Let function f be defined by f(x) = frac{{2x + 1}}{{1 - 3x}} , then {f^{ - 1}}(x) is

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1.Relation and Function

easy

Let the function $f$ be defined by $f(x) = \frac{{2x + 1}}{{1 - 3x}}$, then ${f^{ - 1}}(x)$ is

A

$\frac{{x - 1}}{{3x + 2}}$

B

$\frac{{3x + 2}}{{x - 1}}$

C

$\frac{{x + 1}}{{3x - 2}}$

D

$\frac{{2x + 1}}{{1 - 3x}}$

Solution

(a) Let $y = f(x)$ ==> $y = \frac{{2x + 1}}{{1 – 3x}}$

==> $y – 3xy = 2x + 1$

==> $x = \frac{{y – 1}}{{3y + 2}}$

==> ${f^{ – 1}}(y) = \frac{{y – 1}}{{3y + 2}}$

$\Rightarrow {f^{ – 1}}(x) = \frac{{x – 1}}{{3x + 2}}$.

Standard 12

Mathematics

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