2019 Ga Math (b) MEXT Monbukagakusho

https://www.studyinjapan.go.jp/en/planning/scholarships/mext-scholarships/examination.html

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Create Practice QuestionCreate Practice TestCreate MathGPT VideoSolve the above problem.(6) Sequence

The sequence is

$1,\ 4,\ 10,\ 19,\ 31,\ldots$

Find the differences:

\begin{array}{c|ccccc}

\text{Term} & 1 & 4 & 10 & 19 & 31\\

\hline

\text{First difference} & & 3 & 6 & 9 & 12\\

\text{Second difference} & & & 3 & 3 & 3

\end{array}

Since the second difference is constant, the sequence is quadratic.

The first difference before the first term is 0, so the $n$-th term can be found by adding the differences:

$a_n=1+3(1+2+\cdots +(n-1)).$

Using

$1+2+\cdots +(n-1)=\frac{n(n-1)}{2},$

we get

$a_n=1+\frac{3n(n-1)}{2}.$

Therefore,

$\boxed{a_n=\frac{3n^2-3n+2}{2}}.$

For the sum of the first $n$ terms,

$S_n=\sum_{k=1}^{n}\left(1+\frac{3k(k-1)}{2}\right).$

Thus,

$S_n=n+\frac{3}{2}\sum_{k=1}^{n}k(k-1).$

Since

$\sum_{k=1}^{n}k(k-1)=\frac{n(n+1)(n-1)}{3},$

we have

$S_n=n+\frac{n(n+1)(n-1)}{2}.$

Simplifying,

$\boxed{S_n=\frac{n^3+n}{2}}.$

So the answers for (6) are

$\boxed{[1\text{-}11]=\frac{3n^2-3n+2}{2}}$

and

$\boxed{[1\text{-}12]=\frac{n^3+n}{2}}.$

(7) Minimization

We are given

$\frac{4a+b}{2a}$+$\frac{4a-3b}{b}$,

where a,b>0.

Let

$x=\frac{b}{a}.$

Then

$\frac{4a+b}{2a}$

=2+$\frac{x}{2},$

and

$\frac{4a-3b}{b}$

$=\frac{4}{x}-3.$

Therefore, the expression becomes

$2+\frac{x}{2}+\frac{4}{x}-3$

$=\frac{x}{2}+\frac{4}{x}-1.$

By AM-GM,

$\frac{x}{2}+\frac{4}{x}$

$\geq 2\sqrt{\frac{x}{2}\cdot\frac{4}{x}}$

$=2\sqrt{2}.$

Equality occurs when

$\frac{x}{2}=\frac{4}{x}.$

Hence,

$x^2=8$

and because $x>0,$

$x=2\sqrt{2}.$

Since $ x=\frac ba,$

$\boxed{b=2\sqrt{2}\,a}.$

The minimum value is

$2\sqrt{2}-1.$

Therefore,

$\boxed{[1\text{-}13]=2\sqrt{2}}$

and

$\boxed{[1\text{-}14]=2\sqrt{2}-1}.$

Create Practice QuestionCreate Practice TestCreate MathGPT VideoSolve the above problem.(9) Divisibility of a base-6 number

We have

$\sum_{k=0}^{n}x_k6^k.$

For $ k\geq 1,$

$6^k=1+5(1+6+6^2+\cdots+6^{k-1})

=1+5\sum_{l=0}^{k-1}6^l.$

Therefore,

\begin{aligned}

$\sum_{k=0}^{n}x_k6^k$

$&=x_0+\sum_{k=1}^{n}x_k6^k\\$

$&=x_0+\sum_{k=1}^{n}x_k$

$\left(1+5\sum_{l=0}^{k-1}6^l\right)\\

&=\boxed{x_0}$ $+\boxed{5}\left(\sum_{k=1}^{n}x_k\sum_{l=0}^{k-1}6^l\right).$

\end{aligned}

Thus,

$[1\text{-}21]=\boxed{x_0},

\qquad$

$[1\text{-}22]=\boxed{5}.$

The number is divisible by 5 exactly when its digits add to a multiple of 5, because

$6^k\equiv 1\pmod 5.$

Hence,

$\sum_{k=0}^{n}x_k6^k$

$\equiv \sum_{k=0}^{n}x_k\pmod 5.$

Therefore,

$[1\text{-}23]=\boxed{5}.$

(10) Solving the linear Diophantine equation

We need integer solutions to

253x+256y=1.

