AP EAMCET Engineering 23 Sep 2020 Shift 1 Paper

Given, curve is y=‌

ax−b

(x−1)(x−4)

;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