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Class 12th Continuity and Differentiability

114. If f:[-5,5] →R is a differentiable function and if f'(x) does not vanish anywhere, then prove that f(-5) ≠ f(5).

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alok kumar singh

Contributor-Level 10

114. Solution :
It is given that f: [-5,5]? R is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) f is continuous on [?5, 5].

(b) f is differentiable on (?5, 5).

Therefore, by the Mean Value Theorem, there exists c? (?5, 5) such that

$\begin{array}{l} f^{'} (c) \frac{f (5) ? f (? 5)}{5 ? (? 5)} \\ ? 10 f^{'} (c) = f (5) ? f (? 5) \end{array}$

It is also given that f' (x) does not vanish anywhere.

$\begin{array}{l} ? f^{'} (c) ? 0 \\ ? 10 f^{'} (c) ? 0 \\ ? f (5) ? f (? 5) ? 0 \\ ? f (5) ? f (? 5) \end{array}$

Hence, proved.

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8 months ago

Class 12th Differential Equations

64. $x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + x y$

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Vishal Baghel

Contributor-Level 10

The given D.E. is

$\begin{array}{l} x^{2} \frac{d y}{d x} = x^{2} - 2 y^{2} + + x y \\ \Rightarrow \frac{d y}{d x} = \frac{x^{2} - 2 y^{2} + + x y}{x^{2}} = 1 - \frac{2 y^{2}}{x^{2}} + \frac{y}{x} \\ \Rightarrow \frac{d y}{d x} = 1 - 2 {(\frac{y}{x})}^{2} + \frac{y}{x} = f (\frac{y}{x}) \end{array}$

Hence, the D.E. is homogenous $f x^{n}$

Let, $y = v x . = \frac{y}{x} = v$ so that, $\frac{d y}{d x} = v + x \frac{d v}{d x}$ is the D.E.

Thus, $v + x \frac{d v}{d x} = 1 - 2 v^{2} + v$

$\begin{array}{l} \Rightarrow x \frac{d v}{d x} = 1 - 2 v^{2} \\ \Rightarrow \frac{d v}{1 - 2 v^{2}} = \frac{d x}{x} \end{array}$

Integrating both sides,

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Class 12th Differential Equations

63. $(x^{2} - y^{2}) d x + 2 x y d y = 0$

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Vishal Baghel

Contributor-Level 10

The Given D.E. is

$\begin{array}{l} (x^{2} - y^{2}) d x + 2 x y d y = 0 \\ \Rightarrow 2 * y d y = - (x^{2} - y^{2}) d x \\ = \frac{d y}{d x} = - \frac{(x^{2} - y^{2})}{2 x y} = \frac{(y^{2} - x^{2})}{2 x y} \\ = \frac{1}{2} \frac{(\frac{y^{2} - x^{2}}{x^{2}})}{(\frac{x y}{x^{2}})} \\ = \frac{1}{2} \frac{[{(\frac{y}{x})}^{2} - 1]}{(\frac{y}{x})} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E. is homogenous.

Let, $y = v x \Rightarrow \frac{y}{x} = v, s o t h a t, \frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E

$\begin{array}{l} \Rightarrow v + x \frac{d v}{d x} = \frac{1}{2} [\frac{v^{2} - 1}{v}] \\ \Rightarrow x \frac{d v}{d x} = \frac{v^{2} - 1}{2 v} - v = \frac{v^{2} - 1 - 2 v^{2}}{2 v} = \frac{- 1 - v^{2}}{2 v} \\ \Rightarrow (\frac{2 v}{1 + v^{2}}) d v = - \frac{d x}{x} \end{array}$

Integrating both sides we get,

Putting back $v = \frac{y}{x}$ we get,

$\begin{array}{l} \Rightarrow x [\frac{y^{2}}{x^{2}} + 1] = c \\ \Rightarrow \frac{y^{2} + x^{2}}{x} = c \end{array}$

$\Rightarrow y^{2} + x^{2} = x c$ is the required solution.

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8 months ago

Class 12th Differential Equations

62. $(x - y) d y - (x + y) d x = 0$

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Vishal Baghel

Contributor-Level 10

The Given D.E. is $(x - y) d y - (x + y) d x = 0$

$\begin{array}{l} \Rightarrow (x - y) d y = (x + y) d x \\ \Rightarrow \frac{d y}{d x} = \frac{x + y}{x - y} = \frac{x (1 + \frac{y}{x})}{x (1 - \frac{y}{x})} = \frac{1 + \frac{y}{x}}{1 - \frac{y}{x}} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E. is homogenous.

