Solving Inequality: Find Range of k for ##k\cos^2x-k\cos x+1≥0##

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What is the range of values of the RHS for k > 0?What is the range of values of the RHS for k < 0?What is the range of values of the RHS for k = 0?What are the conditions on k for there to be a RHS value within the LHS range?What are the conditions on k for there to be no RHS value within the LHS range?What are the conditions on k for there to be some RHS values within the LHS range and some outside?What is the combination of cases where the RHS range is completely inside the LHS range?What is the combination of cases where the RHS range is completely outside the LHS range?
  • #1
Saitama
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Homework Statement


The range of k for which the inequality ##k\cos^2x-k\cos x+1≥0## for all x, is
a. k<-1/2
b. k>4
c. -1/2≤k≤4
d. -1/2≤k≤2

Homework Equations


The Attempt at a Solution


I am not sure about how to begin with this. This seems to me a quadratic in cos(x) and here, the discriminant should be less than zero.
##k^2-4k<0##
This gives me ##0<k<4## but this is not given in any of the options.

Any help is appreciated. Thanks!
 
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  • #2
Hey Pranav!

It seems you are not done yet.
Your solution would be right if cos x could take on any value.
However, cos x is limited in range.
This means that the actual solution will have to contain your solution and perhaps have more solution values.

Is there any answer that matches that?

Alternatively, you could try and find the minima and maxima of ##\cos^2x - \cos x##.
Fill those in and solve for k.
But of course that is more work...
 
  • #3
Pranav-Arora said:

Homework Statement


The range of k for which the inequality ##k\cos^2x-k\cos x+1≥0## for all x, is
a. k<-1/2
b. k>4
c. -1/2≤k≤4
d. -1/2≤k≤2

Homework Equations



The Attempt at a Solution


I am not sure about how to begin with this. This seems to me a quadratic in cos(x) and here, the discriminant should be less than zero.
##k^2-4k<0##
This gives me ##0<k<4## but this is not given in any of the options.

Any help is appreciated. Thanks!
For one thing, the discriminant should be non-negative.

Then as I like Serena pointed out, you need be sure that k is such that -1 ≤ cos(x) ≤ 1 .
 
  • #4
SammyS said:
For one thing, the discriminant should be non-negative.
Do you mean non-positive?
 
  • #5
haruspex said:
Do you mean non-positive?

Shall we stick to less than zero? ;p
 
  • #6
I like Serena said:
Shall we stick to less than zero? ;p

No, I think SammyS' point is that it's <= 0.
Pranav, I suggest solving this by rewriting the inequality in the form <square involving cos x> <comparator> <function of k only>, but you'll have to treat different cases according to the sign of k.
 
  • #7
haruspex said:
No, I think SammyS' point is that it's <= 0.
Pranav, I suggest solving this by rewriting the inequality in the form <square involving cos x> <comparator> <function of k only>, but you'll have to treat different cases according to the sign of k.

Sorry, my mistake.
 
  • #8
I like Serena said:
Shall we stick to less than zero? ;p
Yes, I like Serena, you are right !

I misread the problem.

Of course, the discriminant must be less than or equal to zero so that [itex]\ \ k\cos^2(x)-k\cos(x)+1=0\ \ [/itex] has [STRIKE]no[/STRIKE] at most one real root .

Edited above, per haruspex (next post) .
 
Last edited:
  • #9
SammyS said:
Yes, I like Serena, you are right !

I misread the problem.

Of course, the discriminant must be less than zero so that [itex]\ \ k\cos^2(x)-k\cos(x)+1=0\ \ [/itex] has no real roots.
No, it can equal zero. One real root is ok; it just must not have two distinct real roots.
 
  • #10
Pranav-Arora;42410fg19 said:

Homework Statement


The range of k for which the inequality ##k\cos^2x-k\cos x+1≥0## for all x, is
a. k<-1/2
b. k>4
c. -1/2≤k≤4
d. -1/2≤k≤2

Homework Equations


The Attempt at a Solution


I am not sure about how to begin with this. This seems to me a quadratic in cos(x) and here, the discriminant should be less than zero.
##k^2-4k<0##
This gives me ##0<k<4## but this is not given in any of the options.

