license:apache-2.0 size_categories:1K<n<10K format:imagefolder modality:image library:datasets library:mlcroissant region:us
---
license: apache-2.0
---


# Dataset Name

## Dataset Summary

MMSciBench dataset focuses on mathematics and physics that evaluates scientific reasoning capabilities. 

## Dataset Structure

### Data Instances

- `QA_has_img_with_categories.csv`: Contains data of Q&A questions (with images) of the subject indicated by the folder name.
- `QA_no_img_with_categories.csv`: Contains data of text-only Q&A questions of the subject indicated by the folder name.
- `multi_choice_has_img_with_categories.csv`: Contains data of multi-choice questions (with images) of the subject indicated by the folder name.
- `multi_choice_no_img_with_categories.csv`: Contains data of text-only multi-choice of the subject indicated by the folder name.

Example of a data entry in QA_has_img_with_categories.csv:

```csv
question.phasename,question.subjectname,question.hassolution,question.questionid,question.typename,question.difficulty,question.groupdifficulty,question.chaptername,Selection,question.mainkeypoint,question.minorkeypoint,question.content,question.accessories,question.answer,question.solutioncontent,childquestion.parentquestionid,childquestion.childquestionid,childquestion.mainkeypoint,childquestion.minorkeypoint,childquestion.childcontent,childquestion.childaccessories,childquestion.childanswer,childquestion.childsolutioncontent,broken_img,duplicate,content_img,answer_img,msg,ans,is_proof_question,taxonomy_level1,taxonomy_level2,taxonomy_level3
高中,数学,1,1176789,解答,0.7,高,一元二次方程的解集及其根与系数的关系,,,"一元二次方程根的分布问题,直接利用均值不等式求最值","<p>（本小题满分${100}分）</p><p>如图，建立平面直角坐标系<span class=""mathjax-tex"">$$xOy$$</span>，<span class=""mathjax-tex"">$$x$$</span>轴在地平面上，<span class=""mathjax-tex"">$$y$$</span>轴垂直于地平面，单位长度为<span class=""mathjax-tex"">$$1$$</span>千米。某炮位于坐标原点。已知炮弹发射后的轨迹在方程<span class=""mathjax-tex"">$$y=kx-\frac1{20}(1+k^2)x^2(k&gt;0)$$</span>表示的曲线上，其中<span class=""mathjax-tex"">$$k$$</span>与发射方向有关。炮的射程是指炮弹落地点的横坐标。</p><p>（1）求炮的最大射程；</p><p>（2）设在第一象限有一飞行物（忽略其大小），其飞行高度为<span class=""mathjax-tex"">$$3.2$$</span>千米，试问它的横坐标<span class=""mathjax-tex"">$$a$$</span>不超过多少时，炮弹可以击中它？请说明理由。</p><p><img src=""img/13fe0a72fe29dd7.jpg"" /></p>",[],"<p>（1）令<span class=""mathjax-tex"">$$y=0$$</span>，得<span class=""mathjax-tex"">$$kx-\frac1{20}(1+k^2)x^2=0$$</span>，由实际意义和题设条件知<span class=""mathjax-tex"">$$x&gt;0$$</span>，<span class=""mathjax-tex"">$$k&gt;0$$</span>，故<span class=""mathjax-tex"">$$x=\frac{20k}{1+k^2}=\frac{20}{k+\frac1k}\le\frac{20}2=10$$</span>，当且仅当<span class=""mathjax-tex"">$$k=1$$</span>时取等号。所以炮的最大射程为<span class=""mathjax-tex"">$$10$$</span>千米。</p><p>（2）因为<span class=""mathjax-tex"">$$a&gt;0$$</span>，所以炮弹可击中目标<span class=""mathjax-tex"">$$\Leftrightarrow$$</span>存在<span class=""mathjax-tex"">$$k&gt;0$$</span>，使<span class=""mathjax-tex"">$$3.2=ka-\frac1{20}(1+k^2)a^2$$</span>成立<span class=""mathjax-tex"">$$\Leftrightarrow$$</span>关于<span class=""mathjax-tex"">$$k$$</span>的方程<span class=""mathjax-tex"">$$a^2k^2-20ak+a^2+64=0$$</span>有正根<span class=""mathjax-tex"">$$\Leftrightarrow$$</span>判别式<span class=""mathjax-tex"">$$\Delta=(-20a)^2-4a^2(a^2+64)\ge0$$</span><span class=""mathjax-tex"">$$\Leftrightarrow$$</span><span class=""mathjax-tex"">$$a\le6$$</span>。此时，<span class=""mathjax-tex"">$$k=\frac{20a+\sqrt{(-20a)^2-4a^2(a^2+64)}}{2a^2}&gt;0$$</span>（不考虑另一根）。所以当<span class=""mathjax-tex"">$$a$$</span>不超过<span class=""mathjax-tex"">$$6$$</span>千米时，可击中目标。</p>","<p>本题主要考查函数与方程和基本不等式的应用等相关知识。</p><p>（1）求炮的最大射程，即<span class=""mathjax-tex"">$$y=0$$</span>时的一个较大的根，因为含有参数<span class=""mathjax-tex"">$$k$$</span>，所以需根据<span class=""mathjax-tex"">$$k$$</span>的取值范围确定另外一个根的最大值，即为炮的最大射程。</p><p>（2）炮弹能击中目标的含义为炮弹的飞行高度<span class=""mathjax-tex"">$$y=3.2$$</span>时有解。根据二次函数有正根，可得出<span class=""mathjax-tex"">$$a$$</span>的取值范围。</p>",,,,,,,,,0,0.0,['data/math/img_QA/13fe0a72fe29dd7.jpg'],[],问题（解答）：（本小题满分${100}分）如图，建立平面直角坐标系$$xOy$$，$$x$$轴在地平面上，$$y$$轴垂直于地平面，单位长度为$$1$$千米。某炮位于坐标原点。已知炮弹发射后的轨迹在方程$$y=kx-\frac1{20}(1+k^2)x^2(k>0)$$表示的曲线上，其中$$k$$与发射方向有关。炮的射程是指炮弹落地点的横坐标。（1）求炮的最大射程；（2）设在第一象限有一飞行物（忽略其大小），其飞行高度为$$3.2$$千米，试问它的横坐标$$a$$不超过多少时，炮弹可以击中它？请说明理由。,（1）令$$y=0$$，得$$kx-\frac1{20}(1+k^2)x^2=0$$，由实际意义和题设条件知$$x>0$$，$$k>0$$，故$$x=\frac{20k}{1+k^2}=\frac{20}{k+\frac1k}\le\frac{20}2=10$$，当且仅当$$k=1$$时取等号。所以炮的最大射程为$$10$$千米。（2）因为$$a>0$$，所以炮弹可击中目标$$\Leftrightarrow$$存在$$k>0$$，使$$3.2=ka-\frac1{20}(1+k^2)a^2$$成立$$\Leftrightarrow$$关于$$k$$的方程$$a^2k^2-20ak+a^2+64=0$$有正根$$\Leftrightarrow$$判别式$$\Delta=(-20a)^2-4a^2(a^2+64)\ge0$$$$\Leftrightarrow$$$$a\le6$$。此时，$$k=\frac{20a+\sqrt{(-20a)^2-4a^2(a^2+64)}}{2a^2}>0$$（不考虑另一根）。所以当$$a$$不超过$$6$$千米时，可击中目标。,NO,4.几何与5.代数,几何与代数,平面解析几何
```

