Tags Ncert Solutions Maths class 12th

Ncert Solutions Maths class 12th

Get insights from 2.5k questions on Ncert Solutions Maths class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Ncert Solutions Maths class 12th

Follow Ask Question

2.5k

Questions

Discussions

Active Users

Followers

New answer posted

4 months ago

Ncert Solutions Maths class 12th Differential Equations

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

0 Follower 2 Views

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,

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Differential Equations

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

0 Follower 2 Views

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.

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Differential Equations

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

0 Follower 2 Views

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

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

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

0 0

New answer posted

4 months ago

Ncert Solutions Maths 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 $x \in [1, 2]$

0 Follower 2 Views

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 $x \in [1, 2]$ .

less

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Differential Equations

61. Kindly Consider the following

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

0 Follower 2 Views

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,

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

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Continuity and Differentiability

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

0 Follower 2 Views

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.

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Differential Equations

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

0 Follower 2 Views

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.

0 0

New answer posted

4 months ago

Ncert Solutions Maths 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

0 Follower 2 Views

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.

0 0

New answer posted

4 months ago

Ncert Solutions Maths 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.

0 Follower 74 Views

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,

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

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

0 0

New answer posted

4 months ago

Ncert Solutions Maths class 12th Differential Equations

57. In a bank, principal increases continuously at the rate of 5% per year. An amount of Rs 1000 is deposited with this bank, how much will it worth after 10 years

(e^0.5 = 1.645).

0 Follower 20 Views

Vishal Baghel

Contributor-Level 10

Let P and t the principal and time respectively.

Then, increase in principal $\frac{d P}{d t} = P * 5 %$

$\Rightarrow \frac{d P}{P} = \frac{5}{100} d t$

Integrating both sides,

$\begin{array}{l} \Rightarrow \int \frac{d P}{P} = \int \frac{1}{20} d t \\ \Rightarrow l o g P = \frac{t}{20} + c \\ \Rightarrow P = e^{\frac{t}{20} + c} \end{array}$

At, t=0, P=1000

So, $\begin{array}{l} 1000 = e^{\frac{0}{20} + c} \\ \Rightarrow e^{c} = 1000 \end{array}$

And at t=10,

$\begin{array}{l} P = e^{\frac{10}{100} + c} = e^{0.5} . e^{c} \\ \Rightarrow P = 1.648 * 1000 = 1648 \end{array}$

P = ?1648

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
687k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Ncert Solutions Maths class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience