Michael keeps it real

Let $\sin^2{x}=y$ . Note that since $0\le\sin^2{x}\le 1$ . we have $0\le y\le1$ .

Substituting into the equation $\sin^4{x}-\sin^2{x}-k=0$ gives $y^2-y-k=0$ .

Using the quadratic formula gives $y=\dfrac{1\pm\sqrt{1+4k}}{2}$ . Using the fact that $0\le y\le1$ , we see that $0\le\dfrac{1\pm\sqrt{1+4k}}{2}\le 1$ .

We can break the problem into two cases now: $y=\dfrac{1+\sqrt{1+4k}}{2}$ and $y=\dfrac{1-\sqrt{1+4k}}{2}$ .

First case: $y=\dfrac{1+\sqrt{1+4k}}{2}$

We have that $0\le\dfrac{1+\sqrt{1+4k}}{2}\le 1$ . Multiplying both sides by $2$ and then subtracting $1$ gives $-1\le \sqrt{1+4k}\le 1$ . Since $\sqrt{1+4k}\ge 0$ , we have that $0\le \sqrt{1+4k}\le 1$ .

Solving for $k$ gives $-\frac{1}{4}\le k\le 0$ .

Second case: $y=\dfrac{1-\sqrt{1+4k}}{2}$

We have that $0\le\dfrac{1-\sqrt{1+4k}}{2}\le 1$ . Multiplying both sides by $2$ and then subtracting $1$ gives $-1\le -\sqrt{1+4k}\le 1$ . Multiplying by $-1$ and keeping in mind to switch the direction of the inequalities gives $-1\le \sqrt{1+4k}\le 1$ , and we see that we are back at our first case.

We have that in the range $-10\le k\le 0$ , that $-\frac{1}{4}\le k\le 0$ works. This is $\dfrac{\frac{1}{4}}{10}=\dfrac{1}{40}$ of the entire range. Therefore the probability of choosing a $k$ in this range is $\dfrac{1}{40}$ and our desired answer is $1+40=\boxed{41}$ .

Let $\sin^2\theta=X$ We can rewrite the equation as $X^2-X-k=0$ Since this is a quadratic, we can apply a discriminant test. The sine function only gives real values, so therefore X will be real. This means that we need the quadratic to have at least one real root, so $b^2-4ac\ge0$ $(-1)^2-4*1*-k\ge0$ $1+4k\ge0$ $4k\ge-1$ $k\ge\left(\frac{-1}{4}\right)$ This gives us a desired range for k of $0-\left(\frac{-1}{4}\right)=\left(\frac{1}{4}\right)$ This is out of a total possible range of $0-(-10)=10$ The probability that k lies inside this range is therefore $\left(\frac{1}{4}\right)\div10=\left(\frac{1}{40}\right)$ This is in the required form so therefore the answer is $1+10=41$ $\boxed{41}$

Let us transform the original equation a bit:

$\sin^{4}\theta-\sin^{2}\theta-k=0 \Rightarrow \sin^{2}\theta(\sin^{2}\theta-1)=k \Rightarrow -\sin^{2}\theta(1-\sin^{2}\theta)=k \Rightarrow$

$\Rightarrow -\sin^{2}\theta \cos^{2}\theta=k \Rightarrow \sin^{2}\theta \cos^{2}\theta=-k \Rightarrow 4\sin^{2}\theta \cos^{2}\theta=-4k \Rightarrow$

$\Rightarrow sin^{2}(2\theta)=-4k$

Taking into account that $\sin\alpha \in [-1,1]$ , we may note that $\sin^{2}\alpha \in [0,1]$ . For $\alpha = 2\theta$ , we can see that $\sin^{2}(2\theta) \in [0,1]$ .

Therefore, the allowed values for $k$ can be found in the interval between its minimal allowed value and its maximal allowed value.

The minimum is acquired for $\sin^{2}(2\theta)=0$ , thus $-4k_{min}=0 \Rightarrow k_{min}=0$ .

The maximum is acquired for $\sin^{2}(2\theta)=1$ , thus $-4k_{max}=1 \Rightarrow k_{max}=-\frac{1}{4}$ .

At this point it is evident that the allowed values for $k$ are the values in the interval $[-\frac{1}{4},0]$ .

Considering the original domain of $k$ , which is $[-10,0]$ , we may note that $[-\frac{1}{4},0]$ is exactly $\frac{1}{40}$ of it. Therefore, the desired probability is $P=\frac{1}{40}$ .

With that in mind, if $\frac{a}{b}=\frac{1}{40}$ then $a+b=\boxed{41}$ .

PS: This is by FAR the best problem title I have seen so far. Well done!

