Question

If the function $f\left(x\right)=\frac{ax+b}{(x-1)(x-4)}$ has a local maxima at $(2.-1)$, then.

• A. $b=1,a=0$
• B. $a=1,b=0$
• C. $b=-1,a=0$
• D. $a=-1,b=0$

Answer 1

The correct option is B $a=1,b=0$

Taking $log$ of both sides of given equation, we get,

$lnf\left(x\right)=ln(ax+b)-ln(x-1)-ln(x-4)$

Differentiating w.r.t. $x$ we get

$f^{\prime }\left(x\right)=f\left(x\right)[\frac{a}{ax+b}-\frac{1}{x-1}-\frac{1}{x-4}]...........\left(1\right)$

As $(2,-1)$ is local maxima, so

$f\left(2\right)=-1$ and $f^{\prime }\left(2\right)=0$

Putting above values in (1) we get

$\frac{a}{2a+b}-1+\frac{1}{2}=0$

$\Rightarrow 2a=2a+b$

$\Rightarrow b=\mathrm{0.............}\left(2\right)$

Also $x=2$ is local maxima hence $f^{\prime \prime }\left(2\right)<0$

Differentiating (1) we get

$f^{\prime \prime }\left(x\right)=f^{\prime }\left(x\right)[\frac{a}{ax+b}-\frac{1}{x-1}-\frac{1}{x-4}]+f\left(x\right)[\frac{-a^2}{(ax+b{)}^2}+\frac{1}{(x-1{)}^2}+\frac{1}{(x-4{)}^2}]..........\left(1\right)$

Also $f\left(2\right)=-1$ and $f^{\prime }\left(2\right)=0$

Hence $f^{\prime \prime }\left(2\right)=[\frac{-5}{4}+\frac{a^2}{(2a+b{)}^2}]<0$

Putting $b=0$ from (2) we get

$\frac{-5}{4}+\frac{a^2}{4a^2}<0$

Which is true for all real values of $a$, hence $a$ can be any real no.

Also given $f\left(2\right)=-1$

$\Rightarrow -1=\frac{2a}{-2}$

$\Rightarrow a=1$

Hence option B is correct.