If $$f\left( x \right) = \dfrac{{\left[ {2x - 1} \right]}}{{\left[ {x + 5} \right]}}$$, $$x \ne - 5$$ , then $${f^{ - 1}}\left( x \right)$$ is equal to $$\left( 1 \right){\text{ }}\dfrac{{\left[ {x + 5} \right]}}{{\left[ {2x - 1} \right]}}{\text{ }},{\text{ }}x \ne \dfrac{1}{2}$$ $$\left( 2 \right){\text{ }}\dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}}{\text{ }},{\text{ }}x \ne 2$$ $$\left( 3 \right){\text{ }}\dfrac{{\left[ {x - 5} \right]}}{{\left[ {2x + 1} \right]}}{\text{ }},{\text{ }}x \ne \dfrac{1}{2}$$ $$\left( 4 \right){\text{ }}\dfrac{{\left[ {5x - 1} \right]}}{{\left[ {2 - x} \right]}}{\text{ }},{\text{ }}x \ne 2$$

Question

If $$f\left( x \right) = \dfrac{{\left[ {2x - 1} \right]}}{{\left[ {x + 5} \right]}}$$, $$x \ne  - 5$$ , then $${f^{ - 1}}\left( x \right)$$ is equal to $$\left( 1 \right){\text{      }}\dfrac{{\left[ {x + 5} \right]}}{{\left[ {2x - 1} \right]}}{\text{  }},{\text{ }}x \ne \dfrac{1}{2}$$   $$\left( 2 \right){\text{      }}\dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}}{\text{  }},{\text{ }}x \ne 2$$   $$\left( 3 \right){\text{      }}\dfrac{{\left[ {x - 5} \right]}}{{\left[ {2x + 1} \right]}}{\text{  }},{\text{ }}x \ne \dfrac{1}{2}$$   $$\left( 4 \right){\text{      }}\dfrac{{\left[ {5x - 1} \right]}}{{\left[ {2 - x} \right]}}{\text{  }},{\text{ }}x \ne 2$$

Vedantu Content Team · Accepted Answer

Hint: According to the question we have to find the value of $${f^{ - 1}}\left( x \right)$$ .For this first we have to first replace the f(x) with y. Then solve the given function and find the value of x. Then whatever the value of x we found, we have to replace every x variable with a y variable and also replace every y variable with a x variable. Then solve it. After that replace y with  $${f^{ - 1}}\left( x \right)$$ to find the value of an inverse function. And then compare your answer with the given options and select the right one. Complete step by step solution:  Let $$f\left( x \right) = y$$ .Therefore,  $$y = \dfrac{{2x - 1}}{{x + 5}}$$       ------- (i) On multiplying $$\left( {x + 5} \right)$$ on both sides, the equation (i) becomes $$\left( {x + 5} \right)y = \dfrac{{2x - 1}}{{\left( {x + 5} \right)}}.\left( {x + 5} \right)$$  At the right hand side of the equation the $$\left( {x + 5} \right)$$ terms in the numerator and denominator will cancel out each other and we get $$\left( {x + 5} \right)y = 2x - 1$$ Now open the bracket on the left hand side, $$xy + 5y = 2x - 1$$   Now shift $$2x$$ to the left hand side and $$5y$$ to the right hand side, $$xy - 2x =  - 1 - 5y$$   By taking x common from the left hand side we get $$x\left( {y - 2} \right) =  - 1 - 5y$$  Now move $$\left( {y - 2} \right)$$ to the right hand side, $$x = \dfrac{{ - 1 - 5y}}{{\left( {y - 2} \right)}}$$   Now we will find the inverse of the above equation. To find the inverse we will replace x by y and y by x. So we have, $$y = \dfrac{{ - 1 - 5x}}{{x - 2}}$$   By taking minus common from the numerator and the denominator we get $$y = \dfrac{{ - \left( {1 + 5x} \right)}}{{ - \left( {2 - x} \right)}}$$  Now the minus signs will cancel and we get $$y = \dfrac{{1 + 5x}}{{2 - x}}$$   or $${f^{ - 1}}\left( x \right) = \dfrac{{5x + 1}}{{2 - x}}$$   And here x can not be equal to $$2$$ that is $$x \ne 2$$  Hence, the correct option is  $$\left( 2 \right){\text{      }}\dfrac{{\left[ {5x + 1} \right]}}{{\left[ {2 - x} \right]}}{\text{  }},{\text{ }}x \ne 2$$.Note: Remember all the steps to find out the value of any inverse function. While solving mistakes can be made so be careful with all the steps. Don’t forget to replace f(x) by y in the first step as it makes the rest of the process easier. While solving this question there are a couple of steps that we really need to be careful of.