Question

If $\sin^{-1}\frac{\alpha}{17} + \cos^{-1}\frac{4}{5} - \tan^{-1}\frac{77}{36} = 0, 0 < \alpha < 13$, and $A$ is set containing all the natural numbers satisfying the inequality $\cot^{-1}(n^2 - 12n + 33) > \sin\left(\cos^{-1}\sqrt{1 - \frac{\pi^2}{16}}\right)$, then which of the following statement(s) is(are) correct? (consider only the principal values of the inverse trigonometric functions)

• A. Arithmetic mean of all the values in $A$ is smaller than the value of $\alpha$
• B. The smallest element in $A$ is 5
• C. The sum of all the elements in $A$ is 21
• D. The largest element in $A$ is 8

Answer 1

Answer: (A), (B)

A, B

Answer 2

Here's how to solve this problem:

Part 1: Finding the value of $\alpha$

The given equation is $\sin^{-1}\frac{\alpha}{17} + \cos^{-1}\frac{4}{5} - \tan^{-1}\frac{77}{36} = 0$. We rewrite this as:

$$\sin^{-1}\frac{\alpha}{17} + \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{77}{36}$$

Let $X = \sin^{-1}\frac{\alpha}{17}$, $Y = \cos^{-1}\frac{4}{5}$, and $Z = \tan^{-1}\frac{77}{36}$. The equation becomes $X + Y = Z$.

From $X = \sin^{-1}\frac{\alpha}{17}$: Since $0 < \alpha < 13$, we have $0 < \frac{\alpha}{17} < \frac{13}{17} < 1$. Thus, $X$ is in the first quadrant, and:

$$\sin X = \frac{\alpha}{17}$$ $$\cos X = \sqrt{1 - \sin^2 X} = \sqrt{1 - \left(\frac{\alpha}{17}\right)^2} = \frac{\sqrt{289 - \alpha^2}}{17}$$

From $Y = \cos^{-1}\frac{4}{5}$: Since $\frac{4}{5} > 0$, $Y$ is in the first quadrant, and:

$$\cos Y = \frac{4}{5}$$ $$\sin Y = \sqrt{1 - \cos^2 Y} = \sqrt{1 - \left(\frac{4}{5}\right)^2} = \sqrt{1 - \frac{16}{25}} = \sqrt{\frac{9}{25}} = \frac{3}{5}$$

From $Z = \tan^{-1}\frac{77}{36}$: Since $\frac{77}{36} > 0$, $Z$ is in the first quadrant. We can find $\sin Z$ and $\cos Z$. Construct a right triangle with opposite side 77 and adjacent side 36. The hypotenuse will be $\sqrt{77^2 + 36^2} = \sqrt{5929 + 1296} = \sqrt{7225} = 85$.

$$\sin Z = \frac{77}{85}$$ $$\cos Z = \frac{36}{85}$$

Now, take the sine of both sides of $X+Y=Z$:

$$\sin(X+Y) = \sin Z$$

Using the sum formula for sine, $\sin X \cos Y + \cos X \sin Y = \sin Z$:

$$\left(\frac{\alpha}{17}\right)\left(\frac{4}{5}\right) + \left(\frac{\sqrt{289 - \alpha^2}}{17}\right)\left(\frac{3}{5}\right) = \frac{77}{85}$$ $$\frac{4\alpha}{85} + \frac{3\sqrt{289 - \alpha^2}}{85} = \frac{77}{85}$$

Multiply by 85:

$$4\alpha + 3\sqrt{289 - \alpha^2} = 77$$

Isolate the square root term:

$$3\sqrt{289 - \alpha^2} = 77 - 4\alpha$$

For the square root to be well-defined and equal to a non-negative value, we must have $77 - 4\alpha \ge 0$, which implies $4\alpha \le 77$, or $\alpha \le \frac{77}{4} = 19.25$.

