Question

$()=\sqrt[2]{a^{2}+{x^2} }$
PLs find domain and range

Answer 1

Answer:

Step-by-step explanation:

Question 1 :

For domain,

f(x) must be a positive real number( including 0 ), which means √( 2 - 2x - x^2 ) ≥ 0

= > 2 - 2x - x^2 ≥ 0

= > x^2 + 2x - 2 ≤ 0

= > x^2 + 2x + 1 - 1 - 2 ≤ 0

= > ( x + 1 )^2 - 3 ≤ 0

= > ( x + 1 )^2 ≤ 3

= > - √3 ≤ x + 1 ≤ √3

= > - √3 - 1 ≤ x ≤ √3 - 1

Domain of this function is [ - √3 - 1 , √3 - 1 ].

Let, √( 2 - 2x - x^2 ) = a

= > a^2 = 2 - 2x - x^2

= > x^2 + 2x - 2 + a^2 = 0

= > x^2 + 2x - ( 2 - a^2 ) = 0

Since a will always be a positive real number, discriminant of this quadratic equation will always be greater than 0 or 0. It means :

= > 2^2 + 4( 1 )( 2 - a^2 ) ≥ 0

= > 4 + 4( 2 - a^2 ) ≥ 0

= > 1 + 2 - a^2 ≥ 0

= > 3 - a^2 ≥ 0

= > - a^2 ≥ - 3

= > a^2 ≤ 3

= > - √3 ≤ a ≤ √3

Since a can never be negative { a = a real number under square root }

= > 0 ≤ a ≤ √3

Range of the given function is [ 0 , √3 ].

Question 2 :

i)

For domain, f(x) must be a real number, here, x - l x l will always be a real number.

Thus domain is { R }.

For range,

For every real number, l x l ≥ x

= > l x l - x ≥ 0

= > x - l x l ≤ 0

Range is ( - ∞ , 0 ]

ii)

f(x) will always be a real number, so domain is { R }.

For range,

If x = +ve, l x l + x ≥ 0 { +ve no. + +ve no. = another +ve no. or 0 }

If x = - ve, l x l + x = 0 { - x + x = 0 }

Thus, l x l + x ≥ 0

l x l + x will always be a positive number.

Hence, range is [ 0 , ∞ ).

Answer 2

Answer:

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