$f (x) = log ∣ ∣ 4 - x^{2} ∣ ∣$ then domain of $x$ is

x \in R

except

x = \pm 2

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x \in R

except

x = \pm 1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x \in R

except

x = \pm 4

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x = \pm 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $x \in R$ except $x = \pm 2$
Here $f (x)$ is defined for $| 4 - x^{2} | > 0$

Clearly $| 4 - x^{2} |$ is greater than zero for every real $x$ except at $x = \pm 2$

Suggest Corrections

Similar questions

f (x) = \frac{1}{x^{2} - 17 x + 66}

f (\frac{2}{x - 2})

x =

f (x) = \{\begin{matrix} \frac{|x^{2} - x|}{x^{2} - x}, & x \neq 0, 1 \\ 1, & x = 0 \\ - 1, & x = 1 \end{matrix}

f (x) = x | sin x |, x \in R

Let f(x) =

$|x|=⎧⎪ ⎪⎨⎪ ⎪⎩3×(x−1)x2−3x+2 \text{for } ~~~ x\neq 1,2 −3 \text{for } ~~~ x=1 4 \text{for} ~~~ x=2$ . Then f(x) is continuous

f (x) = \frac{x^{2} - 1}{x}, g (x) = \frac{x + 2}{x - 3}

\frac{f (x)}{g (x)}

Join BYJU'S Learning Program