मराठी

Let \[F\Left( X \Right) = \Frac{\Alpha X}{X + 1}, X \Neq - 1\] Then, for What Value of α is \[F \Left( F\Left( X \Right) \Right) = X?\] (A) \[\Sqrt{2}\] (B) \[- \Sqrt{2}\] (C) 1 (D) −1

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प्रश्न

Let  \[f\left( x \right) = \frac{\alpha x}{x + 1}, x \neq - 1\] Then, for what value of α is \[f \left( f\left( x \right) \right) = x?\]

 

पर्याय

  • \[\sqrt{2}\]

  • \[- \sqrt{2}\]

  • 1

  • -1

MCQ
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उत्तर

(d) −1

\[f\left( f\left( x \right) \right) = x\] 
\[ \Rightarrow f\left( \frac{\alpha x}{x + 1} \right) = x\] 
\[ \Rightarrow \frac{\alpha\left( \frac{\alpha x}{x + 1} \right)}{\left( \frac{\alpha x}{x + 1} \right) + 1} = x\] 
\[ \Rightarrow \frac{\alpha^2 x}{\alpha x + x + 1} = x\] 
\[ \Rightarrow \alpha^2 x = \alpha x^2 + x^2 + x\] 
\[ \Rightarrow \alpha^2 x - \alpha x^2 - x^2 - x = 0\] 
\[ \Rightarrow \alpha^2 x - \alpha x^2 - \left( x^2 + x \right) = 0\] 
\[\text{Solving } \text{for } \text{ the } \alpha \text{ we get}, \] 
\[\alpha = \frac{- \left( - x^2 \right) \pm \sqrt{\left( - x^2 \right)^2 - 4 \times x \times \left[ - \left( x^2 + x \right) \right]}}{2x}\] 
\[ = \frac{x^2 \pm \sqrt{x^4 + 4 x^3 + 4 x^2}}{2x}\] 
\[ = x + 1, - 1\] 
\[\text{Here, - 1 is independent of } x, \] 
\[ \therefore for, \alpha = - 1, f\left( f\left( x \right) \right) = x\]

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पाठ 2: Functions - Exercise 2.6 [पृष्ठ ७८]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 2 Functions
Exercise 2.6 | Q 40 | पृष्ठ ७८

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