A function is defined for all positive numbers x as f(x) = ax^(1/2) +

Question

A function is defined for all positive numbers x as  f(x) = a\sqrt{x} + b . What is the value of f(3), if f(4) – f(1) = 2 and f(4) + f(1) = 10?

(A) 1

(B) 2

(C)  2 \sqrt{3}

(D)  2 \sqrt{3} + 2

(E)  2 \sqrt{3} – 2

Archit3110 · Answer

solve for expressions 
we get a=2and b =2
so for f(3) = 2√3+ 2
IMO D

Bunuel · Answer

_________________
Fixed that. Thank you.

NandishSS · Answer

f(4) – f(1) = 2 ==>  a\sqrt{4} + b  -  a\sqrt{1} - b  ==> 2a - a =2 ==> a=2

f(4) + f(1) = 10 ==>  a\sqrt{4} + b  +  a\sqrt{1} + b  ==> 6+2b =10 ==> b=2

Hence D

f(3) =  2 \sqrt{3} + 2

enricosanchez · Answer

How does  a\sqrt{4} + b  +  a\sqrt{1} + b  ==> 6+2b =10??

kaylaquijas · Answer

Bunuel,

Why doesn't it work algebraically to add the two functions together?

f(4)-f(1)=2
+
 f(4)+f(1)=10
-------------------

f(1) cancels out, leaving (2)*f(4) = 12, which reduces to f(4) = 6. When subbing 6 into the second equation, then f(1) = 4.

So I ended up with f(3) = 6( 3) +4

Did I fall into some sort of trap by trying to solve it this way? I need to understand why this doesn't work so I don't do it again in the future.

Prep Toolkit