Since $256=253+3$,

253x+256y=253(x+y)+3y.

Thus,

253(x+y)+3y=1.

Taking the equation modulo 253,

$3y\equiv 1\pmod{253}.$

Because

3(169)=507=2(253)+1,

we have

$y \equiv 169\pmod{253}.$

So write

y=169+253t,

where t is an integer. Substituting into the original equation gives

x=-171-256t.

To minimize \lvert x\rvert, choose t=-1. Then

x=-171+256=85,

and

y=169-253=-84.

Therefore,

$\boxed{[1\text{-}24]=85},

\qquad

\boxed{[1\text{-}25]=-84}.

Check:

253(85)+256(-84)=21,505-21,504=1.$

(11) Translating the parabola

The original function is

$f(x)=2x^2+3x+1.$

Translating the graph 2 units to the right replaces x with x-2:

f(x-2)=2(x-2)^2+3(x-2)+1.

Translating it 3 units downward subtracts 3:

$y=2(x-2)^2+3(x-2)+1-3.$

Expand:

$\begin{aligned}

y

&=2(x^2-4x+4)+3x-6+1-3\\

&=2x^2-8x+8+3x-8\\

&=2x^2-5x.

\end{aligned}$

Thus,

$y=2x^2-5x+0,$

so

$\boxed{[1\text{-}26]=2},

\qquad

\boxed{[1\text{-}27]=-5},

\qquad

\boxed{[1\text{-}28]=0}.$

Create Practice QuestionCreate Practice TestCreate MathGPT VideoSolve the above problem.Retry ResponseSolve the above problem.Retry ResponseSolve the above problem.Retry ResponseSolve the above problem.Problem 2

We are given:

$CD\perp AB$

$\angle BAC=\dfrac{\pi}{12}=15^\circ

AB=2\sqrt{2}$

$AD=\sqrt{6}$

Since D lies on AB,

$DB=AB-AD=2\sqrt{2}-\sqrt{6}.$

(1) Find $\cos\dfrac{\pi}{12}$

Since

$\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4},

use the cosine subtraction formula:

\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)

=\cos\frac{\pi}{3}\cos\frac{\pi}{4}

+

\sin\frac{\pi}{3}\sin\frac{\pi}{4}.

Substitute the familiar values:

\cos\frac{\pi}{3}=\frac12,\qquad

\cos\frac{\pi}{4}=\frac{\sqrt2}{2},

\sin\frac{\pi}{3}=\frac{\sqrt3}{2},\qquad

\sin\frac{\pi}{4}=\frac{\sqrt2}{2}.$

Therefore,

$\cos\frac{\pi}{12}

=\frac{\sqrt2}{4}+\frac{\sqrt6}{4}

=\frac{\sqrt6+\sqrt2}{4}.$

Thus,

$\boxed{[2\text{-}1]=\sqrt6}.$

(2) Find the length AC

In right triangle ACD,

$\cos\frac{\pi}{12}=\frac{AD}{AC}.$

Hence,

$AC=\frac{AD}{\cos(\pi/12)}

=\frac{\sqrt6}{(\sqrt6+\sqrt2)/4}.$

So

$AC=\frac{4\sqrt6}{\sqrt6+\sqrt2}.$

Rationalize the denominator:

$AC

=\frac{4\sqrt6(\sqrt6-\sqrt2)}

{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}.$

The denominator is

6-2=4.

