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

Answer: (B)

$a=1,b=0$

for finding local maxima we have to find derivative of f(x)

so, $f^{\prime }\left(x\right)=\frac{a(x-1)(x-4)-(ax+b)(2x-5)}{\left[\right(x-1\left)\right(x-4){]}^2}$

for local maxima at x=2 , f'(2)=0

so $\frac{a(2-1)(2-4)-(2a+b)(2\ast 2-5)}{\left[\right(2-1\left)\right(2-4){]}^2}$=0

so from this we get $b=0$

and it also satisfies, $f\left(2\right)=-1$

$\frac{2a+0}{\left[\right(2-1\left)\right(2-4\left)\right]}=-1$

so from this we will get $a=1$

so $a=1,b=0$

option${}^{\prime }{B}^{\prime }$ is correct