Any help is appreciated. Thanks!

The restriction that -1≤cosx≤1 can make the range of k wider than [0,4]. There can be roots of the equation ky^2-ky+1=0, but outside the interval (-1,1). The range should contain the interval [0,4] anyway. This is true for which of the given intervals?

ehild
 
Last edited:
  • #11
As I like Serena pointed out, -1 ≤ cos(x) ≤ 1 .

Therefore, there are likely more permissible values of k than just those given by considering the discriminant.

The problem at hand is equivalent to finding the values of k such that [itex]\ \ k\,t^2-k\,t+1\ge0\ \ [/itex] for -1 ≤ t ≤ 1 .
 
  • #12
Thanks everyone for the replies!

Yes, discriminant should be less equal to zero. Sorry about that.
I am not sure if I have understood all the replies properly. Following haruspex' post, I can rewrite the given inequality as
##(\cos x-\frac{1}{2})^2≥\frac{k-4}{4k}##
How should I proceed from here?

@ehild and ILS: The option C includes [0,4] so it should be the answer.
 
  • #13
Pranav-Arora said:
Thanks everyone for the replies!

Yes, discriminant should be less equal to zero. Sorry about that.
I am not sure if I have understood all the replies properly. Following haruspex' post, I can rewrite the given inequality as
##(\cos x-\frac{1}{2})^2≥\frac{k-4}{4k}##
How should I proceed from here?

@ehild and ILS: The option C includes [0,4] so it should be the answer.
See the graph of [itex]\ \displaystyle kt^2-kt+1\,,\ [/itex] with k = -1/4, restricted to -1 ≤ t ≤ 1 .

Because the domain of this function is restricted to [-1, 1], the function's range does not include any negative values.

attachment.php?attachmentid=54993&stc=1&d=1359129189.gif
from WolframAlpha

Added in Edit:

Of course, the graph of [itex]\ \displaystyle k\cos^2(x)-k\cos(x)+1\,,\ [/itex] looks nothing like the above graph, but both graphs have the same range for k = -1/4 .
 

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  • #14
Pranav-Arora said:
Thanks everyone for the replies!

Yes, discriminant should be less equal to zero. Sorry about that.
I am not sure if I have understood all the replies properly. Following haruspex' post, I can rewrite the given inequality as
##(\cos x-\frac{1}{2})^2≥\frac{k-4}{4k}##
How should I proceed from here?

That is not quite true. Depends on the sign of k.

ehild
 
  • #15
Pranav-Arora said:
I can rewrite the given inequality as
##(\cos x-\frac{1}{2})^2≥\frac{k-4}{4k}##
How should I proceed from here?
As I warned, and as ehild has pointed out, you have to split it according to the sign of k. (And k=0 as a third case.) The above is for k > 0.
What is the range of values of the LHS?
 

FAQ: Solving Inequality: Find Range of k for ##k\cos^2x-k\cos x+1≥0##

1. What is the purpose of solving inequalities?

Solving inequalities helps us determine the range of values that satisfy a given equation or expression. It allows us to find the set of values that make the inequality true, rather than just a single solution.

2. What is the first step in solving an inequality?

The first step in solving an inequality is to isolate the variable on one side of the inequality sign. This is done by using inverse operations, such as adding, subtracting, multiplying, or dividing both sides by the same number or expression.

3. How do you know which direction the inequality sign should face?

The direction of the inequality sign depends on the operations used in the expression. If the variable is multiplied or divided by a negative number, the direction of the inequality sign is flipped. Otherwise, the direction remains the same as in the original expression.

4. What is the difference between an open and closed interval?

An open interval does not include the endpoints, while a closed interval includes the endpoints. For example, (0, 5) is an open interval, while [0, 5] is a closed interval.

5. How do you graph the range of values for an inequality on a number line?

To graph the range of values, you can use a number line and plot the values that satisfy the inequality. If the inequality is inclusive of the endpoints, use a closed circle. If the inequality is exclusive of the endpoints, use an open circle. Then, shade the area between the plotted points to represent the range of values.

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