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दिलेल्या वक्रातील सामान्य जर x-अक्षाच्या समांतर असेल तर 

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UP TGT Mathematics 2016 Official Paper
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  1. \(\frac{{dy}}{{dx}} = 0\)
  2. \(\frac{{dy}}{{dx}} = 1\)
  3. \(\frac{{dx}}{{dy}} = 0\)
  4. \(\frac{{dx}}{{dy}} = 1\)

Answer (Detailed Solution Below)

Option 3 : \(\frac{{dx}}{{dy}} = 0\)
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संकल्पना:

सामान्य स्पर्शिकेला लंब आहे, जसे की mt × mn = -1

mt = dy/dx.

गणना:

फल f(x) विचारात घेऊ, जसे की स्पर्शिकेचा उतार mt असेल आणि सामान्य उतार mn असेल.

आता, आपणास माहित आहे की, सामान्य स्पर्शिकेला लंब आहे, जसे की

 mt × mn = -1

mn × dy/dx = -1

⇒ mn = - dx/dy

आता, प्रश्नानुसार, सामान्य हा x-अक्षाच्या समांतर आहे, म्हणून सामान्यचा उतार x-अक्षाच्या उताराइतका आहे. (म्हणजे, शून्य)

∴ mn = 0

⇒ - dx/dy = 0

⇒ dx/dy = 0

∴ ​दिलेल्या वक्रातील सामान्य जर x-अक्षाच्या समांतर असेल तर dx/dy = 0 असेल

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