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Class 12
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The point at which the normal to the curve y = π‘₯ + 1/x,Β  x > 0 is perpendicular to the line 3x – 4y – 7 = 0 is:

(a) (2,Β  5/2)Β  Β  (b) (Β±2, Β 5/2)

(c) (-1/2, 5/2)Β  (d) (1/2 , Β 5/2)

Β 

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

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Question 9 The point at which the normal to the curve y = π‘₯ + 1/π‘₯, x > 0 is perpendicular to the line 3x – 4y – 7 = 0 is: (a) (2, 5/2) (b) (Β±2"," 5/2) (c) (βˆ’1/2 " ," 5/2) (d) (1/2 " ," 5/2) Finding Slope of Normal y = x + 1/π‘₯ Differentiating both sides 𝑑𝑦/𝑑π‘₯ = 1 βˆ’ 1/π‘₯^2 Now, Slope of Normal = (βˆ’πŸ)/(𝟏 βˆ’ 𝟏/𝒙^𝟐 ) Given that Normal is perpendicular to 3x βˆ’ 4y = 7 So, Slope of Normal Γ— Slope of Line = βˆ’1 (βˆ’1)/(1 βˆ’ 1/π‘₯^2 ) Γ— 3/4 = βˆ’1 3/4 = 1 βˆ’ 1/π‘₯^2 1 βˆ’ 1/π‘₯^2 = 3/4 1 βˆ’ 3/4 = 1/π‘₯^2 1/4 = 1/π‘₯^2 x2 = 4 x = Β± 2 Since x > 0 ∴ x = 2 Finding y when x = 2 y = x + 1/π‘₯ y = 2 + 1/2 y = πŸ“/𝟐 Thus, Point at which normal is perpendicular to line = (x, y) = (2, πŸ“/𝟐) So, the correct answer is (a)

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