A derivable function f : R + → R satisfies the condition f x - f y ≥ ln x y + x - y ∀ x , y ∈ R + . If g denotes the derivative of f then compute the value of the sum ∑ n = 1 100 g 1 n .

We have, f x - f y ≥ ln x y + x - y ∀ x , y ∈ R + Now, f x - f y ≥ ln x - ln y + x - y . . . . ( 1 ) Replace y → x & x → y , f ( y ) - f ( x ) ≥ ln y - ln x + y - x ⇒ f ( x ) - f ( y ) ≤ ln x - ln y + x - y . . . . ( 2 ) From 1 and 2 , we get f ( x ) - f ( y ) = ln x - ln y + x - y . . . . . ( 3 ) ⇒ f ( x ) - f ( y ) = ln x + x - ln y + y Hence, f x = ln x + x . Then, g x = f ' x = 1 x + 1 Now, ∑ n = 1 100 g 1 n = g 1 + g 1 2 + . . . + g 1 100 ⇒ ∑ n = 1 100 g 1 n = 1 + 1 + 1 + 2 + 1 + 3 + . . . . + 1 + 100 ⇒ ∑ n = 1 100 g 1 n = 100 + 1 + 2 + 3 + . . . + 100 ⇒ ∑ n = 1 100 g 1 n = 100 + 100 × 101 2 ⇒ ∑ n = 1 100 g 1 n = 100 + 50 × 101 = 5150

A derivable function f : R + → R satisfies the condition f x - f y ≥ ln x y + x - y ∀ x , y ∈ R + . If g denotes the derivative of f then compute the value of the sum ∑ n = 1 100 g 1 n .

We have, f x - f y ≥ ln x y + x - y ∀ x , y ∈ R + Now, f x - f y ≥ ln x - ln y + x - y . . . . ( 1 ) Replace y → x & x → y , f ( y ) - f ( x ) ≥ ln y - ln x + y - x ⇒ f ( x ) - f ( y ) ≤ ln x - ln y + x - y . . . . ( 2 ) From 1 and 2 , we get f ( x ) - f ( y ) = ln x - ln y + x - y . . . . . ( 3 ) ⇒ f ( x ) - f ( y ) = ln x + x - ln y + y Hence, f x = ln x + x . Then, g x = f ' x = 1 x + 1 Now, ∑ n = 1 100 g 1 n = g 1 + g 1 2 + . . . + g 1 100 ⇒ ∑ n = 1 100 g 1 n = 1 + 1 + 1 + 2 + 1 + 3 + . . . . + 1 + 100 ⇒ ∑ n = 1 100 g 1 n = 100 + 1 + 2 + 3 + . . . + 100 ⇒ ∑ n = 1 100 g 1 n = 100 + 100 × 101 2 ⇒ ∑ n = 1 100 g 1 n = 100 + 50 × 101 = 5150

If x < 1 , then prove that 1 - 2 x 1 - x + x 2 + 2 x - 4 x 3 1 - x 2 + x 4 + 4 x 3 - 8 x 7 1 - x 4 + x 8 + … … ∞ = 1 + 2 x 1 + x + x 2 .

Let A = 1 + x + x 2 1 - x + x 2 1 - x 2 + x 4 1 - x 4 + x 8 . . . . . . ∞ ⇒ A = 1 + x 2 + x 4 1 - x 2 + x 4 1 - x 4 + x 8 . . . . . . ∞ ⇒ A = 1 + x 4 + x 8 1 - x 4 + x 8 . . . . . . ∞ ⇒ A = lim n → ∞ 1 + x 2 n + x 4 n Now, x < 1 , so A = lim n → ∞ 1 + x 2 n + x 4 n = 1 Hence, 1 + x + x 2 1 - x + x 2 1 - x 2 + x 4 1 - x 4 + x 8 . . . . . . ∞ = 1 Taking log both sides, we get log e 1 + x + x 2 + log e 1 - x + x 2 + log e 1 - x 2 + x 4 + log e 1 - x 4 + x 8 + . . . . . . ∞ = log e 1 ⇒ log e 1 + x + x 2 + log e 1 - x + x 2 + log e 1 - x 2 + x 4 + log e 1 - x 4 + x 8 + . . . . . . ∞ = 0 Differentiating both sides w.r.t. x , we get 1 + 2 x 1 + x + x 2 + 2 x - 1 1 - x + x 2 + - 2 x + 4 x 3 1 - x 2 + x 4 + 4 x 3 - 8 x 7 1 - x 4 + x 8 + . . . . . . ∞ = 0 ∵ d d x x m = m x m - 1 ⇒ 1 - 2 x 1 - x + x 2 + 4 x 3 - 2 x 1 - x 2 + x 4 + 8 x 7 - 4 x 3 1 - x 4 + x 8 + . . . . . . ∞ = 1 + 2 x 1 + x + x 2

If f n x = e f n - 1 x for all n ∈ N and f 0 x = x then show that d d x f n x = f 1 x . f 2 x . . . . . . f n x .

Given that f n x = e f n - 1 x for all n ∈ N and f 0 x = x . ⇒ f 1 x = e f 0 x = e x and f 2 x = e f 1 x = e e x and f 3 x = e f 2 x = e e e x So, f n x = e f n - 1 x = e e e e . . . . n - 1 times x Now, d d x f n x = e f n - 1 x d d x f n - 1 x = e f n - 1 x d d x e f n - 2 x = e f n - 1 x . e f n - 2 x . . . . . e f 0 x = f n x . f n - 1 x . . . . . f 1 x

If x 2 + y 2 = e arcsin y x 2 + y 2 . Prove that d 2 y d x 2 = 2 x 2 + y 2 x - y 3 , x > 0 .

