If g x is monotonically increasing and f x is monotonically decreasing for x ∈ R and if g o f x is defined for x ∈ R , then prove that g o f x will be monotonically decreasing function. Hence prove that g o f x + 1 ≤ g o f x - 1 .

Since f x is monotonically decreasing & g x is monotonically increasing ⇒ f ' x ≤ 0 , g ' x ≥ 0 . . . . . i Consider g o f x = g f x Now, g ' f x f ' x ≤ 0 , using i ⇒ g o f x is a decreasing function . . . . . i i Also, x + 1 > x - 1 Using i i , g o f x + 1 ≤ g o f x - 1 .

If g x is monotonically increasing and f x is monotonically decreasing for x ∈ R and if g o f x is defined for x ∈ R , then prove that g o f x will be monotonically decreasing function. Hence prove that g o f x + 1 ≤ g o f x - 1 .

Since f x is monotonically decreasing & g x is monotonically increasing ⇒ f ' x ≤ 0 , g ' x ≥ 0 . . . . . i Consider g o f x = g f x Now, g ' f x f ' x ≤ 0 , using i ⇒ g o f x is a decreasing function . . . . . i i Also, x + 1 > x - 1 Using i i , g o f x + 1 ≤ g o f x - 1 .

Check monotonicity of f x at indicated point f x = x 2 + 1 x 2 at x = 1 , 2 .

Given, f x = x 2 + 1 x 2 f ' x = 2 x - 2 x 3 ⇒ f ' x = 2 x 4 - 1 x 3 ⇒ f ' x = 2 x 3 x 2 + 1 x 2 - 1 ⇒ f ' x = 2 x 3 x 2 + 1 x + 1 x - 1 ∙ For - ∞ , - 1 ∪ 1 , ∞ f x is strictly increasing. ∙ For x ∈ - 1 , 0 ∪ 0 , 1 f x is strictly decreasing. So, at x = 2 f x is strictly increasing and at x = 1 . f x is neither increasing nor decreasing.

Prove that sin 1 10 1 10 > sin 1 9 1 9 .

We have sin 1 10 1 10 > sin 1 9 1 9 Let f x = sin x x Differentiating both sides w.r.t. x , we get f ' x = x cos x - 1 · sin x x 2 Let g x = x cos x - sin x ⇒ g ' x = cos x - x sin x - cos x ⇒ g ' x = - x sin x For small angles, x sin x > 0 ⇒ - x sin x < 0 , therefore g ' x < 0 , hence g x is decreasing. For x > 0 ⇒ g x < g 0 ⇒ x cos x - sin x < 0 ⇒ x cos x - sin x x 2 < 0 ⇒ f ' x < 0 Hence, f x is decreasing. So, 1 9 > 1 10 ⇒ sin 1 9 < sin 1 10 ⇒ 10 sin 1 10 > 10 sin 1 9 > 9 sin 1 9 ⇒ sin 1 10 1 10 > sin 1 9 1 9 Hence, proved.

Let f and g be differentiable on R and suppose f 0 = g 0 and f ' x ≤ g ' x for all x ≥ 0 . Then show that f x ≤ g x for all x ≥ 0 .

Consider, h x = g x - f x Now, h ' x = g ' x - f ' x Given, g ' x ≥ f ' x ⇒ g ' x - f ' x ≥ 0 ⇒ h ' x ≥ 0 So, h x is an increasing function. h x ≥ h 0 , for x ∈ [ 0 , ∞ ) ⇒ h x ≥ g 0 - f 0 ⇒ g x - f x ≥ 0 ⇒ g x ≥ f x Hence, proved.

Let f x = 3 - x 0 ≤ x < 1 x 2 + ln b x ≥ 1 . Find the set of values of b such that f x has a local minima at x = 1 .

Given, f x = 3 - x , 0 ≤ x < 1 x 2 + ln b , x ≥ 1 For a function to have local minima at x = a , f a + ≥ f a ≤ f a - Local minima at x = 1 . f 1 = 1 2 + ln b = ln b + 1 Now, f 1 - ≥ ln b + 1 ⇒ 3 - 1 ≥ ln b + 1 ⇒ ln b ≤ 1 ⇒ b ≤ e Also, b > 0 , to define ln b . So, b ∈ ( 0 , e ] .