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 May 29, 2023 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
Learn in your speed, with individual attention  Teachoo Maths 1on1 Class
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)
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.
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