Let f(x)=√ log 2((10 x-4/4-x2))-1. Then sum of all integers in domain of f(x) is

Question

Let f(x)=√ log 2((10 x-4/4-x2))-1. Then sum of all integers in domain of f(x) is (A) - 15 (B) - 16 (C) - 17 (D) - 18

Tardigrade Admin · Accepted Answer

log 2((10 x-4/4-x2)) ≥ 1 ⇒ (10 x-4/4-x2)-2 ≥ 0 ⇒ (10 x-4-8+2 x2/4-x2) ≥ 0 ⇒ (2 x2+10 x-12/x2-4) ≤ 0 ∴ x ∈[-6,-2) ∪[1,2) ∴ x= -6,-5,-4,-3,1 Hence sum =-17

Q. Let f(x)=log2​(4−x210x−4​)−1​. Then sum of all integers in domain of f(x) is

- 15

- 16

- 17

- 18

log2​(4−x210x−4​)≥1⇒4−x210x−4​−2≥0⇒4−x210x−4−8+2x2​≥0 ⇒x2−42x2+10x−12​≤0 ∴x∈[−6,−2)∪[1,2) ∴x={−6,−5,−4,−3,1} Hence sum =−17

Q. Let $f (x) = lo g_{2} (\frac{10 x - 4}{4 - x ^{2}}) - 1$ . Then sum of all integers in domain of $f (x)$ is