Question
If the function $$y = \frac{{ax + b}}{{\left( {x - 1} \right)\left( {x - 4} \right)}}$$ has turning point at $$P\left( {2,\, - 1} \right),$$ then :
A.
$$a = b = 1$$
B.
$$a = b = 0$$
C.
$$a = 1,\,b = 0$$
D.
$$a = b = 2$$
Answer :
$$a = 1,\,b = 0$$
Solution :
$$y = \frac{{ax + b}}{{\left( {x - 1} \right)\left( {x - 4} \right)}} = \frac{{ax + b}}{{{x^2} - 5x + 4}}$$ has turning
point at $$P\left( {2,\, - 1} \right).$$
Thus, $$P\left( {2,\, - 1} \right)$$ lies on the curve.
Therefore $$2a + b = 2.....\left( 1 \right)$$
Also, $$\frac{{dy}}{{dx}} = 0{\text{ at }}P\left( {2,\, - 1} \right)$$
Now, $$\frac{{dy}}{{dx}} = \frac{{a\left( {{x^2} - 5x + 4} \right) - \left( {2x - 5} \right)\left( {ax + b} \right)}}{{{{\left( {{x^2} - 5x + 4} \right)}^2}}}$$
At $$P\left( {2,\, - 1} \right),\,\frac{{dy}}{{dx}} = \frac{{ - 2a + 2a + b}}{4} = 0$$
$${\text{or }}b = 0\,\,{\text{or }}a = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {{\text{From equation }}\left( 1 \right)} \right]$$