A polynomial $p(x)$ satisfies the following:
The minimum degree of such a polynomial is
yes, option $D$ is the correct answer.
Here is how $p(x$) should look like:
Value of $P(x$) will be zero at green circled $(O)$ points,So they will be the roots of the polynomial $P(x)$. Hence, the minimum degree of $P(x)$ will be $4$.
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 .
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.
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
in order to solve this type of problem always go to this link it will never fail
https://math.stackexchange.com/questions/675110/what-is-the-minimum-degree-of-a-polynomial-given-the-initial-conditions/675137#675137
Do by this method , this is more easy .
answer : D
@Dileep kumar M 6
Do one more time
-8 8
16
4th difference same.So, Degree = 4
Gatecse
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