Question
Download Solution PDFComprehension
Consider the following for the two (02) items that follow:
A function f is such that f(xy) = f(x + y) for all real values of x and y, and f(5) = 10
What is f(20)+f(-20) equal to?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFCalculation:
Given,
The function is \( f(xy) = f(x + y) \) for all real values of x and y, and f(5) = 10 .
We are tasked with finding:
\( f(20) + f(-20) \)
Using the given functional equation, we have:
For \( f(0 \cdot 5) = f(0 + 5) \), we get:
\( f(0) = f(5) = 10 \)
For \( f(0 \cdot 20) = f(0 + 20) \), we get:
\( f(0) = f(20) = 10 \)
For \( f(0 \cdot -20) = f(0 + (-20)) \), we get:
\( f(0) = f(-20) = 10 \)
Thus,
\( f(20) + f(-20) = 10 + 10 = 20 \)
Hence, the correct answert is Option 3.
Last updated on Jul 8, 2025
->UPSC NDA Application Correction Window is open from 7th July to 9th July 2025.
->UPSC had extended the UPSC NDA 2 Registration Date till 20th June 2025.
-> A total of 406 vacancies have been announced for NDA 2 Exam 2025.
->The NDA exam date 2025 has been announced. The written examination will be held on 14th September 2025.
-> The selection process for the NDA exam includes a Written Exam and SSB Interview.
-> Candidates who get successful selection under UPSC NDA will get a salary range between Rs. 15,600 to Rs. 39,100.
-> Candidates must go through the NDA previous year question paper. Attempting the NDA mock test is also essential.