Assertion (A): Let f(x) = e^x and g(x) = log x. Then (f + g)(x) = e^x + log x.
Reason (R): Dom(f + g) = Dom(f) ∩ Dom(g).
Find the domain of f(x) = sin⁻¹(-x²).
Answer (2)
The domain of the function f ( x ) = sin − 1 ( − x 2 ) is the interval [ − 1 , 1 ] . This is because the inverse sine function is only defined for inputs between -1 and 1, and in this case, − x 2 meets that criterion when x is within [ − 1 , 1 ] . Therefore, all values of x must lie in this interval for the function to be defined.
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To solve these problems, let's break them down systematically.
Problem 1: Assertion (A) and Reason (R)
Assertion (A): Let f ( x ) = e x and g ( x ) = lo g x . Then ( f + g ) ( x ) = e x + lo g x .
Reason (R): Dom ( f + g ) = Dom ( f ) ∩ Dom ( g ) .
The assertion states that the sum of the two functions f ( x ) = e x and g ( x ) = lo g x is e x + lo g x . This is true according to the definition of adding functions where the functions are simply added together.
The reason states that the domain of the sum of two functions is the intersection of their respective domains. This is also true and is a key concept in function operations.
Now, let's find the domains:
Domain of f ( x ) = e x : The domain of the exponential function e x is all real numbers, x ∈ ( − ∞ , ∞ ) .
Domain of g ( x ) = lo g x : The domain of a logarithmic function lo g x is 0"> x > 0 .
Thus, the domain of ( f + g ) ( x ) = e x + lo g x is the intersection Dom ( f ) ∩ Dom ( g ) = ( 0 , ∞ ) .
Problem 2: Finding the Domain of f ( x ) = sin − 1 ( − x 2 )
The function sin − 1 ( z ) has a domain of − 1 ≤ z ≤ 1 . Here, we have z = − x 2 . For − x 2 to be between -1 and 1, we need:
− 1 ≤ − x 2 ≤ 1
Right Inequality: − x 2 ≤ 1
Rewriting this gives:
x 2 ≥ − 1
This condition is always satisfied since [tex]x^2[/tex] is always non-negative.
Left Inequality: − 1 ≤ − x 2
Rewriting gives:
x 2 ≤ 1
This implies:(
-1 \leq x \leq 1)
Finally, the domain of f ( x ) = sin − 1 ( − x 2 ) is x ∈ [ − 1 , 1 ] .
In summary, the domain where − x 2 fits within the required range of the inverse sine function is x between -1 and 1 inclusive.
