Model Card for EXAONE-3.5-7.8B-Stratos-Ko

Base Model

Dataset

Training

  • 4xA100, 12 hours - Total 48 GPU Hours with A100 PCie

  • Trained with Axolotl, not LLaMA-Factory, but trying to follow SkyThought's training config as much as possible.

  • Difference

    • sample packing: True
    • optimizer: paged_adamw_32bit
Axolotl config
base_model: beomi/EXAONE-3.5-7.8B-Instruct-Llamafied
model_type: AutoModelForCausalLM
tokenizer_config: beomi/EXAONE-3.5-7.8B-Instruct-Llamafied
tokenizer_type: AutoTokenizer

load_in_8bit: false
load_in_4bit: false
strict: false

datasets:
  - path: werty1248/Bespoke-Stratos-17k-KoEnKo-formatted
    field_messages: conversations
    type: chat_template
    chat_template: chatml

dataset_prepared_path: ./data_preparation
output_dir: /workspace/data

hf_use_auth_token: true

sequence_len: 16384
sample_packing: true
pad_to_sequence_len: true

plugins:
  - axolotl.integrations.liger.LigerPlugin
liger_rope: true
liger_rms_norm: true
liger_layer_norm: true
liger_glu_activation: true
liger_fused_linear_cross_entropy: true

wandb_project:
#wandb_entity:
#wandb_watch:
wandb_name:
#wandb_log_model:

gradient_accumulation_steps: 24
micro_batch_size: 1
num_epochs: 3
optimizer: paged_adamw_32bit
lr_scheduler: cosine
learning_rate: 1.0e-5

train_on_inputs: false
group_by_length: false
bf16: auto
fp16: 
tf32: false

gradient_checkpointing: true
early_stopping_patience:
resume_from_checkpoint:
local_rank:
logging_steps: 1
xformers_attention:
flash_attention: true

warmup_ratio: 0.1
eval_table_size:

deepspeed: ./deepspeed_configs/zero3_bf16.json

Evaluation

HAERAE-HUB/HRM8K

  • werty1248/EXAONE-3.5-7.8B-Stratos-Ko: temperature=0.0, max think tokens = 4096
    • In the first stage, the model generates up to 4096 tokens.
    • In the second stage, for answers that fail to generate an <|end_of_thought|> after using the maximum number of tokens, I add \n\n<|end_of_thought|>\n\n<|begin_of_solution|> to end the thought and output the answer.
  • LGAI-EXAONE/EXAONE-3.5-7.8B-Instruct: temperature=0.7, top_p=0.95, max_tokens = 2048
  • Other Models: Reported by OneLineAI - Creator of HRM8K Benchmark
Model GSM8K KSM MATH MMMLU OMNI_MATH Average
GPT-4o 91.21 22.83 74.45 68.72 30.75 57.99
GPT-4o-mini 87.57 19.40 70.68 63.40 26.45 53.50
*EXAONE-3.5-7.8B-Stratos-Ko 83.02 15.97 67.49 **44.68 24.62 49.98
EXAONE-3.5-7.8B-Instruct 81.58 14.71 63.50 ***41.49? 21.69 44.19
Qwen2.5-14B-Instruct 66.34 15.55 53.38 61.49 20.64 43.88
Llama-3.1-8B-Instruct 77.79 7.21 49.01 47.02 15.92 39.39
Qwen2.5-7B-Instruct 58.38 13.10 48.04 48.94 16.55 37.80
EXAONE-3.0-7.8B-Instruct 72.33 7.98 46.79 37.66 15.35 36.02
*Ko-R1-1.5B-preview 43.3 ? 73.1 ? 29.8 ?

* Korean Reasoning Models

** In a 4-option multiple-choice question, both correct answer numbber and answer alphabet are accepted as correct ('A' == 1, 'B' == 2, ...).

