size_categories:n<1K modality:text library:distilabel region:us synthetic distilabel rlaif
---
size_categories: n<1K
dataset_info:
  features:
  - name: problem
    dtype: string
  - name: level
    dtype: string
  - name: solution
    dtype: string
  - name: type
    dtype: string
  - name: generation
    sequence: string
  - name: distilabel_metadata
    list:
    - name: raw_input_text_generation_0
      list:
      - name: content
        dtype: string
      - name: role
        dtype: string
    - name: raw_output_text_generation_0
      dtype: string
    - name: statistics_text_generation_0
      struct:
      - name: input_tokens
        dtype: int64
      - name: output_tokens
        dtype: int64
  - name: model_name
    dtype: string
  splits:
  - name: train
    num_bytes: 222373
    num_examples: 10
  download_size: 118274
  dataset_size: 222373
configs:
- config_name: default
  data_files:
  - split: train
    path: data/train-*
tags:
- synthetic
- distilabel
- rlaif
---

<p align="left">
  <a href="https://github.com/argilla-io/distilabel">
    <img src="https://raw.githubusercontent.com/argilla-io/distilabel/main/docs/assets/distilabel-badge-light.png" alt="Built with Distilabel" width="200" height="32"/>
  </a>
</p>

# Dataset Card for numina-deepseek-r1-qwen-7b

This dataset has been created with [distilabel](https://distilabel.argilla.io/).



## Dataset Summary

This dataset contains a `pipeline.yaml` which can be used to reproduce the pipeline that generated it in distilabel using the `distilabel` CLI:

```console
distilabel pipeline run --config "https://huggingface.co/datasets/gazhitt/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

or explore the configuration:

```console
distilabel pipeline info --config "https://huggingface.co/datasets/gazhitt/numina-deepseek-r1-qwen-7b/raw/main/pipeline.yaml"
```

## Dataset structure

The examples have the following structure per configuration:


<details><summary> Configuration: default </summary><hr>

