1.Relation and Function
easy

જો $y = f(x) = \frac{{x + 2}}{{x - 1}}$, તો $x = $

A

$f(y)$

B

$2f(y)$

C

$\frac{1}{{f(y)}}$

D

એકપણ નહી.

(IIT-1984)

Solution

(a) $y = \frac{{x + 2}}{{x – 1}}\,\,$

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

Standard 12
Mathematics

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