*** In a 4-option multiple-choice question, the correct answer *(not answer number) are accepted as correct.

Example

Q1: From ChuGyouk/GSM8k-Ko

input = tokenizer.apply_chat_template([ { "content": "Your role as an assistant involves thoroughly exploring questions through a systematic long thinking process before providing the final precise and accurate solutions. This requires engaging in a comprehensive cycle of analysis, summarizing, exploration, reassessment, reflection, backtracing, and iteration to develop well-considered thinking process. Please structure your response into two main sections: Thought and Solution. In the Thought section, detail your reasoning process using the specified format: <|begin_of_thought|> {thought with steps separated with '\\n\\n'} <|end_of_thought|> Each step should include detailed considerations such as analisying questions, summarizing relevant findings, brainstorming new ideas, verifying the accuracy of the current steps, refining any errors, and revisiting previous steps. In the Solution section, based on various attempts, explorations, and reflections from the Thought section, systematically present the final solution that you deem correct. The solution should remain a logical, accurate, concise expression style and detail necessary step needed to reach the conclusion, formatted as follows: <|begin_of_solution|> {final formatted, precise, and clear solution written in the same language as the question.} <|end_of_solution|> Now, try to solve the following question through the above guidelines", "role": "system" },
                                       { "content": "Janet의 μ˜€λ¦¬λŠ” ν•˜λ£¨μ— 16개의 달걀을 λ‚³μŠ΅λ‹ˆλ‹€. κ·Έλ…€λŠ” 맀일 μ•„μΉ¨ 3개의 달걀을 λ¨Ήκ³ , μΉœκ΅¬λ“€μ„ μœ„ν•΄ 4개의 λ‹¬κ±€λ‘œ 머핀을 κ΅½μŠ΅λ‹ˆλ‹€. κ·Έλ…€λŠ” 남은 달걀을 농산물 μ‹œμž₯μ—μ„œ μ‹ μ„ ν•œ μ˜€λ¦¬μ•Œ κ°œλ‹Ή 2λ‹¬λŸ¬μ— νŒλ§€ν•©λ‹ˆλ‹€. κ·Έλ…€λŠ” 농산물 μ‹œμž₯μ—μ„œ 맀일 μ–Όλ§ˆλ‚˜ λ§Žμ€ λˆμ„ λ²ŒκΉŒμš”?", "role": "user" },], add_generation_prompt=True, return_tensors='pt')
res = model.generate(input.to('cuda'), max_length = 2048, stop_strings = ['<|end_of_solution|>'], tokenizer=tokenizer)

tokenizer.decode(res[0])
Output
[|assistant|]Okay, let's figure out how much money Janet makes each day from selling her eggs at the market. Let's break it down step by step.

First, Janet's ducks lay 16 eggs every day. So she has 16 eggs each day.

Next, she uses some of these eggs. She eats 3 eggs every morning. So after eating those 3, she has 16 minus 3 eggs left. Let's calculate that: 16 - 3 = 13 eggs left.

Then, she bakes muffins for her friends using 4 eggs. So after making the muffins, she has 13 - 4 eggs left. Let me do that subtraction: 13 - 4 = 9 eggs remaining.

These remaining eggs are the ones she sells at the market. So she has 9 eggs left each day.

Each egg is sold for $2 at the market. Therefore, the amount of money she makes is the number of eggs sold multiplied by the price per egg. So 9 eggs times $2 each equals $18.

Let me double-check to make sure I didn't make a mistake:

16 eggs per day

Eats 3 eggs: 16 - 3 = 13 eggs left

Bakes 4 eggs: 13 - 4 = 9 eggs

9 eggs Γ— $2 each = $18

Yes, that adds up. So Janet sells the eggs at the market and makes $18 each day. That seems straightforward.

Wait, let me think again: she lays 16 eggs, uses 3, bakes 4, leaves 9, so sells 9. 9 times 2 is 18. Yep, that's correct.

Another way to check:

Total eggs each day: 16

Used eggs (eaten + baked): 3 + 4 = 7

So remaining eggs: 16 - 7 = 9

9 eggs Γ— $2 = $18. Yep, same answer.

I don't see any other expenses or other factors here, so the answer should be $18 each day.

**Final Answer**
Janet makes $18 each day at the market.

<|end_of_thought|>

<|begin_of_solution|>

Janet은 맀일 16개의 μ˜€λ¦¬μ•Œμ„λ‚λ‹ˆλ‹€. κ·Έλ…€λŠ” 맀일 μ•„μΉ¨ 3개의 μ•Œμ„ λ¨Ήκ³ , 4개의 μ•Œλ‘œ 머핀을 κ΅½μŠ΅λ‹ˆλ‹€. κ·Έλ…€κ°€ 남은 μ˜€λ¦¬μ•Œμ€ 농산물 μ‹œμž₯μ—μ„œ κ°œλ‹Ή 2λ‹¬λŸ¬μ— νŒλ§€λ©λ‹ˆλ‹€.

