Question

If $a^2+b^2=1$ and $u$ is the minimum value of $\frac{b+1}{a+b-2}$ then find the value of $u^2$.

Answer 1

Answer: 9

Put $a=\cos \theta$
$\text{ and }b=\sin \theta$
$\text{ Hence }E=\frac{b+1}{a+b-2}=\frac{\sin \theta +1}{\sin \theta +\cos \theta -2}=f(\theta )$
$\text{ Formaxima/minima, }{f}^{\prime }(\theta )=0$
$\Rightarrow (\sin \theta +\cos \theta -2)(\cos \theta )-(\sin \theta +1)(\cos \theta -\sin \theta )=0$
$\Rightarrow \sin \theta \cos \theta +{\cos }^2\theta -2\cos \theta -\sin \theta \cos \theta +{\sin }^2\theta -\cos \theta +\sin \theta =0$
$\Rightarrow {\cos }^2\theta -2\cos \theta +{\sin }^2\theta -\cos \theta +\sin \theta =0$
$\Rightarrow 1-3\cos \theta +\sin \theta =0$
$\Rightarrow 1+\sin \theta =3\cos \theta$
$\Rightarrow (1+\sin \theta {)}^2=9{\cos }^2\theta =9-9{\sin }^2\theta$
$\Rightarrow 10{\sin }^2\theta +2\sin \theta -8=0$
$\Rightarrow 5{\sin }^2\theta +\sin \theta -4=0$
$\Rightarrow 5{\sin }^2\theta +5\sin \theta -4\sin \theta -4=0$
$\Rightarrow \left({\sin }^2\theta +1\right)(5\sin \theta -4)=0$
$\Rightarrow \sin \theta =-1\text{ or }\sin \theta =\frac{4}{5}$
Now value of $E$, when $\sin \theta =-1$ is 0 .
and value of $E$ when $\sin \theta =\frac{4}{5}$ is $\frac{\sin \theta +1}{\sin \theta +\cos \theta -2}=\frac{\frac{4}{5}+1}{\frac{4}{5}+\frac{3}{5}-2}$ or $\frac{\frac{4}{5}+1}{\frac{4}{5}-\frac{3}{5}-2}$ i.e $E=\frac{-9}{3}$ or $\frac{9}{-9}$
Hence $E{]}_{\text{minimum }}=-3$ when $\sin \theta =\frac{4}{5}$ and $\cos \theta =\frac{3}{5}\Rightarrow u=-3$
Hence, $u^2=9$