Question

A function given by $f\left(x\right)=\left(\frac{a{x}^2+2bx+c}{A{x}^2+2Bx+C}\right)$ has points of extremum at $x=1$ and $x=-1$ ,

such that $f\left(1\right)=2$, $f(-1)=3$ and $f\left(0\right)=2.5$

Then the Option which holds true is

• A. $a=-2.5A$
• B. $a=2.5A$
• C. $A=-2.5a$
• D. None of these

Answer 1

Answer: (B)

$a=2.5A$

Since $f\left(x\right)$ has point of extremes at $(x+1)$ & $(x-1)$ , value of $f\left(x\right)$ will remain between

$f(-1)$ & $f\left(1\right)$ i.e. 2 & 3.

$\therefore 2\le f\left(x\right)\le 3$

For points $(x+1)$ & $(x-1)$ to be extremes, denominator must be minimum at both these points. Moreover, $f\left(x\right)$ remains positive and is quadratic. Best possible solution is:

Denominator = $k[\left({x-1}^2\right)+\left({x+1}^2\right)]$

$=2k\left({x-1}^2\right)$

Comparing

$2k\left({x-1}^2\right)$ = $A{x}^2+2Bx+C$

We get:

A=2k, B=0 & C=2k

$\therefore A=C\&B=0$

Given $f\left(x\right)=\frac{a{x}^2+2bx+c}{A{x}^2+A}$

$\Rightarrow f\left(0\right)=\frac{a{\left(0\right)}^2+2b\left(0\right)+c}{A{\left(0\right)}^2+A}=\frac{c}{A}=2.5$

$\therefore c=2.5A=2.5C$

$\Rightarrow f\left(1\right)=\frac{a+2b+c}{A+A}=2$

$\therefore a+2b+c=4A$ ... (i)

$\Rightarrow f(-1)=\frac{a-2b+c}{A+A}=3$

$\therefore a-2b+c=6A$ -- (ii)

Adding (i) & (ii), we get:

$2a+2c=10A$

$2a+5A=10A$

$2a=5A$

$a=2.5A$

Hence the answer is (B)