The equation of tangent to the curve y (1 + x 2 ) = 2 – x, where it crosses x-axis is:

(A) x + 5y = 2   (B) x – 5y = 2

(C) 5x – y = 2   (D) 5x + y = 2

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  1. Chapter 6 Class 12 Application of Derivatives (Term 1)
  2. Serial order wise

Transcript

Question 12 The equation of tangent to the curve y (1 + x2) = 2 – x, where it crosses x-axis is: (A) x + 5y = 2 (B) x – 5y = 2 (C) 5x – y = 2 (D) 5x + y = 2 First, let us find the point at which curve cuts π‘₯βˆ’axis Since π’š=𝟎 on π‘₯βˆ’axis ∴ Point =(𝒙,𝟎) For π’™βˆ’ coordinate Putting π’š=𝟎 in 𝑦(1+π‘₯^2 )=2βˆ’π‘₯ 0(1+π‘₯^2 )=2βˆ’π‘₯ 0=2βˆ’π‘₯ 𝒙=𝟐 ∴ Point =(𝟐,𝟎) Now, to find equation of tangent we find slope of tangent at (𝟐, 𝟎) Finding Slope of tangent Given curve 𝑦(1+π‘₯^2 )=2βˆ’π‘₯ Differentiating w.r.t x 𝑑𝑦/𝑑π‘₯ (1+π‘₯^2 )+𝑦 Γ— (0+2π‘₯)=0βˆ’1 𝑑𝑦/𝑑π‘₯ (1+π‘₯^2 )+𝑦(2π‘₯)=βˆ’1 𝑑𝑦/𝑑π‘₯ (1+π‘₯^2 )+2π‘₯𝑦=βˆ’1 𝑑𝑦/𝑑π‘₯ (1+π‘₯^2 )=βˆ’1βˆ’2π‘₯𝑦 π’…π’š/𝒅𝒙=(βˆ’πŸ βˆ’ πŸπ’™π’š" " )/((𝟏 + 𝒙^𝟐 ) ) Since we need Slope at (2, 0) Putting x = 2, y = 0 in 𝑑𝑦/𝑑π‘₯ π’…π’š/𝒅𝒙 =(βˆ’1 βˆ’ (2 Γ— 2 Γ— 0))/(1 + 2^2 ) =(βˆ’1 βˆ’ 0)/(1 + 4) =(βˆ’πŸ)/πŸ“ Finding equation of tangent Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) ∴ Equation of tangent at (2, 0) & Slope (βˆ’1)/5 is (π’šβˆ’πŸŽ)=(βˆ’πŸ)/πŸ“ (π’™βˆ’πŸ) 𝑦=(βˆ’πŸ)/πŸ“ (π‘₯βˆ’2) 5𝑦=βˆ’1(π‘₯βˆ’2) 5𝑦=βˆ’π‘₯+2 5𝑦+π‘₯=2 𝒙+πŸ“π’š=𝟐 Hence, equation of tangent is 𝒙+πŸ“π’š=𝟐 So, the correct answer is (A)

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.