Mathematics - Functions Question with Solution | TestHub

Given $f\left(x\right)={\log }_e\left(\mathrm{s}\mathrm{i}\mathrm{n}x\right),\mathrm{ }\left(0<x<\pi \right)$ and $g\left(x\right)={\sin }^{-1}\left(e^{-x}\right),\mathrm{ }(x\ge 0),$ then we have

$fog\left(x\right)=f\left(g\left(x\right)\right)$

$=f\left({\sin }^{-1}\left(e^{-x}\right)\right)$

$={\log }_e\left(\sin \left({\sin }^{-1}\left(e^{-x}\right)\right)\right)$

$={\log }_e\left(e^{-x}\right)=-x$

$\Rightarrow fog\left(x\right)=-x$

$\Rightarrow b=f\left(g\left(\alpha \right)\right)=-\alpha$

Now ${\left(f\left(g\left(x\right)\right)\right)}^{\text{'}}=-1$

$\Rightarrow a={\left(f\left(g\left(\alpha \right)\right)\right)}^{\text{'}}=-1$

$\therefore a{\alpha }^2-b\alpha -a=\left(-1\right){\alpha }^2-\left(-\alpha \right)\alpha -\left(-1\right)$

$=-{\alpha }^2+{\alpha }^2+1=1.$