Ans.

(i)

Sol. x + 1 > x

f(x + 1) > f(x), f is increasing function

g(f(x + 1)) < g(f(x)), g is decreasing function

(ii)

Sol. f(x) = x 3 – 6x 2 + 15 x + 3

f '(x) = 3x 2 – 12x + 15

3x 2 – 12x + 15 > 0 because {D < 0}

f(x) is increasing function it means

x + 1 > x ⇒ f(x + 1) > f(x)

It function is one-one function. This is

cubic polynomial so range will be (– ∞ , ∞ ) → function is one and onto so it is invertible function

(iii)

Sol. g(x) = f(x) + f(1 – x)

g'(x) = f '(x) – f '(1 – x)

f "(x) < 0 it means f '(x) is decreasing function in 0 ≤ x ≤ 1

so that greatest value of f '(x) in0 ≤ x ≤ 1 at x = 0 and least value at x = 1 so that when 0 < x < 1/2 g'(x) = +ve at x = ½, g'(x) = 0 1/2 < x < 0, g '(x) = –ve it means g(x) is increasing in g(x) is decreasing in