MCQ
If $y = f(x) = \frac{{x + 2}}{{x - 1}}$, then $x = $
  • $f(y)$
  • B
    $2f(y)$
  • C
    $\frac{1}{{f(y)}}$
  • D
    None of these

Answer

Correct option: A.
$f(y)$
a
(a) $y = \frac{{x + 2}}{{x - 1}}\,\,$

$\Rightarrow \,\,x = \frac{3}{{y - 1}} + 1 = \frac{{y + 2}}{{y - 1}} = f(y)$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

If $C = 2\cos \theta $, then the value of the determinant $\Delta = \left| {\,\begin{array}{*{20}{c}}C&1&0\\1&C&1\\6&1&C\end{array}\,} \right|$ is
If ${z^2} = {{{x^{1/2}} + {y^{1/2}}} \over {{x^{1/3}} + {y^{1/3}}}}$ then $x{{\partial z} \over {\partial x}} + y{{\partial z} \over {\partial y}} = $
The magnitude of the vector $\frac{1}{\sqrt{3}} \hat{i}+\frac{1}{\sqrt{3}} \hat{j}-\frac{1}{\sqrt{3}} \hat{k}$ is:
Let $f: R \rightarrow R$ be a function such that $f(x+y)=f(x)+f(y), \quad \forall x, y \in R .$ If $f(x)$ is differentiable at $x=0$, then

$(A)$ $f(x)$ is differentiable only in a finite interval containing zero

$(B)$ $f(x)$ is continuous $\forall x \in R$

$(C)$ $f^{\prime}(x)$ is constant $\forall x \in R$

$(D)$ $f(x)$ is differentiable except at finitely many points

The total revenue in Rupees received from the sale of x units of a product is given by $R(x) 2=3 x^2+36 x+5$. The marginal revenue, when $x = 15$ is:
If $y = sin^{-1 }\left( {x\sqrt {1\,\, - \,\,x} \,\,\, + \,\,\,\sqrt x \,\,\sqrt {1\, - \,{x^2}} } \right) \&\,\, \frac{{dy}}{{dx}}= \frac{1}{{2\,\sqrt {x\,(1\,\, - \,\,x)} }}+ p$, then $p =$
The angle between the lines $2 x=3 y=-z$ and $6 x=-y=-4 z$ is
The area of the parallelogram whose diagonals are $\frac{3}{2}i + \frac{1}{2}j - k$ and $2i - 6j + 8k$ is
If $f(x) = \left\{ {\begin{array}{*{20}{c}}{4x + 3\,,}&{{\rm{if}}}&{1 \le x \le 2}\\{3x + 5\,,}&{{\rm{if}}}&{2 < x \le 4}\end{array}} \right.$ then $\int_1^4 {\,f(x)} \,dx = $
Let $f(x)=x^2$ and $g(x)=2^x$. Then, the solution set of the equation fog $(x)=gof(x)$ is: