Let f x -2-4 x -2 for x - Real numbers- Find out the value of f 1-2001- f 2-2001- - f 2000-2001A- 2002B- 2000C- 1050D- 1000E- 1020

The correct option is D 1000f(1 x) = 2/4^1 x + 2 = f(x) = 2*4^x/4^x + 2 = f(x) = 4^x/4^x + 2 f(x) + f(1 x) = 1 So, the desired sum is 1000. ...

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Quantitative Aptitude

Functions

Let f x =2/4 ...

Question

Let f(x) = $\frac{2}{4^{x} + 2}$ for x ϵ Real numbers. Find out the value of $f (\frac{1}{2001})$ + $f (\frac{2}{2001})$ + ...... + $f (\frac{2000}{2001})$

2000

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2002

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1050

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1020

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1000

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Solution

The correct option is E
1000

f(1 - x) = $\frac{2}{4^{1 - x} + 2}$ = f(x) = $\frac{2 * 4^{x}}{4^{x + 2}}$ = f(x) = $\frac{4^{x}}{4^{x + 2}}$
f(x) + f(1 - x) = 1

So, the desired sum is 1000.

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Let f(x) = 24x+2 for x ϵ Real numbers. Find out the value of f(12001) + f(22001)+ ...... + f(20002001)

The correct option is E 1000f(1 - x) = 241−x+2 = f(x) = 2∗4x4x+2 = f(x) = 4x4x+2 f(x) + f(1 - x) = 1 So, the desired sum is 1000.

Let f(x) = $\frac{2}{4^{x} + 2}$ for x ϵ Real numbers. Find out the value of $f (\frac{1}{2001})$ + $f (\frac{2}{2001})$ + ...... + $f (\frac{2000}{2001})$

The correct option is E
1000

f(1 - x) = $\frac{2}{4^{1 - x} + 2}$ = f(x) = $\frac{2 * 4^{x}}{4^{x + 2}}$ = f(x) = $\frac{4^{x}}{4^{x + 2}}$
f(x) + f(1 - x) = 1

So, the desired sum is 1000.