IOQM mock test 3

 Q1.

Define $n_a!$ for $n$ and $a$ positive to be $n_a! = n(n - a)(n - 2a)(n - 3a) \ldots (n - ka)$ where $k$ is the greatest integer for which $n > ka$. If the quotient $72_8! / 18_2!$ is equal to $2^x$, find $x$.

Solution:
Given number equals:
72.(72-8).(72-16)...(72-64)/18.(18-2).(18-4)...(18-16)
= 4^9 = 2^18

Q2.

 If 

$p, q$ are two real numbers satisfying the relationships $2p^2 - 3p - 1 = 0$ and $q^2 + 3q - 2 = 0$ and $pq \ne 1$, find the value of $\frac{pq + p + 1}{q}$.
Solution:
Clearly p, 1/q are roots of 2x^2 - 3x - 1 = 0.
p + 1/q = 3/2
p/q = -1/2
(pq + p + 1)/q = p + p/q + 1/q = 1 = Answer

Q3.

Convex quadrilateral ABCD has $AB = 9$ and $CD = 12$. Diagonals AC and BD intersect at E, $AC = 14$, and $\triangle AED$ and $\triangle BEC$ have equal areas. What is AE?
Solution:
A trapezium has the property that of the 4 triangles formed by its diagonals - 2 between the parallel lines are similar and the other 2 have equal areas.
Given the information in the question we can assume that AEB and CED are similar with side ratio of 3:4. This gives AE = 6 and EC = 8. Answer = 6.

Q4.

In $\triangle ABC$, $\angle ACB = 60^\circ$, $\angle BAC = 75^\circ$, $AD \perp BC$ at D, $BE \perp AC$ at E, AD intersects BE at H. Find $\angle CHD$ in degrees.
Solution:
Since H is the orthocenter, CH once extended will meet AB as a perpendicular.
Using this information we can find all the angles formed thus.
It will give us angle CHD = 45 degrees.

Q5.

Find the number of rectangles that can be obtained by joining four of the twelve vertices of a 12-sided regular polygon.
Solution:
Every regular polygon has a circumcircle and an incircle.
A rectangle formed using the vertices of this regular 12-gon will have all the angles inscribed on the circumference of the circumcircle.
All the angles of a rectangle are 90 degrees.
It means that both the diagonals of the rectangle are diameters of the circumcircle by Thales' theorem.
A 12-gon has 6 diameters which pass through 2 of its vertices.
For e.g. vertex 1,7 2,8 (i.e. all the vertices with gap of 6 in between).
So such a rectangle can be formed by using 2 of those 6 diameters in 6C2 = 15 ways.
So answer = 15.

What if the question changed from 12-gon to hexagon?
Then it will be 3C2 = 3.
For 10-gon: 5C2 = 10.
For a quadrilateral: 2C2 = 1.

And as is clear, this can be done only in a regular polygon with even number of vertices as in a odd-sided polygon there are no diameters of the circumcircle which pass through 2 of its vertices.

Q6.

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In $\triangle ABC$, $AB = 9$, $BC = 8$ and $AC = 7$. The bisector of $\angle A$ meets $BC$ at $D$. The circle passing through $A$ and touching $BC$ at $D$ cuts $AB$ and $AC$ at $M$ and $N$ respectively. Find $MN$.

Solution:

By the angle bisector theorem,

$$\frac{BD}{DC} = \frac{AB}{AC} = \frac{9}{7}$$

and so

$$BD = 8 \times \frac{9}{9 + 7} = \frac{9}{2}.$$

By the power chord theorem,

$$BD^2 = BM \times BA$$

and so

$$BM = \frac{9}{4}.$$

Finally, $MN$ is parallel to $BC$ since

$$\angle BDM = \angle MAD = \angle DAN = \angle DMN.$$

Let's break the above step into smaller steps.
∠BDM=∠MAD by alternate segment theorem.
∠MAD=∠DAN given in the question.
∠DAN=∠DMN because angles inscribed by the same chord are same. The chord is DN here.
Now since we have ∠BDM=∠DMN it means MN || BC by Alternate Interior Angles Theorem.

Hence we have

$$\frac{AM}{AB} = \frac{MN}{BC},$$

or

$$\frac{9 - \frac{9}{4}}{9} = \frac{MN}{8},$$

so that

$$MN=6.$$

Q7.

