\(x=\frac{8}{3}\).

Solution:
Note \(x^2-1=(x-1)(x+1)\).

Put everything over common denominator \((x^2-1): Left : \frac{x-1}{x^2-1} + \frac{2(x+1)}{x^2-1} = \)

So equation becomes \(\frac{3x+1}{x^2-1} = \frac{9}{x^2-1}\). Multiply both sides by \(x^2-1\) (excluded \(x=\pm1) \):

\(3x+1=9 → 3x=8 → x=\frac{8}{3}\). Check \( x\neq\pm1\). Valid.

Two possibilities: \((a,b)=(3,1)\) or

\((a,b)=(-3,-2)\).

Solution:
Compute \(f(f(x)) = a(ax+b)+b = a^2x + ab + b\)

. Equate coefficients with \(9x+4\):

\(a^2 = 9 → a=3\) or \(a=−3\).

If \(a=3: ab+b = 3b + b = 4b = 4 → b=1\).

If \( a=−3: ab+b=−3b+b=−2b=4 \)

\(→ b=−2\).

But check \( a^2=9 \) yes; however the linear term would be \(−3(−3?)\) We already did. Both pairs produce correct polynomial:

For \(a=−3,b=−2\) we get \(a^2x + ab + b \)

\(=9x+4\).

So both are valid and click one, out of the given choices.

(Short explanation: If a line through C intersects AB at D such that CE ∥ AB, then by intercept theorem the division on AB corresponds to the sides from C to the vertices.)

Correct : \(AD : DB = 3:4 \)

Solution (conceptual/similar triangles):
Because\(CE\parallel AB\), quadrilateral ABEC is a trapezoid with \(AB\parallel CE\). Using similar triangles, triangles that share angle at C produce \(AD:DB = AC:CB = 6:8 = 3:4\)

\(x=3\) or \(x=4\).

Solution:
From \(x+y=7 ⇒ y=7−x\).

Substitute in circle equation:

\(x^2 + (7-x)^2 = 25\)
\(x^2 + 49 -14x + x^2 =25\)
\(2x^2 -14x + 49 -25 =0\)
\(2x^2 -14x +24 =0\).

Divide by \(2: x^2 -7x+12=0\)

Factor: \((x-3)(x-4)=0 ⇒ x=3\) or \(x=4\).

\((x,y)=(2,1)\) or\( (\frac{1}{3},-\frac{2}{3})\)

• Solution
• From \(x−y=1\) we have \(x=y+1\).
• Substitute into the first equation: \(\frac{1}{y+1} + \frac{1}{y} = \frac{3}{2}.\)
• Common denominator \( \frac{y + (y+1)}{y(y+1)} = \frac{3}{2} \quad\Rightarrow\quad \frac{2y+1}{y(y+1)} = \frac{3}{2}.\)
• Cross-multiply: \(2(2y+1) = 3y(y+1).\)
• Left: \(4y+2\). Right: \(3y^2+3y\)
• Bring all terms one side: \(0 = 3y^2 + 3y – 4y – 2 = 3y^2 – y – 2.\)
• Solve quadratic \(3y^2 – y – 2 = 0\). Discriminant \(D = (-1)^2 – 4(3)(-2) = 1 + 24 = 25.\). So \(y = \frac{1 \pm 5}{6}.\).
• Thus \(y = \frac{1+5}{6} = 1\) or \(y = \frac{1-5}{6} = -\frac{4}{6} = -\frac{2}{3}.\)
• Now \( x=y+1\):
• If \( y=1\), \(x=2\).
• If \(y=-\frac{2}{3}\)​, \(x = -\frac{2}{3} + 1 = \frac{1}{3}.\)

\({\frac{5}{2}\text{ or }\frac{17}{4}}\)​​

Solution

We can find the two roots but we can also use symmetric relationships. The quadratic factors:

So roots are 2 and 4.

For each root compute \(x + \frac{1}{x}\):

• If \(x=2, 2 + 1/2 = \frac{4+1}{2}=\frac{5}{2}.\)​
• If \(x=4, 4 + 1/4 = \frac{16+1}{4}=\frac{17}{4}.\)

The question asks “find the value” — the possible values are:

\(\frac{5}{2}\) and \( \frac{17}{4}\)​.

Select the one, out of given choice.

\({3+2i}\)

Solution

If a polynomial P(x) divided by \(x^2+1\) gives remainder \(2x+3\), then for any root r of \(x^2+1\), i.e. \(r=i\) or \(r=−i\), we have \(P(r)=2r+3\).