1. **맀일 λ‚³λŠ” μ•Œ**: \( 16 \)개
2. **맀일 λ¨ΉλŠ” μ•Œ**: \( 3 \)개
3. **맀일 μ‚¬μš©ν•˜λŠ” μ•Œ**: \( 3 + 4 = 7 \)개
4. **남은 μ•Œ**: \( 16 - 7 = 9 \)개

남은 9개의 μ•Œμ— λŒ€ν•œ 판맀 μˆ˜μ΅μ€ λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€.
\[
9 \text{ μ•Œ} \times 2 \text{ λ‹¬λŸ¬/μ•Œ} = 18 \text{ λ‹¬λŸ¬}
\]

κ·ΈλŸ¬λ―€λ‘œ, κ·Έλ…€λŠ” 농산물 μ‹œμž₯μ—μ„œ 맀일 \(\boxed{18}\)λ‹¬λŸ¬λ₯Ό 벌고 μžˆμŠ΅λ‹ˆλ‹€.

<|end_of_solution|>

Q2: From KSM, HAERAE-HUB/HRM8K

input = tokenizer.apply_chat_template([ { "content": "Your role as an assistant involves thoroughly exploring questions through a systematic long thinking process before providing the final precise and accurate solutions. This requires engaging in a comprehensive cycle of analysis, summarizing, exploration, reassessment, reflection, backtracing, and iteration to develop well-considered thinking process. Please structure your response into two main sections: Thought and Solution. In the Thought section, detail your reasoning process using the specified format: <|begin_of_thought|> {thought with steps separated with '\\n\\n'} <|end_of_thought|> Each step should include detailed considerations such as analisying questions, summarizing relevant findings, brainstorming new ideas, verifying the accuracy of the current steps, refining any errors, and revisiting previous steps. In the Solution section, based on various attempts, explorations, and reflections from the Thought section, systematically present the final solution that you deem correct. The solution should remain a logical, accurate, concise expression style and detail necessary step needed to reach the conclusion, formatted as follows: <|begin_of_solution|> {final formatted, precise, and clear solution written in the same language as the question.} <|end_of_solution|> Now, try to solve the following question through the above guidelines", "role": "system" },
                                       { "content": "μ–‘μ˜ μ‹€μˆ˜ t t 에 λŒ€ν•˜μ—¬ ν•¨μˆ˜ f f λ₯Ό \\[ f(t) = \\int_{0}^{\\sqrt{t}} \\int_{y}^{\\sqrt{t}} \\frac{1}{2 + \\sin(x^2)} \\, dx \\, dy \\] 둜 μ •μ˜ν•  λ•Œ, fβ€²left(fracpi2right) f'\\left(\\frac{\\pi}{2}\\right) 의 값을 κ΅¬ν•˜μ‹œμ˜€.", "role": "user" },], add_generation_prompt=True, return_tensors='pt')
res = model.generate(input.to('cuda'), max_length = 8192, stop_strings = ['<|end_of_solution|>'], tokenizer=tokenizer)

tokenizer.decode(res[0])
Output(ν•œκ΅­μ–΄λ‘œ 생각함???)
[|assistant|]

λ¨Όμ € 주어진 ν•¨μˆ˜ \( f(t) \)λ₯Ό μ΄ν•΄ν•˜κΈ° μœ„ν•΄ 이쀑 적뢄을 뢄석해야 ν•©λ‹ˆλ‹€. ν•¨μˆ˜λŠ” 두 개의 μ€‘μ²©λœ μ λΆ„μœΌλ‘œ μ •μ˜λ©λ‹ˆλ‹€:

\[
f(t) = \int_{0}^{\sqrt{t}} \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \, dy
\]

이 ν‘œν˜„μ‹μ€ \( x \)에 λŒ€ν•œ λ‚΄λΆ€ 적뢄 \( \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \)의 결과인 \( y \)에 λŒ€ν•œ μ™ΈλΆ€ 적뢄 \( \int_{0}^{\sqrt{t}} dy \)λ₯Ό λ‚˜νƒ€λƒ…λ‹ˆλ‹€. μ΄λŠ” \( y \)κ°€ \( 0 \)μ—μ„œ \( \sqrt{t} \) μ‚¬μ΄μ˜ μž„μ˜μ˜ 값을 μ·¨ν•˜λ©°, \( x \)λŠ” \( y \)λΆ€ν„° \( \sqrt{t} \)κΉŒμ§€μ˜ 값을 μ·¨ν•˜λŠ” μ˜μ—­ \( R \)을 λ‚˜νƒ€λƒ…λ‹ˆλ‹€. μ˜μ—­ \( R \)의 λ²”μœ„μ™€ 관계λ₯Ό μ΄ν•΄ν•˜λ©΄ 주어진 μ λΆ„μ˜ 의미λ₯Ό 더 잘 νŒŒμ•…ν•  수 μžˆμŠ΅λ‹ˆλ‹€.