Therefore,

$AC=\sqrt6(\sqrt6-\sqrt2)

=6-\sqrt{12}

=6-2\sqrt3.$

Thus,

$\boxed{[2\text{-}2]=6}.$

(3) Find $(BC)^2$

First, use the Pythagorean theorem in right triangle ACD:

$CD^2=AC^2-AD^2.$

We have

$AC=6-2\sqrt3,

so

AC^2=(6-2\sqrt3)^2

=36-24\sqrt3+12

=48-24\sqrt3.$

Also,

$AD^2=(\sqrt6)^2=6.$

Therefore,

$CD^2=(48-24\sqrt3)-6

=42-24\sqrt3.$

Now calculate $DB^2:

DB=2\sqrt2-\sqrt6,$

so

$DB^2=(2\sqrt2-\sqrt6)^2

=8-4\sqrt{12}+6

=14-8\sqrt3.$

In right triangle CDB,

$BC^2=CD^2+DB^2.$

Thus,

$\begin{aligned}

BC^2

&=(42-24\sqrt3)+(14-8\sqrt3)\\

&=56-32\sqrt3.

\end{aligned}$

Therefore,

$\boxed{[2\text{-}3]=56}.$

(4) Find the length BC

We need to simplify

$BC=\sqrt{56-32\sqrt3}.$

Observe that

$56-32\sqrt3

=(4\sqrt2-2\sqrt6)^2.$

Indeed,

$(4\sqrt2-2\sqrt6)^2

=32-16\sqrt{12}+24

=56-32\sqrt3.$

Since $4\sqrt2-2\sqrt6>0,

BC=4\sqrt2-2\sqrt6.$

Therefore,

$\boxed{[2\text{-}4]=4\sqrt2}.$

Final answers

$\boxed{[2\text{-}1]=\sqrt6}

\boxed{[2\text{-}2]=6}

\boxed{[2\text{-}3]=56}

\boxed{[2\text{-}4]=4\sqrt2}$

Create Practice QuestionCreate Practice TestCreate MathGPT Video2019_ga_math_b_ansPDFSolve the above problem.Problem 2

Given:

$CD\perp AB,\qquad \angle BAC=\frac{\pi}{12},\qquad AB=2\sqrt2,\qquad AD=\sqrt6.$

Since D lies on AB, the angle in right triangle ACD is

$\angle CAD=\angle BAC=\frac{\pi}{12}.$

(1) Find $\cos \dfrac{\pi}{12}$

Using

$\frac{\pi}{12}=\frac{\pi}{3}-\frac{\pi}{4},$

we get

\begin{aligned}

$\cos\frac{\pi}{12}$

$&=\cos\left(\frac{\pi}{3}-\frac{\pi}{4}\right)\\$

$&=\cos\frac{\pi}{3}\cos\frac{\pi}{4}$

$+\sin\frac{\pi}{3}\sin\frac{\pi}{4}\\$

$&=\frac12\cdot\frac{\sqrt2}{2}$

$+\frac{\sqrt3}{2}\cdot\frac{\sqrt2}{2}\\$

$&=\frac{\sqrt2+\sqrt6}{4}.$

\end{aligned}

Therefore,

$\boxed{[2\text{-}1]=\sqrt6}.$

(2) Find AC

In the right triangle ACD,

$\cos\frac{\pi}{12}=\frac{AD}{AC}.$

Therefore,

$AC=\frac{AD}{\cos(\pi/12)}

=\frac{\sqrt6}{(\sqrt2+\sqrt6)/4}.$

Thus,

$AC=\frac{4\sqrt6}{\sqrt2+\sqrt6}.$

Rationalizing the denominator,

$\begin{aligned}

AC

&=\frac{4\sqrt6(\sqrt6-\sqrt2)}

{(\sqrt6+\sqrt2)(\sqrt6-\sqrt2)}\\

&=\frac{4\sqrt6(\sqrt6-\sqrt2)}{6-2}\\

&=\sqrt6(\sqrt6-\sqrt2)\\

&=6-\sqrt{12}\\

&=6-2\sqrt3.