As a quadratic in $\sin^2\theta$ , for theta to be real the discriminant $\Delta = b^2-4ac=1+4k$ must be positive. $1+4k>0 \implies k>-\frac{1}{4}$ . Thus, $k$ is in the interval $[-\frac{1}{4},0]$ .

Finally, because of the range of $y = \sin^2x$ is $[0,1]$ , not all values of $k$ in the previous interval $[-\frac{1}{4},0]$ may be valid. By the quadratic formula, $\sin^2\theta = \frac{-b+\sqrt{\Delta}}{2a} = \frac{1+\sqrt{\Delta}}{2}$ . As $\Delta=1+4k$ is between 0 and 1, $\sin^2\theta$ ranges from $[\frac{1}{2},1]\subset [0,1]$ , so there is now no doubt that the previous interval of $[-\frac{1}{4},0]$ is valid for all values of $k$ .

Thus, the probability is $\frac{0-(-\frac{1}{4})}{0-(-10)}=\frac{1}{40}$ and so $a+b=\boxed{41}$ .

(1) Uniform Probability distribution in interval [c,d] has p.d.f

$f(x)=\frac{1}{c-d}$ So in this case, $f(x)= \frac{1}{10}$

(2) $sin^{4}\theta-sin^{2}\theta-k$ is a quadratic in $sin^{2}\theta$

From the quadratic formula:

(3) $sin^{2}\theta=\frac{1±\sqrt{1+4k}}{2}$

For real solutions, $1+4k>0 \implies 4k>-\frac{1}{4}$ So k in interval $-\frac{1}{4}<k<0$

Using p.d.f,

(4) $Pr(\frac{-1}{4}<k<0)=\int^{0}_{\frac{-1}{4}} \frac{1}{10} dx = \frac{1}{40}$

(5) So $a=1, b=40, a+b=41$

I started solving it by substituting $\sin^2 \theta$ with $x$ . Thus this makes the equation: $x^2 - x - k = 0$ Using the quadratic formula, $x = \frac{1 \pm \sqrt{1+4k}}{2}$

Thus, the maximum possible value of $k$ is 0 and the minimum possible value of $\frac{-1}{4}$ . The distribution of $\frac{-1}{4}$ to 0 is $\frac{\frac{-1}{4}-0}{-10-0}$ which gives the fraction as $\frac{1}{40}$

Thus $a+b = 41$

Let $x = \sin^{2} \theta$

Now, we can rewrite the above equation as a simple quadratic:

$x^{2} - x - k = 0$

Let's calculate the probability for which there are NO real solutions to this equation. In order for this to be true, the discriminant, $(b^{2} - 4ac)$ has to be less than 0.

Substituting the given coefficients, we get:

$4k + 1 < 0$

$k < \frac{-1}{4}$

Since $k$ can only be chosen from the interval $[-10, 0]$ , it means that the probability of choosing a $k$ such that the equation has no solution is:

$\frac{-10 - \frac{-1}{4}}{-10} = \frac{39}{40}$

Hence, the probability that the equation has at least one real solution $\theta$ is:

$1 - \frac{39}{40} = \frac{1}{40}$ , where $a = 1$ and $b = 40$ .

Therefore, the value of $a + b = \boxed{41}$ .

Letting $u = sin^2\theta$ and using quadratic formula, we have:

$u = \frac{1\pm{\sqrt{1+4k}}}{2}$

Thus:

$\sin{\theta} = \sqrt{\frac{1\pm{\sqrt{1+4k}}}{2} }$ . Solving for $\theta$ :

$\theta = \sin^{-1}({\sqrt{\frac{1\pm{\sqrt{1+4k}}}{2} } })$

For $\theta$ to be real, $\sqrt{\frac{1\pm{\sqrt{1+4k}}}{2} }$ must be between $-1$ and $1$ inclusive.

Because $\sqrt{\frac{1\pm{\sqrt{1+4k}}}{2} }$ is always positive, we only check for $k$ that makes that value less than $1$ .

Squaring the radical:

$\frac{1\pm{\sqrt{1+4k}}}{2} \le 1$

Simplifying:

$\pm{\sqrt{1+4k}} \le 1$ , or just $k \le 0$

Also notice from the original equation where we isolated $\theta$ , for $\theta$ to be real, $\sqrt{4k+1}$ must be non-negative. So, $k \ge \frac{-1}{4}$ . The range $[\frac{-1}{4}, 0]$ is $\frac{\frac{1}{4}}{10}$ or $\frac{1}{40}$ of the total range. So, the probability is $\frac{1}{40}$ . The desired result is thus $40+1 =$ $41$ .

