Find the interval to which b may belong so that the function f x = 1 - 21 - 4 b - b 2 b + 1 x 3 + 5 x + 6 is increasing at every point of its domain.

Given, f x = 1 - 21 - 4 b - b 2 b + 1 x 3 + 5 x + 6 ⇒ f ' x = 1 - 21 - 4 b - b 2 b + 1 3 x 2 + 5 Given that the function f ( x ) is increasing for all real values of x . ∴ f ' x ≥ 0 ⇒ 3 1 - 21 - 4 b - b 2 b + 1 x 2 + 5 ≥ 0 ⇒ 1 - 21 - 4 b - b 2 b + 1 ≥ 0 ⇒ 1 ≥ 21 - 4 b - b 2 b + 1 ⇒ 1 2 ≥ 21 - 4 b - b 2 b + 1 2 ⇒ b + 1 2 - 21 + 4 b + b 2 b + 1 2 ≥ 0 ⇒ b 2 + 3 b - 10 b + 1 2 ≥ 0 ⇒ b + 5 b - 2 b + 1 2 ≥ 0 , here b ≠ - 1 b ∈ [ - 7 , - 1 ) ∪ [ 2 , 3 ]

Find the interval to which b may belong so that the function f x = 1 - 21 - 4 b - b 2 b + 1 x 3 + 5 x + 6 is increasing at every point of its domain.

Given, f x = 1 - 21 - 4 b - b 2 b + 1 x 3 + 5 x + 6 ⇒ f ' x = 1 - 21 - 4 b - b 2 b + 1 3 x 2 + 5 Given that the function f ( x ) is increasing for all real values of x . ∴ f ' x ≥ 0 ⇒ 3 1 - 21 - 4 b - b 2 b + 1 x 2 + 5 ≥ 0 ⇒ 1 - 21 - 4 b - b 2 b + 1 ≥ 0 ⇒ 1 ≥ 21 - 4 b - b 2 b + 1 ⇒ 1 2 ≥ 21 - 4 b - b 2 b + 1 2 ⇒ b + 1 2 - 21 + 4 b + b 2 b + 1 2 ≥ 0 ⇒ b 2 + 3 b - 10 b + 1 2 ≥ 0 ⇒ b + 5 b - 2 b + 1 2 ≥ 0 , here b ≠ - 1 b ∈ [ - 7 , - 1 ) ∪ [ 2 , 3 ]

If 0 < x < 1 prove that y = x ln x - x 2 2 + 1 2 is a function such that d 2 y d x 2 > 0 . Deduce that x ln x > x 2 2 - 1 2 .

Given: y = x ln x - x 2 2 + 1 2 Differentiating above equation w.r.t. x , we get d y d x = x d d x ln x + ln x d x d x - 2 x 2 + 0 ⇒ d y d x = x 1 x + ln x - x ⇒ d y d x = 1 + ln x - x . . . i Again differentiating the above equation w.r.t. x , we get d 2 y d x 2 = 1 x - 1 ∵ 0 < x < 1 ⇒ ∞ > 1 x > 1 ⇒ ∞ > 1 x - 1 > 0 ⇒ ∞ > d 2 y d x 2 > 0 ∴ d 2 y d x 2 > 0 Now, we will prove for x ∈ 0 , 1 , x ln x > x 2 2 - 1 2 x ln x > x 2 2 - 1 2 ⇒ x ln x - x 2 2 + 1 2 > 0 ⇒ y > 0 So we will prove that for x ∈ 0 , 1 , y > 0 From equation i , d y d x = 1 + ln x - x ⇒ d y d x = ln x + 1 - x So from the graph, we can see that for x ∈ 0 , 1 ln x ∈ - ∞ , 0 and 1 - x ∈ 0 , 1 So ln x is more dominating than 1 - x ⇒ d y d x < 0 ⇒ y = x ln x - x 2 2 + 1 2 is a decreasing function in x ∈ 0 , 1 So we can say that in the interval x ∈ 0 , 1 , the function has its minimum value at x = 1 and maximum value at x = 0 So at x = 1 , y min = 1 ln 1 - 1 2 + 1 2 ⇒ y min = 0 ⇒ y > 0 , x ∈ 0 , 1 ⇒ x ln x - x 2 2 + 1 2 > 0 ⇒ x ln x > x 2 2 - 1 2 Hence proved.

