Question

Consider $\phi(a, b, t)=a^4-5 a^2+b^2+5 t^2-4 b t-2 t+\frac{33}{4}$ where $a, b, t \in R$. Given that $f(t)$ and $g(b)$ are the minimum values of $\phi( a , b , t )$.
$ \int \frac{ dx }{ f ( x )}$ is
[Note: Where C is the constant of integration]

• A. $\tan ^{-1}( x -1)+ C$
• B. $\frac{1}{2} \tan ^{-1}\left(\frac{ x -1}{2}\right)+ C$
• C. $\ln \left( x -1+\sqrt{ x ^2-2 x +2}\right)+ C$
• D. $\frac{1}{2} \ln \left|\frac{ x -1}{ x +1}\right|+ C$

Answer 1

Answer: (A)

$ \phi( a , b , t )=\left( a ^2-\frac{5}{2}\right)^2+( b -2 t )^2+( t -1)^2+1$
Now consider $E=(b-2 t)^2+(t-1)^2+1=5 t^2-2 t(2 b+1)+b^2+2$
Now $g(b)=\frac{-D}{4 a}=\frac{b^2-4 b+9}{5}=\frac{(b-2)^2+5}{5}$
Hence $g ( b )=\left.\phi( a , b , t )\right|_{\min .}=\frac{1}{5}( b -2)^2+1$
$\| l y E=b^2-4 b t+5 t^2-2 t+2$
Hence $f(t)=\left.\phi(a, b, t)\right|_{\min .} =\frac{-D}{4 a}=\frac{4 \cdot 1\left(5 t^2-2 t+2\right)-16 t^2}{4} $
$ =5 t^2-2 t+2-4 t^2=t^2-2 t+2=(t-1)^2+1$
(i) $ \int \frac{ dx }{( x -1)^2+1}=\tan ^{-1}( x -1)+ C$