Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa

Let f : R\rightarrow R be a function which statisfies  f\left ( x+y \right ) = f(x) + f(y) ,x,y \epsilon R , if f (1) =2 ,and g(n) = \sum_{k=1}^{n-1} f(k), n \epsilon N then the value of n for which g(n) = 20, is :
Option: 1 4
Option: 2 5
Option: 3 9
Option: 4 20

Answers (1)

best_answer

f(x+y)=f(x)+f(y)

f(1)=2

g(x)=\sum_{k=1}^{n-1} f(k) \quad n \in N

g(n)=20

\\f(1)=f(1)+f(0) \\\\ f(2)=2 f(2)=4 \\\\ f(3)=8 f(2)+f(2)=6 \\\\ f(4)=8

\\f(5)=f(3)+f(2)=6+4=10 \\ \\f(x)=2x\\\\\Rightarrow 2 \times \frac{n(n+1)}{2}=20 \\ \\\Rightarrow 4 \times 5=20 \Rightarrow n=4

Posted by

Suraj Bhandari

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

JEE Main 2024 session 2 paper 2 answer key objection window closes today; fees
anam-khan

JEE Main 2024 Topper: Farmer’s son Gajare Nilkrishna says ‘keep questioning until you understand’
anam-khan

NIT Puducherry Cut-off 2024: Computer Science and Engineering had highest closing rank last year

Latest Question