The point at which the normal to the curve y = 𝑥 + 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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Transcript

Question 9 The point at which the normal to the curve y = 𝑥 + 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) Finding Slope of Normal y = x + 1/𝑥 Differentiating both sides 𝑑𝑦/𝑑𝑥 = 1 − 1/𝑥^2 Now, Slope of Normal = (−𝟏)/(𝟏 − 𝟏/𝒙^𝟐 ) Given that Normal is perpendicular to 3x − 4y = 7 So, Slope of Normal × Slope of Line = −1 (−1)/(1 − 1/𝑥^2 ) × 3/4 = −1 3/4 = 1 − 1/𝑥^2 1 − 1/𝑥^2 = 3/4 1 − 3/4 = 1/𝑥^2 1/4 = 1/𝑥^2 x2 = 4 x = ± 2 Since x > 0 ∴ x = 2 Finding y when x = 2 y = x + 1/𝑥 y = 2 + 1/2 y = 𝟓/𝟐 Thus, Point at which normal is perpendicular to line = (x, y) = (2, 𝟓/𝟐) So, the correct answer is (a)

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.