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Ex 6.3, 12 - Find equations of all lines having slope 0, tangent - Ex 6.3

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  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise
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Ex 6.3,12 Find the equations of all lines having slope 0 which are tangent to the curve 𝑦 = ﷐1﷮𝑥2 −2𝑥 + 3﷯ Given Curve is 𝑦 = ﷐1﷮𝑥2 −2𝑥 + 3﷯ Slope of tangent is ﷐𝑑𝑦﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐𝑑﷐﷐1﷮𝑥2 − 2𝑥 + 3﷯﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐𝑑﷐﷐𝑥2 − 2𝑥 + 3﷯﷮−1﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=−1﷐﷐𝑥2−2𝑥+3﷯﷮−2﷯ . ﷐𝑑﷐﷐𝑥﷮2﷯− 2𝑥 + 3﷯﷮𝑑𝑥﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=−﷐﷐𝑥2−2𝑥+3﷯﷮−2﷯﷐2𝑥−2﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=−2﷐﷐𝑥2−2𝑥+3﷯﷮−2﷯﷐𝑥−1﷯ ﷐𝑑𝑦﷮𝑑𝑥﷯=﷐−2﷐𝑥 − 1﷯﷮﷐﷐𝑥2 − 2𝑥 + 3﷯﷮2﷯﷯ Hence Slope of tangent is ﷐−2﷐𝑥 − 1﷯﷮﷐﷐𝑥2 − 2𝑥 + 3﷯﷮2﷯﷯ Given Slope of tangent is 0 ⇒ ﷐𝑑𝑦﷮𝑑𝑥﷯=0 ﷐−2﷐𝑥 − 1﷯﷮﷐﷐𝑥2 − 2𝑥 + 3﷯﷮2﷯﷯=0 −2﷐𝑥 − 1﷯=0 ×﷐﷐𝑥2 − 2𝑥 + 3﷯﷮2﷯ −2﷐𝑥 − 1﷯=0 ﷐𝑥−1﷯=0 𝑥=1 Finding y when 𝑥=1 𝑦=﷐1﷮𝑥2 − 2𝑥 + 3﷯ 𝑦=﷐1﷮﷐﷐1﷯﷮2﷯ − 2﷐1﷯ + 3﷯ 𝑦=﷐1﷮1 − 2 + 3﷯ 𝑦=﷐1﷮2 ﷯ Point is ﷐𝟏 , ﷐𝟏﷮𝟐﷯﷯ Thus , tangent passes through ﷐1 , ﷐1﷮2﷯﷯ Equation of tangent at ﷐1 , ﷐1﷮2﷯﷯ & having Slope zero is ﷐𝑦 −﷐1﷮2﷯﷯=0﷐𝑥−1﷯ 𝑦 −﷐1﷮2﷯=0 𝑦=﷐1﷮2﷯ Hence Equation of tangent is 𝒚=﷐𝟏﷮𝟐﷯

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