Given: x 2 + y 2 = e sin - 1 y x 2 + y 2 Take log both sides, we get 1 2 log x 2 + y 2 = sin - 1 y x 2 + y 2 ⇒ 1 2 log x 2 + y 2 = tan - 1 y x Differentiating both sides w.r.t. x , we get 2 x + y y ' 2 x 2 + y 2 = 1 1 + y 2 x 2 x y ' - y × 1 x 2 ⇒ x + y y ' x 2 + y 2 = x y ' - y x 2 + y 2 ⇒ x + y = y ' ( x - y ) ⇒ y ' = x + y x - y ⇒ d y d x = x + y x - y ⇒ d 2 y d x 2 = ( x - y ) 1 + y ' - ( x + y ) 1 - y ' ( x - y ) 2 ⇒ d 2 y d x 2 = x - y - x - y + y ' x - y + x + y ( x - y ) 2 ⇒ d 2 y d x 2 = 2 x y ' - 2 y ( x - y ) 2 ⇒ d 2 y d x 2 = 2 x x + y x - y - 2 y ( x - y ) 2 ⇒ d 2 y d x 2 = 2 x 2 + 2 x y - 2 x y + 2 y 2 ( x - y ) 3 ⇒ d 2 y d x 2 = 2 x 2 + y 2 ( x - y ) 3

If x = sec θ - cos θ and y = sec n θ - cos n θ , then show that x 2 + 4 d y d x 2 = n 2 y 2 + 4

Given: x = sec θ - cos θ . . . 1 and y = sec n θ - cos n θ . . . 2 Differentiate equation 1 w.r.t. θ , we get d x d θ = d d θ sec θ − cos θ ⇒ d x d θ = sec θ tan θ + sin θ ⇒ d x d θ = sin θ cos θ ( sec θ + cos θ ) ⇒ d x d θ = tan θ sec θ + cos θ . . . 3 Differentiate equation 2 w.r.t θ , we get d y d θ = n sec n − 1 θ ⋅ d d θ sec θ − n cos n − 1 θ ⋅ d d θ cos θ [ Using chain rule ] ⇒ d y d θ = n sec n − 1 θ ⋅ sec θ tan θ − n cos n − 1 θ ⋅ − sin θ ⇒ d y d θ = n sec n θ 1 sec θ ⋅ sec θ tan θ − n cos n θ 1 cos θ ⋅ − sin θ ⇒ d y d θ = n sec n θ tan θ + n cos n θ tan θ ⇒ d y d θ = n tan θ sec n θ + cos n θ . . . 4 ∴ d y d x = d y d θ d x d θ ⇒ d y d x = n tan θ sec n x + cos n θ tan θ sec θ + cos θ {from eq 3 and 4 } ⇒ d y d x 2 = n 2 sec n θ + cos n θ 2 ( sec θ + cos θ ) 2 ⇒ d y d x 2 = n 2 sec n θ − cos n θ 2 + 4 sec θ − cos θ 2 + 4 ∵ a + b 2 = a − b 2 + 4 a b ⇒ d y d x 2 = n 2 y 2 + 4 x 2 + 4 ⇒ x 2 + 4 d y d x 2 = n 2 y 2 + 4 Hence proved

If x = 1 z and y = f x , show that: d 2 f d x 2 = 2 z 3 d y d z + z 4 d 2 y d z 2

Given that x = 1 z and y = f x ⇒ d x d z = - 1 z 2 Now, d d x f x = d y d x = d y d z . d z d x = - z 2 d y d z ⇒ d 2 f d x 2 = d d x - z 2 d y d z = d d z - z 2 d y d z d z d x = - z 2 d d z - z 2 d y d z = z 2 2 z 2 - 1 d y d z + z 2 d 2 y d z 2 = 2 z 3 d y d z + z 4 d 2 y d z 2

Prove that if a 1 sin x + a 2 sin 2 x + . . . . + a n sin n x ≤ sin x for x ∈ R , then a 1 + 2 a 2 + 3 a 3 + . . . . + n a n ≤ 1

Given that a 1 sin x + a 2 sin 2 x + . . . . + a n sin n x ≤ sin x for x ∈ R Let f x = a 1 sin x + a 2 sin 2 x + . . . . + a n sin n x So, f x ≤ sin x ⇒ f 0 ≤ 0 And f ' x = a 1 cos x + 2 a 2 cos 2 x + 3 a 3 cos 3 x + . . . . + n a n cos n x We have to prove that f ' 0 ≤ 1 Now, f ' 0 = lim h → 0 f h - f 0 h ⇒ f ' 0 ≤ lim h → 0 f h h = lim h → 0 sin h h = 1

Show that the substitution z = ln tan x 2 changes the equation d 2 y d x 2 + cot x d y d x + 4 y cosec 2 x = 0 to d 2 y d z 2 + 4 y = 0 .

Given that z = ln tan x 2 ⇒ d z d x = 1 tan x 2 × 1 2 sec 2 x 2 = 1 2 sin x 2 cos x 2 = 1 sin x = cosec x And d 2 y d x 2 + cot x d y d x + 4 y cosec 2 x = 0 … 1 ⇒ d y d x = d y d z . d z d x = cosec x d y d z ⇒ d 2 y d x 2 = d d x d y d x = d d z d y d x . d z d x = cosec x d d z cosec x d y d z = cosec x cosec x d 2 y d z 2 + d y d z × d d x cosec x d x d z = cosec 2 x d 2 y d z 2 - cosec x . cot x d y d z From equation 1 , we get cosec 2 x d 2 y d z 2 - cosec x . cot x . d y d z + cot x cosec x d y d z + 4 y cosec 2 x = 0 ⇒ d 2 y d z 2 + 4 y = 0 Which is required equation.