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  • Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 9 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Function Exercise 2.4 Question 9 Maths Textbook Solution.

Answers (1)

Answer:

fis invertible with the function.

                f^{-1}\left ( x \right )=\frac{\sqrt{x-6}-1}{3}

Given:

                f:R_{+}\rightarrow \left [ 5,\infty \right ] given by f\left ( x \right )=9x^{2}+6x-5

Hint:

Bijection function should fulfil the injection and surjection condition.

Solution:

Let  x,y be two elements of domain\left ( R^{+} \right ), such thatf\left ( x \right )=f\left ( y \right ).

                \! \! \! \! \! \! \! \! 9x^{2}+6x-5=9y^{2}+6y-5\\\\9x^{2}+6x=9y^{2}+6y\\\\x=y\left ( as,x,y\: \epsilon \: R^{+} \right )

                fis one-one.

Subjectivity of f:

Let y is in the co-domain\left ( Q \right ) such that f\left ( x \right )=y

\! \! \! \! \! \! \! \! \! 9x^{2}+6x-5=y\\\\9x^{2}+6x=y+5

9x^{2}+6x+1=y+6                                                                                 [Adding 1 on both sides]
   \left ( 3x+1 \right )^{2}=6+y\\\\3x+1=\sqrt{y+6}\\\\3x=\sqrt{y+6}-1\\\\x=\frac{\sqrt{y+6}1}{3}\: \epsilon \: R^{+}\left ( domain \right )

                f is onto.

So, f is a bijection and hence, it is invertible.

f^{-1}\left ( x \right )=y                                                                                                                                       … (i)

\! \! \! \! \! \! \! \! x=f\left ( y \right )\\\\x=9y^{2}+6y-5\\\\x+5=9y^{2}+6y

x+6=\left ( 3y+1 \right )^{2}                                                                                 [Adding  on both sides]

 

 

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

f^{-1}\left ( x \right )=\frac{\sqrt{x+6}-1}{3}                                                                                                         [From (i)]

 

 

Posted by

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