Question

Find the domain and range

$f(x) = \sqrt{2x - 1}$
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Answer 1

f(x)=√2x-1

Domain : It is defined as a set of all points for which the function is real and defined

Now,the given function f(x) is a square root function and is real and defined for all the positive numbers.So,the domain of the given function f(x) is given by 2x-1 > 0.

↪x > 1/2

Domain : All real number numbers are greater than or equal to 1/2

Range : It is defined as a set of all values of dependent variable for which the function is defined.

Now, To find the range of the given function

The range consists of all the value greater than 0 or equal to 0.

↪Range : All real numbers are greater than 0 or equal to 0.

Thus, the domain is all real numbers greater than or equal to 0, and the range is all real numbers greater than or equal to 0.

Answer 2

Answer:

v = 1/2l √T/m Write the dimensions on either sides, we have L.H.S. = v =[T-1] = [M0L0T-1] R.H.S. = 1/2l = √T/m (As 1/2 has no dimensions) Tension is force and m is mass per unit length,so Read more on Sarthaks.com - https://www.sarthaks.com/563226/check-the-accuracy-of-the-relation-v-1-2l-t-m-where-v-is-the-frequency-l-is-the-length....''''''