\( y \)κ°€ 0λΆ€ν„° \( \sqrt{t} \)κΉŒμ§€, 그리고 \( x \)κ°€ \( y \)λΆ€ν„° \( \sqrt{t} \)κΉŒμ§€ λ³€ν•  λ•Œ, μ˜μ—­ \( R \)은 μ§μ‚¬κ°ν˜•μ˜ 일뢀가 될 수 μžˆμŠ΅λ‹ˆλ‹€. \( y \)κ°€ 0인 경우, \( x \)λŠ” 0μ—μ„œ \( \sqrt{t} \)κΉŒμ§€ μ¦κ°€ν•©λ‹ˆλ‹€. \( y \)κ°€ \( \sqrt{t} \)인 경우, \( x \)λŠ” \( \sqrt{t} \)μ—μ„œ λ‹€μ‹œ \( \sqrt{t} \)둜 λŒμ•„μ˜΅λ‹ˆλ‹€. \( y \)κ°€ \( 0 \)κ³Ό \( \sqrt{t} \) μ‚¬μ΄μ˜ μ–΄λ–€ κ°’ \( y_0 \)인 경우, \( x \)λŠ” \( y_0 \)μ—μ„œ \( \sqrt{t} \)κΉŒμ§€ μ¦κ°€ν•©λ‹ˆλ‹€. λ”°λΌμ„œ, μ˜μ—­ \( R \)은 λ‹€μŒκ³Ό 같은 λ„€ 개의 μ‚¬κ°ν˜•μœΌλ‘œ ꡬ성될 수 μžˆμŠ΅λ‹ˆλ‹€:
- \( y = 0 \)λΆ€ν„° \( y = a \)κΉŒμ§€ (단, \( a \leq \sqrt{t} \))의 μ‚¬κ°ν˜•: μ΄λ•Œ \( x \)λŠ” \( 0 \)κ³Ό \( \sqrt{t} \) 사이.
- \( y = a \)λΆ€ν„° \( y = \sqrt{t} \)κΉŒμ§€ (단, \( a \leq y \leq \sqrt{t} \))의 μ‚¬κ°ν˜•: μ΄λ•Œ \( x \)λŠ” \( a \)λΆ€ν„° \( \sqrt{t} \) 사이.

이 μ‚¬κ°ν˜•λ“€μ€ \( y = a \)μ—μ„œ μ ‘ν•©λ‹ˆλ‹€. λ”°λΌμ„œ, \( y \)의 λ²”μœ„λŠ” \( 0 \)μ—μ„œ \( \sqrt{t} \) 사이에 있고, \( a \)λŠ” 이전에 μ •μ˜λœ νŠΉμ • κ°’, λ˜λŠ” \( t \)에 μ˜ν•΄ κ²°μ •λ˜λŠ” κ°’μž…λ‹ˆλ‹€.

λ¬Έμ œλŠ” \( f'(t) \)λ₯Ό κ²°μ •ν•˜λŠ” κ²ƒμž…λ‹ˆλ‹€. μ΄λŠ” \( t \)에 λŒ€ν•œ \( f(t) \)의 λ„ν•¨μˆ˜λ‘œ, Leibniz 적뢄 κ·œμΉ™μ„ μ‚¬μš©ν•˜μ—¬ ꡬ할 수 μžˆμŠ΅λ‹ˆλ‹€. 이쀑 적뢄에 λŒ€ν•œ Leibniz 적뢄 κ·œμΉ™μ€ λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€:

\[
\frac{d}{dt} \int_{a(t)}^{b(t)} \int_{c(t)}^{d(t)} f(x, y) \, dx \, dy = \int_{a(t)}^{b(t)} \frac{\partial}{\partial t} \left[ \int_{c(t)}^{d(t)} f(x, y) \, dx \right] \, dy + \int_{c(t)}^{d(t)} \frac{\partial}{\partial t} \left[ \int_{a(t)}^{d(t)} f(x, y) \, dx \right] \, dy
\]

주어진 ν•¨μˆ˜ \( f(t) \)λŠ” λ‹€μŒκ³Ό 같이 μ“Έ 수 μžˆμŠ΅λ‹ˆλ‹€:

\[
f(t) = \int_{0}^{\sqrt{t}} \left( \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \right) dy
\]

λ”°λΌμ„œ, Leibniz κ·œμΉ™μ„ μ μš©ν•˜λ©΄, \( f'(t) \)λŠ” λ‹€μŒ 두 개의 적뢄을 ν¬ν•¨ν•©λ‹ˆλ‹€:

1. \( y \)에 λŒ€ν•œ μ™ΈλΆ€ 적뢄: \( \int_{0}^{\sqrt{t}} \frac{\partial}{\partial t} \left( \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \right) dy \)
2. \( x \)에 λŒ€ν•œ λ‚΄λΆ€ 적뢄: \( \int_{0}^{\sqrt{t}} \frac{\partial}{\partial t} \left( \int_{0}^{x} \frac{1}{2 + \sin(y^2)} \, dy \right) dx \)

이 두 개λ₯Ό λͺ¨λ‘ 계산해야 ν•©λ‹ˆλ‹€.

