CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
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CBSE Class 12 Sample Paper for 2022 Boards (MCQ Based - for Term 1)
Last updated at August 14, 2026 by Teachoo
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 = š„ + 1/š„, x > 0 is perpendicular to the line 3x ā 4y ā 7 = 0 is: (a) (2, 5/2) (b) (Ā±2"," 5/2) (c) (ā1/2 " ," 5/2) (d) (1/2 " ," 5/2) Finding Slope of Normal y = x + 1/š„ Differentiating both sides šš¦/šš„ = 1 ā 1/š„^2 Now, Slope of Normal = (āš)/(š ā š/š^š ) Given that Normal is perpendicular to 3x ā 4y = 7 So, Slope of Normal Ć Slope of Line = ā1 (ā1)/(1 ā 1/š„^2 ) Ć 3/4 = ā1 3/4 = 1 ā 1/š„^2 1 ā 1/š„^2 = 3/4 1 ā 3/4 = 1/š„^2 1/4 = 1/š„^2 x2 = 4 x = Ā± 2 Since x > 0 ā“ x = 2 Finding y when x = 2 y = x + 1/š„ y = 2 + 1/2 y = š/š Thus, Point at which normal is perpendicular to line = (x, y) = (2, š/š) So, the correct answer is (a)
