Question 5 - Examples - Chapter 6 Class 12 Application of Derivatives
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
Question 5 Find the equation of the tangent to the curve y = (𝑥 − 7)/((𝑥 − 2)(𝑥 − 3)) at the point where it cuts the x-axis.Slope of the tangent to the curve is 𝑑𝑦/𝑑𝑥 = ((𝑥 − 7)^′ [(𝑥 − 2) (𝑥 − 3)]− (𝑥 − 7) [(𝑥 − 3) (𝑥 − 2)]^′)/((𝑥 − 2)^2 (𝑥 − 3)^2 ) 𝑑𝑦/𝑑𝑥 = (1 × (𝑥 − 2) (𝑥 − 3) − (𝑥 − 7)[(𝑥 − 3)^′ (𝑥 − 2) + (𝑥 − 3) (𝑥 − 2)^′ ])/((𝑥 − 2)^2 (𝑥 − 3)^2 ) 𝑑𝑦/𝑑𝑥 = ((1) (𝑥 − 2) (𝑥 − 3) − (𝑥 − 7)[1 × (𝑥 − 2) + (𝑥 − 3) × 1])/((𝑥 − 2)^2 (𝑥 − 3)^2 ) 𝑑𝑦/𝑑𝑥 = ((𝑥 − 2) (𝑥 − 3) − (𝑥 − 7)(2𝑥 − 5))/((𝑥 − 2)^2 (𝑥 − 3)^2 ) 𝑑𝑦/𝑑𝑥 = ((𝑥 − 2) (𝑥 − 3) )/((𝑥 − 2)^2 (𝑥 − 3)^2 )−(𝑥 − 7)(2𝑥 − 5)/((𝑥 − 2)^2 (𝑥 − 3)^2 ) 𝑑𝑦/𝑑𝑥 = (1 )/((𝑥 − 2) (𝑥 − 3) )−((𝑥 − 7))/((𝑥 − 2) (𝑥 − 3) ) × ((2𝑥 − 5))/((𝑥 − 2) (𝑥 − 3) ) 𝑑𝑦/𝑑𝑥 = (1 )/((𝑥 − 2) (𝑥 − 3) )−𝑦 × ((2𝑥 − 5))/((𝑥 − 2) (𝑥 − 3) ) 𝒅𝒚/𝒅𝒙 = (𝟏 − 𝒚(𝟐𝒙 − 𝟓))/((𝒙 − 𝟐) (𝒙 − 𝟑) ) We need to find Equation of tangent at the point where the curve cuts the x axis, Thus, y = 0 We need to find value of x Putting y = 0 in equation of curve 0 = (𝑥 − 7)/(𝑥 − 2)(𝑥 − 3) ∴ x = 7 Thus, curve cuts the x-axis at point (7, 0) Finding Slope at point (7, 0) 𝑑𝑦/𝑑𝑥 = (1 − 𝑦(2𝑥 − 5))/((𝑥 − 2) (𝑥 − 3) ) Putting x = 7, y = 0 = (1 − 0[2(7)−5])/((7 − 2) (7 − 3) ) = 1/(5 × 4) = 𝟏/𝟐𝟎 Now, Equation of the tangent at point (7, 0) with slope 1/20 is 𝑦 −𝑦_1= 𝑚 (𝑥 − 𝑥1) 𝑦−0= 1/20 (𝑥−7) 20𝑦=𝑥−7 𝟐𝟎𝒚−𝒙+𝟕=𝟎
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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo