Question

If $f\left(x\right)=\frac{4^x}{4^x+2},f\left(x\right)+f(1-x)=a$ and $f\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)+\cdots +f\left(\frac{96}{97}\right)=b$ then order pair $(a,b)$ is

• A. $(2,20)$
• B. $(1,48)$
• C. $(1,24)$
• D. $(92,96)$

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A $(2,20)$
B $(1,48)$
C $(1,24)$
D $(92,96)$

$f\left(x\right)=\frac{4^x}{4^x+2}f\left(x\right)+f(1-x)=af\left(\frac{1}{97}\right)+f\left(\frac{2}{97}\right)........+f\left(\frac{96}{97}\right)=b\left[f\right(\frac{1}{97})+f(\frac{96}{97}\left)\right]+\left[f\right(\frac{2}{97})+f(\frac{95}{97}\left)\right]......=b$

Total pair$=48$

$a+a+a+....a=b48a=bifa=1,b=48$

So,pair of $(a,b)$ is $(1,48)$