```json
{
    "distilabel_metadata": [
        {
            "raw_input_text_generation_0": [
                {
                    "content": "Let \\[f(x) = \\left\\{\n\\begin{array}{cl} ax+3, \u0026\\text{ if }x\u003e2, \\\\\nx-5 \u0026\\text{ if } -2 \\le x \\le 2, \\\\\n2x-b \u0026\\text{ if } x \u003c-2.\n\\end{array}\n\\right.\\]Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper).",
                    "role": "user"
                }
            ],
            "raw_output_text_generation_0": "Okay, so I have this piecewise function, and I need to find the value of \\( a + b \\) such that the function is continuous. Hmm, continuity for piecewise functions generally means that there are no jumps or breaks at the points where the function changes its definition. So, in this case, the function changes at \\( x = -2 \\) and \\( x = 2 \\). It must be continuous at both of those points. That means the left-hand limit and the right-hand limit at those points should both exist and be equal to the function\u0027s value at those points.\n\nAlright, let me break it down. There are three pieces:\n\n1. For \\( x \u003e 2 \\), \\( f(x) = ax + 3 \\)\n2. For \\( -2 \\leq x \\leq 2 \\), \\( f(x) = x - 5 \\)\n3. For \\( x \u003c -2 \\), \\( f(x) = 2x - b \\)\n\nSo, the points to check are at \\( x = -2 \\) and \\( x = 2 \\), right?\n\nStarting with \\( x = 2 \\). For continuity there, the limit from the left (as \\( x \\) approaches 2 from below) should equal the limit from the right (as \\( x \\) approaches 2 from above), and both should equal \\( f(2) \\).\n\nCalculating the left limit first: when \\( x \\) approaches 2 from the left, which is using the middle piece, \\( f(x) = x - 5 \\). So, plugging in 2, we get \\( 2 - 5 = -3 \\).\n\nCalculating the right limit: when \\( x \\) approaches 2 from the right, which is using the first piece, \\( f(x) = ax + 3 \\). Plugging in 2, that\u0027s \\( 2a + 3 \\).\n\nSo, for continuity at \\( x = 2 \\), we set the left limit equal to the right limit: \\( -3 = 2a + 3 \\).\n\nWait, let me write that down more clearly:\n\n\\[ \\lim_{x \\to 2^-} f(x) = 2 - 5 = -3 \\]\n\\[ \\lim_{x \\to 2^+} f(x) = a(2) + 3 = 2a + 3 \\]\n\nSet them equal: \\( -3 = 2a + 3 \\)\n\nSolving for \\( a \\):\n\nSubtract 3 from both sides: \\( -3 - 3 = 2a \\) =\u003e \\( -6 = 2a \\)\n\nDivide both sides by 2: \\( a = -3 \\). \n\nOkay, so found \\( a = -3 \\). Now, moving on to \\( x = -2 \\).\n\nAgain, checking continuity at \\( x = -2 \\): the left limit (from the middle piece) and the right limit (from the third piece) should both equal \\( f(-2) \\) which is calculated using the middle piece, so \\( f(-2) = (-2) - 5 = -7 \\).\n\nWait, no, sorry. Let me correct that:\n\nWait, actually, for \\( x = -2 \\), in the middle piece, \\( f(-2) = (-2) - 5 = -7 \\). But for the left and right limits approaching \\( x = -2 \\), we use the other defined pieces.\n\nSo, first, compute the right limit: as \\( x \\) approaches \\( -2 \\) from the right (i.e., from values greater than -2 but less than or equal to 2), we are using the middle piece, which is \\( f(x) = x - 5 \\). So plugging in \\( x = -2 \\):\n\n\\( f(-2^+) = -2 - 5 = -7 \\).\n\nNow, the left limit: as \\( x \\) approaches \\( -2 \\) from the left (values less than or equal to -2), we use the third piece, \\( f(x) = 2x - b \\). Substituting \\( x = -2 \\):\n\n\\( f(-2^-) = 2(-2) - b = -4 - b \\).\n\nFor continuity at \\( x = -2 \\), the left limit must equal the right limit (which is \\( -7 \\)) and equal \\( f(-2) = -7 \\). \n\nSo, setting the left limit equal to \\( -7 \\):\n\n\\( -4 - b = -7 \\)\n\nSolving for \\( b \\):\n\nAdd 4 to both sides: \\( -b = -7 + 4 \\) =\u003e \\( -b = -3 \\)\n\nMultiply both sides by -1: \\( b = 3 \\)\n\nSo, found \\( b = 3 \\). Hmm.\n\nTherefore, \\( a = -3 \\) and \\( b = 3 \\), so \\( a + b = -3 + 3 = 0 \\). Wait, is that right? That seems surprisingly simple. Let me double-check.