For any acute angled Δ A B C , find the maximum value of sin A A + sin B B + sin C C .

Let f x = sin x x f ' x = x cos x - sin x x 2 = cos x x - tan x x 2 < 0 , ∀ x ∈ 0 , π 2 ; ∵ tan x > x f " x = - x 2 sin x - 2 x cos x + 2 sin x x 4 Let g x = - x 2 sin x - 2 x cos x + 2 sin x g ' x = - x 2 cos x < 0 , ∀ x ∈ 0 , π 2 So, g x will be a decreasing function. For x > 0 , we have g x < g 0 i.e., g x < 0 . ∴ f ' x < 0 and f " x < 0 ∀ ∈ 0 , π 2 . So, graph of function f x will be concave downwards. ⇒ f A + B + C 3 > f A + f B + f C 3 ⇒ sin A + B + C 3 A + B + C 3 ≥ sin A A + sin B B + sin C C 3 ⇒ sin π 3 π 3 ≥ sin A A + sin B B + sin C C 3 ⇒ 3 3 2 π ≥ sin A A + sin B B + sin C C 3 ⇒ sin A A + sin B B + sin C C ≤ 9 3 2 π

For what real values of ' a ' and ' b ' all the extrema of the function f x = 5 a 2 3 x 3 + 2 a x 2 - 9 x + b are positive and the maximum is at the point x 0 = - 5 9

Given, f x = 5 a 2 3 x 3 + 2 a x 2 - 9 x + b ⇒ f ' x = 5 a 2 x 2 + 4 a x - 9 ⇒ f ' x = 5 a 2 x 2 + 9 a x - 5 a x - 9 ⇒ f ' x = a x - 1 5 a x + 9 ⇒ f ' x = 0 at x = 1 a & x = - 9 5 a Given that the function is positive at the point of extrema x = - 5 9 . Case No i : When x = 1 a = - 5 9 ⇒ a = - 9 5 ⇒ f - 5 9 = 5 3 - 9 5 2 - 5 9 3 + 2 - 9 5 - 5 9 2 - 9 - 5 9 + b > 0 ⇒ - 25 27 - 10 9 + 5 + b > 0 ⇒ b > - 80 27 Case No i i : When x = - 9 5 a = - 5 9 ⇒ a = 81 25 ⇒ f - 5 9 = 5 3 81 25 2 - 5 9 3 + 2 81 25 - 5 9 2 - 9 - 5 9 + b > 0 ⇒ f - 5 9 = - 3 + 2 + 5 + b > 0 ⇒ b > - 4

