Show that tan x > x for all x ∈ 0 , π 2

Let f ( x ) = tan x - x ....(i) Now, f ' ( x ) = sec 2 x - 1 We know that, a function is always increasing, when f ' x > 0 . Now, f ' ( x ) = sec 2 x - 1 > 0 for every x ∈ 0 , π 2 ∴ f ( x ) is increasing for every x ∈ 0 , π 2 Now, f ( 0 ) = tan 0 - 0 = 0 - 0 = 0 (substitute in equation (i)) ∴ tan x - x > 0 as f x is increasing. ⇒ tan x > x for every x ∈ 0 , π 2 Hence, showed

If x ∈ ( 0 , π 2 ) then show that 2 x π < sin x < x .

Given that, x > 0 , π 2 To prove: 2 x π < sin x < x Let f ( x ) = x - sin x ⇒ f ' ( x ) = 1 - cos x Here, f ' x > 0 , as the cos x < 1 in the given interval 0 , π 2 ⇒ f ' ( x ) > 0 ⇒ f is strictly increasing function ⇒ x - sin x > 0 ⇒ x > sin x .......(i) Now let us prove 2 x π < sin x ⇒ 2 π < sin x x .......(ii) Let h x = sin x x .......(iii) h ' ( x ) = x cos x - sin x x 2 ...(iv) In the interval 0 , π 2 lim x → 0 sin x x = 1 and h π 2 = 2 π let g ( x ) = x cos x - sin x ⇒ g ' ( x ) = cos x - x sin x - cos x = - x sin x < 0 ⇒ g ' x < 0 Therefore, we can say that h x is a decreasing function ⇒ sin x x < 0 ....(v) From (ii) and (v) ⇒ sin x x > 2 π ⇒ sin x > 2 x π ......(vi) From (i) and (vi) 2 x π < sin x < x in the interval 0 , π 2 Hence, showed.

If x ∈ ( 0 , 1 ) then show that 2 x < ln ( 1 + x ) ( x - 1 ) < 2 x 1 + x 2 2 1 - x 2 .

Given that, x ∈ ( 0 , 1 ) To prove: 2 x < ln ( 1 + x ) ( x - 1 ) < 2 x 1 + x 2 2 1 - x 2 Let f x = ln 1 + x x - 1 - 2 x .......(i) f ' ( x ) = 1 1 + x x - 1 x - 1 - 1 - x x - 1 2 - 2 Let us take x = 0 . 5 ∈ 0 , 1 f ' x = - 0 . 5 1 . 5 - 2 - 0 . 5 2 - 2 = 2 . 66 - 2 > 0 ⇒ f ' x > 0 in the interval 0 , 1 . Therefore, we can say that f x is strictly increasing function. ⇒ f ( x ) > 0 ⇒ ln 1 + x x - 1 - 2 x > 0 ⇒ ln 1 + x x - 1 > 2 x .....................(i) Now, Let us take g x = 2 x 1 + x 2 2 1 - x 2 - ln ( 1 + x ) ( x - 1 ) ⇒ g ' x = 2 1 1 + x 2 2 1 - x 2 + 2 x 0 + 2 x 2 1 - x 2 - x 2 2 - 2 x 2 1 - x 2 2 - x - 1 1 + x 1 x - 1 - ( 1 + x ) 1 x - 1 2 = 2 1 + 0 . 5 2 2 1 - 0 . 25 + 2 x 2 × 0 . 5 2 1 - 0 . 5 2 - 0 . 5 2 - 4 × 0 . 5 2 1 - 0 . 5 2 2 - 0 . 5 - 1 1 + 0 . 5 - 2 0 . 5 - 1 2 = 2 1 . 09 + 2 × 0 . 5 × 1 1 . 5 - 0 . 5 2 . 25 - - 0 . 5 1 . 5 - 2 - 0 . 5 2 > 0 ∵ d d x u v = u ' v - v ' u v 2 Let us take x = 0 . 5 ∈ 0 , 1 , in this interval g ' x = positive value - negative value > 0 ⇒ g ' ( x ) > 0 Therefore, we can say that g x is strictly increasing function. ⇒ g ( x ) > 0 ⇒ 2 x 1 + x 2 2 1 - x 2 - ln ( 1 + x ) ( x - 1 ) > 0 ⇒ 2 x 1 + x 2 2 1 - x 2 > ln ( 1 + x ) ( x - 1 ) .......(ii) From (i) and (ii), 2 x < ln ( 1 + x ) ( x - 1 ) < 2 x 1 + x 2 2 1 - x 2 Hence, showed.