We have,

$sin^4\theta - sin^2\theta - k = 0$

$\Rightarrow$ $sin^2\theta(sin^2\theta -1)-k=0$

$\Rightarrow$ $sin^2\theta cos^2\theta + k = 0$

$\Rightarrow$ $sin^2 2\theta = -4k$ $.$

Since the value of $sin^2 2\theta$ lies between $0$ and $1$ so the possible values of $k$ lies between $0$ and $-0.25$ $.$

So, the probability is,

$0.25/10 = 1/40$

And the answer is $40 + 1 = \boxed{41}$

The first thing to notice is that we are required to calculate the probability of finding a solution $\theta$ to the equation. This gives us our first clue: there are going to be a range of values of $k$ in the given interval that satisfy the given conditions. If there were only one or two select numbers, the probability would be infinite.

The next, rather obvious thing to notice that the equation is a quadratic in $\sin^2\theta$ . We can express it so, and write its solution out in the quadratic formula:

$\sin^2\theta = \frac{1\pm \sqrt{1 + 4k}}{2}$

Now, lets look for constraints on $k$ . The first constraint is given to us: as $k$ is in the interval $[-10, 0], -10\leq k \leq 0$ . But lets narrow it down a bit more.

As we are looking for real solutions, the discriminant, $1+4k \geq 0 \implies k \geq -\frac{1}{4}$ .

Also, $\sin^2\theta\leq 1 \implies \frac{1\pm \sqrt{1 + 4k}}{2} \leq 1 \implies k \geq 0$ . So, we conclude that $k$ has to be in the interval, $[\frac{-1}{4}, 0]$ whose "size" is exactly one-fortieth of the interval given. The probability that a number will be chosen from this interval therefore is $\frac{1}{40}$ where $1$ and $40$ are co-prime. So, we get the required answer $1 + 40 = \boxed{41}$ .

Using the discriminant, let x = sin(theta), k must be greater than or equal to -1/4 in order to have at least one real solution.

Given the interval [-10, 0], k can only be in the interval [-1/4, 0].

Finding the probability, we find the distances for these intervals. The interval given has distance of 10 and the k has distance of 1/4.

P(at least one real solution) = (1/4)/10 = 1/40. Then a + b = 1 + 40 = 41. The probability

Let sin^2(theta)=y. We want to find the probability that there is a real solution of y^2-y-k=0. That means that 1+4k>=0. So, k>=1/4. Since k is chosen uniformly in the interval [-10, 0] we get probability 1/40.

the expression is (sin² Θ)² -(sin² Θ) -k =0

---> put x=sin² Θ ,& we get ==> x² - x -k =0 ,

here since x=sin² Θ --->x must lie in range [0,1] as sinΘ lies in range [-1,1]

quadratic roots for x=(1± √1+4k)/2 & discriminant,d=1+4k

OR

let the roots be represented as

♦x1 =(1+ I√d I)/2 ♦x2 =(1- I√d I)/2

===>x=x1 or x2 discriminant, d = 1+4k ≥ 0 ----for real soln.(s)

==> k ≥ -1/4

===>the range of k ε [-0.25,0] for real soln.(s)

at k= -0.25 or -1/4 --->discriminant,d =0

as k increases to its provided maximum k=0 ----->discriminant,d= 1

so as -1/4 ≤ k ≤0 -----> 0≤ d≤ 1 ------> 0≤ I√d I ≤ 1

x1 =(1+ I√d I)/2 --------> 1/2 ≤ x1 ≤ 1 (due to reason given above ↑)

x2 =(1- I√d I)/2 --------> 0 ≤ x2 ≤ 1/2 (since 0≤ I√d I ≤ 1)

hence ,x= x1 or x2 ===> 0≤x≤1 throughout the range k ε [-0.25,0]

so our initial condition (x ε [0,1] ) is satisfied

so probability of having real roots = distance in number line for [-0.25,0] to that of [-10,0]

==> a/b= I(0)- (-0.25)I / I (0) - (-10)I = 1/40

==>a+b =41

sin x ^ 4 - sin x ^ 2 will always be a -ve number [-1, 0] .. hence loose bound for k is [-1, 0] however, the minima of y^2 - y (if sin x ^ 2 is called y) occurs at y = 0.5 .. hence the max diff can only be [-0.25, 0] which is the range of k .. hence prob = 0.25/10 or 1/40 => a+b = 41

This has real solution only and only if k>=-0.25 and probability is given by (upper limit of - lower limit)/(max value - lowest value) i.e. (0-(-1/4))/(0-(-10)) solve this and you will get the right answer.