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For

$$\frac{2^2(1^2 + 2^2 + 3^2 + \ldots + n^2)}{1^2 + 3^2 + 5^2 + \ldots + (2n - 1)^2} > 1.04,$$

where 'n' is a natural number, then maximum value of n is =

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Solution:
Prerequisite: sum of squares of first n natural, odd, even numbers.
Numerator  = sum of squares of first n even numbers = 2n.(n+1).(2n+1)/3
Denominator = odd squares sum = (2n-1).n.(2n+1)/3
=> (2n+2)/(2n-1) > 104/100 = 52/50
=> 100n + 100 > 104n - 52
=> 4n < 152
=> n < 38
=> n = 37 = Answer.

Q8.

A regular octagon is formed by cutting congruent isosceles right angled triangles from the corners of a square. If the square has side-length 1 and the side-length of the octagon is 

$\sqrt{a} - b$ then $a + b =$"

Solution:

2x + x.sqrt(2) = 1
=> x = 1/(2 + sqrt(2))
=> x.sqrt(2) = 1/(sqrt(2) + 1) = sqrt(2) - 1
=> a = 2, b = 1
Answer = 3

Q9.

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There are a few integers n such that $n^2 + n + 1$ divides $n^{2022} + 61$. Find the sum of the squares of these integers.

Correct Answer
62

Solution
Prerequisite: Modulo arithmetic

Since $n^3 - 1 = (n - 1)(n^2 + n + 1)$, we know that $n^2 + n + 1$ divides $n^3 - 1$.

Also, since $n^{2022} - 1 = (n^3)^{674} - 1$, we also know that $n^2 + n + 1$ divides $n^{2022} - 1$.

As $n^{2022} + 61 = n^{2022} - 1 + 62$,

we must have that $n^2 + n + 1$ divides $n^{2022} + 61$ if and only if $n^2 + n + 1$ divides 62.

Case (i) : If $n^2 + n + 1 = 1$, then $n = 0, -1$

Case (ii) : If $n^2 + n + 1 = 2$, there is no integer solution for $n$.

Case (iii) : If $n^2 + n + 1 = 31$, then $n = 6, -5$

Case (iv) : If $n^2 + n + 1 = 62$, there is no integer solution for $n$.

Thus, all the integer values of $n$ are $0, -1, 6, -5$. Hence the sum of squares is $1 + 36 + 25 = 62$

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Q10.

"Determine the number of real solutions of ⌊x/2⌋ + ⌊2x/3⌋ = x, where ⌊x⌋ is greatest integer function of x."

Solution:
First remember that greatest integer function is the Floor function not the ceiling one.
For e.g. floor(3.5) = 3, ceiling(3.5) = 4.
Greatest means, the greatest integer smaller than x. Now:

$$\left\lfloor \frac{x}{2}\right\rfloor+\left\lfloor \frac{2x}{3}\right\rfloor=x$$

The left side is an integer, so $x$ must be an integer.

Write $x=6k+r$ with $r\in\{0,1,2,3,4,5\}$. Then

$$\left\lfloor \frac{x}{2}\right\rfloor =3k+\left\lfloor \frac{r}{2}\right\rfloor,\qquad \left\lfloor \frac{2x}{3}\right\rfloor =4k+\left\lfloor \frac{2r}{3}\right\rfloor.$$

Plugging into the equation gives

$$3k+\left\lfloor \frac{r}{2}\right\rfloor+4k+\left\lfloor \frac{2r}{3}\right\rfloor =6k+r \quad\Longrightarrow\quad k+\left\lfloor \frac{r}{2}\right\rfloor+\left\lfloor \frac{2r}{3}\right\rfloor=r.$$

Compute for $r=0,1,2,3,4,5$:

$$\begin{array}{c|cccccc} r & 0 & 1 & 2 & 3 & 4 & 5\\ \hline \lfloor r/2\rfloor & 0 & 0 & 1 & 1 & 2 & 2\\ \lfloor 2r/3\rfloor & 0 & 0 & 1 & 2 & 2 & 3\\ k=r-\lfloor r/2\rfloor-\lfloor 2r/3\rfloor & 0 & 1 & 0 & 0 & 0 & 0 \end{array}$$

Thus $x=6k+r$ yields $x\in\{0,2,3,4,5,7\}$.
Therefore, there are 6 real solutions.

