;P=(2,−1) Since, P(2−1) lies on the above curve. −1=‌
a(2)−b
(2−1)(2−4)
−1=‌
2a−b
−2
⇒‌‌2a−b=2....(i) y=‌
ax−b
(x−1)(x−4)
y=‌
ax−b
x2−5x+4
differentiate w.r.t. ' x ' on both sides, y′=‌
a(2)(x2−5x+4)−(ax−b)(2x−5)
(x2−5x+4)2
Since, P(2,−1) is a turning point then yP(2−1)′=0 a[22−5(2)+4]−(2a−b)(2(2)−5)=0 −2a−(2a−b)(−1)=0 b=0 Substitute b=0 in Eq. (i) 2a−0‌‌=2 a‌‌=1