Let g(x)=x2−4x+9 ‌D<0 ‌g(x)>0‌ for ‌x∈R
∴[
f(−5)=0
f(3)=0
. So, f(x) is many-one. again, ‌yx2−4xy+9y=x2+2x−15 ‌x2(y−1)−2x(2y+1)+(9y+15)=0 ‌‌ for ‌∀x∈R⇒D≥0 ‌D=4(2y+1)2−4(y−1)(9y+15)≥0 ‌5y2+2y+16≤0 ‌(5y−8)(y+2)≤0
y∈[−2,‌
8
5
]‌ range ‌ Note : If function is defined from f:R⟶R then only correct answer is option (3) ⇒ Bonus