Now, as an exercise swap the floor function with ceiling function and find the answers.
You should get 6 solutions again: {0, -2, -3, -4, -5, -7}

Q11.

"A function $f$ is such that $f : \mathbb{R} \to \mathbb{R}$ where
$f(xy + 1) = f(x)f(y) - f(y) - x + 2$
for all $x, y \in \mathbb{R}$. Find $f(2006) + f(0)$."

Solution:
f(x).f(y) - f(y) - x + 2 = f(y).f(x) - f(x) - y + 2
=> f(x) - f(y) = x - y
=> f(2006) - f(0) = 2006
Put x = y = 0 in the original definition
f(1) = f(0)^2 - f(0) + 2
But f(1) = f(0) + 1
=> f(0)^2 - 2.f(0) + 1 = 0
=> (f(0) - 1)^2 = 0
=> f(0) = 1
=> f(2006) = 2007
=> f(0) + f(2006) = 2008 = answer.

Q12.

"Given a set $S = \{1, 2, 3, \ldots, 20\}$. The subset $A = \{a, b, c\}$ is said to be 'nice', if $a, b, c$ are in A.P. Then how many 'nice' subsets does $S$ have?"

Solution:
One way is enumeration with conditions.
If a, a+d,a+2d are the 3 terms then a+2d <= 20 => d <= 10 - a/2
a = 1 => d = 1 to 9
a = 2 => d = 1 to 9
a = 3,4 => d = 1 to 8
...
a = 15,16 => d = 1 to 2
a = 17,18 => d = 1
So total such nice subsets = 2.(1 + 2 .. 9) = 90

Another way:
If a,b,c are in A.P. then a + c = 2b
=> Both a,c have same parity(both are even or both are odd).
Choosing a,c fixes b.
There are 10 even and 10 odd numbers.
So 2.10C2 = 90

Q13.

"The 12 numbers $a_1, a_2, \ldots, a_{12}$ are in arithmetical progression. The sum of all these numbers is 354. Let $P = a_2 + a_4 + \cdots + a_{12}$ and $Q = a_1 + a_3 + \cdots + a_{11}$. If the ratio $P : Q$ is $32 : 27$, the common difference of the progression is"

Solution:

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Let the progression be $a, a + d, \ldots, a + 11d$.

$$a_1 + a_2 + \cdots + a_{12} = 6(2a + 11d) = 354$$ $$P = a_2 + a_4 + \cdots + a_{12} = 3(2a + 12d)$$ $$Q = a_1 + a_3 + \cdots + a_{11} = 3(2a + 10d)$$

Hence we have:

$$2a + 11d = 59$$ $$2a + 12d : 2a + 10d = 32 : 27 \Rightarrow 5a = 2d$$

Solving for $d$, we obtain $d = 5$.

Q14.

"In $\triangle ABC$, $AB = AC = \sqrt{3}$ and D is a point on BC such that $AD = 1$. Find the value of $BD \times DC$."

Solution:
One way is to construct such a triangle.
Let AD be median falling on BC.
Then by Pythagoras theorem, BD = DC = sqrt(2).
BC = 2.sqrt(2)
It's a valid triangle since sum of any 2 sides is more than the third.
So BD x DC = 2 = answer.

More formal way:

Construct a circle with $A$ as the center and $AB = AC = \sqrt{3}$ as the radius. Extend $AD$ to meet the circumference at $M$ and $N$ as shown.

Using the Intersecting Chord Theorem:

$$BD\cdot DC=MD\cdot ND=(\sqrt{3}-1)(\sqrt{3}+1)=2.$$
Q15.

"If $\sqrt{x+1} + \sqrt{y} + \sqrt{z-4} = \frac{x + y + z}{2}$, then the value of $x + y + z$ is"

Solution:

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We have,

$$2\sqrt{x+1} + 2\sqrt{y} + 2\sqrt{z - 4} = x + y + z$$ $$\Rightarrow 2\sqrt{x+1} + 2\sqrt{y} + 2\sqrt{z - 4} = (x + 1) + y + (z - 4) + 3$$ $$\Rightarrow \{(x + 1) + 1 - 2\sqrt{x+1} \} + \{y + 1 - 2\sqrt{y} \} + \{(z - 4) + 1 - 2\sqrt{z - 4} \} = 0$$ $$\Rightarrow (\sqrt{x+1} - 1)^2 + (\sqrt{y} - 1)^2 + (\sqrt{z - 4} - 1)^2 = 0$$ $$\Rightarrow \sqrt{x+1} = 1 \Rightarrow x = 0,\quad \sqrt{y} = 1 \Rightarrow y = 1,\quad \text{and} \quad z - 4 = 1 \Rightarrow z = 5$$ $$\therefore x+y+z=6$$
Q16.

