Answer:
A
Step-by-step explanation:
[tex]\frac{x^2+x-6}{x-7} \leq 0\\[/tex]
hence numerator and denominator are of opposite sign.
case1 .
let x²+x-6≥ 0
and x-7<0
x²+x≥6
x²+x+1/4≥6+1/4
(x+1/2)²≥25/4
(x+1/2)²≥(5/2)²
|x+1/2|≥5/2
x+1/2≥5/2
x≥5/2-1/2
x≥2
or
x+1/2≤-5/2
x≤-5/2-1/2
x≤-3
also x-7<0
x<7
combining
x≤-3 or 2≤x<7
case 2.
x²+x-6<0
and x-7≥0
x²+x<6
x²+x+1/4<6+1/4
(x+1/2)²<25/4
|x+1/2|<5/2
so -5/2 <x+1/2<5/2
-5/2-1/2<x<5/2-1/2
-3<x<2
also x-7≥0
so x≥7
combining
we get no vlaue satisfying both the values.
Hence rejected.
