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 April 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
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)
Made by
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
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