(b,c,d)

f(x) = x cos , x ≥ 1

⇒ f ′ (x) = sin + cos

⇒ f ′′ (x) = – cos

Now f ′ (x) = 0 + 1 = 1 ⇒ option ‘B’ is correct

x ∈ [1, ∞ ) ⇒ ∈ (0, 1]

⇒ f ′′ (x) < 0 ⇒ option ‘D’ is correct

As f ′ (1) = sin 1 + cos 1 > 1

f ′ (x) is strictly decreasing and f ′ (x) = 1

so graph of f ′ (x) is as below

Now in [x, x + 2], x ∈ [1, ∞ ), f(x) is continuous and differentiable

so by LMVT, f ′ (x) =

as f ′ (x) > 1 for all x ∈ [1, ∞ )

⇒ > 1 ⇒ f(x + 2) – f(x) > 2

for all x ∈ [1, ∞ )

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