## 4 Answers

### 10 Comments

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. ]

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) = ax^{2}+bx+c and should satisfy p(0) = 5, p(1) = 4, p(2) = 9 and p(3) = 20. . and so on .

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.

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

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 $D$