Let, $y = v x = \frac{y}{x} = v, s o, \frac{d y}{d x} = v + x \frac{d v}{d x}$ in the D.E

Then, $\begin{array}{l} v + x \frac{d v}{d x} = \frac{1 + v}{1 - v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 + v}{1 - v} - v = \frac{1 + v - 1 + v^{2}}{1 - v} = \frac{1 + v^{2}}{1 - v} \\ \Rightarrow [\frac{1 - v}{1 + v^{2}}] d v = \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{1 - v}{1 + v^{2}} d v = \int \frac{d x}{x} \\ \Rightarrow \int \frac{1}{1 + v^{2}} d v - \frac{1}{2} \int \frac{2 v}{1 + v^{2}} d v = l o g x + c \\ \Rightarrow t a n^{- 1} v - \frac{1}{2} l o g (1 + v^{2}) = l o g x + c \end{array}$

Putting back $v = \frac{y}{x}, w e g e t,$

$\begin{array}{l} = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g | 1 + \frac{y^{2}}{x^{2}} | = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g | \frac{x^{2} + y^{2}}{x^{2}} | = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} [l o g (x^{2} + y^{2}) - l o g x^{2}] = l o g x + c \end{array}$

$\begin{array}{l} = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g (x^{2} + y^{2}) + \frac{1}{2} l o g x^{2} = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g (x^{2} + y^{2}) + l o g {(x^{2})}^{\frac{1}{2}} = l o g x + c \\ = t a n^{- 1} \frac{y}{x} - \frac{1}{2} l o g (x^{2} + y^{2}) + l o g x = l o g + c \\ = t a n^{- 1} \frac{y}{x} = \frac{1}{2} l o g (x^{2} + y^{2}) + c \end{array}$

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8 months ago

Class 12th Continuity and Differentiability

113. Examine if Rolle's Theorem is applicable to any of the following functions. Can you say something about the converse of Rolle's Theorem from these examples?

$f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

$f (x) = x^{2} - 1$ for $x \in [1, 2]$

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alok kumar singh

Contributor-Level 10

113. Solution:

By Rolle's Theorem, for a function $f [a, b] \to R,$ if

f is continuous on $[a, b]$

f is differentiable on $(a, b)$

f(a)= f(b)

then, there exists some $c \in (a, b)$ such that $f^{'} (c) = 0$

therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f(x) is not continuous at every integral point.

In particular, f(x) is not continuous at x=5 and x=9

$\Rightarrow$ f(x) is not continuous in $[5, 9]$

Also, $\begin{array}{l} f (5) = [5] = 5, a n d, f (9) = [9] = 9 \\ ∴ f (5) \neq f (9) \end{array}$

The differentiability of f in $(5, 9)$ is checked as follows.

Let n be an integer such that $n \in (5, 9)$ .

The left hand limit of f at x

...more

113. Solution:

By Rolle's Theorem, for a function $f [a, b] \to R,$ if

f is continuous on $[a, b]$

f is differentiable on $(a, b)$

f(a)= f(b)

then, there exists some $c \in (a, b)$ such that $f^{'} (c) = 0$

therefore, Rolle's Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

$f (x) = [x]$ for $x \in [5, 9]$

It is evident that the given function f(x) is not continuous at every integral point.

In particular, f(x) is not continuous at x=5 and x=9

$\Rightarrow$ f(x) is not continuous in $[5, 9]$

Also, $\begin{array}{l} f (5) = [5] = 5, a n d, f (9) = [9] = 9 \\ ∴ f (5) \neq f (9) \end{array}$

The differentiability of f in $(5, 9)$ is checked as follows.

Let n be an integer such that $n \in (5, 9)$ .

The left hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - 1 - n}{h} = \underset{h \to 0}{l i m} \frac{- 1}{h} = \infty$

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left and right hand limits of f at x=n are not equal, f is not differentiable at x=n

$∴$ f is not differentiable in (5,9).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for $f (x) = [x]$ for $x \in [5, 9]$

$f (x) = [x]$ for $x \in [- 2, 2]$

It is evident that the given function f(x) is not continuous at every integral point.

In particular, f(x) is not continuous at x=-2 and x=2

$\Rightarrow$ f(x) is not continuous in $[- 2, 2]$ .

Also, $\begin{array}{l} f (- 2) = [- 2] = - 2, a n d, f (2) = [2] = 2 \\ ∴ f (- 2) \neq f (2) \end{array}$

The differentiability of f in $(- 2, 2)$ is checked as follows.

Let n be an integer such that $n \in (- 2, 2)$ .

