Question

Let f(x)=√log(x^2+kx+k+1) /x^2+k
If f(x) is continuous for all x€R ,then the range of k is

Answer 1

The surd is just over numerator and not over denominator term

Answer 2

For this function to be continous we just have to analyze the denominator and the function inside the logarithm.
So, first of all, we will create the following body of equations:
\[x^2+kx+k+1 > 0\]
\[x^2+k \ne 0\]

Answer 3

Done! I did the same ..

Answer 4

it is a guess only.
\[x^2+kx+\frac{ k^2 }{ 4 }-\frac{ k^2 }{ 4 }+k+1>0\]
\[\left( x+\frac{ k }{ 2 } \right)^2>\frac{ k^2 }{ 4 }-k-1\]
\[\left( x+\frac{ k }{ 2 } \right)^2>\left( \frac{ k^2-4k+4-4 }{ 4 } \right)-1\]
\[\left( x+\frac{ k }{ 2 } \right)^2>\left( \frac{ k-2 }{ 2 } \right)^2-2>-2\]
i don't know if i am correct.

Answer 5

Alright! Bt what shall we do with this result??

Answer 6

@michele_laino

Answer 7

is your function, like this:
\[f\left( x \right) = \sqrt {\log \left( {\frac{{{x^2} + kx + k}}{{{x^2} + k}}} \right)} \]

Answer 8

No , sir
The surd is just over numerator

Answer 9

X^2+kx+k+1 is numerator and it is under surd sign...

Answer 10

like this one?
\[f\left( x \right) = \frac{{\sqrt {\log \left( {{x^2} + kx + k} \right)} }}{{{x^2} + k}}\]

Answer 11

+1 is also there in numerator

Answer 12

Under surd sign in addition with that term u wrote

Answer 13

here it it then:
\[f\left( x \right) = \frac{{\sqrt {\log \left( {{x^2} + kx + k + 1} \right)} }}{{{x^2} + k}}\]

Answer 14

Ryt!!!

Answer 15

oops..here it is then...

Answer 16

Np :)

Answer 17

we have to request these conditions:
\[\begin{gathered}
\log \left( {{x^2} + kx + k + 1} \right) \geqslant 0 \hfill \\
\hfill \\
{x^2} + k \ne 0 \hfill \\
\end{gathered} \]

Answer 18

from the second condition, we get:
\(k>0\)

Answer 19

from the first condition, we get:
\[{x^2} + kx + k + 1 \geqslant 1\]

Answer 20

now, if:
\[{x^2} + kx + k + 1 \geqslant 1\]
then
\[{x^2} + kx + k \geqslant 0\]

Answer 21

that above inequality, is equivalent to this one:
\[\large \left\{ {x - \left( {\frac{k}{2} + \sqrt {\frac{{{k^2}}}{4} - k} } \right)} \right\}\left\{ {x - \left( {\frac{k}{2} - \sqrt {\frac{{{k^2}}}{4} - k} } \right)} \right\} \geqslant 0\]

Answer 22

oops.. I made a typo:
\[\large \left\{ {x - \left( {\frac{{ - k}}{2} + \sqrt {\frac{{{k^2}}}{4} - k} } \right)} \right\}\left\{ {x - \left( {\frac{{ - k}}{2} - \sqrt {\frac{{{k^2}}}{4} - k} } \right)} \right\} \geqslant 0\]

Answer 23

we have the further condition:
\[\frac{{{k^2}}}{4} - k \geqslant 0 \Rightarrow k \leqslant 0 \cup k \geqslant 4\]
which gives \(k \geq 4\), since k has to be positive

Answer 24

Sir bt ans is different

Answer 25

Look at Let f(x)=√log(x^2+kx+k+1) /(x^2+k) again. I am assuming that parentheses enclose x^2+k. If this assumption is incorrect, please take a screen shot or find some other way to communicate this function to the rest of us.

Since division by zero is not defined, x^2 + k must never be zero. Note that x=0 and k = 0 is not permissible, because x^2 + k would then be zero.

