Question
Download Solution PDFवक्र y = 2x3 - 5x2 + x - 2 पर बिंदु (1, -1) पर अभिलंब की ढलान क्या है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFअवधारणा:
वक्र y = f(x) के स्पर्श रेखा की ढलान m = \(\rm dy\over dx\)
अभिलंब की ढलान = \(\rm -{1\over m}\) = \(\rm -{1\over {dy\over dx}}\)
गणना:
दिया गया वक्र y = 2x3 - 5x2 + x - 2
समीकरण को x के सापेक्ष अवकलन करने पर
\(\rm dy\over dx\) = 6x2 - 10x + 1
बिंदु (1, -1) पर ढलान
\(\rm dy\over dx\) x = 1 पर = 6(1)2 - 10(1) + 1
\(\rm dy\over dx\) = 6 - 10 + 1
\(\rm dy\over dx\) = -3
अभिलंब की ढलान (m') = \(\rm -{1\over {dy\over dx}}\)
m' = \(\boldsymbol{\rm -{1 \over -3}}\) = \(\boldsymbol{\rm {1 \over 3}}\)
बिंदु (1, -1) पर वक्र के अभिलंब की ढलान 1/3 है।
∴ विकल्प 2 सही है।Last updated on Jul 16, 2025
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