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joy francis

Q1)If f(x) satisfies the relation f(x+y)=f(x)+f(y) for all x

R and f(1)=5, then find _{[ n=1]}

^{[m ]}f(n).

Q2)a function fx is defined for all x

R and satisfies f(x+y)=fx+2y^2+kxy

x,y

R where k is a given constant. If f(1)=2 and f(2)=8 find f(x) and show that

f(x+y).f(1/x+y)=k, x+y

0 .

Q3)let n be a positive integer and f(x)=1!+2!+3!+4!...........+n!. Find polynomials P(x) and Q(x) such that f(n+2)=P(n).f(n+1) + Q(n).f(n) for all n

N.

HIMANSHU JAIN

1) put f(x) = tx

f(1) = 5

=> t = 5

:. ans is 5m(m+1)/2

waterdemon

I am solving the third question

I don't have much time so it's a bit short try understanding it

We have to find the value of "f(x)" and "Q(x)"

Given:

f(n+2) = Q(n).f(n) + P(n).f(n+1)

[1! + 2! + ........+ n! + (n+1)! + (n+2)!]

=Q(n). [ 1! + 2! + ................+ n! ] + P(n). [ 1! + 2! + ........+ n! + (n+1)! ]

Now

(1! + 2! + ........+ n!) + [ (n+1)! + (n+2)! ]

=Q(n). [1! + 2! + ........+ n!] + P(n). (1! + 2! + ........+ n!) + P(n).(n+1)!

On solving we will get

P(n) + Q(n) = 1 ...................@ 1 equation

and then on substituting the value of Q(n) = 1 - P(n) in the equation we get

P(n) = n+3

Therefore

P(x) = x+3

For Q(x) substitute P(x) value in equation 1

We get

Q(x) = -(x+2)

plZ rate me for my efforts.

Cheers !!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Prabhas Bajaj

(1)

using given prop,

f(1)=5

f(2)=10 [f(2)= f(1) + f(1)]

f(3)=15 [f(3)= f(2) + f(1)]

f(4)= 20
.
.
.
and so on...

=> The series of functions forms an AP with a=5, d=5 -------(1)

Use Sum of an AP formula upto 'm' terms-

S= m (2a+(m-1)d)
2

Substitute values of a and d from (1)

Ans-

S= 5m (m+1)
2

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