RD Chapter 17 Increasing and Decreasing Functions Ex 17.2 Solutions
Question - 11 : - Show that f(x) =cos2 x is a decreasing function on (0, π/2).
Answer - 11 : -
Given f (x) = cos2 x
⇒ 
⇒ f’(x)= 2 cos x (–sin x)
⇒ f’(x)= –2 sin (x) cos (x)
⇒ f’(x)= –sin2x
Now, as given x belongs to (0, π/2).
⇒ 2x ∈ (0, π)
⇒ Sin(2x)> 0
⇒ –Sin(2x) < 0
⇒ f’(x)< 0
Hence, condition for f(x) to be decreasing
Thus f(x) is decreasing on interval (0, π/2).
Hence proved
Question - 12 : - Show that f(x) =sin x is an increasing function on (–π/2, π/2).
Answer - 12 : -
Given f (x) = sin x
⇒ 
⇒ f’(x)= cos x
Now, as given x ∈ (–π/2,π/2).
That is 4th quadrant, where
⇒ Cosx> 0
⇒ f’(x)> 0
Hence, condition for f(x) to be increasing
Thus f(x) is increasing on interval (–π/2, π/2).
Question - 13 : - Show that f(x) =cos x is a decreasing function on (0, π), increasing in (–π, 0) and neitherincreasing nor decreasing in (–π, π).
Answer - 13 : -
Given f(x) = cos x
⇒ 
⇒ f’(x)= –sin x
Taking different region from 0 to 2π
Let x ∈ (0,π).
⇒ Sin(x)> 0
⇒ –sinx < 0
⇒ f’(x)< 0
Thus f(x) is decreasing in (0, π)
Let x ∈ (–π,o).
⇒ Sin(x) < 0
⇒ –sinx > 0
⇒ f’(x)> 0
Thus f(x) is increasing in (–π, 0).
Therefore, from above condition we find that
⇒ f(x) is decreasing in (0, π) and increasing in (–π, 0).
Hence, condition for f(x) neither increasing nor decreasing in(–π, π)
Question - 14 : - Show that f(x) =tan x is an increasing function on (–π/2, π/2).
Answer - 14 : -
Given f (x) = tan x
⇒ 
⇒ f’(x)= sec2x
Now, as given
x ∈ (–π/2,π/2).
That is 4th quadrant, where
⇒ sec2x> 0
⇒ f’(x)> 0
Hence, Condition for f(x) to be increasing
Thus f(x) is increasing on interval (–π/2, π/2).
Question - 15 : - Show that f(x) =tan–1 (sin x + cos x) is a decreasing function on the interval(π/4, π /2).
Answer - 15 : -
Question - 16 : - Show that the function f (x) = sin (2x + π/4) isdecreasing on (3π/8, 5π/8).
Answer - 16 : -
Thus f (x) is decreasing on the interval (3π/8, 5π/8).
Show that thefunction f(x) = cot–1 (sin x + cos x) is decreasing on (0, π/4)and increasing on (π/4, π/2).
Answer - 17 : -
Given f(x) = cot–1 (sin x + cos x)
Question - 18 : - Show that f(x) = (x– 1) ex + 1 is an increasing function for all x > 0.
Answer - 18 : -
Given f (x) = (x – 1) ex + 1
Now differentiating the given equation with respect to x, we get
⇒ 
⇒ f’(x)= ex + (x – 1) ex
⇒ f’(x)= ex(1+ x – 1)
⇒ f’(x)= x ex
As given x > 0
⇒ ex >0
⇒ xex > 0
⇒ f’(x)> 0
Hence, condition for f(x) to be increasing
Thus f(x) is increasing on interval x > 0
Question - 19 : - Show that the function x2 – x + 1 is neither increasing nor decreasing on (0, 1).
Answer - 19 : -
Given f(x) = x2 – x + 1
Now by differentiating the given equation with respect to x, weget
⇒ 
⇒ f’(x)= 2x – 1
Taking different region from (0, 1)
Let x ∈ (0, ½)
⇒ 2x– 1 < 0
⇒ f’(x)< 0
Thus f(x) is decreasing in (0, ½)
Let x ∈ (½, 1)
⇒ 2x– 1 > 0
⇒ f’(x)> 0
Thus f(x) is increasing in (½, 1)
Therefore, from above condition we find that
⇒ f(x) is decreasing in (0, ½) and increasing in (½, 1)
Hence, condition for f(x) neither increasing nor decreasing in(0, 1)
Question - 20 : - Show that f(x) = x9 + 4x7 +11 is an increasing function for all x ϵ R.
Answer - 20 : -
Given f (x) = x9 + 4x7 + 11
Now by differentiating above equation with respect to x, we get
⇒ 
⇒ f’(x)= 9x8 + 28x6
⇒ f’(x)= x6(9x2 + 28)
As given x ϵ R
⇒ x6 >0 and 9x2 + 28 > 0
⇒ x6 (9x2 +28) > 0
⇒ f’(x)> 0
Hence, condition for f(x) to be increasing
Thus f(x) is increasing on interval x ∈ R