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
yes, option D is the correct answer.
Here is how p(x) should look like:
Value of P(x) will be zwro 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
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