λ¨Όμ €, \( y \)에 λŒ€ν•œ μ™ΈλΆ€ 적뢄을 λ‹¨μˆœν™”ν•©λ‹ˆλ‹€. \( \frac{\partial}{\partial t} \left( \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \right) \)λ₯Ό κ³„μ‚°ν•©λ‹ˆλ‹€. μ΄λŠ” Leibniz κ·œμΉ™μ„ μ‚¬μš©ν•œ λ‚΄λΆ€ 적뢄에 λŒ€ν•œ \( t \)에 λŒ€ν•œ λ„ν•¨μˆ˜λ‘œ μ“Έ 수 μžˆμŠ΅λ‹ˆλ‹€:

\[
\frac{\partial}{\partial t} \left( \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \right) = \frac{1}{2 + \sin((\sqrt{t})^2)} \cdot \frac{d}{dt}(\sqrt{t}) = \frac{1}{2 + \sin(t)} \cdot \frac{1}{2\sqrt{t}}
\]

λ”°λΌμ„œ, 첫 번째 λΆ€λΆ„ 적뢄은 λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€:

\[
\int_{0}^{\sqrt{t}} \frac{1}{2 + \sin(t)} \cdot \frac{1}{2\sqrt{t}} dy = \frac{1}{2\sqrt{t}(2 + \sin(t))} \int_{0}^{\sqrt{t}} dy = \frac{1}{2\sqrt{t}(2 + \sin(t))} \cdot \sqrt{t} = \frac{1}{2(2 + \sin(t))}
\]

λ‹€μŒμœΌλ‘œ, \( x \)에 λŒ€ν•œ λ‚΄λΆ€ 적뢄을 λ‹¨μˆœν™”ν•©λ‹ˆλ‹€. \( \frac{\partial}{\partial t} \left( \int_{0}^{x} \frac{1}{2 + \sin(y^2)} \, dy \right) \)λ₯Ό κ³„μ‚°ν•©λ‹ˆλ‹€. μ΄λŠ” \( y = x \)μ—μ„œ \( t \)에 λŒ€ν•œ λ„ν•¨μˆ˜λ‘œ μ“Έ 수 μžˆμŠ΅λ‹ˆλ‹€:

\[
\frac{\partial}{\partial t} \left( \int_{0}^{x} \frac{1}{2 + \sin(y^2)} \, dy \right) = \frac{\partial}{\partial t} \left( \int_{0}^{x} f(y) \, dy \right) = f(x) = \frac{1}{2 + \sin(x^2)}
\]

λ”°λΌμ„œ, 두 번째 λΆ€λΆ„ 적뢄은 λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€:

\[
\int_{0}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx
\]

이것을 κ²°ν•©ν•˜λ©΄, \( f'(t) \)의 전체 식은 λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€:

\[
f'(t) = \frac{1}{2(2 + \sin(t))} + \int_{0}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx
\]

\( f'(t) \)의 각 뢀뢄을 λ‹¨μˆœν™”ν•˜λ©΄ λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€:

1. 첫 번째 ν•­ \( \frac{1}{2(2 + \sin(t))} = \frac{1}{4 + 2\sin(t)} \)
2. 두 번째 항은 μ λΆ„λ˜μ—ˆμ§€λ§Œ, 주어진 ν˜•νƒœλ‘œ λ‹¨μˆœν™”ν•  수 μ—†μŠ΅λ‹ˆλ‹€ (더 λ³΅μž‘ν•¨). κ·ΈλŸ¬λ‚˜ \( t_{\text{max}} \)λ₯Ό ꡬ할 λ•Œ λ¬Έμ œκ°€ ν•„μš”λ‘œ ν•©λ‹ˆλ‹€.

λ”°λΌμ„œ, \( f'(t) \)λŠ” λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€:

\[
f'(t) = \frac{1}{4 + 2\sin(t)} + \int_{0}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx
\]

\( t = \frac{\pi}{2} \)인 경우, μš°λ¦¬λŠ” \( f'\left( \frac{\pi}{2} \right) \)λ₯Ό ν•„μš”λ‘œ ν•©λ‹ˆλ‹€.

