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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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