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Bunuel
If \(\frac{f(x)}{f(x−1)} = \frac{(x−2)}{(x+1)}\), for all \(x ≥ 0\) and \(f(6) = 81\), what is the value of \(f(4)\)?

A. 282
B. 282.5
C. 283
D. 283.5
E. 284


Solution


    • We have \(\frac{f(x)}{f(x-1)} = \frac{(x – 2)}{(x+1)}\)
Substituting x = 6 in the above equation, we get,
    • \(\frac{f(6)}{f(5)} =\frac{4}{7} …………..Eq.(i)\)
Substituting x = 5 in the the given expression, we get,
    • \(\frac{f(5)}{f(4)} =\frac{3}{6} …………..Eq.(ii)\)
Now, from Eq.(i) and Eq.(ii), we can write,
    • \(\frac{f(6)}{f(5)}* \frac{f(5)}{f(4)} =\frac{4}{7}*\frac{3}{6}\)
      \(⟹\frac{f(6)}{f(4)} = \frac{2}{7}\)
      \(⟹f(4) = f(6)* \frac{7}{2} = 81*\frac{7}{2} =283.5\)
Thus, the correct answer is Option D.
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If f(x)/f(x−1)=(x−2)/(x+1), for all x≥0x≥0 and f(6)=81, what is the value of f(4)?

A. 282
B. 282.5
C. 283
D. 283.5
E. 284

f(6)/f(5) = 4/7

similarly, f(5)/f(4) = 3/6 = 1/2


Multiplying both, f(6)/f(4) = 2/7 => f(4) = 7 * 81/2 = 283.5

So D
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Bunuel
If \(\frac{f(x)}{f(x−1)} = \frac{(x−2)}{(x+1)}\), for all \(x ≥ 0\) and \(f(6) = 81\), what is the value of \(f(4)\)?

A. 282
B. 282.5
C. 283
D. 283.5
E. 284


First, letting x = 6, we have:

f(6)/f(5) = 4/7

81/f(5) = 4/7

4 * f(5) = 567

f(5) = 567/4

Now, letting x = 5, we have:

f(5)/f(4) = 3/6

(567/4)/f(4) = 1/2

f(4) = 2 * 567/4

f(4) = 567/2 = 283.5

Answer: D
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