Question

Consider $\varphi (a,b,t)=a^4-5a^2+b^2+5t^2-4bt-2t+\frac{33}{4}$ where $a,b,t\in R$. Given that $f(t)$ and $g(b)$ are the minimum values of $\varphi (a,b,t)$.
$\int \frac{dx}{f(x)}$ 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. $\ln \left(x-1+\sqrt{x^2-2x+2}\right)+C$
• D. $\frac{1}{2}\ln \left|\frac{x-1}{x+1}\right|+C$

Answer 1

Answer: (A)

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