For all x∈ (0,1)

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For all x&#8712; (0,1)

Accepted Answer

Note Inequation based upon uncompatible function. This type of inequation can be solved by calulus only.
Part (a) f(x) = e^x - 1 - x
then f(x) = e^x - 1 - x
⇒ f(x) increases in (0,1)
⇒ f(x) > f(0) for 0 < x < 1
⇒ e^x - 1 - x > 0 or e^x > 1 + x for 0 < x <1
Part (b) Next, let g (x) log_e (1 + x) - x, 0 < x < 1
g′

<0 for 0 < x < 1
⇒ g(x) decreases for 0 < x <1
⇒ g(x) < g (0) for 0 < x <1
⇒ log_e (1 + x) - x < 0 for 0 < x < 1
or log_e (1 + x) < x for 0 < x < 1
Therefore, (b) is the answer.
Part (c) sin x > x
let h(x) = sin x - x
h′(x) = cos x -1
for x ∈ (0,1), cos x -1 < 0 ⇒ h(x) is decreasing function.
⇒ h(x) < h (0)
⇒ sin x - x < 0
⇒ sin x < x, which is not true,
Part (d) p(x) = log x - x
p′(x)

- 1 0, ∀ x ∈ (0,1)
Therefore, p′(x) is an increasing function
⇒ p (0) < p (x)< p (1
⇒ - ∞ < log x - x < - 1
⇒ log x - x < 0 ⇒ log x < x
Therefore, (d) is not the answer,

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