The domain of f(x)=(x/16 - x2)+(l o g)2(x3 - 2 x) is

Question

The domain of f(x)=(x/16 - x2)+(l o g)2(x3 - 2 x) is (A) (- √2 , 0)∪(√2 , ∈ fty) (B) (- √2 , 0)∪(√2 , 4) (C) (- √2 , 0)∪(4 , ∈ fty) (D) (- √2 , 0)∪(√2 , 4)∪(4 , ∈ fty)

Tardigrade Admin · Accepted Answer

16-x2≠ 0 and x3-2x>0 x≠ 4,-4 ldots ..(i) and x(x2 - 2)>0 x(x + √2)(x - √2)>0 <img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/m-dkdddevfjwka.png" /> x∈ (- √2 , 0)∪(√2 , ∈ fty) ldots (i i) Taking intersection of (i) &(i i) , we get, x∈ (- √2 , 0)∪(√2 , 4)∪(4 , ∈ fty)

Q. The domain of $f (x) = \frac{x}{16 - x ^{2}} + (l o g)_{2} (x^{3} - 2 x)$ is

$(- 2, 0) \cup (2, \in f t y)$

$(- 2, 0) \cup (2, 4)$

$(- 2, 0) \cup (4, \in f t y)$

$(- 2, 0) \cup (2, 4) \cup (4, \in f t y)$

$16 - x^{2} \neq = 0$ and $x^{3} - 2 x > 0$
$x \neq = 4, - 4 \dots .. (i)$
and $x (x^{2} - 2) > 0$
$x (x + 2) (x - 2) > 0$

$x \in (- 2, 0) \cup (2, \in f t y) \dots (ii)$
Taking intersection of $(i) & (ii)$ , we get,
$x \in (- 2, 0) \cup (2, 4) \cup (4, \in f t y)$

Q. The domain of f(x)=16−x2x​+(log)2​(x3−2x) is

(−2​,0)∪(2​,∈fty)

(−2​,0)∪(2​,4)

(−2​,0)∪(4,∈fty)

(−2​,0)∪(2​,4)∪(4,∈fty)