Example of a data entry in QA_no_img_with_categories.csv:

```csv
question.phasename,question.subjectname,question.questionid,question.typename,question.difficulty,question.content,question.accessories,question.answer,question.solutioncontent,childquestion.childquestionid,childquestion.childcontent,childquestion.childaccessories,childquestion.childanswer,childquestion.childsolutioncontent,msg,ans,is_proof_question,taxonomy_level1,taxonomy_level2,taxonomy_level3
高中,数学,1161468,解答,0.7,"<p>（本小题满分13分）</p><p>将数列<span class=""mathjax-tex"">$$\{a_n\}$$</span>中的所有项按每一行比上一行多两项的规则排成如下数表：</p><p><span class=""mathjax-tex"">$$a_1$$</span></p><p><span class=""mathjax-tex"">$$a_2\ \ \ a_3\ \ \ a_4$$</span></p><p><span class=""mathjax-tex"">$$a_5\ \ \ a_6\ \ \ a_7\ \ \ a_8\ \ \ a_9$$</span></p><p><span class=""mathjax-tex"">$$\cdots\cdots$$</span></p><p>已知表中的第一列数<span class=""mathjax-tex"">$$a_1$$</span>，<span class=""mathjax-tex"">$$a_2$$</span>，<span class=""mathjax-tex"">$$a_5$$</span>，<span class=""mathjax-tex"">$$\cdots$$</span>构成一个等差数列，记为<span class=""mathjax-tex"">$$\{b_n\}$$</span>，且<span class=""mathjax-tex"">$$b_2=4$$</span>，<span class=""mathjax-tex"">$$b_5=10$$</span>，表中每一行正中间的一个数<span class=""mathjax-tex"">$$a_1$$</span>，<span class=""mathjax-tex"">$$a_3$$</span>，<span class=""mathjax-tex"">$$a_7$$</span>，<span class=""mathjax-tex"">$$\cdots$$</span>构成数列<span class=""mathjax-tex"">$$\{C_n\}$$</span>，其前<span class=""mathjax-tex"">$$n$$</span>项和为<span class=""mathjax-tex"">$$S_n$$</span>。</p><p>（1）求数列<span class=""mathjax-tex"">$$\{b_n\}$$</span>的通项公式；</p><p>（2）若上表中，从第二行起，每一行中的数按从左到右的顺序均构成等比数列，公比为同一个正数，且<span class=""mathjax-tex"">$$a_{13}=1$$</span>。</p><p>①求<span class=""mathjax-tex"">$$S_n$$</span>；</p><p>②记<span class=""mathjax-tex"">$$M=\{n|\ (n+1)C_n\geqslant\lambda$$</span>，<span class=""mathjax-tex"">$$n\in\Bbb N^*\}$$</span>。若集合<span class=""mathjax-tex"">$$M$$</span>的元素个数为<span class=""mathjax-tex"">$$3$$</span>，求<span class=""mathjax-tex"">$$\lambda$$</span>的取值范围。</p>",[],"<p>（1）因为<span class=""mathjax-tex"">$$\{b_n\}$$</span>为等差数列，且已知<span class=""mathjax-tex"">$$b_2=4$$</span>，<span class=""mathjax-tex"">$$b_5=10$$</span>，</p><p>设其公差为<span class=""mathjax-tex"">$$d$$</span>，有<span class=""mathjax-tex"">$$\cases{ b_1+d=4\cr b_1+4d=10}$$</span>，</p><p>解得<span class=""mathjax-tex"">$$\cases{ b_1=2\cr d=2}$$</span>，所以数列<span class=""mathjax-tex"">$$\{b_n\}$$</span>的通项公式<span class=""mathjax-tex"">$$b_n=2n$$</span>。      ......3分</p><p>（2）①设每一行组成的等比数列公比为<span class=""mathjax-tex"">$$q$$</span>，且前<span class=""mathjax-tex"">$$n$$</span>行共有<span class=""mathjax-tex"">$$1+3+5+\cdots $$</span><span class=""mathjax-tex"">$$+(2n-1)=n^2$$</span>个数，且<span class=""mathjax-tex"">$$3^2&lt;13&lt;4^2$$</span>，所以<span class=""mathjax-tex"">$$a_{10}=b_4=8$$</span>，<span class=""mathjax-tex"">$$a_{13}=a_{10}q^3=8q^3=1$$</span>，得<span class=""mathjax-tex"">$$q={1\over2}$$</span>，</p><p>因此<span class=""mathjax-tex"">$$C_n=2n\cdot({1\over2})^{n-1}$$</span><span class=""mathjax-tex"">$$={n\over2^{n-2}}$$</span>。      ......5分</p><p>所以<span class=""mathjax-tex"">$$S_n={1\over2^{-1}}+{2\over2^0}+{3\over2^1}+\cdots $$</span><span class=""mathjax-tex"">$$+{n-1\over2^{n-3}}+{n\over2^{n-2}}$$</span>，</p><p><span class=""mathjax-tex"">$${1\over2}S_n={1\over2^0}+{2\over2^1}+{3\over2^2}+\cdots $$</span><span class=""mathjax-tex"">$$+{n-1\over2^{n-2}}+{n\over2^{n-1}}$$</span>，</p><p>两式相减得<span class=""mathjax-tex"">$${1\over2}S_n={1\over2^{-1}}+{1\over2^0}+{1\over2^1}+\cdots $$</span><span class=""mathjax-tex"">$$+{1\over2^{n-2}}-{n\over2^{n-1}}$$</span><span class=""mathjax-tex"">$$=4-{n+2\over2^{n-1}}$$</span>，</p><p>得<span class=""mathjax-tex"">$$S_n=8-{n+2\over2^{n-2}}$$</span>。      ......9分</p><p>②由①知<span class=""mathjax-tex"">$$C_n={n\over2^{n-2}}$$</span>，所以<span class=""mathjax-tex"">$$\lambda\leqslant{n(n+1)\over2^{n-2}}$$</span>。</p><p>设<span class=""mathjax-tex"">$$f(n)={n(n+1)\over2^{n-2}}$$</span>，计算得<span class=""mathjax-tex"">$$f(1)=4$$</span>，<span class=""mathjax-tex"">$$f(2)=f(3)=6$$</span>，<span class=""mathjax-tex"">$$f(4)=5$$</span>，<span class=""mathjax-tex"">$$f(5)={15\over4}$$</span>，</p><p>且<span class=""mathjax-tex"">$$f(n+1)-f(n) $$</span><span class=""mathjax-tex"">$$={(n+1)(2-n)\over2^{n-1}}$$</span>，</p><p>所以当<span class=""mathjax-tex"">$$n\geqslant3$$</span>时，<span class=""mathjax-tex"">$$f(n+1)&lt;f(n)$$</span>。      ......11分</p><p>因为<span class=""mathjax-tex"">$$M$$</span>的元素个数为<span class=""mathjax-tex"">$$3$$</span>，所以<span class=""mathjax-tex"">$$\lambda\in(4,5]$$</span>。      ......13分</p>","<p>本题主要考查等差数列和等比数列。