λ¨Όμ €, 첫 번째 항을 ν‰κ°€ν•©λ‹ˆλ‹€:

\[
\frac{1}{4 + 2\sin\left( \frac{\pi}{2} \right)} = \frac{1}{4 + 2 \cdot 1} = \frac{1}{6}
\]

두 번째 항을 ν‰κ°€ν•˜λ €λ©΄, \( \sin\left( x^2 \right) \)에 λŒ€ν•œ \( x \)에 λŒ€ν•œ 적뢄을 \( \frac{\pi}{2} \)κΉŒμ§€ \( \sqrt{t} \)κΉŒμ§€ κ³ λ €ν•΄μ•Ό ν•©λ‹ˆλ‹€. ν•˜μ§€λ§Œ \( t = \frac{\pi}{2} \)μ΄λ―€λ‘œ, 이 적뢄은 \( x \)κ°€ 0μ—μ„œ \( \sqrt{\frac{\pi}{2}} \)κΉŒμ§€μ˜ λ²”μœ„λ₯Ό κ°–κ²Œ λ©λ‹ˆλ‹€. κ·ΈλŸ¬λ‚˜ μš°λ¦¬κ°€ μ•Œκ³  μžˆλŠ” 점은 \( \sin(x^2) \)의 적뢄에 λŒ€ν•œ λ‹«νžŒ ν˜•νƒœκ°€ μ‘΄μž¬ν•˜μ§€ μ•ŠμœΌλ©°, λ”°λΌμ„œ 적뢄은 μ§μ ‘μ μœΌλ‘œ 평가할 수 μ—†λ‹€λŠ” κ²ƒμž…λ‹ˆλ‹€. 이 적뢄은 주어진 λ²”μœ„μ—μ„œ \(\sin(x^2)\)의 주기적이고 λ³΅μž‘ν•œ 행동 λ•Œλ¬Έμ— 평가할 수 μ—†μŠ΅λ‹ˆλ‹€.

κ·ΈλŸ¬λ‚˜ 문제의 λ§₯λ½μ—μ„œ 주어진 값을 λ°”νƒ•μœΌλ‘œ κ°€λŠ₯ν•œ 간단화 λ˜λŠ” νŠΉμ • κ°’μ˜ ν™œμš©μ„ κ³ λ €ν•  수 μžˆμŠ΅λ‹ˆλ‹€. 주어진 선택지와 계산 κ²°κ³Όλ₯Ό 톡해, \( f'\left( \frac{\pi}{2} \right) \)의 값은 첫 번째 ν•­ \( \frac{1}{6} \)κ³Ό 두 번째 ν•­ \(\int_{0}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx\)의 ν•©μœΌλ‘œλΆ€ν„° μΆ”μ •λ©λ‹ˆλ‹€. μ λΆ„μ˜ μ„±μ§ˆκ³Ό \( x = 0 \)μ—μ„œ \( x = \sqrt{\frac{\pi}{2}} \)κΉŒμ§€ \( 2 + \sin(x^2) \)에 λŒ€ν•œ λŒ€μΉ­μ„±μ„ κ³ λ €ν•  λ•Œ, 첫 번째 ν•­κ³Ό 두 번째 ν•­ λͺ¨λ‘ \( \frac{1}{6} \)에 κ°€κΉŒμšΈ μˆ˜λ„ μžˆμ§€λ§Œ, μ‹€μ œλ‘œ \( t = \frac{\pi}{2} \)λ₯Ό μ‚¬μš©ν•˜λŠ” 문제의 λ§₯λ½μ—μ„œ 이 μ λΆ„μ˜ 값은 κ°„λ‹¨ν™”ν•˜κΈ° μ–΄λ ΅μŠ΅λ‹ˆλ‹€.

κ·ΈλŸ¬λ‚˜, 문제의 λ§₯락과 주어진 닡지 쀑 κ°€μž₯ κ°€λŠ₯μ„± μžˆλŠ” 값을 μ„ νƒν•˜κΈ° μœ„ν•΄, \( f'\left( \frac{\pi}{2} \right) \)에 λŒ€ν•œ λ‹¨μˆœν™”λœ μ ‘κ·Ό 방식과 κ°€λŠ₯ν•œ λ‹¨μˆœν™”λ₯Ό κ³ λ €ν•΄μ•Ό ν•©λ‹ˆλ‹€. κ·ΈλŸ¬λ‚˜ 두 번째 뢀뢄을 μ œμ™Έν•˜κ³ λŠ” \( f'\left( \frac{\pi}{2} \right) = \frac{1}{6} \)μž…λ‹ˆλ‹€.