Next, focus on the \[\sqrt{\log (x^2+kx+k+1)}\]

Answer 26

Look at Let f(x)=√log(x^2+kx+k+1) /(x^2+k) again. I am assuming that parentheses enclose x^2+k. If this assumption is incorrect, please take a screen shot or find some other way to communicate this function to the rest of us.

Since division by zero is not defined, x^2 + k must never be zero. Note that x=0 and k = 0 is not permissible, because x^2 + k would then be zero.

Next, focus on the \[\sqrt{\log (x^2+kx+k+1)}\]

Answer 27

Your assumption is correct parantheses has that term (x^2+k) excluded from the surd sign

Answer 28

x^2+kx+k+1>0 for log to be defined

Answer 29

I think that the subsequent conditions hold:
\[\left\{ \begin{gathered}
\log \left( {{x^2} + kx + k + 1} \right) \geqslant 0 \hfill \\
\hfill \\
{x^2} + kx + k + 1 > 0 \hfill \\
\hfill \\
{x^2} + k \ne 0 \hfill \\
\end{gathered} \right.\]

Answer 30

Regarding your post of 5 minutes ago: Yes.. x^2+kx+k+1>0 has to be greater than 0 for that log to be defined. However, the log itself (I am not speaking of x^2+kx+k+1) could be negative, which would produce a negative argument for the sqrt function and thus would not be permissible.

This is not the answer to your question, of course, but I hope this additional info to consider will be helpful.

Answer 31

Yes that would give the equation to be same as that Michele sir gave

Answer 32

Log(x^2+kx+k+1)>=0 which gives us the same equation as he gave ...
X^2+kx+k+1>=1

Answer 33

we have the subsequent logic implication:
\[\huge {x^2} + k \ne 0 \Rightarrow k > 0\]

Answer 34

Yes sir this part Is definitely true BT what about that k>=4 that u came up with

Answer 35

since we can write this:
\[\Large \log \left( {{x^2} + kx + k + 1} \right) \geqslant 0 \Rightarrow {x^2} + kx + k + 1 \geqslant 1\]

Answer 36

Sure!

Answer 37

provided that the base of logarithm is greter than \(1\)

Answer 38

greater*

Answer 39

Well! That's not mentioned in the question though...
If it's greater or less than 1

Answer 40

so, I ask, what is the base of logarithm?

Answer 41

OK sir shall I show u my work now...?

Answer 42

please wait a moment I consider the case where the base of logaritm is less than 1 and greater than zero

Answer 43

I'm doing some computations...

Answer 44

x^2+kx+k+1>=1
x^2+kx+k>=0
That means that this quadratic must have Its DETERMINANT AS <=0
So that means k^2-4k<0
That means k €[0,4]

Answer 45

Sorry open at 0

Answer 46

yes! correct!

Answer 47

I have base to be greater than 1 and most preferably it should be greater than 1 only bcoz until unless we r mentioned we have been taught that consider that case only

Answer 48

Considered*

Answer 49

Michele: The base of the log here is 10; that's a given.
So there's no point in wondering what would happen if the base were between 0 and 1.

Answer 50

ok! thanks! @mathmale

Answer 51

No, @mathmale anyone can wonder that becoz question never says that...

Answer 52

Base is greater than 1 or not

Answer 53

if base is less than 1, we get:
\[\begin{gathered}
{x^2} + kx + k + 1 \leqslant 1 \hfill \\
\hfill \\
{x^2} + kx + k \leqslant 0 \hfill \\
\end{gathered} \]

Answer 54

BT sir we can do it considering base greater than 1 becoz it's the more preferable technique and I solved the question by that and got the ans

Answer 55

more precisely:
\[\begin{gathered}
0 < {x^2} + kx + k + 1 \leqslant 1 \hfill \\
\hfill \\
- 1 < {x^2} + kx + k \leqslant 0 \hfill \\
\end{gathered} \]

Answer 56

Thanks all! I got the ans ....

Answer 57

please what is the answer?