Consider $f:R_{+} \rightarrow [-5,\infty)$ given by $f(x)=9x^{2}+6x-5$. Show that $f$ is invertible with $f^{-1}(y)=\dfrac{\sqrt{y+6}-1}{3}$. Hence find (i) $f^{-1}(10)$ and (ii) $y$ if $f^{-1}(y)=\dfrac{4}{3}$. (Here $R_{+}$ is the set of all non-negative real numbers.)

CBSE Class 12 · 2017 · Relations And Functions XIILong Answer

Consider

f : R_{+} \to [- 5, \infty)

given by

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

. Show that

f

is invertible with

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

. Hence find
(i)

f^{- 1} (10)

and
(ii)

y

f^{- 1} (y) = \frac{4}{3}

. (Here

R_{+}

is the set of all non-negative real numbers.)

Answer: $f^{- 1} (10) = 1$ ; $y = 19$

Complete the square:

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

.
For

x \geq 0

3 x + 1 > 0

, so take the positive root:

(3 x + 1)^{2} = y + 6 \Rightarrow x = \frac{y + 6 - 1}{3}

This is single-valued on

[- 5, \infty)

, confirming

f

is invertible with

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

.
(i)

f^{- 1} (10) = \frac{10 + 6 - 1}{3} = \frac{16 - 1}{3} = \frac{4 - 1}{3} = 1

.
(ii) Set

\frac{y + 6 - 1}{3} = \frac{4}{3} \Rightarrow y + 6 - 1 = 4 \Rightarrow y + 6 = 5 \Rightarrow y + 6 = 25

f^{- 1} (10) = 1, y = 19

CBSE Class 12 Board · 2017 · Subjective