Thus \(P(i)=2i+3.\)

0

Solution

Distance from origin to line

\(y = mx + 2 \) (line in form \(mx – y +2 =0)\) is

So

Square: \(m^2+1 =1 → m^2=0\).

Thus \(m^2 = 0\).

(Interpretation: the tangent line is horizontal y=2, tangent to circle radius 2 centered at origin.)

1

Solution

If \(f \) is an involution, then \( f(f(x))=x\).

For Möbius transformations of form \(f(x)=\frac{ax+b}{cx+d}\), being its own inverse implies matrix \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\) is proportional to its inverse.

One property: fixed-point relation — easier route: if \(f(1)=2\) then applying f again: \(f(2)= f(f(1)) = 1. \)

Because \(f(f(1)) = 1\) by involution property.

Thus \(f(2)=1. \)

(exact).

Solution

If \( \angle A = 90^\circ\),

then the legs are \( AB=13\) and \(AC=15\).

Hypotenuse \( BC = \sqrt{13^2 + 15^2}\)​.

Compute:

• \(13^2 = 169\).
• \(15^2 = 225. \)
  Sum \(169 + 225 = 394\).

So hypotenuse length is \(\sqrt{394}\)​.

Altitude h from right angle to hypotenuse in a right triangle satisfies

\( h = \frac{(\text{product of legs})}{\text{hypotenuse}} = \frac{13\cdot 15}{\sqrt{394}}\).

Compute numerator \(13\cdot 15 = 195\).

So \(h = \frac{195}{\sqrt{394}}.\)

We can rationalize or simplify: note \( 394 = 2\cdot197 \), no square factors.

So final simplified exact form is \(195/\sqrt{394}\)​.

If desired, express as \(\frac{195\sqrt{394}}{394}\)​​ but original is fine.

\( \frac{9}{25}\).

Solution

Total number of 4-digit numbers with these digits, no repeat and first \(digit ≠ 0\):

• Choose first digit: 5 choices (1–5).
• Remaining 3 digits: choose any 3 from the remaining 5 digits: that’s permutations \( 5\cdot 4\cdot 3\) for the last three places.
  So total count =\( 5 \times 5 \times 4 \times 3\). Wait carefully: after choosing first digit (5 choices), for second digit there are 5 remaining digits (including 0), then 4, then 3: so total \(= 5 \cdot 5 \cdot 4 \cdot 3 = 300\).

Now count favorable numbers divisible by 5. A number is divisible by 5 if its last digit is 0 or 5.

Case 1: Last digit = 0.

• Last digit fixed (0).
• First digit: cannot be 0; choose from {1,2,3,4,5} but not 0. However 0 is already used as last, so first has 5 choices still \( (1–5) \). But if last is 0, first digit choices = 5.
• Second digit: choose from remaining digits (after choosing first and last), count = 4.
• Third digit: remaining = 3.
  So count case \(1 = 5 \cdot 4 \cdot 3 = 60\).

Case 2: Last digit = 5.

• Last digit fixed (5).
• First digit cannot be 0 and cannot be 5. So first digit choices = digits {1,2,3,4}= 4 choices.
• Second digit: choose from remaining (including 0 now) — after picking first and last, remaining digits count = 4.
• Third digit: remaining = 3.
  So count case \( 2 = 4 \cdot 4 \cdot 3 = 48\).

Total favorable \( 60 + 48 = 108\).

Probability \(= \frac{108}{300}\)

Simplify divide numerator and denominator by [/latex]12: 108/12=9, 300/12=25 [/latex].

So probability \(= \frac{9}{25}\).

\({15}\).

Solution

means \( 7 \le \sqrt{n} < 8.\)

Square inequalities:

\(7^2 \le n < 8^2 \quad\Rightarrow\quad 49 \le n < 64\).

Integers n satisfying this are \(49,50,\dots,63 \).

Count them: from 49 to 63 inclusive: \(63-49+1 = 15 \).

All these are within \(0\le n\le100\), so the count is \(15\).

or

Solution (step-by-step)

1. Factor the denominator: \(x^2 – 3x + 2 = (x-1)(x-2) \). So \(2x\ne 1,2 \).
2. Multiply both sides by the denominator (valid for \(x\ne1,2 \)):

3. Expand the right side:

4. Bring all terms to the left:

\(x^3−6x^2+7x+2=0. \).

5. Try rational roots among divisors of 2: \( \pm1,\pm2\).
6. \(x=1\) gives \(1 – 6 + 7 + 2 = 4 \neq 0\).
7. \(x=2\) gives \(8 – 24 + 14 + 2 = 0\). So \(x=2\) is a root — but recall \(x=2\) is excluded (denominator zero), so it is an extraneous root for the original equation.
8. Divide the cubic by \((x-2)\) (synthetic division or polynomial long division):
   • Synthetic division yields quotient\(x^2 – 4x – 1\).
     So

\(x^3 – 6x^2 + 7x + 2 = (x-2)(x^2 – 4x – 1)\).