</p><p>（1）由于<span class=""mathjax-tex"">$$\{b_n\}$$</span>为等差数列，则可根据已知解得<span class=""mathjax-tex"">$$b_1$$</span>、<span class=""mathjax-tex"">$$d$$</span>，从而得到数列<span class=""mathjax-tex"">$$\{b_n\}$$</span>的通项公式；</p><p>（2）①由于<span class=""mathjax-tex"">$$a_{10}=b_4=8$$</span>，又由已知得<span class=""mathjax-tex"">$$a_{13}=1$$</span>，从而得到<span class=""mathjax-tex"">$$q$$</span>，则可得到数列<span class=""mathjax-tex"">$$\{C_n\}$$</span>的通项公式，可以看出<span class=""mathjax-tex"">$$\{C_n\}$$</span>为等比数列和等差数列乘积的形式，利用错位相减法即可得到其前<span class=""mathjax-tex"">$$n$$</span>项和<span class=""mathjax-tex"">$$S_n$$</span>；</p><p>②由已知可得<span class=""mathjax-tex"">$$\lambda\leqslant{n(n+1)\over2^{n-2}}$$</span>，设<span class=""mathjax-tex"">$$f(n)={n(n+1)\over2^{n-2}}$$</span>，根据<span class=""mathjax-tex"">$$f(n)$$</span>的单调性即可得到<span class=""mathjax-tex"">$$\lambda$$</span>的取值范围。</p>",,,,,,问题（解答）：（本小题满分13分）将数列$$\{a_n\}$$中的所有项按每一行比上一行多两项的规则排成如下数表：$$a_1$$$$a_2\ \ \ a_3\ \ \ a_4$$$$a_5\ \ \ a_6\ \ \ a_7\ \ \ a_8\ \ \ a_9$$$$\cdots\cdots$$已知表中的第一列数$$a_1$$，$$a_2$$，$$a_5$$，$$\cdots$$构成一个等差数列，记为$$\{b_n\}$$，且$$b_2=4$$，$$b_5=10$$，表中每一行正中间的一个数$$a_1$$，$$a_3$$，$$a_7$$，$$\cdots$$构成数列$$\{C_n\}$$，其前$$n$$项和为$$S_n$$。（1）求数列$$\{b_n\}$$的通项公式；（2）若上表中，从第二行起，每一行中的数按从左到右的顺序均构成等比数列，公比为同一个正数，且$$a_{13}=1$$。①求$$S_n$$；②记$$M=\{n|\ (n+1)C_n\geqslant\lambda$$，$$n\in\Bbb N^*\}$$。若集合$$M$$的元素个数为$$3$$，求$$\lambda$$的取值范围。,"（1）因为$$\{b_n\}$$为等差数列，且已知$$b_2=4$$，$$b_5=10$$，设其公差为$$d$$，有$$\cases{ b_1+d=4\cr b_1+4d=10}$$，解得$$\cases{ b_1=2\cr d=2}$$，所以数列$$\{b_n\}$$的通项公式$$b_n=2n$$。      ......3分（2）①设每一行组成的等比数列公比为$$q$$，且前$$n$$行共有$$1+3+5+\cdots $$$$+(2n-1)=n^2$$个数，且$$3^2<13<4^2$$，所以$$a_{10}=b_4=8$$，$$a_{13}=a_{10}q^3=8q^3=1$$，得$$q={1\over2}$$，因此$$C_n=2n\cdot({1\over2})^{n-1}$$$$={n\over2^{n-2}}$$。      ......5分所以$$S_n={1\over2^{-1}}+{2\over2^0}+{3\over2^1}+\cdots $$$$+{n-1\over2^{n-3}}+{n\over2^{n-2}}$$，$${1\over2}S_n={1\over2^0}+{2\over2^1}+{3\over2^2}+\cdots $$$$+{n-1\over2^{n-2}}+{n\over2^{n-1}}$$，两式相减得$${1\over2}S_n={1\over2^{-1}}+{1\over2^0}+{1\over2^1}+\cdots $$$$+{1\over2^{n-2}}-{n\over2^{n-1}}$$$$=4-{n+2\over2^{n-1}}$$，得$$S_n=8-{n+2\over2^{n-2}}$$。      ......9分②由①知$$C_n={n\over2^{n-2}}$$，所以$$\lambda\leqslant{n(n+1)\over2^{n-2}}$$。设$$f(n)={n(n+1)\over2^{n-2}}$$，计算得$$f(1)=4$$，$$f(2)=f(3)=6$$，$$f(4)=5$$，$$f(5)={15\over4}$$，且$$f(n+1)-f(n) $$$$={(n+1)(2-n)\over2^{n-1}}$$，所以当$$n\geqslant3$$时，$$f(n+1)<f(n)$$。      ......11分因为$$M$$的元素个数为$$3$$，所以$$\lambda\in(4,5]$$。      ......13分",NO,3.函数,函数,数列
```

