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

Class 12
Solutions of Sample Papers and Past Year Papers - for Class 12 Boards

## (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 β 4y β 7 = 0 is: (a) (2, 5/2) (b) (Β±2"," 5/2) (c) (β1/2 " ," 5/2) (d) (1/2 " ," 5/2) Finding Slope of Normal y = x + 1/π₯ Differentiating both sides ππ¦/ππ₯ = 1 β 1/π₯^2 Now, Slope of Normal = (βπ)/(π β π/π^π ) Given that Normal is perpendicular to 3x β 4y = 7 So, Slope of Normal Γ Slope of Line = β1 (β1)/(1 β 1/π₯^2 ) Γ 3/4 = β1 3/4 = 1 β 1/π₯^2 1 β 1/π₯^2 = 3/4 1 β 3/4 = 1/π₯^2 1/4 = 1/π₯^2 x2 = 4 x = Β± 2 Since x > 0 β΄ x = 2 Finding y when x = 2 y = x + 1/π₯ y = 2 + 1/2 y = π/π Thus, Point at which normal is perpendicular to line = (x, y) = (2, π/π) So, the correct answer is (a)