Skip to main content
Back to FUNCTIONS

JEE MATHEMATICS MCQ: FUNCTIONS

1 Medium
1 of 1
Application +4 | -1
If f(2x2+y28,2x2−y28)=xyf\left(2 x^2+\frac{y^2}{8}, 2 x^2-\frac{y^2}{8}\right)=x y, then the value of f(60,48)f(60,48) +f(80,48)+f(13,5)+f(80,48)+f(13,5) must be:
View answer and explanation
Correct option: A
112
The correct answer is A (112). The value is obtained by substituting appropriate values of x and y into the given function and summing the results.

More Questions from FUNCTIONS

Practice All Chapter MCQs →
Type at least 2 characters or jump directly:
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.