The left hand limit of f at x=n is,

The right hand limit of f at x=n is,

$\underset{h \to 0}{l i m} \frac{f (n + h) - f (n)}{h} = \underset{h \to 0}{l i m} \frac{[n + h] - [n]}{h} = \underset{h \to 0}{l i m} \frac{n - n}{h} = \underset{h \to 0}{l i m} 0 = 0$

Since the left- and right-hand limits of f at x=n are not equal, f is not differentiable at x=n

$∴$ f is not differentiable in (-2,2).

It is observed that f does not satisfy all the conditions of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for $f (x) = [x]$ for $x \in [- 2, 2]$

$f (x) = x^{2} - 1$ for $x \in [1, 2]$

It is evident that f, being a polynomial function, is continuous in $[1, 2]$ and is differentiable on (1,2).

$\begin{array}{l} f (1) = 1^{2} = 0 \\ f (2) = 2^{2} - 1 = 3 \\ ∴ f (1) \neq f (2) \end{array}$

It is observed that f does not satisfy a condition of the hypothesis of Rolle's Theorem.

Hence, Rolle's Theorem is not applicable for $f (x) = x^{2} - 1$ for $x \in [1, 2]$ .

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New answer posted

8 months ago

Class 12th Differential Equations

61. Kindly Consider the following

$y' = \frac{x + y}{x}$

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Vishal Baghel

Contributor-Level 10

The given D.E. is

$\begin{array}{l} y^{1} = \frac{x + y}{x} \\ = \frac{d y}{d x} = 1 + \frac{y}{x} = f (\frac{y}{x}) \end{array}$

Hence, the given D.E is homogenous.

Let, $y = v x \to \frac{y}{x} = v, s o t h a t, \frac{d y}{d x} = v + \frac{x d v}{d x}$

So, the D.E. becomes

$\begin{array}{l} v + \frac{x d v}{d x} = 1 + v \\ \Rightarrow \frac{x d v}{d x} = 1 \\ \Rightarrow d v = \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int d v = \int \frac{d x}{x} \\ v = l o g | x | + c \end{array}$

Putting $v = \frac{y}{x}$ back we get,

$\frac{y}{x} = l o g | x | + c$

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Class 12th Continuity and Differentiability

112. Verify Rolle's Theorem for the function $f (x) = x^{2} + 2 x - 8, x \in [- 4, 2]$

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alok kumar singh

Contributor-Level 10

112. Given, $f (x) = x^{2} + 2 x - 8$ , being polynomial function is continuous in $[- 4, 2]$ and also differentiable in $(- 4, 2)$ .

$\begin{array}{l} f (- 4) = {(- 4)}^{2} + 2 (- 4) - 8 \\ = 16 - 8 - 8 \\ = 0 \\ f (2) = {(2)}^{2} + 2 * 2 - 8 \\ = 4 + 4 - 8 \\ = 0 \end{array}$

Therefore, $f (- 4) = f (2) = 0$

The value of $f (x)$ at -4 and 2 coincides.

Rolle's Theorem states that there is a point $c \in (- 4, 2)$ such that $f^{'} (c) = 0$ $f (x) = x^{2} + 2 x - 8$

Therefore, $f^{'} (x) = 2 x + 2$

Hence,

$\begin{array}{l} f^{'} (c) = 0 \\ 2 c + 2 = 0 \\ c = - 1 \end{array}$

Thus, $c = - 1 \in (- 4, 2)$

Hence, Rolle's Theorem is verified.

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8 months ago

Class 12th Differential Equations

60. $(x^{2} + x y) d y = (x^{2} + y^{2}) d x$

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Vishal Baghel

Contributor-Level 10

The given D.E. is

$\begin{array}{l} \frac{d y}{d x} = \frac{x^{2} + y^{2}}{x^{2} + x + y} = \frac{x^{2} (1 + \frac{y^{2}}{x^{2}})}{x^{2} (1 + \frac{y}{x})} = F (x, y) \\ T h e n, F (λ x, λ y) = \frac{λ^{2} x^{2}}{λ^{2} x^{2}} [\frac{1 + \frac{λ^{2} y^{2}}{λ^{2} x^{2}}}{1 + \frac{λ y}{λ x}}] = λ^{0} F (x, y) \\ = λ^{2} F (x, y) \end{array}$

Hence, $F (x, y)$ is a homogenous $f x^{n}$ of degree 2.

