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Question 9 - CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1) - Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

Last updated at Dec. 16, 2024 by Teachoo

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

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

    3. CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)

    CBSE Class 12 Sample Paper for 2021 Boards →
CBSE Class 12 Sample Paper for 2021 Boards →
Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

About the Author

Davneet Singh

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