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13 Jan 2023, 10:00
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (02:30) correct
33%
(02:38)
wrong
based on 206
sessions
History
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The function f(x) is defined as \(f(x)=\frac{1}{(x+1)}\), where x is not equal to -1. If a is not equal to -1 or 0, which of the following expressions must be true for all values of a?
I. \(f(a−1)=f(a)–f(1)\)
II. \(\frac{2}{f(a−1)}=2/f(a)−1/f(1)\)
III. \(f(a−1)=f(a)*f(1)\)
A. I only
B. II only
C. III only
D. I, II and III
E. None of the above
I. \(f(a−1)=f(a)–f(1)\)
II. \(\frac{2}{f(a−1)}=2/f(a)−1/f(1)\)
III. \(f(a−1)=f(a)*f(1)\)
A. I only
B. II only
C. III only
D. I, II and III
E. None of the above
14 Jan 2023, 10:04
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Given that \(f(x)=\frac{1}{(x+1)}\), where x is not equal to -1 and a is not equal to -1 or 0. We need to find which of the following expressions must be true for all values of a?
I. \(f(a−1)=f(a)–f(1)\)
\(f(a-1)=\frac{1}{(a-1+1)}\) = \(\frac{1}{a}\)
f(a)–f(1) = \(\frac{1}{(a+1)}\) - \(\frac{1}{(1+1)}\) = \(\frac{2 - a - 1}{2*(a+1)}\) = \(\frac{1-a}{2*(a+1)}\)
Clearly, \(f(a−1)≠f(a)–f(1)\)
=> FALSE
II. \(\frac{2}{f(a−1)}=2/f(a)−1/f(1)\)
\(f(x)=\frac{1}{(x+1)}\)
=> \(\frac{1}{f(x)}=x+1\)
=> \(\frac{2}{f(a-1)}=2*(a-1+1)\) = 2a
\(\frac{2}{f(a)}=2*(a+1)\) = 2a + 2
\(\frac{1}{f(1)}=(1+1)\) = 2
=> 2/f(a)−1/f(1) = 2a + 2 - 2 = 2a = 2/f(a−1)
=> TRUE
III. \(f(a−1)=f(a)*f(1)\)
\(f(a-1)=\frac{1}{(a-1+1)}\) = \(\frac{1}{a}\)
\(f(a)=\frac{1}{(a+1)}\)
\(f(1)=\frac{1}{(1+1)}\) = \(\frac{1}{2}\)
=> f(a)*f(1) = \(\frac{1}{(a+1)} * \frac{1}{2}\) = \(\frac{1}{2*(a+1)}\)
Clearly, \(f(a−1)≠f(a)*f(1)\)
=> FALSE
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
I. \(f(a−1)=f(a)–f(1)\)
\(f(a-1)=\frac{1}{(a-1+1)}\) = \(\frac{1}{a}\)
f(a)–f(1) = \(\frac{1}{(a+1)}\) - \(\frac{1}{(1+1)}\) = \(\frac{2 - a - 1}{2*(a+1)}\) = \(\frac{1-a}{2*(a+1)}\)
Clearly, \(f(a−1)≠f(a)–f(1)\)
=> FALSE
II. \(\frac{2}{f(a−1)}=2/f(a)−1/f(1)\)
\(f(x)=\frac{1}{(x+1)}\)
=> \(\frac{1}{f(x)}=x+1\)
=> \(\frac{2}{f(a-1)}=2*(a-1+1)\) = 2a
\(\frac{2}{f(a)}=2*(a+1)\) = 2a + 2
\(\frac{1}{f(1)}=(1+1)\) = 2
=> 2/f(a)−1/f(1) = 2a + 2 - 2 = 2a = 2/f(a−1)
=> TRUE
III. \(f(a−1)=f(a)*f(1)\)
\(f(a-1)=\frac{1}{(a-1+1)}\) = \(\frac{1}{a}\)
\(f(a)=\frac{1}{(a+1)}\)
\(f(1)=\frac{1}{(1+1)}\) = \(\frac{1}{2}\)
=> f(a)*f(1) = \(\frac{1}{(a+1)} * \frac{1}{2}\) = \(\frac{1}{2*(a+1)}\)
Clearly, \(f(a−1)≠f(a)*f(1)\)
=> FALSE
So, Answer will be B
Hope it helps!
Watch the following video to learn the Basics of Functions and Custom Characters
10 Dec 2024, 01:27
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It is given that a≠0,-1 then take a as any other value like 1 and check all the options,this will be quick way to solve this.
