1.Relation and Function
easy

જો $f(x) = 3x - 5$, તો ${f^{ - 1}}(x) =$

A

$\frac{1}{{3x - 5}}$

B

$\frac{{x + 5}}{3}$

C

અસ્તિત્વ ન ધરાવે કારણ કે $f$ એ એક-એક નથી.

D

અસ્તિત્વ ન ધરાવે કારણ કે $f$ એ વ્યાપ્ત નથી.

(IIT-1998)

Solution

(b) Let $f(x) = y\,\, \Rightarrow \,\,x = {f^{ – 1}}(y).$

Hence$f(x) = y = 3x – 5\,\, $

$\Rightarrow \,\,x = \frac{{y + 5}}{3}\, \Rightarrow \,{f^{ – 1}}(y) = x = \frac{{y + 5}}{3}$

$\therefore \,\,{f^{ – 1}}(x) = \frac{{x + 5}}{3}$

Also $f$ is one-one and onto,

so ${f^{ – 1}}$ exists and is given by ${f^{ – 1}}(x) = \frac{{x + 5}}{3}$.

Standard 12
Mathematics

Similar Questions

Start a Free Trial Now

Confusing about what to choose? Our team will schedule a demo shortly.