Let $f:N \rightarrow N$ be a function defined as $f(x)=9x^{2}+6x-5$. Show that $f:N \rightarrow S$, where $S$ is the range of $f$, is invertible. Find the inverse of $f$ and hence find $f^{-1}(43)$ and $f^{-1}(163)$.

$f^{-1}(x)=\dfrac{\sqrt{x+6}-1}{3}$; $f^{-1}(43)=2$, $f^{-1}(163)=4$

CBSE Class 12 · 2016 · Relations And Functions XIILong Answer

Let

f : N \to N

be a function defined as

f (x) = 9 x^{2} + 6 x - 5

. Show that

f : N \to S

, where

S

is the range of

f

, is invertible. Find the inverse of

f

and hence find

f^{- 1} (43)

and

f^{- 1} (163)

Answer: $f^{- 1} (x) = \frac{x + 6 - 1}{3}$ ; $f^{- 1} (43) = 2$ , $f^{- 1} (163) = 4$

Restricting the codomain to the range

S

makes

f

onto by construction; we show it is one-one, hence invertible.
Complete the square:

y = 9 x^{2} + 6 x - 5 = (3 x + 1)^{2} - 6

(3 x + 1)^{2} = y + 6 \Rightarrow 3 x + 1 = y + 6 (positive root, since x \in N)

\Rightarrow x = \frac{y + 6 - 1}{3}

This gives a well-defined single-valued inverse, so

f

is invertible with

f^{- 1} (x) = \frac{x + 6 - 1}{3}

f^{- 1} (43) = \frac{49 - 1}{3} = \frac{7 - 1}{3} = 2

and

f^{- 1} (163) = \frac{169 - 1}{3} = \frac{13 - 1}{3} = 4

f^{- 1} (x) = \frac{x + 6 - 1}{3}, f^{- 1} (43) = 2, f^{- 1} (163) = 4

CBSE Class 12 Board · 2016 · Subjective