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

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, $u^2=9$