\end{aligned}$

Hence,

$\boxed{[2\text{-}2]=6}.$

(3) Find (BC)^2

First calculate CD^2. By the Pythagorean theorem in \triangle ACD,

$CD^2=AC^2-AD^2.$

Now,

$\begin{aligned}

AC^2

&=(6-2\sqrt3)^2\\

&=36-24\sqrt3+12\\

&=48-24\sqrt3.

\end{aligned}$

Therefore,

$CD^2=(48-24\sqrt3)-6

=42-24\sqrt3.$

Also,

$DB=AB-AD=2\sqrt2-\sqrt6,$

so

$\begin{aligned}

DB^2

&=(2\sqrt2-\sqrt6)^2\\

&=8-4\sqrt{12}+6\\

&=14-8\sqrt3.

\end{aligned}$

In the right triangle $CDB$,

$BC^2=CD^2+DB^2.$

Thus,

$\begin{aligned}

BC^2

&=(42-24\sqrt3)+(14-8\sqrt3)\\

&=56-32\sqrt3.

\end{aligned}$

Therefore,

$\boxed{[2\text{-}3]=56}.$

(4) Find BC

We need to simplify

$BC=\sqrt{56-32\sqrt3}.$

Notice that

$56-32\sqrt3=(4\sqrt2-2\sqrt6)^2,$

because

$(4\sqrt2-2\sqrt6)^2

=32-16\sqrt{12}+24

=56-32\sqrt3.$

Hence,

$BC=4\sqrt2-2\sqrt6.$

Therefore,

$\boxed{[2\text{-}4]=4\sqrt2}.$

Final answers

$\boxed{

[2\text{-}1]=\sqrt6,\quad

[2\text{-}2]=6,\quad

[2\text{-}3]=56,\quad

[2\text{-}4]=4\sqrt2

}$

Create Practice QuestionCreate Practice TestCreate MathGPT VideoSolve the above problem.Worked with your files

Problem 1

(1) Number and sum of the positive divisors of 2,019

Factor 2,019 :

$2,019=3\cdot 673.$

Since 673 is prime, the positive divisors are

1,\ 3,\ 673,\ 2,019.

Therefore, the number of positive divisors is

$\boxed{4}.$

Their sum is

$1+3+673+2,019=\boxed{2,696}.$

(2) Equation of the angle bisector of \angle AOB

The points are

$O(0,0),\qquad A(3,0),\qquad B(3,4).$

The ray $OA$ has unit direction vector

(1,0).

The vector from $O$ to $B$ is $(3,4)$, whose length is $5$, so its unit direction vector is

$\left(\frac35,\frac45\right).$

The internal angle-bisector direction is the sum of the two unit vectors:

$(1,0)+\left(\frac35,\frac45\right)

=\left(\frac85,\frac45\right).$

Its slope is

$\frac{4/5}{8/5}=\frac12.$

Since the angle bisector passes through O(0,0), its equation is

$\boxed{y=\frac12x}.$

(3) Tangent to y=x^2

The slope of the line through (-1,1) and (3,9) is

$\frac{9-1}{3-(-1)}=\frac84=2.$

Therefore, the required tangent line must have slope 2.

For

$y=x^2,$

the derivative is

$\frac{dy}{dx}=2x.$

At the point of tangency, the tangent slope is 2, so

$2x=2 \implies x=1.$

The corresponding y-coordinate is

$y=1^2=1.$

Thus, the tangent point is (1,1). The tangent line is

y-1=2(x-1),

so

y=2x-1.

Therefore,

$\boxed{y=2x-1}$

and the point of tangency is

$\boxed{(1,1)}.$

(4) Intersection of the line and the circle

The line is

y=m(x-5)+3=mx+3-5m.

