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Question

Let f:RR be a function which satisfies f(x+y)=f(x)+f(y)x,yR. If f(1)=2 and g(n)=f(k)k=1n-1, nN then the value of n for which g(n)=20is:


A

9

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B

5

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C

4

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D

20

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Solution

The correct option is B

5


Determine the value of n for which g(n)=20

We have,

f(x+y)=f(x)+f(y)x,yR

g(n)=f(k)k=1n-1

We know that: f(1)=2

Letx=y=1f(1+1)=f(1)+f(1)f(2)=2f(1)f(2)=4

Similarly,

Letx=1,y=2f(1+2)=f(1)+f(2)f(3)=2+4f(3)=6

g(n)=f(1)+f(2)+f(3)+....+f(n-1)=2+4+6+....2(n-1)=2(1+2+3...(n-1))=2.(n-1)n2=n(n-1)=n2-n

Since we know, g(n)=20, placing the value we get,

n2-n=20n2-n-20=0(n-5)(n+4)=0n=-4,5ncannotbenegativen=5

Therefore, the value of n for which g(n)=20is: 5.

Hence, the correct answer is Option (B).


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