The point at which the normal to the curve y = ๐‘ฅ + 1/x,  x > 0 is perpendicular to the line 3x – 4y – 7 = 0 is:

(a) (2,  5/2)    (b) (±2,  5/2)

(c) (-1/2, 5/2)  (d) (1/2 ,  5/2)

 

This question is inspired from Question 22 - CBSE Class 12 Sample Paper for 2021 Boards

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  1. Class 12
  2. Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Transcript

Question 9 The point at which the normal to the curve y = ๐‘ฅ + 1/๐‘ฅ, x > 0 is perpendicular to the line 3x โ€“ 4y โ€“ 7 = 0 is: (a) (2, 5/2) (b) (ยฑ2"," 5/2) (c) (โˆ’1/2 " ," 5/2) (d) (1/2 " ," 5/2) Finding Slope of Normal y = x + 1/๐‘ฅ Differentiating both sides ๐‘‘๐‘ฆ/๐‘‘๐‘ฅ = 1 โˆ’ 1/๐‘ฅ^2 Now, Slope of Normal = (โˆ’๐Ÿ)/(๐Ÿ โˆ’ ๐Ÿ/๐’™^๐Ÿ ) Given that Normal is perpendicular to 3x โˆ’ 4y = 7 So, Slope of Normal ร— Slope of Line = โˆ’1 (โˆ’1)/(1 โˆ’ 1/๐‘ฅ^2 ) ร— 3/4 = โˆ’1 3/4 = 1 โˆ’ 1/๐‘ฅ^2 1 โˆ’ 1/๐‘ฅ^2 = 3/4 1 โˆ’ 3/4 = 1/๐‘ฅ^2 1/4 = 1/๐‘ฅ^2 x2 = 4 x = ยฑ 2 Since x > 0 โˆด x = 2 Finding y when x = 2 y = x + 1/๐‘ฅ y = 2 + 1/2 y = ๐Ÿ“/๐Ÿ Thus, Point at which normal is perpendicular to line = (x, y) = (2, ๐Ÿ“/๐Ÿ) 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.