Continuity and DifferentiabilityHard
Question
Let g(x) = 
; 0 < x < 2, m and n are integers, m ≠ 0, n > 0, and let p be the left hand derivative of |x - 1| at x = 1. If 
g(x) = p, then
; 0 < x < 2, m and n are integers, m ≠ 0, n > 0, and let p be the left hand derivative of |x - 1| at x = 1. If g(x) = p, thenOptions
A.n = 1, m = 1
B.n = 1, m = - 1
C.n = 2, m = 2
D.n > 2, m = n
Solution
From graph, p = - 1
⇒
g(x) = - 1⇒
g(1 + h) = - 1⇒
= - 1⇒
= - 1, which holds if n = m = 2.Create a free account to view solution
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