Home
Class 12
MATHS
A continuous function f(x)onR->R satisfi...

A continuous function `f(x)onR->R` satisfies the relation `f(x)+f(2x+y)+5x y=f(3x-y)+2x^2+1forAAx ,y in R ` Then find `f(x)dot`

Text Solution

Verified by Experts

We have,
`f(x) +f(2x+y)+5xy=f(3x-y)+2x^(2)+1 AA x,y in R` …(1)
To get function we should eliminate `f(2x+y)` and `f(3x-y).`
For that let `2x+y=3x-y`
`implies y=(x)/(2)`
So, putting `y=(x)/(2),` we get
`f(x)+f((5x)/(2))+(5x^(2))/(2)=f((5x)/(2))+2x^(2)+1`
`implies f(x)=1-(x^(2))/(2)`
Promotional Banner

Topper's Solved these Questions

  • RELATIONS AND FUNCTIONS

    CENGAGE ENGLISH|Exercise Solved Examples|15 Videos
  • RELATIONS AND FUNCTIONS

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 1.1|15 Videos
  • PROPERTIES AND SOLUTIONS OF TRIANGLE

    CENGAGE ENGLISH|Exercise Archives (Numerical Value Type)|3 Videos
  • SCALER TRIPLE PRODUCTS

    CENGAGE ENGLISH|Exercise DPP 2.3|11 Videos

Similar Questions

Explore conceptually related problems

A function f(x) satisfies the relation f(x+y) = f(x) + f(y) + xy(x+y), AA x, y in R . If f'(0) = - 1, then

A function f: R -> R satisfy the equation f (x)f(y) - f (xy)= x+y for all x, y in R and f(y) > 0 , then

A function f : R→R satisfies the equation f(x)f(y) - f(xy) = x + y ∀ x, y ∈ R and f (1)>0 , then

If a real valued function f(x) satisfies the equation f(x +y)=f(x)+f (y) for all x,y in R then f(x) is

If the function / satisfies the relation f(x+y)+f(x-y)=2f(x),f(y)AAx , y in R and f(0)!=0 , then

A function f: R->R satisfies that equation f(x+y)=f(x)f(y) for all x ,\ y in R , f(x)!=0 . Suppose that the function f(x) is differentiable at x=0 and f^(prime)(0)=2 . Prove that f^(prime)(x)=2\ f(x) .

A function f: R->R satisfies that equation f(x+y)=f(x)f(y) for all x ,\ y in R , f(x)!=0 . Suppose that the function f(x) is differentiable at x=0 and f^(prime)(0)=2 . Prove that f^(prime)(x)=2\ f(x) .

Suppose the function f(x) satisfies the relation f(x+y^3)=f(x)+f(y^3)dotAAx ,y in R and is differentiable for all xdot Statement 1: If f^(prime)(2)=a ,t h e nf^(prime)(-2)=a Statement 2: f(x) is an odd function.

If f be decreasing continuous function satisfying f(x + y) = f(x) + f(y)-f(x) f(y) Y x, y epsilon R ,f'(0)=1 then int_0^1 f(x)dx is

If a function 'f' satisfies the relation f(x)f^('')(x)-f(x)f^(')(x) -f^(')(x)^(2)=0 AA x in R and f(0)=1=f^(')(0) . Then find f(x) .