The domain of fx - a2-x2 a - 0 is

The correct option is B [ a, a]f(x) = √(a^2 x^2). For f(x) to be defined nbsp;a2 x2≥0⇒x2≤a2⇒ a≤x≤a∴x∈[ a, nbsp;a] nbsp; ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Domain

The domain of...

Question

The domain of $f (x) = \sqrt{a^{2} - x^{2}}$ (a > 0) is

(-a, a)

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

[-a, a]

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

[0, a]

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

(-a, 0]

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

Open in App

Solution

Suggest Corrections

108

Similar questions

f (x) = \sqrt{a^{2} - x^{2}}, a > 0

f (x) = \frac{cos m x}{x^{^{2}} + a^{2}}

∣ ∣ ∣ \int_{0}^{\infty} \frac{cos m x d x}{x^{2} + a^{2}} ∣ ∣ ∣ \leq A

A

f (x) = \frac{a + \sqrt{a^{2} - x^{2}} + x}{\sqrt{a^{2} - x^{2}} + a - x}

a > 0

x < a

f^{^{'}} (0)

{(1 + a x + b x^{2})}^{4} = a_{0} + a_{1} x + a_{2} x^{2} + . . . . . . . . . + a_{8} x^{8}

a, b, a_{0}, a_{1}, . . . . . . . a_{8} \in R

a_{0} + a_{1} + a_{2} \neq 0

∣ ∣ ∣ ∣ \begin{matrix} a_{0} & a_{1} & a_{2} a_{1} & a_{2} & a_{0} a_{2} & a_{0} & a_{1} \end{matrix} ∣ ∣ ∣ ∣ = 0

\frac{5 a}{b}

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{x^{2}}{a} - a, i f 0 < x < a 0, i f x = 0 a - \frac{a^{2}}{x^{2}}, i f x > a \end{matrix}

lim x \to a f (x)

Join BYJU'S Learning Program