Question

x² 3x² + axt2
xtl
R=R
R.
x-2
R=R2
RitR2=5​

Answer 1

y=

x

2

−3x+2

x−p

yx

2

−3xy+2y=x−p

yx

2

+∗(−3y−1)x+2y+p=θ

For x to be real i.e. x∈R

⇒ Discriminat ≥0

⇒ (−3y−1)

2

≥4y(2y+b)

9y

2

+1+6y ≤8y

2

+4py

y

2

+(6+4p)y+1 ≤0

For this equation to be always true, its discriminat has to be ≥0

⇒ (+64b)

2

≤4

36+16p

2

+48p≤4

16p

2

+48p+32≤0

p

2

+3p+2≤0

(p+1)(p+2)≤0

⇒ p≥1 & p≤2 OR p=≤1 & p≥2 Notpossible

∴ p∈[1,2]