If all roots of the quadratic equation 

$(k^2 - 1)x^2 - 6(3k - 1)x + 72 = 0$ are positive integers, and the sum of all possible values of the constant $k$ is $\frac{a}{5}$, then find $a$.

Solution:
Product of roots = 72/(k^2 - 1) = positive integer. But it doesn't tell us much as k could be any rational number, not necessarily an integer.
Sum of roots = 6.(3k - 1)/(k^2 - 1) = positive integer. Again hard to figure out anything.
Let's compute the discriminant:
D = 36.(k-3)^2
Easy to take its square root, so let's compute the roots.
Roots = {6.(3k - 1) -+ 6.(k - 3)}/2.(k^2 - 1) = 12/(k + 1), 6/(k - 1)
Let 12/(k + 1) = m and 6/(k - 1) = n
=> k = 12/m - 1 = 6/n + 1
=> m = 6n/(n + 3) and n = 3m/(6 - m)
m = 6n/(n + 3) doesn't seem of much use.
But n = 3m/(6 - m) is quite useful as it restricts m.
For n to be positive, 1 <= m <= 5
3 value of m yield positive integer value of n.
m = 3, n = 3
m = 4, n = 6
m = 5, n = 15
These values give 3,2,7/5 as values of k.
So a/5 = 32/5 => a = 32.

Q17.

Consider a set of 6 non-overlapping triangles in a plane such that no three points in the plane is collinear. If all the possible triangles are drawn taking vertices of these triangle such that not more than one point is selected from a triangle. If total number of triangle hence drawn in N, find 

$\frac{N}{10}$?

Solution:
Choosing 3 triangles: 6C3
Choosing 1 vertex from each: 3C1 * 3C1 * 3C1
Total: 20 * 3 * 3 * 3 = 540

One wrong way to solve:
Choosing first vertex: 18
Choosing second vertex: 15
Choosing third vertex: 12
Total: 18 * 15 * 12 = 6 * 540

So it's more by a factor of 6.
What's wrong?
We are repeating the combinations.
For e.g. when the first vertex is 1 then the remaining 2 can be 2,3.
But when the first vertex is 2 the remaining 2 can be 1,3 again.
So we have to divide this by 3! to get the correct answer.

Q18.

Let $f(n)$ denote the square of the sum of the digits of natural number $x$, where
$f^2(x)$ denote $f(f(x))$,
$f^3(x)$ denote $f(f(f(x)))$ and so on.

Then value of

$$\frac{f^{2019}(2011)-f^{2018}(2011)}{f^{2021}(2011)-f^{2020}(2011)}$$
Solution:

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$$f(2011) = (2 + 0 + 1 + 1)^2 = 16$$ $$f^2(2011) = (1 + 6)^2 = 49$$ $$f^3(2011) = (4 + 9)^2 = 169$$ $$f^4(2011) = (1 + 6 + 9)^2 = 256$$ $$f^5(2011) = (2 + 5 + 6)^2 = 169$$

Hence, $f(x)$ is periodic.

$$\therefore f^{2n}(2011) = 256 \quad \& \quad f^{2n+1}(2011) = 169$$ $$\therefore \frac{f^{2019}(2011)-f^{2018}(2011)}{f^{2021}(2011)-f^{2020}(2011)}=\frac{169-256}{169-256}=01$$
Q19.

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The number of distinct primes dividing $12! + 13! + 14!$ is (where $n!$ = factorial n)

Solution:

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$$12! + 13! + 14! = 12!(1 + 13 + 14 \times 13) = 12! \times 14^2$$

The prime factors of $12!$ are:

$$2, 3, 5, 7, 11$$

∴ There are 5 distinct primes dividing $12! + 13! + 14!$.

Q20.