To solve it, let

$\begin{array}{l} y = v x, s o t h a t, \frac{d y}{d x} = v . \frac{d x}{d x} + x \frac{d v}{d x} \\ v = \frac{y}{x} \Rightarrow \frac{d y}{d x} = v + x \frac{d v}{d x} \end{array}$

The D.E. now becomes,

$\begin{array}{l} v + x \frac{d v}{d x} = \frac{1 + v^{2}}{1 + v} \\ \Rightarrow x \frac{d v}{d x} = \frac{1 + v^{2}}{1 + v} - v = \frac{1 + v^{2} - v - v^{2}}{1 + v} = \frac{1 - v}{1 + v} \\ \Rightarrow (\frac{1 + v}{1 - v}) d v = \frac{d x}{x} \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{1 + v}{1 - v} d v = \int \frac{d x}{x} \\ \Rightarrow \int \frac{1 + v}{1 - v} d v = l o g x + c \\ \Rightarrow \int \frac{2 - (1 - v)}{1 - v} d v = l o g x + c \\ \Rightarrow \int \frac{2}{1 - v} d v - \int 1 d v = l o g x + c \\ \Rightarrow \frac{2 l o g | 1 - v |}{- 1} - v = l o g x + c \\ \Rightarrow l o g {(1 - v)}^{2} + v = - l o g x - c \end{array}$

Put $v = \frac{y}{x},$

$\begin{array}{l} l o g {(1 - \frac{y}{x})}^{2} + \frac{y}{2} = - l o g x + c \\ \Rightarrow l o g {(\frac{x - y}{x})}^{2} + l o g x = - \frac{y}{x} - c \end{array}$

$\begin{array}{l} \Rightarrow l o g [{(\frac{x - y}{x})}^{2} * x] = - \frac{y}{x} - c \\ \Rightarrow \frac{x - y}{x} = e^{- \frac{y}{x} - c} = c_{1} e^{- \frac{y}{x} - c}, w h e r e, c_{1} e \end{array}$

$\Rightarrow {(x - y)}^{2} = c_{1} . x e^{- \frac{y}{x}}$ is the required solution of the D.E.

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Class 12th Differential Equations

59. The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is:

(A) e^x + e^-y = C

(B) e^x + e^y = C

(C) e^-x + e^y = C

(D) e^x + e^-y = C

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Vishal Baghel

Contributor-Level 10

The given D.E. is

$\begin{array}{l} \frac{d y}{d x} = e^{x + y} \\ \Rightarrow \frac{d y}{d x} = e^{x} . e^{y} \\ \Rightarrow \frac{d y}{e^{y}} = e^{x} d x \end{array}$

Integrating both sides,

$\begin{array}{l} \int \frac{d y}{e^{y}} = \int e^{x} d x \\ \Rightarrow \frac{e^{- y}}{- 1} = e^{x} + c_{1} \\ \Rightarrow e^{- y} = - e^{x} - c_{1} \\ \Rightarrow e^{- y} + e^{x} = c, w h e r e, c = - c_{1} \end{array}$

$∴$ Option (A) is correct.

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8 months ago

Class 12th Differential Equations

58. In a culture the bacteria count is 1,00,000. The number is increased by 10% in 2 hours. In how many hours will the count reach 2,00,000, if the rate of growth of bacteria is proportional to the number present.

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Vishal Baghel

Contributor-Level 10

Let 'x' be the number of bacteria present in instantaneous time t.

Then, $\frac{d x}{d t} \propto x$

$\Rightarrow \frac{d x}{d t} = k x, w h e r e, k =$ constant of proportionality.

$\Rightarrow \frac{d x}{x} = k d t$

Integrating both sides,

$\begin{array}{l} \int \frac{d x}{x} = \int k d t \\ \Rightarrow l o g x = k t + c \end{array}$

Given, at $t = 0, x = x_{0} (s a y) t h e n,$

$l o g x_{0} = c (I n i t i a l, x_{0} = 100000)$

So, the differential equation is

$\begin{array}{l} l o g x = k t + l o g x_{0} \\ \Rightarrow l o g x - l o g x_{0} = k t \\ \Rightarrow l o g \frac{x}{x_{0}} = k t \end{array}$

As the bacteria number increased by 10% in 2 hours.

The number of bacteria increased in 2hours $= 10 % * 100000 = 10000$

Hence, at t=2,

$x = 100000 + 10000 = 110000$

So, $\begin{array}{l} l o g \frac{110000}{100000} = 2 k \\ \Rightarrow k = \frac{1}{2} l o g (\frac{11}{10}) \end{array}$

Hence, $l o g \frac{x}{x_{0}} = [\frac{1}{2} l o g \frac{11}{10}] * t$

$w h e n, x = 200000,$ then we get,

$l o g \frac{200000}{100000} = \frac{1}{2} l o g \frac{11}{10} * t$

$\begin{array}{l} \Rightarrow 2 l o g 2 = l o g (\frac{11}{10}) * t \\ \Rightarrow t = \frac{2 l o g 2}{l o g (\frac{11}{10})} h o u r s \end{array}$

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