\[
f'\left( \frac{\pi}{2} \right) = \frac{1}{6} + \int_{0}^{\sqrt{\pi/2}} \frac{1}{2 + \sin(x^2)} \, dx
\]

이것은 μ‹€μ œλ‘œ 더 λ³΅μž‘ν•œ 평가λ₯Ό ν•„μš”λ‘œ ν•˜λ©°, 일반적인 μˆ˜ν•™ 문제 ν•΄κ²° λ°©λ²•μœΌλ‘œλŠ” λ‹¨μˆœν™”λ˜μ§€ μ•ŠμŠ΅λ‹ˆλ‹€. κ·ΈλŸ¬λ‚˜ 문제의 λ§₯락과 주어진 ν•΄λ‹΅ 선택지에 따라, 이 적뢄은 λ¬΄μ‹œλ˜κ³  \( \frac{1}{6} \)만 λ‚¨κ²Œ λ©λ‹ˆλ‹€. μ΄λŠ” 문제λ₯Ό λ‹¨μˆœν™”ν•˜κ±°λ‚˜ νŠΉμ • λŒ€μΉ­μ„±μ„ μ‚¬μš©ν•˜κΈ° λ•Œλ¬Έμ— κ°€λŠ₯ν•©λ‹ˆλ‹€.

\[
\boxed{\frac{1}{6}}
\]

<|end_of_thought|>

<|begin_of_solution|>

ν•¨μˆ˜ \( f(t) \)λŠ” λ‹€μŒκ³Ό 같이 μ •μ˜λ©λ‹ˆλ‹€.
\[ f(t) = \int_{0}^{\sqrt{t}} \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \, dy \]

\( t \)에 λŒ€ν•œ λ„ν•¨μˆ˜ \( f'\left(t\right) \)λ₯Ό κ΅¬ν•˜κΈ° μœ„ν•΄ Leibniz 적뢄 κ·œμΉ™μ„ μ‚¬μš©ν•©λ‹ˆλ‹€.
\( f'(t) \)λŠ” \( y \)에 λŒ€ν•œ μ™ΈλΆ€ 적뢄과 \( x \)에 λŒ€ν•œ λ‚΄λΆ€ μ λΆ„μ˜ λ„ν•¨μˆ˜λ₯Ό ν¬ν•¨ν•©λ‹ˆλ‹€.

1. **\( y \)에 λŒ€ν•œ μ™ΈλΆ€ 적뢄:**
   \[
   \frac{\partial}{\partial t} \left( \int_{y}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx \right) = \frac{1}{2 + \sin(t)} \cdot \frac{1}{2\sqrt{t}}
   \]
   이 κ²°κ³Όλ₯Ό 0μ—μ„œ \( \sqrt{t} \)κΉŒμ§€ μ λΆ„ν•˜λ©΄,
   \[
   \frac{1}{2\sqrt{t}(2 + \sin(t))} \int_{0}^{\sqrt{t}} dy = \frac{1}{2(2 + \sin(t))}
   \]

2. **\( x \)에 λŒ€ν•œ λ‚΄λΆ€ 적뢄:**
   \[
   \frac{\partial}{\partial t} \left( \int_{0}^{x} \frac{1}{2 + \sin(y^2)} \, dy \right) = \frac{1}{2 + \sin(x^2)}
   \]
   이 κ²°κ³Όλ₯Ό 0μ—μ„œ \( \sqrt{t} \)κΉŒμ§€ μ λΆ„ν•˜λ©΄,
   \[
   \int_{0}^{\sqrt{t}} \frac{1}{2 + \sin(x^2)} \, dx
   \]

\( t = \frac{\pi}{2} \)λ₯Ό \( f'(t) \)에 λŒ€μž…ν•˜λ©΄,
\[
f'\left(\frac{\pi}{2}\right) = \frac{1}{2(2 + \sin(\pi/2))} + \int_{0}^{\sqrt{\pi/2}} \frac{1}{2 + \sin(x^2)} \, dx
\]

\( \sin(\pi/2) = 1 \)μ΄λ―€λ‘œ,
\[
\frac{1}{2(2 + 1)} = \frac{1}{6}
\]

두 번째 항은 μ •ν™•ν•˜κ²Œ ν‰κ°€λ˜μ§€ μ•ŠμœΌλ©°, 문제의 λ§₯λ½μ—μ„œ κ³ λ €ν•˜μ§€ μ•Šκ³ , μ΅œμ’… 닡은 λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€.
\[
\boxed{\frac{1}{6}}
\]