Consider the function 

$f(x) = \frac{1}{3^x + \sqrt{3}}$. Find the value of

$$\sqrt{3}[f(-5)+f(-4)+f(-3)+f(-2)+f(-1)+f(0)+f(1)+f(2)+f(3)+f(4)+f(5)+f(6)]$$
Solution:

$$f(x) + f(1 - x) = \frac{1}{3^x + \sqrt{3}} + \frac{1}{3^{1 - x} + \sqrt{3}} = \frac{\sqrt{3}}{3^x \sqrt{3} + 3} + \frac{3^x}{3 + 3^x \sqrt{3}} = \frac{3^x + \sqrt{3}}{3 + 3^x \sqrt{3}} = \frac{1}{\sqrt{3}}$$

Therefore,

$$\sqrt{3}[f(-5) + f(6)] + \sqrt{3}[f(-4) + f(5)] + \sqrt{3}[f(-3) + f(4)] + \sqrt{3}[f(-2) + f(3)] + \sqrt{3}[f(-1) + f(2)] + \sqrt{3}[f(0) + f(1)]$$ $$=\sqrt{3}(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}})=6$$
Q21.

Let $P(x)$ be a polynomial of degree 4 such that

$$P(n) = \frac{120}{n} \quad \text{for } n = 1, 2, 3, 4, 5.$$

Determine the value of $P(6)$.
Solution:

Let $Q(x) = xP(x) - 120$,
where $Q(x)$ is a 5th-degree polynomial having 1, 2, 3, 4, 5 as its roots.

∴

$$xP(x) - 120 = C(x - 1)(x - 2)(x - 3)(x - 4)(x - 5) \quad \text{(where 'C' is constant)}$$

Put $x = 0$:

$$-120 = C(-1)(-2)(-3)(-4)(-5)$$ $$-120 = -120C \Rightarrow C = 1$$

So,

$$xP(x) - 120 = (x - 1)(x - 2)(x - 3)(x - 4)(x - 5)$$

Now put $x = 6$:

$$6P(6) - 120 = 5 × 4 × 3 × 2 × 1$$ $$6P(6) = 120 + 120 = 240$$ $$P(6) = \frac{240}{6} = 40$$

∴ $\boxed{{\displaystyle P(6)=40}}$

Q22.

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Let

$$S = \left\lfloor \sqrt{1} \right\rfloor + \left\lfloor \sqrt{2} \right\rfloor + \cdots + \left\lfloor \sqrt{2020} \right\rfloor$$

Find the sum of the digits of

$$\left\lfloor \sqrt{S} \right\rfloor$$

where $\left\lfloor x \right\rfloor$ is the greatest integer function of $x$.

Solution:

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For any positive integer $k$ and $x$, the following relations are equivalent:

$$\left\lfloor \sqrt{x} \right\rfloor = k \Leftrightarrow k^2 \leq x < (k+1)^2 \Leftrightarrow x \in [k^2, k^2 + 2k]$$

So $2k + 1$ values of $x$ satisfy the relation.

Since $44^2 = 1936 < 2020 < 2025 = 45^2$, and
$2020 - 1936 + 1 = 85$,

$$S = 1(3) + 2(5) + 3(7) + \ldots + 43(87) + 44(85)$$ $$\therefore \sum_{k=1}^{43} k(2k + 1) = \sum_{k=1}^{43} (2k^2 + k) = 2 \sum_{k=1}^{43} k^2 + \sum_{k=1}^{43} k$$ $$= \frac{43 \cdot 44 \cdot 87}{3} + \frac{43 \cdot 44}{2} + 3740 = 54868 + 946 + 3740 = 59554$$

Since $244^2 = 59536 < 59554 < 60025 = 245^2$,

$$\left\lfloor \sqrt{S} \right\rfloor = 244$$

Therefore, required sum is $2 + 4 + 4 = \boxed{10}$.
A note on how to compute the square root above quickly:

Use nearby perfect–square bounds and the “next odd jump.”

General trick
If $m^2\le x$, then $(m+1)^2=m^2+(2m+1)$. So after you guess an $m$, just compare
$\Delta=x-m^2$ with $2m+1$:

• If $\Delta<2m+1$, then $\lfloor\sqrt{x}\rfloor=m$.

• Otherwise it’s $m+1$.

For $x=59554$:

1. Rough bound: $240^2=57600 < x < 250^2=62500$ ⇒ $240<\sqrt{x}<250$.

2. Try the middle: $244^2=(240+4)^2=57600+1920+16=59536$.

3. Compare the jump to the next square:
   $(244+1)^2-244^2=2\cdot244+1=489$.
   $\Delta=x-244^2=59554-59536=18<489$.

Hence $\lfloor\sqrt{59554}\rfloor=244$.
Sum of digits $=2+4+4=10$.

