Let us consider page of size 16 bytes and process address space 32 pages and physical address space of 64 frames.
Calculate following.
A) Size of logical address i.e number of bits needed to uniquely identify a page in this address
Space of 16 pages.
Answers
Answer:
\begin{gathered}\\\;\underbrace{\underline{\sf{Understanding\;the\;Question\;:-}}}\end{gathered}
UnderstandingtheQuestion:−
Here the Concept of Quadratic Equations has been used. Here we see that we are given a quadratic equation. We know that a Quadratic Equation has 2 roots either can be equal, imaginary or distinct but here we have to go with equal case since, we need to find the value of a such that there are only one roots that is only one value of x.
Let's do it !!
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★ Formula Used :-
\begin{gathered}\\\;\boxed{\sf{For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}}\end{gathered}
Forequalroots:b
2
−4ac=0
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★ Solution :-
Given,
✒ ax² - 5x + 2 = 0
Here, a = a, b = -5 and c = 2.
We know that,
\begin{gathered}\\\;\;\sf{:\mapsto\;\;For\;equal\;roots\;:\;b^{2}\;-\;4ac\;=\;\bf{0}}\end{gathered}
:↦Forequalroots:b
2
−4ac=0
\begin{gathered}\\\;\;\sf{:\mapsto\;\;b^{2}\;-\;4ac\;=\;\bf{0}}\end{gathered}
:↦b
2
−4ac=0
By applying values, we get,
\begin{gathered}\\\;\;\sf{:\mapsto\;\;(-\:5)^{2}\;-\;4\;\times\;a\;\times\;2\;=\;\bf{0}}\end{gathered}
:↦(−5)
2
−4×a×2=0
\begin{gathered}\\\;\;\sf{:\mapsto\;\;25\;-\;8\;\times\;a\;=\;\bf{0}}\end{gathered}
:↦25−8×a=0
\begin{gathered}\\\;\;\sf{:\mapsto\;\;\;-\;8\;\times\;a\;=\;\bf{-\:25}}\end{gathered}
:↦−8×a=−25
\begin{gathered}\\\;\;\sf{:\mapsto\;\;\;8\;\times\;a\;=\;\bf{25}}\end{gathered}
:↦8×a=25
\begin{gathered}\\\;\;\underline{\underline{\bf{:\mapsto\;\;\;a\;=\;\bf{\dfrac{25}{8}}}}}\end{gathered}
:↦a=
8
25
Also, when a = 0 , we get,
✒ ax² - 5x + 2 = 0
✒ 0x² - 5x + 2 = 0
✒ -5x = -2
✒ 5x = 2
✒ x = 2/5
This also, gives single root of equation since it forms linear equation. Thus,
a = 0
is also correct answer.
\begin{gathered}\\\;\underline{\boxed{\tt{Hence\;\;for\;\;single\;\;roots,\;a\;=\;\bf{\dfrac{25}{8}\;\;\tt{or}\;\;\bf{0}}}}}\end{gathered}
Henceforsingleroots,a=
8
25
or0
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★ More to know :-
• Verification ::
We need to verify, if the equation has single roots or not for the value of a we got. Then,
✒ ax² - 5x + 2 = 0
✒ (25/8)x² - 5x + 2 = 0
✒ 25x² - 40x + 8 = 0
Now using the Quadratic Formula, we get,
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:b\;\pm\;\sqrt{b^{2}\;-\;4ac}}{2a}}\end{gathered}
⇒x=
2a
−b±
b
2
−4ac
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:40\;\pm\;\sqrt{(-\:40)^{2}\;-\;4\;\times\;25\;\times\;8}}{2\;\times\;25}}\end{gathered}
⇒x=
2×25
−40±
(−40)
2
−4×25×8
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{-\:(-\:40)\;\pm\;\sqrt{1600\;-\;1600}}{50}}\end{gathered}
⇒x=
50
−(−40)±
1600−1600
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40\;\pm\;\sqrt{0}}{50}}\end{gathered}
⇒x=
50
40±
0
\begin{gathered}\\\;\tt{\Rightarrow\;\;x\;=\;\dfrac{40}{50}\;=\;\dfrac{4}{5}}\end{gathered}
⇒x=
50
40
=
5
4
Clearly,
\begin{gathered}\\\;\bf{\Rightarrow\;\;x\;=\;\dfrac{4}{5}\;or\;\dfrac{4}{5}}\end{gathered}
⇒x=
5
4
or
5
4
This is because, this is Quadratic equation so it must have two equal roots.
This gives that this equation has only single root that is 4/5 when a = 25/8
So our answer is correct.
Hence, Verified.
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★ Supplementary Counsel :-
\begin{gathered}\\\;\sf{\leadsto\;\;For\;imaginary\;roots\;:\;b^{2}\;-\;4ac\; < \;0}\end{gathered}
⇝Forimaginaryroots:b
2
−4ac<0
\begin{gathered}\\\;\sf{\leadsto\;\;For\;distinct\;roots\;:\;b^{2}\;-\;4ac\; > \;0}\end{gathered}
⇝Fordistinctroots:b
2
−4ac>0
Explanation:
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