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 $\varphi (a,b,t)$.
If $\int e^{x}g(x)dx==e^x(A{x}^2+Bx+C)+D$, where $D$ is constant of integration, then $(A+B+C)$ is equal to

• A. 2
• B. $\frac{14}{5}$
• C. 5
• D. $\frac{19}{5}$

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 $f(t)={\varphi (a,b,t)|}_{\min .}=\frac{-D}{4a}=\frac{4\cdot 1\left(5t^2-2t+2\right)-16t^2}{4}$
$=5t^2-2t+2-4t^2=t^2-2t+2=(t-1{)}^2+1$
(ii) $\frac{1}{5}\int e^x\left(x^2-4x+9\right)dx=e^x\left(A{x}^2+Bx+C\right)+D$
Differentiating both sides w.r.t. $x$
$\frac{1}{5}{e}^x\left(x^2-4x+9\right)=e^x(2Ax+B)+\left(A{x}^2+Bx+C\right)e^x$
$A=\frac{1}{5},B=\frac{-6}{5},C=3$
$A+B+C=\frac{1}{5}-\frac{6}{5}+3=2$