At what points the slopes of the tangents to the curve y = x 3 6 - 3 x 2 2 + 11 x 2 + 12 increases?

Equation of the curve is y = x 3 6 - 3 2 x 2 + 11 x 2 + 12 . ⇒ d y d x = 3 x 2 6 - 3 2 · 2 x + 11 2 = x 2 2 - 3 x + 11 2 Slope m = x 2 2 - 3 x + 11 2 ∵ d y d x = m Now, differentiating with respect to x , we get ⇒ d m d x = 2 x 2 - 3 = x - 3 Now, Slope m increasing if d m d x > 0 (Given that, slope of tangent to the curve is increasing) ⇒ x - 3 > 0 ⇒ x > 3 Therefore, the Slope is increasing on ( 3 , ∞ ) .

Show that the functions ln ( 1 + x ) x and x ( 1 + x ) ln ( 1 + x ) are decreasing on ( 0 , ∞ ) .

Let f x = ln ( 1 + x ) x ⇒ f ' x = x 1 1 + x - ln ( 1 + x ) x 2 ∵ d d x u v = u ' v - v ' u v 2 We know that, a function f x is always decreasing, when f ' x < 0 . Now, x - ( 1 + x ) ln ( 1 + x ) x 2 ( 1 + x ) < 0 When x is taken as negative values then f ' x > 0 and if x is taken as positive values then f ' x < 0 . Here, the function is decreasing on 0 , ∞ ∴ f ( x ) is a decreasing when x ∈ ( 0 , ∞ ) Let g x = x 1 + x ln 1 + x ⇒ g ' x = 1 . 1 + x ln 1 + x - x ( 1 + x ) . 1 x + ln 1 + x 0 + 1 1 + x ln 1 + x 2 ∵ d d x u v = u ' v - v ' u v 2 and d dx uv = uv ' + vu ' We know that, a function g x is always decreasing, when g ' x < 0 . Now, 1 . 1 + x ln 1 + x - x ( 1 + x ) . 1 x + ln 1 + x 0 + 1 1 + x ln 1 + x 2 < 0 When x is taken as negative values then g ' x > 0 and if x is taken as positive values then g ' x < 0 . Here, the function is decreasing on 0 , ∞ Hence, both the functions are decreasing functions on 0 , ∞ .

Find the intervals in which the function f ( x ) = sin 4 x + cos 4 x ∀ x ∈ [ 0 , π / 2 ] is increasing and decreasing.

f ( x ) = sin 4 x + cos 4 x f ( x ) = sin 2 x 2 + cos 2 x 2 = sin 2 x + cos 2 x - 2 sin 2 x cos 2 x = 1 - 1 2 sin 2 2 x f ' ( x ) = - 1 2 × 2 sin 2 x · cos 2 x × 2 = - 2 sin 2 x · cos 2 x = - sin 4 x ∵ 2 sin A cos A = sin 2 A As x is in the interval 0 , π 2 hence 4 x is in the interval 0 , 2 π . sin 4 x ≥ 0 for x ∈ 0 , π 4 sin 4 x ≤ 0 for x ∈ π 4 , π 2 Here, f ' ( x ) < 0 for 0 < x < π 4 ∴ f ( x ) is decreasing on 0 , π 4 . Now, f ' ( x ) > 0 for π 4 < x < π 2 . ∴ f ( x ) is increasing on π 4 , π 2 .

E M B I B E

Intermediate First Year Mathematics Paper 1B>Chapter 10 - Applications of Derivatives>Exercise 10(g)

Telangana Board Solutions for Chapter: Applications of Derivatives, Exercise 7: Exercise 10(g)

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Chapter: Applications of Derivatives, Exercise 7: Exercise 10(g)

Attempt the free practice questions on Chapter 10: Applications of Derivatives, Exercise 7: Exercise 10(g) with hints and solutions to strengthen your understanding. Intermediate First Year Mathematics Paper 1B solutions are prepared by Experienced Embibe Experts.