Question

Let $f\left(x\right)=\frac{1}{x^2}$ for $x\ge 1$, and $g\left(x\right)$ is its reflection in the line mirror $y=x$, then function $h\left(x\right)=\{\begin{array}{cc}f\left(x\right)& x\ge 1\\ g\left(x\right)& 0<x<1\end{array}$, is

• A. derivable at $x=1$
• B. continuous at $x=1$
• C. not derivable at $x=1$
• D. not continuous at $x=1$

Answer 1

The correct option is A derivable at $x=1$

Let $(t,\frac{1}{t^2})$ be a point on f(x)

$\therefore$ reflection in line mirror y = x

$\therefore t=\gamma$

$\therefore \frac{\gamma -\frac{1}{t^2}}{1}=\frac{x-t}{-1}=-2\frac{(-t+\frac{1}{t^2})}{2}$

$\therefore \gamma =\frac{1}{t^2}+t-\frac{1}{t^2}$

$\therefore \gamma =t$

$\therefore x=\frac{1}{t^2}$

$\therefore x=\frac{1}{y^2}$

$\therefore x{y}^2=1$ $=9\left(x\right)$

$\therefore h^{\prime }\left(x\right)$ at $x=1$ for $x\ge 1=f^{\prime }\left(x\right)$

$=\frac{-2}{x^3}=\frac{-2}{1}=-2$

$\therefore h^{\prime }\left(x\right)$ at x = 1 for $x\le 1=9^{\prime }\left(x\right)$

$=\frac{-2}{13}=-2$

derivable at $x=1$.