Example of a data entry in multi_choice_has_img_with_categories.csv:

```csv
question.phasename,question.subjectname,question.hassolution,question.questionid,question.typename,question.difficulty,question.groupdifficulty,question.chaptername,question.mainkeypoint,question.minorkeypoint,question.content,question.accessories,question.answer,question.solutioncontent,childquestion.parentquestionid,childquestion.childquestionid,childquestion.mainkeypoint,childquestion.minorkeypoint,childquestion.childcontent,childquestion.childaccessories,childquestion.childanswer,childquestion.childsolutioncontent,broken_img,img,msg,accessories,sol,taxonomy_level1,taxonomy_level2,taxonomy_level3
高中,数学,1.0,612946.0,单选,0.7,高,指数函数,,"指数函数与分式函数复合,指数函数的图象及变换","<p>函数<span class=""mathjax-tex"">$$y=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$</span>的图象大致为（  ）。</p><p><img src=""img/142fe60383e5939.jpg"" /></p>","[{""type"": 102, ""options"": [""A"", ""B"", ""C"", ""D""]}]","{""choice"":0,""markable"":true,""type"":304}","<p>本题主要考查指数函数的性质和图象。</p><p>由函数定义可得<span class=""mathjax-tex"">$$y=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$</span>的定义域为<span class=""mathjax-tex"">$$\{x|x\ne0\}$$</span>，故C项，D项不符合题意，化简函数表达式得<span class=""mathjax-tex"">$$y=\frac2{e^{2x}-1}+1$$</span>，所以函数在<span class=""mathjax-tex"">$$(0,+\infty)$$</span>上为减函数，故B项不符合题意。</p><p>故本题正确答案为A。</p>",,,,,,,,,0.0,['data/math/img_multiple_choices/142fe60383e5939.jpg'],"问题（单选）：函数$$y=\frac{e^x+e^{-x}}{e^x-e^{-x}}$$的图象大致为（  ）。 选项：A, B, C, D","选项：A, B, C, D",A,3.函数,函数,函数概念与性质
```

