Question

Question number 6,7,8,9,10 on
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Answer 1

Answer on Question #56500 – Math – Algebra

6. The sequence is defined as follows

$$a _ {1} = \frac {1}{2}, \quad a _ {n + 1} = a _ {n} ^ {2} + a _ {n},$$ $$S = \frac {1}{a _ {1} + 1} + \frac {1}{a _ {2} + 1} + \dots + \frac {1}{a _ {1 0 0} + 1}$$

then find $[S]$ ($[\cdot]$ is G.I. function).

Solution. Let $b_{n} = \frac{1}{a_{n} + 1}$ and therefore $S = \sum_{n=1}^{100} b_{n}$. We have

$$b _ {1} = \frac {1}{a _ {1} + 1} = \frac {1}{\frac {1}{2} + 1} = \frac {2}{3}, b _ {2} = \frac {1}{a _ {2} + 1} = \frac {1}{\frac {1}{2} + \frac {1}{4} + 1} = \frac {4}{7}$$

and $S > b_{1} + b_{2} = \frac{2}{3} +\frac{4}{7} = \frac{26}{21} >1.$

In different ways,

$$b _ {1} = \frac {1}{a _ {1} + 1} = \frac {1}{1 + \frac {1}{2}}, b _ {2} = \frac {1}{a _ {2} + 1} = \frac {1}{\frac {1}{2} + \frac {1}{4} + 1} = \frac {1}{1 + \frac {3}{4}} = \frac {1}{1 + \frac {1}{1 + \frac {1}{3}}},$$ $$b _ {3} = \frac {1}{a _ {3} + 1} = \frac {1}{\frac {3}{4} + \frac {9}{16} + 1} = \frac {1}{2 + \frac {5}{16}} = \frac {1}{2 + \frac {1}{3 + \frac {1}{5}}}, \dots$$

It follows that $\lim_{n\to \infty}\sum_{k = 1}^{n}b_k = 2$ (by the properties of continued fractions), but $\sum_{n = 1}^{100}b_n < 2$. Thus, $1 < S < 2$ and

$$[ S ] = 1.$$

Answer: (A): 1.

7: Consider the equation $(x + iy)^{2002} = x - iy$. If the number of ordered pairs $(x, y)$ is $N$ satisfying the given equation then find the sum of digits of $N$.

Solution. We rewrite the complex number in trigonometric form

$$x + i y = z = r (\cos t + i \sin t),$$

where $t = \operatorname{Arg} z = \arctan \frac{y}{x}, r = |z| = \sqrt{x^2 + y^2}$. Then $x - iy = \bar{z} = r(\cos t - i \sin t)$, and

$$\begin{array}{l} (x + i y) ^ {2 0 0 2} = r ^ {2 0 0 2} (\cos t + i \sin t) ^ {2 0 0 2} = r ^ {2 0 0 2} (\cos 2 0 0 2 t + i \sin 2 0 0 2 t) = \\ = r (\cos t - i \sin t) \end{array}$$

or $r^{2001}(\cos 2002t + i\sin 2002t) = \cos t - i\sin t$. Because $|\cos(\cdot)| \leq 1, |\sin(\cdot)| \leq 1$ the number $r$ is bound to be equal to 1, and we have the next equivalent system

$$\cos (2 0 0 2 t) = \cos t, \quad \sin (2 0 0 2 t) = - \sin t$$

The last system has 1001 solutions, for the pairs $(x,y)$ we have 2002 solutions and so on $N = 2002$.

Answer: (A): 4.

8: If in an equation $|x| + |y| + |z| = 10$, $x, y, z \in I$. Then number of solutions is

Solution. If $x, y, z \in \mathbb{Z} \backslash \{0\}$ then number of solutions is equal to

$$\bar {P} _ {9} (2, 7) \cdot \overline {{A _ {2} ^ {3}}} = 3 6 \cdot 8 = 2 8 8$$

Let only one of the variables $x, y, z$, is equal to zero. Then number of solutions is

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$$3 \cdot \overline{A_2^2} \cdot \overline{P_9}(1,8) = 12 \cdot 9 = 108.$$

Let two of the variables $x, y, z$, is equal to zero. Then number of solutions is

$$3 \cdot 2 = 6.$$

It follows that the general number of solutions of the given equation is

$$288 + 108 + 6 = 402.$$

Answer: (B) 402.

9: It $\cot^2 x + 8 \cot x + 3 = 0, x \in [0, 2\pi]$. Then sum of all the solutions is?

**Solution.** We rewrite the given equation

$$t^2 + 8t + 3 = 0$$

where $t = \cot x$. Then $t_1 = -4 - \sqrt{13}$ and $t_2 = -4 + \sqrt{13}$. It follows that the solutions of this equation, which belong to the segment $[0, 2\pi]$, are

$x = \pi - \operatorname{arccot}(4 + \sqrt{13})$ or $x = 2\pi - \operatorname{arccot}(4 + \sqrt{13})$ or $x = \pi - \operatorname{arccot}(4 - \sqrt{13})$ or $x = 2\pi - \operatorname{arccot}(4 - \sqrt{13})$

and the sum of all the solutions is $6\pi - 2\operatorname{arccot}(4 - \sqrt{13}) - 2\operatorname{arccot}(4 + \sqrt{13})$.

Answer: (D) None of these.

10: In any $\triangle ABC$, line joining circumcentre and Incentre is parallel to $AC$ then $OI$ is equal to ($R$ is circumradius of $\triangle ABC$)?

**Solution.** If $AO = BO = CO = R$ and $OI \parallel AC$ then $\angle IOA = \angle OAC = \angle CAO$. By the law of sines we have that

$$\frac{IO}{\sin \angle ICO} = \frac{CO}{\sin \frac{C}{2}}$$

and

$$\frac{IO}{\sin \angle IAO} = \frac{AO}{\sin \frac{A}{2}}$$

Because $\angle IAO = \frac{A}{2} - \angle IOA$ and

$$\begin{array}{l} \angle ICO = \angle IOA - \frac{C}{2}, \text{ then} \\ \frac{IO}{\sin \left(\angle IOA - \frac{C}{2}\right)} = \frac{R}{\sin \frac{C}{2}}, \\ \frac{IO}{\sin \left(\frac{A}{2} - \angle IOA\right)} = \frac{R}{\sin \frac{A}{2}}, \end{array}$$

It follows that $\frac{\sin\left(\frac{A}{2} - \angle IOA\right)}{\sin\frac{A}{2}} = \frac{\sin\left(\angle IOA - \sin\frac{C}{2}\right)}{\sin\frac{C}{2}}$

and by the tangent law we obtain that $\frac{\sin\left(\frac{A}{2} - \angle IOA\right)}{\sin\frac{A}{2}} = \tan \left(\frac{A}{2} - \frac{C}{2}\right)$. Thus, $IO = R\left|\tan \left(\frac{A - C}{2}\right)\right|$.

Answer: (A) $R\left|\tan \left(\frac{A - C}{2}\right)\right|$.

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