\n\nFor \\( x = 2 \\):\n\nLeft limit: 2 - 5 = -3\nRight limit: 2a + 3\nSo -3 = 2a + 3 =\u003e subtract 3: -6 = 2a =\u003e a = -3. That seems correct.\n\nFor \\( x = -2 \\):\n\nRight limit: approaches from the right, x = -2, middle piece: -2 - 5 = -7\nLeft limit: approaches from left, x = -2, third piece: 2(-2) - b = -4 - b\nSet = -7: -4 - b = -7 =\u003e -b = -3 =\u003e b = 3. That also seems correct.\n\nTherefore, adding \\( a + b \\) gives -3 + 3 = 0. So, 0 is the answer.\n\nBut let me just visualize the function to make sure I didn\u0027t make mistakes.\n\nFor x \u003e 2, it\u0027s a line with slope a, which we found to be -3, so f(x) = -3x + 3. As x increases beyond 2, the function would decrease since the slope is negative.\n\nFor -2 \u003c= x \u003c= 2, it\u0027s f(x) = x - 5. That\u0027s a line with slope 1, so it\u0027s increasing from x = -2 to x = 2. At x = -2, f(x) = -2 - 5 = -7; at x = 2, f(x) = 2 - 5 = -3. Wait, hold on. If x = -2, f(x) = -7 and x = 2, f(x) = -3. So between -2 and 2, it\u0027s going from -7 to -3, which is an increase of 4 over an increase of 4 units in x (from -2 to 2 is 4 units), so the slope is indeed 1.\n\nNow, for x \u003c -2, f(x) = 2x - b. So, f(x) is a line with slope 2. Since b is 3, at x = -2, f(-2) = 2*(-2) - 3 = -4 -3 = -7. So both the middle piece and the left piece meet at x = -2 with f(-2) = -7.\n\nAnd for x \u003e 2, f(3) would be 3a + 3, where a is -3, so f(3) = 3*(-3) + 3 = -9 + 3 = -6. Let\u0027s check continuity as x approaches 2 from the right: f(2) is -3, right? It\u0027s the value from the middle piece, so it\u0027s as defined there. However, if we approach 2 from above, using the right-hand side, f(x) = 2a + 3 = 2*(-3) + 3 = -6 + 3 = -3. So, the function at x=2 is -3, and as x approaches 2 from above, it\u0027s approaching -3 as well.\n\nSimilarly, approaching from x = -2, from the left, we get from -4 - b with b=3 equals -7, which is the value in the middle piece, so that\u0027s continuous.\n\nThus, all three pieces connect smoothly at x = -2 and x = 2, so the function is continuous everywhere.\n\nSo, putting it all together, since a is -3 and b is 3, adding them together, we get 0. Hmm, 0 seems quite clean. Maybe that\u0027s just how the constants work here.\n\nAll right, so after going through each step, making sure the limits match the function values, both at x = 2 and x = -2, I\u0027m confident that a = -3 and b = 3, so their sum is 0. I think that\u0027s the right answer.\n\n**Final Answer**\n\\boxed{0}\n\u003c/think\u003e\n\nTo ensure the piecewise function is continuous, we need to check the continuity at the points \\( x = -2 \\) and \\( x = 2 \\).\n\n1. **Continuity at \\( x = 2 \\)**:\n   - Left limit (from the middle piece): \\( f(2) = 2 - 5 = -3 \\)\n   - Right limit (from the first piece): \\( f(2) = 2a + 3 \\)\n   - Setting the left and right limits equal: \\( -3 = 2a + 3 \\)\n   - Solving for \\( a \\): \\( -6 = 2a \\) \u27f9 \\( a = -3 \\)\n\n2. **Continuity at \\( x = -2 \\)**:\n   - Right limit (from the middle piece): \\( f(-2) = -2 - 5 = -7 \\)\n   - Left limit (from the third piece): \\( f(-2) = 2(-2) - b = -4 - b \\)\n   - Setting the right and left limits equal: \\( -4 - b = -7 \\)\n   - Solving for \\( b \\): \\( -b = -3 \\) \u27f9 \\( b = 3 \\)\n\nThus, \\( a = -3 \\) and \\( b = 3 \\). Adding these values together, we get \\( a + b = -3 + 3 = 0 \\).\n\n\\[\n\\boxed{0}\n\\]",
            "statistics_text_generation_0": {
                "input_tokens": 109,
                "output_tokens": 2226
            }
        }
    ],
    "generation": [
