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Let f: \mathbb{N} \rightarrow R: f(x)=4 x^{2}+12 x+15 . Show that f: N \rightarrow range (f) is invertible. Find f^{-1}
CBSE | Class 12 | Exercise 2D | Functions | Maths | RS Aggarwal
Let f: \mathbb{N} \rightarrow R: f(x)=4 x^{2}+12 x+15 . Show that f: N \rightarrow range (f) is invertible. Find f^{-1}

Solution:

\mathrm{f}(\mathrm{x})=4 \mathrm{x} 2+12 \mathrm{x}+15 \quad \text { (as given) }
\mathrm{f}(\mathrm{x}) is invertible if \mathrm{f}(\mathrm{x}) is a bijection (i.e one-one onto function)
One-One function
Suppose \mathrm{p}, \mathrm{q} be two arbitrary elements in \mathrm{R}
Therefore, f(p)=f(q)
\begin{array}{l} \Rightarrow 4 \mathrm{p}_{2}+12 \mathrm{p}+15=4 \mathrm{q}_{2}+12 \mathrm{q}+15 \\ \Rightarrow 4 \mathrm{p}_{2}+12 \mathrm{p}=4 \mathrm{q}_{2}+12 \mathrm{q} \\ \Rightarrow 4 \mathrm{p}_{2}-4 \mathrm{q}_{2}+12 \mathrm{p}-12 \mathrm{q}=0 \\ \Rightarrow(\mathrm{p}-\mathrm{q})(\mathrm{p}+\mathrm{q})+3(\mathrm{p}-\mathrm{q})=0 \\ \Rightarrow(\mathrm{p}-\mathrm{q})[\mathrm{p}+\mathrm{q}+3]=0 \\ \Rightarrow \mathrm{p}-\mathrm{q}=0 \quad \text { or } \quad \mathrm{p}+\mathrm{q}+3=0 \\ \Rightarrow \mathrm{p}=\mathrm{q} \end{array}
When f(p)=f(q), p=q
So, \mathrm{f}(\mathrm{x}) is one-one function.
Onto function
Suppose \mathrm{v} be an arbitrary element of \mathrm{R} (Co-domain)
Therefore, f(x)=v
4 \mathrm{x} 2+12 \mathrm{x}+15=\mathrm{v}
On Solving, we obtain
\mathrm{x}=\frac{-3+\sqrt{v-6}}{2}
As \mathrm{v} \in \mathrm{R} \rightarrow[6, \infty]
\Rightarrow \frac{-3+\sqrt{v-6}}{2} \in \mathrm{R} \rightarrow[6, \infty]
Thus, Range of f(x)= co-domain of f(x)=[6, \infty] Thus, \mathrm{f}(\mathrm{x}) is onto function.
Thus, \mathrm{f}(\mathrm{x}) is invertible.
Now we need to find \mathrm{f}_{-1},
Suppose \mathrm{f}(\mathrm{x})=\mathrm{y}
y=4 x 2+12 x+15
Now, replace all \mathrm{x} with \mathrm{y} and all \mathrm{y} with \mathrm{x}.
Now, solve for y
\Rightarrow x=4 y_{2}+12 y+15
On Solving, we obtain
\mathrm{y}=\frac{-3+\sqrt{v-6}}{2}
Now replace y with f_{-1}(x)
As a result, \mathrm{f}_{-1}(\mathrm{x})=\frac{-3+\sqrt{v-6}}{2}

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