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1.Relation and Function
easy
ધારો કે, $f: N \rightarrow Y $ એ $f(x)=4 x+3$ દ્વારા વ્યાખ્યાયિત વિધેય છે, જ્યાં $Y =\{y \in N :$ કોઈક $x \in N$ માટે $y=4 x+3$ $\} $. સાબિત કરો કે $f$ વ્યસ્તસંપન્ન છે. આ વિધેયનું પ્રતિવિધેય શોધો.
Option A
Option B
Option C
Option D
Solution
Consider an arbitrary element $y$ of $Y$. By the definition of $Y, y=4 x+3$ for some $x$ in the domain $N$. This shows that $x=\frac{(y-3)}{4} .$ Define $g: Y \rightarrow N$ by $g(y)=\frac{(y-3)}{4} .$ Now, $gof\,(x)=g(f(x))=g(4 x+3)$ $=\frac{(4 x+3-3)}{4}=x$ and $fog\,(y)=f(g(y))=f\left(\frac{(y-3)}{4}\right)$ $=\frac{4(y-3)}{4}+3=y-3+3=y .$ This shows that $gof= I _{ N }$ and $f o g=I_{Y}$, which implies that $f$ is invertible and $g$ is the inverse of $f$
Standard 12
Mathematics
