Question

if x > 100, then \( \sqrt{x+16} \) is between which of the two given limits?
options are ------
A) \( \sqrt{x} \) and \( \sqrt{x} \) +1
B) \( \sqrt{x} \) +1 and \( \sqrt{x} \) +2
C) \( \sqrt{x} \) +2 and \( \sqrt{x} \) +3
D) \( \sqrt{x} \) +3 and \( \sqrt{x} \) +4

Answer 1

try it with a number and see

Answer 2

ok

Answer 3

can u tell me in detail

Answer 4

Let us take a trial with x = 105

\( \mathsf {\sqrt{x+16} = \sqrt{\color{blue}{105} + 16}\ = \sqrt{121} = \bf{11} }\)

Take this into consideration now:

\(\mathsf{\sqrt{x} = \sqrt{\color{blue}{105}}} \\
\text{Will be between 10 and 11, quite close to 10 though.} \)

Clearly, since, \(\sqrt{x+16}\) for \(x=105\) has come out to be 11, and therefore we will require limit having 11 in it.

Use your logical reasoning now, from the options, pick out the option which will contain \(11\) for \(x = 105\) .

For me, seems quite logical to say that square root of 105 will nearly lie from 10 to 10.5

With this point, you can strongly conclude the correct answer as with \(\sqrt{x} \approx 10 \ \text{to} \ 10.5 \), there is only one option suitable for it.

Answer 5

To make it easier, I'll suggest that you go up for any perfect square as x.

Like, take \(x = 121\)

\(\sqrt{x+16} = \sqrt{121+16} = \sqrt{137} \approx 11.5 \ \text{to} \ 12 \)
(as 144 is quite close to 137 )

\(\sqrt{x} = \sqrt{121} = 11\)
and thus, \(\sqrt{x+16}\) (which is somewhat between 11.5 to 12) will surely lie between \(\sqrt{x}\) (=11) to \(\sqrt{x}+ 1\) (=12)