Mixing
Value of $a$ for which g(x) is continuous
$begingroup$
Consider $f(x) = 4 sin(2x) $
and
$
g(x)= left{
begin{array}{ll}
ax-f(x), space x > 0 \
3a space cos(x-pi), space x leq 0 \
end{array}
right.
$
, which is continuous in $mathbb R$, $a in mathbb R $.
Which is the value of $a$ ?
$
(A) space 1 space
(B) space 2 space
(C) space 3 space
(D) space 4 space
$
So we should have
$ lim_{xto 0^+} g(x) = lim_{xto 0^-} g(x) $
But in my resolution gives
$ lim_{xto 0^+} g(x) = a times 0 space - space sin(2 times 0) = 0 - 0 = 0 $
and
$ lim_{xto 0^-} g(x) = 3a times cos (0- pi) = 3a times cos (- pi) = 3a times (-1) = -3a $
Therefore we should have $ -3a = 0 Leftrightarrow a = 0$
Which is none of the options! Where is my reasoning wrong? I'd deeply appreciate your help.
Thank you.
continuity
edited Jan 27 at 0:39
Andrei
13.2k21230
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
$endgroup$
add a comment |
$begingroup$
Consider $f(x) = 4 sin(2x) $
and
$
g(x)= left{
begin{array}{ll}
ax-f(x), space x > 0 \
3a space cos(x-pi), space x leq 0 \
end{array}
right.
$
, which is continuous in $mathbb R$, $a in mathbb R $.
Which is the value of $a$ ?
$
(A) space 1 space
(B) space 2 space
(C) space 3 space
(D) space 4 space
$
So we should have
$ lim_{xto 0^+} g(x) = lim_{xto 0^-} g(x) $
But in my resolution gives
$ lim_{xto 0^+} g(x) = a times 0 space - space sin(2 times 0) = 0 - 0 = 0 $
and
$ lim_{xto 0^-} g(x) = 3a times cos (0- pi) = 3a times cos (- pi) = 3a times (-1) = -3a $
Therefore we should have $ -3a = 0 Leftrightarrow a = 0$
Which is none of the options! Where is my reasoning wrong? I'd deeply appreciate your help.
Thank you.
continuity
edited Jan 27 at 0:39
Andrei
13.2k21230
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
$endgroup$
3
$begingroup$
There is no mistake in your answer. There might be one in the question?
$endgroup$
– Stan Tendijck
Jan 27 at 0:24
add a comment |
$begingroup$
Consider $f(x) = 4 sin(2x) $
and
$
g(x)= left{
begin{array}{ll}
ax-f(x), space x > 0 \
3a space cos(x-pi), space x leq 0 \
end{array}
right.
$
, which is continuous in $mathbb R$, $a in mathbb R $.
Which is the value of $a$ ?
$
(A) space 1 space
(B) space 2 space
(C) space 3 space
(D) space 4 space
$
So we should have
$ lim_{xto 0^+} g(x) = lim_{xto 0^-} g(x) $
But in my resolution gives
$ lim_{xto 0^+} g(x) = a times 0 space - space sin(2 times 0) = 0 - 0 = 0 $
and
$ lim_{xto 0^-} g(x) = 3a times cos (0- pi) = 3a times cos (- pi) = 3a times (-1) = -3a $
Therefore we should have $ -3a = 0 Leftrightarrow a = 0$
Which is none of the options! Where is my reasoning wrong? I'd deeply appreciate your help.
Thank you.
continuity
edited Jan 27 at 0:39
Andrei
13.2k21230
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
$endgroup$
Consider $f(x) = 4 sin(2x) $
and
$
g(x)= left{
begin{array}{ll}
ax-f(x), space x > 0 \
3a space cos(x-pi), space x leq 0 \
end{array}
right.
$
, which is continuous in $mathbb R$, $a in mathbb R $.
Which is the value of $a$ ?
$
(A) space 1 space
(B) space 2 space
(C) space 3 space
(D) space 4 space
$
So we should have
$ lim_{xto 0^+} g(x) = lim_{xto 0^-} g(x) $
But in my resolution gives
$ lim_{xto 0^+} g(x) = a times 0 space - space sin(2 times 0) = 0 - 0 = 0 $
and
$ lim_{xto 0^-} g(x) = 3a times cos (0- pi) = 3a times cos (- pi) = 3a times (-1) = -3a $
Therefore we should have $ -3a = 0 Leftrightarrow a = 0$
Which is none of the options! Where is my reasoning wrong? I'd deeply appreciate your help.
Thank you.
continuity
continuity
edited Jan 27 at 0:39
Andrei
13.2k21230
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
edited Jan 27 at 0:39
Andrei
13.2k21230
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
edited Jan 27 at 0:39
Andrei
13.2k21230
edited Jan 27 at 0:39
Andrei
13.2k21230
edited Jan 27 at 0:39
Andrei
13.2k21230
13.2k21230
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
asked Jan 27 at 0:21
BrunoCAfonsoBrunoCAfonso
756
756
3
$begingroup$
There is no mistake in your answer. There might be one in the question?
$endgroup$
– Stan Tendijck
Jan 27 at 0:24
add a comment |
3
$begingroup$
There is no mistake in your answer. There might be one in the question?
$endgroup$
– Stan Tendijck
Jan 27 at 0:24
3
3
$begingroup$
There is no mistake in your answer. There might be one in the question?
$endgroup$
– Stan Tendijck
Jan 27 at 0:24
$begingroup$
There is no mistake in your answer. There might be one in the question?
$endgroup$
– Stan Tendijck
Jan 27 at 0:24
add a comment |
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$begingroup$
There is no mistake in your answer. There might be one in the question?
$endgroup$
– Stan Tendijck
Jan 27 at 0:24