Question

Let $f:R\to R$ be an invertible and a differentiable function defined by $f\left(x\right)=\{\begin{array}{cc}{x}^2+ax-6,& x\le 2\\ \alpha x^2+\beta ,& x>2\end{array}$ where $a,\beta \in Z,\alpha \in R$ and $a\ge -5.$ If $f^{-1}$ denotes the inverse function of $f,$ then

• A. $a+4\alpha +\beta =-17$
• B. $f^{-1}(-12)=3$
• C. $\left(f^{-1}{)}^{\prime }\right(0)=-\frac{1}{7}$
• D. $\left(f^{-1}\right)(-12)=2$

Answer 1

Answer: (A), (C), (D)

$\left(f^{-1}\right)(-12)=2$

$f$ is continuous at $x=2.$
$\therefore 4+2a-6=4\alpha +\beta$
$\Rightarrow 2a-2=4\alpha +\beta \dots \left(1\right)$
$f$ is differentiable at $x=2.$
$4+a=4\alpha \dots \left(2\right)$

Also, $f$ is one-one.
$\therefore \frac{-a}{2\times 1}\ge 2$
$\Rightarrow a\le -4$
$\Rightarrow a=-5$ or $a=-4$
From equation $\left(2\right),$ at $a=-4,$ $\alpha =0$ and $f\left(x\right)$ becomes many-one. So, rejected.
Hence, $a=-5$
Solving $\left(1\right)$ and $\left(2\right),$
$\alpha =\frac{-1}{4},\beta =-11$
$\Rightarrow f\left(x\right)=\{\begin{array}{cc}{x}^2-5x-6,& x\le 2\\ \frac{-x^2}{4}-11,& x>2\end{array}$

Now, let $x^2-5x-6=-12$
$\Rightarrow x=2$ or $x=3$ (rejected)
$\therefore x=2$
Let $-\frac{x^2}{4}-11=-12$
$\Rightarrow x=\pm 2$ (rejected)
$f^{-1}(-12)=2$

Let $g\left(x\right)=f^{-1}\left(x\right)$
Since $g^{\prime }\left(f\right(x\left)\right)=\frac{1}{f^{\prime }\left(x\right)}$ and $f(-1)=0,$
$\therefore g^{\prime }\left(f\right(-1\left)\right)=\frac{1}{f^{\prime }(-1)}$
$\Rightarrow \left(f^{-1}{)}^{\prime }\right(0)=\frac{1}{f^{\prime }(-1)}=\frac{-1}{7}$