Prove that the function f x = x 2 sin 2 1 x for x ≠ 0 0 for x = 0 has a minimum at the point x 0 = 0 (not a strict minimum).

f x = x 2 sin 2 1 x , x ≠ 0 0 , x = 0 For x ≠ 0 , f x = x sin 1 x 2 > 0 So, lim x → 0 + f x > f 0 & lim x → 0 - f x > f 0 ⇒ x 0 = 0 is point of minima. f ' x = 2 x sin 2 1 x + x 2 2 sin 1 x cos 1 x - 1 x 2 , x ≠ 0 ⇒ f ' x = 2 sin 1 x sin 1 x - x cos 1 x , x ≠ 0 For x = 1 n π , n ∈ I , f ' x = 0 Also, f 1 n π = 0 So, graph of f x will touch x -axis at the points x = 1 n π . Here, f x ≥ 0 So, possible graph's nature at x = 1 n π will be Function will have minima at more than one point, so x 0 = 0 is not a strict minima.

For what choice of the parameter h Does the "curve of probabilities" y = n π e - h 2 x 2 h > 0 Have points of inflection x = ± σ ?

Given equation is: y = n π e - h 2 x 2 h > 0 Now, the first derivative is ⇒ y = n π e - h 2 x 2 ⇒ y ' = n π e - h 2 x 2 - 2 h 2 x ⇒ y ' = - 2 h 2 x n π e - h 2 x 2 Now, the second derivative is ⇒ y ' ' = - n h 2 π e - h 2 x 2 . x . - 2 h 2 x ⇒ y ' ' = - n h 2 π e - h 2 x 2 . x . - 2 h 2 x - h 2 n π e - h 2 x 2 ⇒ y ' ' = 2 n h 4 x 2 π e - h 2 x 2 - h 2 n π e - h 2 x 2 Given point of inflection is x = ± σ Case 1 : when x = + σ ⇒ y '' σ = 0 ⇒ 2 n h 4 x 2 π e - h 2 x 2 - h 2 n π e - h 2 x 2 = 0 ⇒ 2 n h 4 σ 2 π e - h 2 σ 2 - h 2 n π e - h 2 σ 2 = 0 ⇒ 2 h 2 σ 2 e - h 2 σ 2 = e - h 2 σ 2 ⇒ h = ± 1 σ 2 ∴ h > 0 ∵ h = 1 σ 2 Case 2 : When x = - σ ⇒ y ' ' - σ = 0 ⇒ 2 n h 4 x 2 π e - h 2 x 2 - h 2 n π e - h 2 x 2 = 0 ⇒ 2 n h 4 - σ 2 π e - h 2 - σ 2 - h 2 n π e - h 2 - σ 2 = 0 ⇒ 2 n h 4 σ 2 π e - h 2 σ 2 - h 2 n π e - h 2 σ 2 = 0 ⇒ 2 h 2 σ 2 e - h 2 σ 2 = e - h 2 σ 2 ⇒ h = ± 1 σ 2 ∴ h > 0 ∵ h = 1 σ 2 Hence, the desired value of h is 1 σ 2 .

HARD

Mathematics

IMPORTANT

Earn 100

Prove that between two maxima (minima) of a continuous function there is a minimum (maximum) of this function.

Important Questions on Application of Differential Calculus to Investigation of Functions

HARD

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.27.

Prove that the function

f (x) = \{\begin{matrix} x^{2} \sin^{2} (\frac{1}{x}) & for x \neq 0 \\ 0 & for x = 0 \end{matrix}

has a minimum at the point

x_{0} = 0

(not a strict minimum).

HARD

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.28.

Prove that if at the point of a minimum there exists a right-side derivative, then it is non-negative, and if there exists a left-side derivative, then it is non-positive.

HARD

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.29.

Show that the function

$y = \{\begin{cases} \frac{1}{x^{2}} (x > 0) \\ 3 x^{2} (x \leq 0) \end{cases}$

has a minimum at the point $x = 0$ , though its first derivative does not change sign when passing through this point.

MEDIUM

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.30.

Let

x_{0}

be the abscissa of the point of inflection on the curve

y = f (x)

. Will the point

x_{0}

be a point of extremum for the function

y = f^{'} (x)

MEDIUM

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.31.

Sketch the function $y = f (x)$ in the neighbourhood of the point $x = - 1$ if

$f (- 1) = 2, f^{'} (- 1) = - 1, f^{''} (- 1) = 0, f^{'''} (x) > 0$ .

HARD

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.32.

For what choice of the parameter

h

Does the "curve of probabilities"

y = \frac{n}{\sqrt{π}} e^{- h^{2} x^{2}} (h > 0)

Have points of inflection

x = \pm σ

MEDIUM

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.33.

Show that any twice continuously differentiable function has at least one abscissa of the point of inflection on the graph of the function between two points of extremum.

HARD

Mathematics

IMPORTANT

PROBLEMS IN CALCULUS OF ONE VARIABLE>Chapter 3 - Application of Differential Calculus to Investigation of Functions>Additional Problems>Q 3.13.34.

Taking the function

y = x^{4} + 8 x^{3} + 18 x^{2} + 8

as an example, ascertain that there any be no points of extremum between the abscissas of the points of inflection on the graph of a function.

Prove that between two maxima (minima) of a continuous function there is a minimum (maximum) of this function.

Important Questions on Application of Differential Calculus to Investigation of Functions

Learn and Prepare for any exam you want !