Writing it in standard form,

mx-y+3-5m=0.

The distance from the origin to this line is

$d=\frac{|3-5m|}{\sqrt{m^2+1}}.$

The line intersects the circle

$x^2+y^2=r^2$

if and only if the distance from the origin to the line is at most r:

$\frac{|3-5m|}{\sqrt{m^2+1}}\le r.$

The problem states that the allowable values have the form

$0\le m\le \text{constant}.$

At m=0, the line is y=3. This must be tangent to the circle, so

r=3.

Now substitute r=3 :

$\frac{|3-5m|}{\sqrt{m^2+1}}\le 3.$

Squaring both sides,

$(3-5m)^2\le 9(m^2+1).$

Expand:

$9-30m+25m^2\le 9+9m^2.$

Thus,

$16m^2-30m\le 0,$

# Calculus and Discrete Mathematics Study Guide

## 🔢 Inductive Sequences and Pascal's Triangle

- **Definition:** The function $I(m, n)$ follows the recurrence relation $I(m+1, n) + I(m, n+1) = I(m+1, n+1)$ with boundary conditions $I(m, 1) = I(1, n) = 1$.

- **Pattern Recognition:** This recurrence is identical to the construction of Pascal's Triangle, where $I(m, n) = \binom{m+n-2}{m-1} = \binom{m+n-2}{n-1}$.

- **Specific Values:**

  - $I(2, n) = \binom{n+2-2}{2-1} = \binom{n}{1} = n$.

  - $I(3, n) = \binom{n+3-2}{3-1} = \binom{n+1}{2} = \frac{n(n+1)}{2}$.

  - $I(5, 3) = \binom{5+3-2}{5-1} = \binom{6}{4} = \binom{6}{2} = 15$.

## 📈 Taylor Series and Function Comparison

- **Function Definitions:** 

  - $f(x) = e^x$

  - $g(x) = 1 + x$

  - $h(x) = 1 + x + \frac{1}{2}x^2$

- **Derivatives for $x < 0$:**

  - $f'(x) = e^x$

  - $g'(x) = 1$

  - $h'(x) = 1 + x$

  - Comparison: Since $e^x < 1+x < 1$ for $x < 0$, the order is $f'(x) < h'(x) < g'(x)$.

- **Function Values for $x < 0$:**

  - Using Taylor expansion, $e^x = 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \dots$

  - Since $\frac{x^3}{6} < 0$ for $x < 0$, $f(x) < h(x)$.

  - Since $\frac{x^2}{2} > 0$ for $x < 0$, $h(x) > g(x)$.

  - Order: $f(x) < g(x) < h(x)$ is incorrect; rather, $f(x) < g(x) < h(x)$ depends on the specific interval, but generally $f(x) < g(x) < h(x)$ holds for small negative $x$.

## 📐 Definite Integrals of Absolute Differences

- **Integral $I_1$:** $\int_{-1}^{0} |e^x - (1+x)| dx$.

  - Since $e^x \ge 1+x$ for all $x$, $|e^x - (1+x)| = e^x - 1 - x$.

  - $I_1 = [e^x - x - \frac{1}{2}x^2]_{-1}^{0} = (1 - 0 - 0) - (e^{-1} + 1 - \frac{1}{2}) = \frac{1}{2} - \frac{1}{e}$.

- **Integral $I_2$:** $\int_{-1}^{0} |e^x - (1+x+\frac{1}{2}x^2)| dx$.

  - Since $e^x \le 1+x+\frac{1}{2}x^2$ for $x \in [-1, 0]$, $|f(x) - h(x)| = 1 + x + \frac{1}{2}x^2 - e^x$.

  - $I_2 = [x + \frac{1}{2}x^2 + \frac{1}{6}x^3 - e^x]_{-1}^{0} = (0 + 0 + 0 - 1) - (-1 + \frac{1}{2} - \frac{1}{6} - e^{-1}) = e^{-1} - \frac{2}{3}$.

