GATE CSE 1997 | Question: 4.4

Question

A polynomial $p(x)$ is such that $p(0) = 5, p(1) = 4, p(2) = 9$ and $p(3) = 20$. The minimum degree it should have is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Comments

Easier Method

https://math.stackexchange.com/questions/675110/what-is-the-minimum-degree-of-a-polynomial-given-the-initial-conditions/675137#675137

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First read  Sachin Mittal  sir's comment that is written below the best answer, too good approach. Although best answer that is written by Arjun sir is also good but it is like formal approach.

Answer 1

Lets take $p(x) =  ax +b$
Now, $p(0) = 5 \implies b = 5.$

$p(1) = 4 \implies a + b = 4, a = -1$
$p(2) = 9 \implies 2a + b = 9 \implies -2 + 5 = 9$, which is false. So, degree $1$ is not possible.

Let $p(x) =$ $ax^{2}$ $+$ $bx$ $+$ $c$

$p(0) = 5 \implies c = 5$
$p(1) = 4 \implies a + b + c = 4 \implies a + b = -1 \quad \to(1)$
$p(2) = 9 \implies 4a + 2b + c = 9 \implies 2a + b = 2 \quad \to (2)$

$(2) - (1) \implies a = 3, b = -1 - 3 = -4$

$p(3) = 20 \implies 9a + 3b + c = 20, 27 -12 + 5 = 20$, equation holds.
So, degree $2$ also will suffice.

Correct Answer: $B$

Comments

One more very nice explanation:

http://math.stackexchange.com/a/675137/309722

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 your case would be this:
 

0    1    2    3   
5    4    9    20

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

0    1    2    3   
5    4    9    20
  -1    5   11

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 differences:
 

0    1    2    3   
5    4    9    20
  -1    5   11
     6    6
For a quadratic function, these second differences will all be the same. 
Your values actually all the same, so your function can be expressed as a quadratic.

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nice explanation !!! :)

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Thanks @SachinBhai

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@SachinMIttal1 this solution fails for https://gateoverflow.in/2245/gate1997_4-4#viewbutton

IS there any special case for this to work

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awesome.......

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@Arjun Sir Why degree 1 is not possible ?

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he explained the same case that you just decalred "failed"

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awesome explanation @sachin sir

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one calculation mistake in P(2) of 1 degree. P(2) = 2a+b = 9 => -2+5 =9 false.

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nice one Thanks.. !

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https://brilliant.org/wiki/method-of-differences/

 

This might help

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is this still in gate syllabus for 2021 ?

Answer 2

Since value of P(x) seems to be decreasing b/w 0 and 1 and then 1 to 2 it is increasing. So, there must be atleast two roots (may be both real and both imaginary). Because polynomials are smooth and continous on real numbers. So, there has to be one U-turn.

Answer 3

(b) is the answer. we can take a generalized quadratic equation ax2 +bx+c and verify.

Answer 4

.