Question

Let g(x) =x^6+ax^5+bx^4+cx^3+dx^2+ex+f be a polynomial such that g(1)=1 , g(2)=2 ,g(3)=3, g(4)=4 g(5)=5 , and g(6) = 6 then the value of g(7) = ( the answer is not 7 )
A.0
B.7
C.14
D.727
please answer fast​

Answer 1

In the above Question , the following information is given -

g ( x ) = x⁶ + ax⁵ + bx⁴ + cx³ + dx² + ex + f .

g ( 1 ) = 1

g ( 2 ) = 2

g ( 3 ) = 3

g ( 4 ) = 4

g ( 5 ) = 5

g ( 6 ) = 6

To find -

g ( 7 ) = ?

Solution -

Here , we can observe that , g ( x ) = x for values of x as 1, 2, 3, 4, 5 and 6 .

We will use this to solve this question .

So ,we can use the division algorithm and solve this -

Thus , this polynomial can be expressed as -

f ( x ) - x = k ( x - 1 )( x - 2)( x - 3)( x - 4)( x - 5)( x - 6 )

Here , k is the leading coefficient.

However , here the leading coefficient is 1 .

So , the new polynomial becomes -

f ( x ) - x = ( x - 1 )( x - 2)( x - 3)( x - 4)( x - 5)( x - 6 )

We have 6 data points and we can substitute any one of them.

If x = 7

=> f ( 7 ) - 7 = 6 × 5 × 4 × 3 × 2 × 1

=> f ( 7 ) - 7 = 720

=> f ( 7 ) = 727 .

Thus , the required value of f ( 7 ) = 727

Sum of the digits of n

=> 7 + 2 + 7

=> 16

Last digit of n

=> 7

Thus , Options A and C are correct .

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Answer 2

Answer:Given P(1)=1,P(2)=2,P(3)=3,P(4)=4,P(5)=5,P(6)=6

P(x)−x will be a polynomial of degree 6 & with roots as 1,2,3,4,5 & 6.

on factorizing P(x)−x=C(x−1)(x−2)(x−3)(x−4)(x−5)(x−6)-------(1)

where C is constant

when x=0⇒P(0)−0=C×6!

C=f/6!-------(2)

P(7)=7+[P(7)−7]=7+  

6!

f

​

(7−1)(7−2)(7−3)(7−4)(7−5)(7−6)

                                         =7+f

As coefficient of x  

6

 is 1⇒c=1

Then constant term f in P(x)=6!=720

P(7)=7+720=727

Step-by-step explanation: