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 $\phi( a , b , t )$.
If $\int e^x g(x)dx = = e^x(Ax^2 + Bx + C) + D$, 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)

$ \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$
(ii) $\frac{1}{5} \int e ^{ x }\left( x ^2-4 x +9\right) dx = e ^{ x }\left( Ax ^2+ Bx + C \right)+ D$
Differentiating both sides w.r.t. $x$
$\frac{1}{5} e ^{ x }\left( x ^2-4 x +9\right)= e ^{ x }(2 Ax + 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 $