Application of DerivativeHard
Question
Let f and g be increasing and decresing functions respectively from [0, ∞) to [0, ∞) Let h(x) = f (g(x)). If h(0) = 0, then h(x) - h(1)
Options
A.always negative
B.always positive
C.strictly increasing
D.None of these
Solution
Let F(x) = h(x) - h(1)
= f(g(x)) - f(g(1))
∴ F′(x) = f′(g(x)).g′(x) = (+)(-) = - ve
[Since, f(x) is increasing function f′(g(x)) is + ve and g(x) is decreasing function g′(f(x)) is - ve]
Sinnce, fg′(x) is - ve
∴ f(x) is decreasing function
∴ when 0 ≤ x < 1
⇒ h(x) - h(1) = + ve
When x ≥ 1
⇒ h(x) - h(1) = - ve
Hence, for x > 0
h(x) - h(1) is neither always positive nor always negative, so it is not strictly increasing throughout.
Therefore, (d) is the answer.
= f(g(x)) - f(g(1))
∴ F′(x) = f′(g(x)).g′(x) = (+)(-) = - ve
[Since, f(x) is increasing function f′(g(x)) is + ve and g(x) is decreasing function g′(f(x)) is - ve]
Sinnce, fg′(x) is - ve
∴ f(x) is decreasing function
∴ when 0 ≤ x < 1
⇒ h(x) - h(1) = + ve
When x ≥ 1
⇒ h(x) - h(1) = - ve
Hence, for x > 0
h(x) - h(1) is neither always positive nor always negative, so it is not strictly increasing throughout.
Therefore, (d) is the answer.
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