Question
Download Solution PDFउस बिंदु को ज्ञात कीजिए जिसपर वक्र y = x3 – 3x2 + 3x - 6 के लिए स्पर्श रेखा x - अक्ष के समानांतर है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
वक्र का ढलान = dy/dx
गणना:
माना कि वक्र पर बिंदु (x, y) है।
दिया गया है: वक्र का समीकरण y = x3 – 3x2 + 3x - 6 ---(1)
x के संबंध में अवकलन करने पर, हमें निम्न प्राप्त होता है
\( \Rightarrow \frac{{{\rm{dy}}}}{{{\rm{dx}}}} = 3{{\rm{x}}^2} - 6{\rm{x}} + 3\)
चूँकि, स्पर्श रेखा अक्ष के समानांतर है,
\(\therefore \frac{{{\rm{dy}}}}{{{\rm{dx}}}}{\rm{\;}} = {\rm{\;}}0\)
⇒ 3x2 – 6x + 3 = 0
⇒ 3 (x2 – 2x + 1) = 0
⇒ (x – 1)2 = 0
∴ x = 1
समीकरण 1 में x का मान रखने पर,
⇒ y = 1 – 3 + 3 – 6 = -5
अतः आवश्यक बिंदु (1, -5) है।
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