Memorizing squares up to $50^2$ (or at least up to $25^2$) makes step (1) instant, and steps (2)–(3) are just a couple of quick additions.

Q23.

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Let $2\lfloor x \rfloor$, $6\{x\}$, and $3x$ be the first three terms of a G.P., where $x \in \mathbb{R}^+$ and
$\lfloor \cdot \rfloor$ and $\{ \cdot \}$ represent the greatest integer and fractional part functions respectively.

If there is another G.P. whose terms are the square of reciprocals of the terms of the given G.P., and the sum of infinite terms of this G.P. is $S$, then find the value of $100S$.

Solution:
Let x = p + q where p,q > 0 (if any of them is 0, the G.P. won't be valid). p is the integer part and q is the fractional part.
36.q^2 = 2p.3(p + q) = 6(p^2 + pq)
=> 6q^2 = p^2 + pq
=> p^2 + pq - 6q^2 = 0
Factorize to get:
(p - 2q)(p + 3q) = 0
p = 2q,-3q
But p,q are both positive => p  = 2q
Now q is the fractional part and when multiplied by 2 it gives p which is an integer..
Only solution is p = 1 and q = 1/2.
x = 1 + 1/2 = 3/2
Common ratio r = 3q/p = 3/2

S = 1/4p^2 * 1/(1 - 1/r^2) = 1/4 * 1/(1 - 4/9) = 9/20
100S = 45 = Answer

Q24.

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Let $a$ and $b$ be integers, and $a + b$ is a root of the equation

$$x^2 + ax + b = 0$$

If the smallest possible value of $ab$ is $k$, find $|k|$.

Solution:
Putting x = a + b in the equation gives:
b^2 + b(3a + 1) + 2a^2 = 0
Since b is integer, D = p.s. of integer
D = (3a + 1)^2 - 8a^2 = a^2 + 6a + 1 = m^2
=> (a + 3)^2 - m^2 = 8
=> (|a + 3| + |m|) (|a + 3| - |m|) = 8
=> |a + 3| = (8 + 1)/2 or (2 + 4)/2
=> |a+3| = 3 as a is integer.
=> a+3 = -+3
=> a = 0,-6
a = 0 => b^2 + b = 0 => b = 0,-1
a = -6 => b = 8,9
Smallest value of ab = -54 => |k| = 54 = answer.


Q26. Evaluate

$$12 \sum_{r=1}^{\infty} \frac{12r^2 + 1}{64r^6 - 48r^4 + 12r^2 - 1}$$

Solution:

$$S=12\sum_{r=1}^{\infty}\frac{12r^2+1}{64r^6-48r^4+12r^2-1}$$

Note that

$$64r^6-48r^4+12r^2-1=(4r^2-1)^3=((2r-1)(2r+1))^3 .$$

Also,

$$\frac{12r^2+1}{(4r^2-1)^3} =\frac12\!\left(\frac{1}{(2r-1)^3}-\frac{1}{(2r+1)^3}\right)$$

since

$$(2r+1)^3-(2r-1)^3=24r^2+2=2(12r^2+1).$$

Hence

$$S=12\sum_{r=1}^{\infty}\frac12\Big(\frac{1}{(2r-1)^3}-\frac{1}{(2r+1)^3}\Big) =6\sum_{r=1}^{\infty}\Big(\frac{1}{(2r-1)^3}-\frac{1}{(2r+1)^3}\Big),$$

which telescopes:

$$6\left(1-\frac{1}{3^3}+\frac{1}{3^3}-\frac{1}{5^3}+\cdots\right) =6\lim_{N\to\infty}\left(1-\frac{1}{(2N+1)^3}\right)=6.$$ $$\boxed{{\displaystyle 6}}$$
Q27.

