TS EAMCET 6-Aug-2021 Shift 2 Question Paper

Given equations of lines are
x+2y−5‌‌=0 . . . (i)
‌‌ and ‌‌3x−4y+5‌‌=0 . . . (ii)
‌‌ and let ‌‌P‌≡[(a−1)2,a] From Eq. (i), origin O and P are on the same side.
(−5)(a−1)2+2a−5)>0
⇒‌‌a2+1−2a+2a−5<0
⇒‌‌a2<4
⇒‌‌−2<a<2 From Eq. (ii), we get
‌‌ (5) ‌[3(a−1)2−4a+5]>0
⇒‌3(a2+1−2a)−4a+5>0
⇒‌3a2−6a+3−4a+5>0
⇒‌3a2−10a+8>0
⇒‌(a−2)(3a−4)>0
⇒‌a<‌

4

3

‌ or ‌a>2 Now, Eqs. (iii) ∩ (iv),
a∈(−2,4∕3)
∴‌‌a={−1,0,1}
Hence, there are 3 points.