?
Seems like some property it follows