Answer: 2. (1) $I(2,n)=n, I(3,n)=\frac{n(n+1)}{2}$

(2) $I(5,3)=15$

3. (1) $h'(x) < f'(x) < g'(x)$

(2) $g(x) < h(x) < f(x)$

(3) $I_1 = \frac{1}{2} - \frac{1}{e}, I_2 = \frac{1}{3} - \frac{1}{e}$ 

Explanation: Hello! I am happy to help you work through these problems. Let's break them down step by step, starting with the function $I(m, n)$ and then moving to the calculus problems.

### Part 2: Inductive Function $I(m,n)$

**Key Concepts:**

- This is a recurrence relation similar to Pascal's Triangle. 

- Given $I(m, 1) = 1$ and $I(1, n) = 1$, and the rule $I(m+1, n+1) = I(m+1, n) + I(m, n+1)$, we can observe that $I(m, n) = \binom{m+n-2}{m-1}$.

**Step 1: Find $I(2, n)$ and $I(3, n)$.**

- For $I(2, n)$: $I(2, 1)=1, I(2, 2)=I(2, 1)+I(1, 2)=1+1=2, I(2, 3)=I(2, 2)+I(1, 3)=2+1=3$.

- Pattern: $I(2, n) = n$.

- For $I(3, n)$: $I(3, 1)=1, I(3, 2)=I(3, 1)+I(2, 2)=1+2=3, I(3, 3)=I(3, 2)+I(2, 3)=3+3=6$.

- Pattern: $I(3, n) = \frac{n(n+1)}{2}$.

**Step 2: Find $I(5, 3)$.**

- Using the formula $I(m, n) = \binom{m+n-2}{m-1}$:

- $I(5, 3) = \binom{5+3-2}{5-1} = \binom{6}{4} = \binom{6}{2} = \frac{6 \times 5}{2} = 15$.

---

### Part 3: Calculus Functions

**Key Concepts:**

- $f(x) = e^x$, $g(x) = 1+x$, $h(x) = 1+x+\frac{1}{2}x^2$.

- Derivatives: $f'(x) = e^x$, $g'(x) = 1$, $h'(x) = 1+x$.

**Step 1: Arrange $f'(x), g'(x), h'(x)$ for $x < 0$.**

- For $x < 0$: $h'(x) = 1+x < 1$. Also, $e^x$ is between $0$ and $1$.

- Specifically, $e^x > 1+x$ for $x < 0$. Thus, $h'(x) < f'(x) < g'(x)$.

**Step 2: Arrange $f(x), g(x), h(x)$ for $x < 0$.**

- Using Taylor series: $e^x = 1+x+\frac{1}{2}x^2 + \frac{1}{6}x^3 + \dots$

- For $x < 0$, $f(x) > h(x) > g(x)$ is not correct; rather, $g(x) < h(x) < f(x)$ because the higher-order terms of $e^x$ are positive/negative alternating.

- Actually, for $x \in (-1, 0)$, $g(x) < h(x) < f(x)$.

**Step 3: Compute $I_1$ and $I_2$.**

- $I_1 = \int_{-1}^{0} (e^x - (1+x)) dx = [e^x - x - \frac{1}{2}x^2]_{-1}^{0} = (1 - 0 - 0) - (e^{-1} + 1 - \frac{1}{2}) = 1 - e^{-1} - 0.5 = 0.5 - \frac{1}{e}$.

- $I_2 = \int_{-1}^{0} (e^x - (1+x+\frac{1}{2}x^2)) dx = [e^x - x - \frac{1}{2}x^2 - \frac{1}{6}x^3]_{-1}^{0} = (1) - (e^{-1} + 1 - \frac{1}{2} + \frac{1}{6}) = 1 - e^{-1} - \frac{2}{3} = \frac{1}{3} - \frac{1}{e}$.