Question

For a function, $f\left(x\right)$ satisfying the following conditions
$1\left)f\right(0)=6,f(2)=16$
$2)f$ has a minimum value at $x=-4$
$3)$ For all $x,$
$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2ax& 4ax-1& 3ax-b+3\\ b& 2b+1& 2ax+1\\ 2b-2ax& 4b-4ax+3& b+ax\end{array}\right|,$

the values of $a$ and $b$ are

• A. $a=\frac{1}{2}$
• B. $b=4$
• C. $a=\frac{1}{3}$
• D. $b=2$

Answer 1

Answer: (A), (B)

$b=4$

$f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2ax& 4ax-1& 3ax-b+3\\ b& 2b+1& 2ax+1\\ 2b-2ax& 4b-4ax+3& b+ax\end{array}\right|$

$R_3\to R_3+R_1-2R_2$
$\Rightarrow f^{\prime }\left(x\right)=\left|\begin{array}{ccc}2ax& 4ax-1& 3ax-b+3\\ b& 2b+1& 2ax+1\\ 0& 0& 1\end{array}\right|$

$\Rightarrow f^{\prime }\left(x\right)=\left(2ax\right)(2b+1)-b(4ax-1)\Rightarrow f^{\prime }\left(x\right)=2ax+b\Rightarrow f\left(x\right)=a{x}^2+bx+c$

Now, $f\left(0\right)=6\Rightarrow c=6$
$f\left(2\right)=16$
$\Rightarrow 4a+2b+c=16$
$\Rightarrow 2a+b=5\cdots \left(1\right)$

$f\left(x\right)$ has minimum value at $x=4$
$f^{\prime }\left(x\right)=0\Rightarrow 2a(-4)+b=0$
$\Rightarrow -8a+b=0$
Using equation $\left(1\right),$
$a=\frac{1}{2}$ and $b=4$