Example of a data entry in multi_choice_no_img_with_categories.csv:

```csv
question.phasename,question.subjectname,question.questionid,question.typename,question.difficulty,question.content,question.accessories,question.answer,question.solutioncontent,childquestion.childquestionid,childquestion.childcontent,childquestion.childaccessories,childquestion.childanswer,childquestion.childsolutioncontent,msg,accessories,sol,taxonomy_level1,taxonomy_level2,taxonomy_level3
高中,数学,2041442,单选,0.7,"<p>已知<span class=""mathjax-tex"">$$\overrightarrow{{a}\mathstrut}=(1,2)$$</span>，<span class=""mathjax-tex"">$$\overrightarrow{{b}\mathstrut}=(3,4)$$</span>，<span class=""mathjax-tex"">$$(\overrightarrow{{a}\mathstrut}+2\overrightarrow{{b}\mathstrut})\bot (\lambda \overrightarrow{{a}\mathstrut}-\overrightarrow{{b}\mathstrut})$$</span>，则<span class=""mathjax-tex"">$$\lambda =$$</span>（  ）。</p>","[{""options"":[""<p><span class=\""mathjax-tex\"">$$-\\frac{61}{27}$$</span></p>"",""<p><span class=\""mathjax-tex\"">$$\\frac{61}{27}$$</span></p>"",""<p><span class=\""mathjax-tex\"">$$-\\frac12$$</span></p>"",""<p><span class=\""mathjax-tex\"">$$\\frac12$$</span></p>""],""type"":102}]","{""choice"":1,""markable"":true,""type"":304}","<p>本题主要考查平面向量基本定理及坐标表示和平面向量的数量积。</p><p>由题意知<span class=""mathjax-tex"">$$\overrightarrow{{a}\mathstrut}+2\overrightarrow{{b}\mathstrut}$$</span><span class=""mathjax-tex"">$$=(1+3\times2,2+4\times2)=(7,10)$$</span>，</p><p><span class=""mathjax-tex"">$$\lambda \overrightarrow{{a}\mathstrut}-\overrightarrow{{b}\mathstrut}$$</span><span class=""mathjax-tex"">$$=(\lambda -3,2\lambda -4)$$</span>。</p><p>若<span class=""mathjax-tex"">$$(\overrightarrow{{a}\mathstrut}+2\overrightarrow{{b}\mathstrut})\bot (\lambda \overrightarrow{{a}\mathstrut}-\overrightarrow{{b}\mathstrut})$$</span>，</p><p>则<span class=""mathjax-tex"">$$(\overrightarrow{{a}\mathstrut}+2\overrightarrow{{b}\mathstrut})\cdot (\lambda \overrightarrow{{a}\mathstrut}-\overrightarrow{{b}\mathstrut})=0$$</span>，</p><p>即<span class=""mathjax-tex"">$$7(\lambda -3)+10(2\lambda -4)=0$$</span>。</p><p>整理得<span class=""mathjax-tex"">$$27\lambda -61=0$$</span>，解得<span class=""mathjax-tex"">$$\lambda =\frac{61}{27}$$</span>。</p><p>故本题正确答案为B。</p>",,,,,,"问题（单选）：已知$$\overrightarrow{{a}\mathstrut}=(1,2)$$，$$\overrightarrow{{b}\mathstrut}=(3,4)$$，$$(\overrightarrow{{a}\mathstrut}+2\overrightarrow{{b}\mathstrut})\bot (\lambda \overrightarrow{{a}\mathstrut}-\overrightarrow{{b}\mathstrut})$$，则$$\lambda =$$（  ）。 选项：A. $$-\frac{61}{27}$$, B. $$\frac{61}{27}$$, C. $$-\frac12$$, D. $$\frac12$$","选项：A. $$-\frac{61}{27}$$, B. $$\frac{61}{27}$$, C. $$-\frac12$$, D. $$\frac12$$",B,4.几何与5.代数,几何与代数,平面向量及其应用
```

