Question

The function $y=\frac{\mathrm{ax}+b}{\left(x-1\right)\left(x-4\right)}$ has a turning point or a point of inflexion at P(2, -1). Then, which of the following is(are) true about a and b?

• A. a = 0
• B. a = 1
• C. b = 1
• D. b = 0

Answer 1

Answer: (B), (D)

b = 0

$y=\frac{\mathrm{ax}+b}{\left(x-1\right)\left(x-4\right)}=\frac{\mathrm{ax}+b}{x^2-5x+4}\mathrm{has}a\mathrm{turning}\mathrm{point}\mathrm{at}P\left(2,-1\right).\mathrm{Thus},P\left(2,-1\right)\mathrm{lies}\mathrm{on}\mathrm{the}\mathrm{curve}.\therefore 2a+b=2\mathrm{Also},\frac{\mathrm{dy}}{\mathrm{dx}}=0\mathrm{at}P\left(2,-1\right).\mathrm{Now},\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{a\left(x^2-5x+4\right)-\left(2x-5\right)\left(\mathrm{ax}+b\right)}{{\left(x^2-5x+4\right)}^2}\mathrm{At}P\left(2,-1\right),\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-2a+2a+b}{4}=0\mathrm{or},a=1\mathrm{and}b=0.$