Matrices and Determinants 2 Question 32
36. Suppose, $f(x)$ is a function satisfying the following conditions
(a) $f(0)=2, f(1)=1$
(b) $f$ has a minimum value at $x=5 / 2$, and
(c) for all $x, f^{\prime}(x)$
$=\begin{vmatrix} 2 a x & 2 a x-1 & 2 a x+ b+1\\ b & b+1 & -1 \\ 2(a x+b) & 2 a x+2 b+1 & 2 a x+b \end{vmatrix}$
where $a, b$ are some constants. Determine the constants $a, b$ and the function $f(x)$.
(1998, 3M)
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Answer:
Correct Answer: 36.
$ a=\frac{1}{4}, b=-\frac{5}{4} $
$ f(x)=\frac{1}{4} x^{2}-\frac{5}{4} x+2 $
Solution:
- Given, $f^{\prime}(x)=\begin{vmatrix}2 a x & 2 a x-1 & 2 a x+b+1 \\ b & b+1 & -1 \\ 2(a x+b) & 2 a x+2 b+1 & 2 a x+b\end{vmatrix}$
Applying $R _3 \rightarrow R _3-R _1-2 R _2$, we get
$ \begin{aligned} & f^{\prime}(x)=\begin{vmatrix} 2 a x & 2 a x-1 & 2 a x+b+1 \\ b & b+1 & -1 \\ 0 & 0 & 1 \end{vmatrix} \\ &=\begin{vmatrix} 2 a x & 2 a x-1 \\ b & b+1 \end{vmatrix} \end{aligned} $
$=\begin{vmatrix} 2 a x & -1 \\ b & 1 \end{vmatrix}$
$f^{\prime}(x) =2ax +b$
On integrating, we get $f(x)=a x^{2}+b x+c$, where $c$ is an arbitrary constant.
Since, $f$ has maximum at $x=5 / 2$.
$ \begin{aligned} & \Rightarrow \quad f^{\prime}(5 / 2)=0 \Rightarrow 5 a+b=0 \quad…(i)\\ & \text { Also, } \quad f(0)=2 \Rightarrow c=2 \text { and } f(1)=1 \\ & \Rightarrow \quad a+b+c=1 \quad…(ii) \end{aligned} $
On solving Eqs. (i) and (ii) for $a, b$, we get
$ a=\frac{1}{4}, b=-\frac{5}{4} $
Thus,
$ f(x)=\frac{1}{4} x^{2}-\frac{5}{4} x+2 $
