MCQ
If $f(x) = 3x - 5$, then ${f^{ - 1}}(x)$
- AIs given by $\frac{1}{{3x - 5}}$
- ✓Is given by $\frac{{x + 5}}{3}$
- CDoes not exist because $f$ is not one-one
- DDoes not exist because $f$ is not onto
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 ${f^{ - 1}}(x) = \frac{{x + 5}}{3}$.
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$(A)$ $(f(c))^2+3 f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$
$(B)$ $(f(c))^2+f(c)=(g(c))^2+3 g(c)$ for some $c \in[0,1]$
$(C)$ $(f(c))^2+3 f(c)=(g(c))^2+g(c)$ for some $c \in[0,1]$
$(D)$ $(f(c))^2=(g(c))^2$ for some $c \in[0,1]$