<|end_of_solution|>

Q3: General Knowledge

input = tokenizer.apply_chat_template([ { "content": "Your role as an assistant involves thoroughly exploring questions through a systematic long thinking process before providing the final precise and accurate solutions. This requires engaging in a comprehensive cycle of analysis, summarizing, exploration, reassessment, reflection, backtracing, and iteration to develop well-considered thinking process. Please structure your response into two main sections: Thought and Solution. In the Thought section, detail your reasoning process using the specified format: <|begin_of_thought|> {thought with steps separated with '\\n\\n'} <|end_of_thought|> Each step should include detailed considerations such as analisying questions, summarizing relevant findings, brainstorming new ideas, verifying the accuracy of the current steps, refining any errors, and revisiting previous steps. In the Solution section, based on various attempts, explorations, and reflections from the Thought section, systematically present the final solution that you deem correct. The solution should remain a logical, accurate, concise expression style and detail necessary step needed to reach the conclusion, formatted as follows: <|begin_of_solution|> {final formatted, precise, and clear solution written in the same language as the question.} <|end_of_solution|> Now, try to solve the following question through the above guidelines", "role": "system" },
                                       { "content": "μ„Έκ³„μ—μ„œ μ„Έ 번째둜 높은 산은?", "role": "user" },], add_generation_prompt=True, return_tensors='pt')
res = model.generate(input.to('cuda'), max_length = 2048, temperature = 0.5, stop_strings = ['<|end_of_solution|>'], tokenizer=tokenizer)

tokenizer.decode(res[0])
  • Normal temperature: hallucination (unstable answer)
  • Low temperature: repitition
Output
[|assistant|]

Okay, so I need to figure out the third highest mountain in the world. Hmm, I know that Mount Everest is the highest, so that's first. But where does the third one come in?

First, I should confirm the order of the highest mountains. Let me think. I think there are a few peaks in the Himalayas, right? Because Mount Everest is in the Himalayas. So maybe the next ones are also in that range.

Let me try to recall the order. The highest is Mount Everest, then K2, then Kangchenjunga, then Lhotse, then Makalu, then Cho Oyu, then Dhaulagiri, then Manaslu, then Nanga Parbat. Wait, is that correct?

Wait, I might be mixing up some names. Let me check. Mount Everest is the highest, then K2 is the second highest. So the third highest is Kangchenjunga? Or is it Lhotse?

Wait, I'm a bit confused here. Maybe I need to look this up to be sure.

Upon checking, the correct order is:

1. Mount Everest
2. K2
3. Kangchenjunga
4. Lhotse
5. Makalu
6. Cho Oyu
7. Dhaulagiri
8. Manaslu
9. Nanga Parbat

So, yes, Kangchenjunga is the third highest mountain in the world.

But wait, is there any debate about this? I recall that there have been some discussions about the exact heights of some of these peaks, but generally, the order is accepted as above.

Kangchenjunga is located in the Himalayas on the border between Nepal and India. It has an elevation of 8,586 meters (28,169 feet). It's a sacred mountain in the region and is known for its challenging climbing routes.

So, to answer the question: the third highest mountain in the world is Kangchenjunga.

**Final Answer**
The third highest mountain in the world is Kangchenjunga.

<|end_of_thought|>

<|begin_of_solution|>

μ„Έκ³„μ—μ„œ μ„Έ 번째둜 높은 산은 λ¬΄μ—‡μΌκΉŒμš”?

1. **μ—λ² λ ˆμŠ€νŠΈ**κ°€ κ°€μž₯ λ†’λ‹€λŠ” 것을 μ•Œκ³  μžˆμŠ΅λ‹ˆλ‹€.
2. κ·Έ λ‹€μŒμ€ **K2**μž…λ‹ˆλ‹€.
3. **Kangchenjunga**λŠ” μ„Έ 번째둜 λ†’μŠ΅λ‹ˆλ‹€.

KangchenjungaλŠ” λ„€νŒ”κ³Ό 인도 μ‚¬μ΄μ˜ νžˆλ§λΌμ•Όμ— μœ„μΉ˜ν•˜λ©°, λ†’μ΄λŠ” μ•½ 8,586λ―Έν„°(28,169ν”ΌνŠΈ)μž…λ‹ˆλ‹€.

λ”°λΌμ„œ μ΅œμ’… 닡변은 λ‹€μŒκ³Ό κ°™μŠ΅λ‹ˆλ‹€.

\boxed{Kangchenjunga}

<|end_of_solution|>
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