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JEE Mains · Maths · STD 12 - 1. relation and function

અહી \(f: R -\{3\} \rightarrow R -\{1\}\) એ \(f(x)=\frac{x-2}{x-3} \) દ્વારા આપેલ છે. અને  \(g: R \rightarrow R\) એ \(g ( x )=2 x -3\) દ્વારા આપેલ છે. તો \(x\) ની બધીજ કિમતોનો સરવાળો મેળવો કે જેથી  \(f^{-1}( x )+ g ^{-1}( x )=\frac{13}{2}\) થાય.

  1. A \(7\)
  2. B \(2\)
  3. C \(5\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(5\)

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Detailed explanation

\(f(x)=y=\frac{x-2}{x-3}\) \(\therefore x=\frac{3 y-2}{y-1}\) \(\therefore f^{-1}(x)=\frac{3 x-2}{x-1}\) \( g(x)=y=2 x-3\) \(\therefore x=\frac{y+3}{2}\) \(\therefore g^{-1}(x)=\frac{x+3}{2}\) \(\because f^{-1}(x)+g^{-1}(x)=\frac{13}{2}\) \(\therefore x^{2}-5 x+6=0\)…
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