### Data Fields

- **question.phasename**: The phase name of the question (All rows are "高中" for high school).
- **question.subjectname**: The subject of the question (e.g., "数学" for Mathematics).
- **question.hassolution**: Indicates whether the question has a solution (1 for yes, 0 for no).
- **question.questionid**: Unique identifier for the question.
- **question.typename**: The type of the question (e.g., "解答" for Q&A).
- **question.difficulty**: The difficulty level of the question (e.g., 0.7).
- **question.groupdifficulty**: Group difficulty level of the question (e.g., "高" for high).
- **question.chaptername**: The chapter name to which the question belongs.
- **Selection**: Optional field that may contain additional selection details. (Ignored)
- **question.mainkeypoint**: Main key point(s) for the question (e.g., "一元二次方程的解集及其根与系数的关系").
- **question.minorkeypoint**: Minor key point(s) for the question.
- **question.content**: The main content of the question, including text and LaTeX math formulas, in HTML.
- **question.accessories**: Accessories for MCQs (empty for Q&A).
- **question.answer**: The answer(s) to the question in HTML.
- **question.solutioncontent**: Detailed explanation or solution content for the question.
- **childquestion.parentquestionid**: The parent question ID for this child question.
- **childquestion.childquestionid**: Unique identifier for the child question.
- **childquestion.mainkeypoint**: Main key point for the child question.
- **childquestion.minorkeypoint**: Minor key point for the child question.
- **childquestion.childcontent**: The content of the child question in HTML.
- **childquestion.childaccessories**: Accessories for the child MCQ.
- **childquestion.childanswer**: The answer to the child question in HTML.
- **childquestion.childsolutioncontent**: Solution content for the child question.
- **broken_img**: Indicates whether there is a broken image (1 for yes, 0 for no).
- **duplicate**: Indicates if the question is a duplicate (1 for yes, 0 for no).
- **content_img**: Image in the Q&A question content (a path to the image).
- **answer_img**: Image(s) in the answer.
- **img**: Image in the MCQ content.
- **msg**: The question content.
- **ans**: The correct answer for the question.
- **is_proof_question**: Indicates whether this is a proof question (1 for yes, 0 for no).
- **taxonomy_level1**: The first level of taxonomy related to the question.
- **taxonomy_level2**: The second level of taxonomy related to the question.
- **taxonomy_level3**: The third level of taxonomy related to the question.

## Licensing Information

This dataset is licensed under the Apache-2.0 license.