        "Okay, so I have this piecewise function, and I need to find the value of \\( a + b \\) such that the function is continuous. Hmm, continuity for piecewise functions generally means that there are no jumps or breaks at the points where the function changes its definition. So, in this case, the function changes at \\( x = -2 \\) and \\( x = 2 \\). It must be continuous at both of those points. That means the left-hand limit and the right-hand limit at those points should both exist and be equal to the function\u0027s value at those points.\n\nAlright, let me break it down. There are three pieces:\n\n1. For \\( x \u003e 2 \\), \\( f(x) = ax + 3 \\)\n2. For \\( -2 \\leq x \\leq 2 \\), \\( f(x) = x - 5 \\)\n3. For \\( x \u003c -2 \\), \\( f(x) = 2x - b \\)\n\nSo, the points to check are at \\( x = -2 \\) and \\( x = 2 \\), right?\n\nStarting with \\( x = 2 \\). For continuity there, the limit from the left (as \\( x \\) approaches 2 from below) should equal the limit from the right (as \\( x \\) approaches 2 from above), and both should equal \\( f(2) \\).\n\nCalculating the left limit first: when \\( x \\) approaches 2 from the left, which is using the middle piece, \\( f(x) = x - 5 \\). So, plugging in 2, we get \\( 2 - 5 = -3 \\).\n\nCalculating the right limit: when \\( x \\) approaches 2 from the right, which is using the first piece, \\( f(x) = ax + 3 \\). Plugging in 2, that\u0027s \\( 2a + 3 \\).\n\nSo, for continuity at \\( x = 2 \\), we set the left limit equal to the right limit: \\( -3 = 2a + 3 \\).\n\nWait, let me write that down more clearly:\n\n\\[ \\lim_{x \\to 2^-} f(x) = 2 - 5 = -3 \\]\n\\[ \\lim_{x \\to 2^+} f(x) = a(2) + 3 = 2a + 3 \\]\n\nSet them equal: \\( -3 = 2a + 3 \\)\n\nSolving for \\( a \\):\n\nSubtract 3 from both sides: \\( -3 - 3 = 2a \\) =\u003e \\( -6 = 2a \\)\n\nDivide both sides by 2: \\( a = -3 \\). \n\nOkay, so found \\( a = -3 \\). Now, moving on to \\( x = -2 \\).\n\nAgain, checking continuity at \\( x = -2 \\): the left limit (from the middle piece) and the right limit (from the third piece) should both equal \\( f(-2) \\) which is calculated using the middle piece, so \\( f(-2) = (-2) - 5 = -7 \\).\n\nWait, no, sorry. Let me correct that:\n\nWait, actually, for \\( x = -2 \\), in the middle piece, \\( f(-2) = (-2) - 5 = -7 \\). But for the left and right limits approaching \\( x = -2 \\), we use the other defined pieces.\n\nSo, first, compute the right limit: as \\( x \\) approaches \\( -2 \\) from the right (i.e., from values greater than -2 but less than or equal to 2), we are using the middle piece, which is \\( f(x) = x - 5 \\). So plugging in \\( x = -2 \\):\n\n\\( f(-2^+) = -2 - 5 = -7 \\).\n\nNow, the left limit: as \\( x \\) approaches \\( -2 \\) from the left (values less than or equal to -2), we use the third piece, \\( f(x) = 2x - b \\). Substituting \\( x = -2 \\):\n\n\\( f(-2^-) = 2(-2) - b = -4 - b \\).\n\nFor continuity at \\( x = -2 \\), the left limit must equal the right limit (which is \\( -7 \\)) and equal \\( f(-2) = -7 \\). \n\nSo, setting the left limit equal to \\( -7 \\):\n\n\\( -4 - b = -7 \\)\n\nSolving for \\( b \\):\n\nAdd 4 to both sides: \\( -b = -7 + 4 \\) =\u003e \\( -b = -3 \\)\n\nMultiply both sides by -1: \\( b = 3 \\)\n\nSo, found \\( b = 3 \\). Hmm.\n\nTherefore, \\( a = -3 \\) and \\( b = 3 \\), so \\( a + b = -3 + 3 = 0 \\). Wait, is that right? That seems surprisingly simple. Let me double-check.\n\nFor \\( x = 2 \\):\n\nLeft limit: 2 - 5 = -3\nRight limit: 2a + 3\nSo -3 = 2a + 3 =\u003e subtract 3: -6 = 2a =\u003e a = -3. That seems correct.\n\nFor \\( x = -2 \\):\n\nRight limit: approaches from the right, x = -2, middle piece: -2 - 5 = -7\nLeft limit: approaches from left, x = -2, third piece: 2(-2) - b = -4 - b\nSet = -7: -4 - b = -7 =\u003e -b = -3 =\u003e b = 3. That also seems correct.\n\nTherefore, adding \\( a + b \\) gives -3 + 3 = 0. So, 0 is the answer.\n\nBut let me just visualize the function to make sure I didn\u0027t make mistakes.\n\nFor x \u003e 2, it\u0027s a line with slope a, which we found to be -3, so f(x) = -3x + 3. As x increases beyond 2, the function would decrease since the slope is negative.