Square both sides of the equation:

$$9(289 - \alpha^2) = (77 - 4\alpha)^2$$ $$2601 - 9\alpha^2 = 5929 - 616\alpha + 16\alpha^2$$

Rearrange into a quadratic equation:

$$25\alpha^2 - 616\alpha + 3328 = 0$$

Solve using the quadratic formula $\alpha = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:

$$\alpha = \frac{616 \pm \sqrt{(-616)^2 - 4(25)(3328)}}{2(25)}$$ $$\alpha = \frac{616 \pm \sqrt{379456 - 332800}}{50}$$ $$\alpha = \frac{616 \pm \sqrt{46656}}{50}$$

We find that $\sqrt{46656} = 216$.

$$\alpha = \frac{616 \pm 216}{50}$$

Two possible values for $\alpha$:

$$\alpha_1 = \frac{616 + 216}{50} = \frac{832}{50} = 16.64$$ $$\alpha_2 = \frac{616 - 216}{50} = \frac{400}{50} = 8$$

The problem states $0 < \alpha < 13$. $\alpha_1 = 16.64$ does not satisfy this condition. $\alpha_2 = 8$ satisfies $0 < 8 < 13$. Also, for $\alpha=8$, $77 - 4(8) = 77 - 32 = 45 \ge 0$, so it's a valid solution. Thus, $\alpha = 8$.

Part 2: Finding the set A

The inequality is $\cot^{-1}(n^2 - 12n + 33) > \sin\left(\cos^{-1}\sqrt{1 - \frac{\pi^2}{16}}\right)$.

First, evaluate the Right Hand Side (RHS): Let $K = \cos^{-1}\sqrt{1 - \frac{\pi^2}{16}}$.

For $K$ to be defined, we need $-1 \le \sqrt{1 - \frac{\pi^2}{16}} \le 1$. Since $\pi \approx 3.14$, $\pi^2 \approx 9.86$. $1 - \frac{\pi^2}{16} \approx 1 - \frac{9.86}{16} = 1 - 0.616 = 0.384$. Since $0 < 0.384 < 1$, $\sqrt{1 - \frac{\pi^2}{16}}$ is real and between 0 and 1, so $K$ is well-defined and $K \in [0, \frac{\pi}{2}]$.

$\cos K = \sqrt{1 - \frac{\pi^2}{16}}$. We need to find $\sin K$:

$$\sin K = \sqrt{1 - \cos^2 K} = \sqrt{1 - \left(1 - \frac{\pi^2}{16}\right)} = \sqrt{\frac{\pi^2}{16}} = \frac{\pi}{4}$$

Since $K \in [0, \frac{\pi}{2}]$, $\sin K \ge 0$, so we take the positive root. So, the RHS of the inequality is $\frac{\pi}{4}$.

The inequality becomes:

$$\cot^{-1}(n^2 - 12n + 33) > \frac{\pi}{4}$$

The function $\cot^{-1}x$ is a strictly decreasing function. Therefore, if $\cot^{-1}(f(n)) > \cot^{-1}(1)$ (since $\cot^{-1}(1) = \frac{\pi}{4}$), then $f(n) < 1$.

$$n^2 - 12n + 33 < 1$$ $$n^2 - 12n + 32 < 0$$

Factor the quadratic expression:

$$(n-4)(n-8) < 0$$

This inequality holds when $n$ is between the roots 4 and 8.

$$4 < n < 8$$

The set $A$ contains all natural numbers satisfying this inequality. Natural numbers are $\{1, 2, 3, \dots\}$. The natural numbers $n$ such that $4 < n < 8$ are $5, 6, 7$. So, $A = \{5, 6, 7\}$.

Part 3: Evaluating the statements

We have $\alpha = 8$ and $A = \{5, 6, 7\}$.

A. Arithmetic mean of all the values in $A$ is smaller than the value of $\alpha$. Arithmetic mean of $A = \frac{5+6+7}{3} = \frac{18}{3} = 6$. Is $6 < 8$? Yes. Statement A is correct.

B. The smallest element in $A$ is 5. The elements of $A$ are $\{5, 6, 7\}$. The smallest element is 5. Statement B is correct.

C. The sum of all the elements in $A$ is 21. Sum of elements in $A = 5+6+7 = 18$. The statement says the sum is 21, which is incorrect. Statement C is incorrect.

D. The largest element in $A$ is 8. The elements of $A$ are $\{5, 6, 7\}$. The largest element is 7. The statement says the largest element is 8, which is incorrect. Statement D is incorrect.

Therefore, statements A and B are correct.