7. Solve \(x^2 – 4x – 1 = 0\) using the quadratic formula:

\(x = \frac{4 \pm \sqrt{(-4)^2 – 4(1)(-1)}}{2}\) = \(\frac{4 \pm \sqrt{16 + 4}}{2} \) = \( \frac{4 \pm \sqrt{20}}{2}\)​​.

\(x = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}\)​.

8. Check domain: \(2 \pm \sqrt{5}\)​ are neither 1 nor 2, so both are valid real solutions.

All integer pairs \((m,n) \))

with \(\{( \pm 8,\pm 1),\; (\pm 1,\pm 8),\; (\pm 7,\pm 4),\; (\pm 4,\pm 7)\}. \)

Solution (step-by-step)

1. Observe \(m^2 \) and \(n^2 \) are nonnegative integers. List perfect squares \(\le 65 \):

\(0,1,4,9,16,25,36,49,64.0,1,4,9,16,25,36\),\(49,64.0,1,4,9,16,25,36,49,64. \)

2. We need two squares from that list whose sum is 65. Check candidate pairs (order matters for sign and swapping \(m,n \)):
   • \(64 + 1 = 65 \) → corresponds to \((m^2,n^2) = (64,1) \).
   • \(49 + 16 = 65 \) → corresponds to \((m^2,n^2) = (49,16) \).
   • Check other combinations quickly: \(36+25=61, 36+16=52 \), etc. No other pairs sum to 65.
3. Convert squares back to integers (considering signs and order):
   • From \(64 + 1 \): \(|m| = 8,\; \). So the integer pairs are
     \((\pm 8,\pm 1) \) and swapped \((\pm 1,\pm 8) \).
   • From \(49 + 16 \): \(|m| = 7,\; |n| = 4 \). So the integer pairs are
     \((\pm 7,\pm 4) \) and swapped \((\pm 4,\pm 7) \).
4. List all distinct integer solutions (showing signs and order):

\(\quad (7,4),(7,−4),(−7,4),(−7,−4) \).

Solution

We use Newton’s identities (power-sum relations) for the symmetric polynomials of the three numbers.
Define the power sums \(p_k = x^k + y^k + z^k. \)

The elementary symmetric sums are

\(e_1 = x+y+z = 3\),

\(\qquad e_2 = xy+yz+zx = 3\),

For a cubic, Newton’s identities give for \(k\ge 1 \): \(p_k – e_1 p_{k-1} + e_2 p_{k-2} – e_3 p_{k-3} = 0, \)

with the convention \(p_0 = 3 \) (since \(x^0+y^0+z^0=3 \)).

Compute sequentially:

• \(p_1 = x+y+z = e_1 = 3. \)
• For \(k=2 \): \(p_2 – e_1 p_1 + 2e_2 = 0 \)

(equivalently the same identity rearranged),

so using the standard form above, \( p_2 – e_1 p_1 + 2e_2 = 0 ⟹ p_2 = e_1 p_1 – 2e_2\) \( = 3\cdot3 – 2\cdot3 = 9 – 6 = 3.\)

For \(k=3 \):

\(p_3 – e_1 p_2 + e_2 p_1 – 3 e_3 = 0 \) \(⟹ p_3 = e_1 p_2 – e_2 p_1 + 3 e_3. \)​

Substitute values: \(p_3 = 3\cdot3 – 3\cdot3 + 3\cdot1 = 9 – 9 + 3 = 3. \)

• For \(k=4 \):

\(p_4 – e_1 p_3 + e_2 p_2 – e_3 p_1 = 0 \) \(⟹ p_4 = e_1 p_3 – e_2 p_2 + e_3 p_1. \)

Substitute: \(p_4 = 3\cdot3 – 3\cdot3 + 1\cdot3 = 9 – 9 + 3 = 3. \)

• For \(k=5 \):

\(p_5 – e_1 p_4 + e_2 p_3 – e_3 p_2 = 0 \) \(⟹ p_5 = e_1 p_4 – e_2 p_3 + e_3 p_2. \)

Substitute: \(p_5 = 3\cdot3 – 3\cdot3 + 1\cdot3 = 9 – 9 + 3 = 3. \)

Therefore

(Remark: one quick check:

satisfies the three given symmetric conditions, and indeed;

The Newton-sum method shows this value is forced for any triple meeting the conditions.)