Question

Let f(x)=x+a/x+b such that f(f(1)=0 & f(2)=-3 then a&b resctivily are.

Answer 1

Answer:

The value of a = - 1  ,   And b = $\frac{ - 7 }{3}$    

Step-by-step explanation:

Given as :

Function f(x) = $\frac{(x + a)}{(x + b)}$

And f(f(1)) = 0     And f(2) = - 3

Now For , x = 2 , y = - 3

I.e  f(2) = $\frac{(2 + a)}{(2 + b)}$

or,  - 3 = $\frac{(2 + a)}{(2 + b)}$

I.e 2 + a = - 6 - 3b

Or, a + 3b = - 8               ....... 1

Again  f(f(1)) = 0

So,  $\frac{(1 + a)}{(1 + b)}$ = 0

Or,  1 + a = 0

∴    a = - 1

So , put htis value of a i n eq 1 , we get value of  b

So , - 1 + 3b = - 8

Or,   3b = - 7

∴       b = $\frac{ - 7 }{3}$

Hence The value of a = - 1  ,   And b = $\frac{ - 7 }{3}$    Answer