Mathematics - Application of Derivative Question with Solution | TestHub
Match the following List: Column I: Equation of tangent
(A) to at the point where the curve achieves local maxima
(B) Common to and
(C) to at the point where its slope is the abscissa of the point
(D) to , , which tangent is parallel to the x-axis
Column II:
(p)
(q)
(r)
(s)
Options:
Answer:
Solution:
(A) For x>0, y = x e^(-x); y' = e^(-x)(1-x) = 0 at x = 1, a local maximum with y = 1/e; the tangent there is y = 1/e, so (A) matches (r). (B) A line y = mx + c tangent to y^2 = 8x has c = 2/m. For it to touch xy = -1, m x^2 + (2/m) x + 1 = 0 must have a double root: (2/m)^2 = 4m, so m = 1 and the common tangent is y = x + 2; (B) matches (p). (C) For y = x^2 the slope is 2x; slope equal to the abscissa gives 2x = x, so x = 0 and the tangent at (0,0) is y = 0; (C) matches (s). (D) Differentiate (1+x^2)y = 2-x: 2xy + (1+x^2) y' = -1. With y' = 0, 2xy = -1, so y = -1/(2x). Substituting back: -(1+x^2)/(2x) = 2-x, giving x^2 - 4x - 1 = 0 and x = 2 + sqrt(5) (since x > 0). Then y = (2 - sqrt(5))/2, i.e. the tangent is 2y = 2 - sqrt(5); (D) matches (q). Answer: A-r, B-p, C-s, D-q (option D).