Let

$$f(x) = \frac{15}{x+1} + \frac{16}{x^2 + 1} - \frac{17}{x^3 + 1}$$

Find the value of

$$f(\tan 15^\circ) + f(\tan 30^\circ) + f(\tan 45^\circ) + f(\tan 60^\circ) + f(\tan 75^\circ)$$

Solution:
Prerequisite: Reciprocal trick.

$$f(x)=\frac{15}{x+1}+\frac{16}{x^2+1}-\frac{17}{x^3+1}$$

Notice the reciprocal symmetry:

$$f(x)+f\!\left(\frac1x\right) = \frac{15}{x+1}+\frac{16}{x^2+1}-\frac{17}{x^3+1} +\frac{15}{1+\frac1x}+\frac{16}{1+\frac1{x^2}}-\frac{17}{1+\frac1{x^3}} =14.$$

The angles pair by tangent reciprocals:

$$\tan75^\circ=\frac1{\tan15^\circ},\quad \tan60^\circ=\frac1{\tan30^\circ},\quad \tan45^\circ=1.$$

Hence

$$f(\tan15^\circ)+f(\tan75^\circ)=14,\qquad f(\tan30^\circ)+f(\tan60^\circ)=14,$$

and

$$f(\tan45^\circ)=f(1)=\frac{15}{2}+\frac{16}{2}-\frac{17}{2}=7.$$

Summing:

$$14+14+7=35.$$ $$\boxed{{\displaystyle 35}}$$

Q28.

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Let $f(x) = x^6 - x^5 - x^3 - x^2 - x$ and $g(x) = x^4 - x^3 - x^2 - 1$. If a, b, c, d are the four roots to the equation $g(x) = 0$, find the value of
$f(a) + f(b) + f(c) + f(d)$.

Solution:
g(a) = 0 => a^4 = a^3 + a^2 + 1
Multiply it with and simplify to get:
a^5 = 2a^3 + a^2 + a + 1
Do it again to get:
a^6 = 3a^3 + 3a^2 + a + 2

Put all this in f(a) to get:
f(a) = a^2 - a + 1
f(a) + f(b) + f(c) + f(d)  = (a^2 + b^2 + c^2 + d^2) - (a + b + c + d) + 4

Using Vitae's formulae on g(x):
a + b + c + d = 1
And a^2 + b^2 + c^2 + d^2 = (a + b + c + d)^2 -2.(-1) = 3
Answer = 4 + 3 - 1 = 6

Q29.

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PQR is a triangle with PQ = 15, QR = 25, RP = 30. A, B are points on PQ and PR respectively such that $\angle PBA = \angle PQR$. The perimeter of the triangle PAB is 28, then the length of AB is _____.

Solution:

Triangles PAB and PQR are similar. Thus

$$\frac{AB}{QR} = \frac{PB}{PQ} = \frac{PA}{PR} = \frac{AB + PB + PA}{QR + PQ + PR} = \frac{28}{70}$$

and $AB = QR \times \frac{2}{5} = 10$.

Q30.

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Let $\overline{CH}$ be an altitude of $\triangle ABC$. Let R and S be the points where the circles inscribed in triangles ACH and BCH are tangent to $\overline{CH}$. If AB = 2021, AC = 2020, and BC = 2019, then RS can be expressed as $\frac{m}{n}$, where m and n are relatively prime positive integers. Find

$$\left\lfloor \frac{n - m}{11} \right\rfloor,$$

where $\left\lfloor x \right\rfloor$ represents the greatest integer function of x.

Solution:

Let r1,r2 be the inradius of ACH,BCH.
CH perpendicular to AB => OR perpendicular to OT.
=> RS = |r1 - r2|
So let's work towards finding r1 - r2.
OR = OT = r1.
AC = AP + PC
AP = AT = AH - r1
PC = CR = CH - r1
AC = AH + CH  - 2r1
=> r1 = (AH + CH - AC)/2
Similarly,
r2 = (BH + CH - BC)/2
r1 - r2 = (AH - BH + BC - AC)/2
AC,BC are already known.
Let's work on AH - BH.
CH^2 = AC^2 - AH^2 = BC^2 - BH^2
=> AC^2 - BC^2 = AH^2 - BH^2
=> AH - BH = (AC^2 - BC^2)/(AH + BH) = (AC^2 - BC^2)/(AB)
=> r1 - r2 = 1/2[(AC^2 - BC^2)/(AB) + BC - AC]
= (AC - BC)/2[(AC + BC)/AB - 1]
= 1/2[AC + BC - AB]/AB = 2018/(2*2021) = 1009/2021
Check if 2021,1009 are relative prime.
Euclid's gcd algorithm gives:
gcd(1009,2021) = gcd(3,1009) = gcd(1,3) = 1
So yes, they are relative prime.
n - m = 2021 - 1009 = 1012
1012/11 = 92 = Answer

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