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:

  1. 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 $



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