Application of Derivatives 4 Question 43
####45. Let $f(x)$ is a function satisfying the following conditions
(i) $f(0)=2, f(1)=1$
(ii) $f(x)$ has a minimum value at $x=5 / 2$ and
(iii) For all $x, f^{\prime}(x)=\left|\begin{array}{ccc}2 a x & 2 a x-1 & 2 a x+b+1 \\ b & b+1 & -1 \\ 2(a x+b) & 2 a x+2 b+1 & 2 a x+b\end{array}\right|$
where, $a$ and $b$ are some constants. Determine the constants $a, b$ and the function $f(x)$.
$(1998,8$ M)
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Answer:
Correct Answer: 45. $\frac{4 \sqrt{3}}{9}$
Solution:
- Given, $f^{\prime}(x)=\left|\begin{array}{ccc}2 a x & 2 a x-1 & 2 a x+b+1 \\ b & b+1 & -1 \\ 2(a x+b) & 2 a x+2 b+1 & 2 a x+b\end{array}\right|$
Applying $R_{3} \rightarrow R_{3}-R_{1}-2 R_{2}$, we get
$ f^{\prime}(x)=\left|\begin{array}{ccc} 2 a x & 2 a x-1 & 2 a x+b+1 \\ b & b+1 & -1 \\ 0 & 0 & 1 \end{array}\right| $
$ \Rightarrow \quad f^{\prime}(x)=2 a x+b $
On integrating both sides, we get
$ f(x)=a x^{2}+b x+c $
Since, maximum at $x=5 / 2 \Rightarrow f^{\prime}(5 / 2)=0$
$ \Rightarrow \quad 5 a+b=0 $
Also, $\quad f(0)=2 \quad \Rightarrow \quad c=2$
and $\quad f(1)=1 \quad \Rightarrow \quad a+b+c=1$
On solving Eqs. (i), (ii) and (iii), we get
$ a=\frac{1}{4}, b=-\frac{5}{4}, c=2 $
Thus,
$ f(x)=\frac{1}{4} x^{2}-\frac{5}{4} x+2 $