\n\nFor -2 \u003c= x \u003c= 2, it\u0027s f(x) = x - 5. That\u0027s a line with slope 1, so it\u0027s increasing from x = -2 to x = 2. At x = -2, f(x) = -2 - 5 = -7; at x = 2, f(x) = 2 - 5 = -3. Wait, hold on. If x = -2, f(x) = -7 and x = 2, f(x) = -3. So between -2 and 2, it\u0027s going from -7 to -3, which is an increase of 4 over an increase of 4 units in x (from -2 to 2 is 4 units), so the slope is indeed 1.\n\nNow, for x \u003c -2, f(x) = 2x - b. So, f(x) is a line with slope 2. Since b is 3, at x = -2, f(-2) = 2*(-2) - 3 = -4 -3 = -7. So both the middle piece and the left piece meet at x = -2 with f(-2) = -7.\n\nAnd for x \u003e 2, f(3) would be 3a + 3, where a is -3, so f(3) = 3*(-3) + 3 = -9 + 3 = -6. Let\u0027s check continuity as x approaches 2 from the right: f(2) is -3, right? It\u0027s the value from the middle piece, so it\u0027s as defined there. However, if we approach 2 from above, using the right-hand side, f(x) = 2a + 3 = 2*(-3) + 3 = -6 + 3 = -3. So, the function at x=2 is -3, and as x approaches 2 from above, it\u0027s approaching -3 as well.\n\nSimilarly, approaching from x = -2, from the left, we get from -4 - b with b=3 equals -7, which is the value in the middle piece, so that\u0027s continuous.\n\nThus, all three pieces connect smoothly at x = -2 and x = 2, so the function is continuous everywhere.\n\nSo, putting it all together, since a is -3 and b is 3, adding them together, we get 0. Hmm, 0 seems quite clean. Maybe that\u0027s just how the constants work here.\n\nAll right, so after going through each step, making sure the limits match the function values, both at x = 2 and x = -2, I\u0027m confident that a = -3 and b = 3, so their sum is 0. I think that\u0027s the right answer.\n\n**Final Answer**\n\\boxed{0}\n\u003c/think\u003e\n\nTo ensure the piecewise function is continuous, we need to check the continuity at the points \\( x = -2 \\) and \\( x = 2 \\).\n\n1. **Continuity at \\( x = 2 \\)**:\n   - Left limit (from the middle piece): \\( f(2) = 2 - 5 = -3 \\)\n   - Right limit (from the first piece): \\( f(2) = 2a + 3 \\)\n   - Setting the left and right limits equal: \\( -3 = 2a + 3 \\)\n   - Solving for \\( a \\): \\( -6 = 2a \\) \u27f9 \\( a = -3 \\)\n\n2. **Continuity at \\( x = -2 \\)**:\n   - Right limit (from the middle piece): \\( f(-2) = -2 - 5 = -7 \\)\n   - Left limit (from the third piece): \\( f(-2) = 2(-2) - b = -4 - b \\)\n   - Setting the right and left limits equal: \\( -4 - b = -7 \\)\n   - Solving for \\( b \\): \\( -b = -3 \\) \u27f9 \\( b = 3 \\)\n\nThus, \\( a = -3 \\) and \\( b = 3 \\). Adding these values together, we get \\( a + b = -3 + 3 = 0 \\).\n\n\\[\n\\boxed{0}\n\\]"
    ],
    "level": "Level 5",
    "model_name": "deepseek-ai/DeepSeek-R1-Distill-Qwen-1.5B",
    "problem": "Let \\[f(x) = \\left\\{\n\\begin{array}{cl} ax+3, \u0026\\text{ if }x\u003e2, \\\\\nx-5 \u0026\\text{ if } -2 \\le x \\le 2, \\\\\n2x-b \u0026\\text{ if } x \u003c-2.\n\\end{array}\n\\right.\\]Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper).",
    "solution": "For the piecewise function to be continuous, the cases must \"meet\" at $2$ and $-2$. For example, $ax+3$ and $x-5$ must be equal when $x=2$. This implies $a(2)+3=2-5$, which we solve to get $2a=-6 \\Rightarrow a=-3$. Similarly, $x-5$ and $2x-b$ must be equal when $x=-2$. Substituting, we get $-2-5=2(-2)-b$, which implies $b=3$. So $a+b=-3+3=\\boxed{0}$.",
    "type": "Algebra"
}
```

This subset can be loaded as:

```python
from datasets import load_dataset

ds = load_dataset("gazhitt/numina-deepseek-r1-qwen-7b", "default")
```

Or simply as it follows, since there's only one configuration and is named `default`: 

```python
from datasets import load_dataset

ds = load_dataset("gazhitt/numina-deepseek-r1-qwen-7b")
```


</details>




