Match the following List: Column I (curve)
(A) $f (x) = s g n (x)$
(B) $f (x) = sin (\frac{1}{x})$
(C) $f (x) = \frac{∣ x - 2∣}{x ^{2}}$
(D) $f (x) = \frac{1}{x ^{2} - x}$ Column II (Number of critical points) (p) 2 (q) 1 (r) 0 (s) infinite

Question

Match the following List: Column I (curve)

(A) $f (x) = s g n (x)$

(B) $f (x) = sin (\frac{1}{x})$

(C) $f (x) = \frac{∣ x - 2∣}{x ^{2}}$

(D) $f (x) = \frac{1}{x ^{2} - x}$ Column II (Number of critical points) (p) 2 (q) 1 (r) 0 (s) infinite

TestHub · Accepted Answer

(A) f(x) = sgn(x): f'(x) = 0 for every x != 0 and the graph has a vertical tangent at x = 0, so f'(0) does not exist; infinitely many critical points, so (A) matches (s). (B) f(x) = sin(1/x): f'(x) = -(1/x^2) cos(1/x), which vanishes whenever 1/x = (2n+1) pi/2 for an integer n, and f'(0) does not exist; infinitely many critical points, so (B) matches (s). (C) f(x) = |x-2|/x^2, with x = 0 excluded from the domain: for x > 2, f'(x) = (4-x)/x^3, and for x < 2, f'(x) = (x-4)/x^3. Hence f'(x) = 0 at x = 4 and f'(2) does not exist; two critical points, so (C) matches (p). (D) f(x) = 1/(x^2-x) = 1/(x-1) - 1/x: f'(x) = (1-2x)/(x^2 (x-1)^2), which vanishes at x = 1/2; one critical point, so (D) matches (q). Answer: A-s, B-s, C-p, D-q (option C).

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