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Mark the Correct Alternative in the Following Question: Let F : R − { 3 5 } → R Be Defined by F(X) = 3 X + 2 5 X − 3 Then,(A) F-1 (X) = F (X) (B) F − 1 ( X ) = − F ( X ) (C) Fo F(X) = - X(D) F − 1 - Mathematics

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प्रश्न

Mark the correct alternative in the following question:
Let f :  \[-\] \[\left\{ \frac{3}{5} \right\}\] \[\to\]  R be defined by f(x) = \[\frac{3x + 2}{5x - 3}\] Then,

 

विकल्प

  • f-1 (x) = f (x)

  • `f^-1 (x) = - f(x)`

  • fo f(x) = - x 

  • `f^-1(x) = 1/19f(x)`

MCQ

उत्तर

We have,

 f :  \[-\] \[\left\{ \frac{3}{5} \right\}\] \[\to\]  R be defined by f(x) = \[\frac{3x + 2}{5x - 3}\]

\[fof\left( x \right) = f\left( f\left( x \right) \right)\] 
\[ = f\left( \frac{3x + 2}{5x - 3} \right)\] 
\[ = \frac{3\left( \frac{3x + 2}{5x - 3} \right) + 2}{5\left( \frac{3x + 2}{5x - 3} \right) - 3}\] 
\[ = \frac{\left( \frac{9x + 6}{5x - 3} \right) + 2}{\left( \frac{15x + 10}{5x - 3} \right) - 3}\] 
\[ = \frac{\left( \frac{9x + 6 + 10x - 6}{5x - 3} \right)}{\left( \frac{15x + 10 - 15x + 9}{5x - 3} \right)}\] 
\[ = \frac{19x}{19}\] 
\[ = x\] 

\[\text{Let } y = \frac{3x + 2}{5x - 3}\] 
\[ \Rightarrow 5xy - 3y = 3x + 2\] 
\[ \Rightarrow 5xy - 3x = 3y + 2\] 
\[ \Rightarrow x\left( 5y - 3 \right) = 3y + 2\] 
\[ \Rightarrow x = \frac{3y + 2}{5y - 3}\] 
\[ \Rightarrow f^{- 1} \left( y \right) = \frac{3y + 2}{5y - 3}\] 

\[So, f^{- 1} \left( x \right) = \frac{3x + 2}{5x - 3} = f\left( x \right)\]

Hence, the correct alternative is option (a).

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अध्याय 2: Functions - Exercise 2.6 [पृष्ठ ७९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 2 Functions
Exercise 2.6 | Q 55 | पृष्ठ ७९

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