Find the minimum value of f ( x ) = 8 x + 8 - x - 4 4 x + 4 - x ∀ x ∈ R

Given, f ( x ) = 8 x + 8 - x - 4 4 x + 4 - x ∀ x ∈ R ⇒ f ' ( x ) = 8 x ln 8 - 8 - x ln 8 - 4 4 x ln 4 - 4 - x ln 4 ⇒ f ' ( x ) = 2 x 3 ln 2 3 - 1 2 x 3 ln 2 3 - 4 2 x 2 ln 2 2 - 1 2 x 2 ln 2 2 ⇒ f ' ( x ) = 3 2 x 3 - 1 2 x 3 - 8 2 x 2 - 1 2 x 2 ln 2 ⇒ f ' ( x ) = 3 2 x - 1 2 x 2 x 2 + 1 2 x 2 + 1 - 8 2 x - 1 2 x 2 x + 1 2 x ln 2 ⇒ f ' ( x ) = 3 2 x 2 + 1 2 x 2 + 2 - 1 - 8 2 x + 1 2 x 2 x - 1 2 x ln 2 ⇒ f ' ( x ) = 3 2 x + 1 2 x 2 - 8 2 x + 1 2 x - 3 2 x - 1 2 x + 1 2 x ln 2 ⇒ f ' ( x ) = 3 2 x + 1 2 x + 1 2 x + 1 2 x - 3 2 x - 1 2 x + 1 2 x ln 2 ⇒ f ' ( x ) = 3 2 x + 1 2 x + 1 2 x 2 - 3 2 x + 1 2 x - 1 2 x + 1 2 2 x ln 2 ⇒ f ' ( x ) = 3 2 x + 1 2 x + 1 2 x + 1 2 2 x ln 2 2 x - 3 - 5 2 2 x - 3 + 5 2 2 x - 1 Here, f ' x changes its sign from negative to positive about x = 0 , log 2 3 + 5 2 i.e. the given function has two points of local minima at x = 0 & x = log 2 3 + 5 2 . ⇒ f ( 0 ) = 8 0 + 8 - 0 - 4 4 0 + 4 - 0 & f ln 3 + 5 2 = 8 log 2 3 + 5 2 + 8 - log 2 3 + 5 2 - 4 4 log 2 3 + 5 2 + 4 - log 2 3 + 5 2 ⇒ f ( 0 ) = - 6 & f ln 3 + 5 2 = 3 + 5 2 3 + 3 - 5 2 3 - 4 3 + 5 2 2 + 3 - 5 2 2 ⇒ f ( 0 ) = - 6 & f ln 3 + 5 2 = 3 + 5 2 + 3 - 5 2 9 + 5 + 6 5 4 + 9 + 5 + 6 5 4 - 1 - 4 9 + 5 + 6 5 4 + 9 + 5 + 6 5 4 ⇒ f ( 0 ) = - 6 & f ln 3 + 5 2 = - 10 Hence, the minimum value of the given function is equal to - 10 .

E M B I B E

Alpha Question Bank for Engineering: Mathematics>Chapter 28 - Application of Derivatives>Exercise-4>Q 20

MEDIUM

JEE Main/Advance

IMPORTANT

Earn 100

Find the interval to which $b$ may belong so that the function $f (x) = (1 - \frac{\sqrt{21 - 4 b - b^{2}}}{b + 1}) x^{3} + 5 x + \sqrt{6}$ is increasing at every point of its domain.

Important Questions on Application of Derivatives

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 28 - Application of Derivatives>Exercise-4>Q 21

If

0 < x < 1

prove that

y = x \ln x - \frac{x^{2}}{2} + \frac{1}{2}

is a function such that

\frac{d^{2} y}{d x^{2}} > 0 .

Deduce that

x \ln x > \frac{x^{2}}{2} - \frac{1}{2} .

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 28 - Application of Derivatives>Exercise-4>Q 22

Find positive real numbers

' a'

and

' b'

such that

f (x) = a x - b x^{3}

has four extrema on

[- 1, 1]

at each of which

| f (x) | = 1

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 28 - Application of Derivatives>Exercise-4>Q 23

For any acute angled

Δ A B C

, find the maximum value of

\frac{\sin A}{A} + \frac{\sin B}{B} + \frac{\sin C}{C}

.

HARD

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 28 - Application of Derivatives>Exercise-4>Q 24

Suppose

p, q, r and s

are fixed real numbers such that a quadrilateral can be formed with sides

p, q, r and s

in clockwise order. Prove that the vertices of the quadrilateral of maximum area lie on a circle.

MEDIUM

JEE Main/Advance

IMPORTANT

Alpha Question Bank for Engineering: Mathematics>Chapter 28 - Application of Derivatives>Exercise-4>Q 25

For what real values of $' a'$ and $' b'$ all the extrema of the function $f (x) = \frac{5 a^{2}}{3} x^{3} + 2 a x^{2} - 9 x + b$ are positive and the maximum is at the point $x_{0} = \frac{- 5}{9}$