Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Class 12th

Find gof and fog , if:

(i) f(x) = |x| and g(x) = |5x - 2|

(ii) f(x) = 8x³ and g(x) = x^1/3

10 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Kindly go through the solution

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New answer posted•a year ago

Ncert Solutions Maths class 12th Limits and Derivatives

17. Kindly consider the following

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

17. Kindly go through the solution

$=$ $= \underset{x \to 0}{l i m} \frac{2 s i n^{2} x}{2 s i n^{2} \frac{x}{2}}$

$= \frac{\underset{x \to 0}{l i m} {(\frac{s i n x}{x})}^{2} \times x^{2}}{\underset{x \to 0}{l i m} {(\frac{s i n^{} \frac{x}{2}}{\frac{x}{2}})}^{2} {\frac{x}{2}}^{2}}$

$= \frac{{(1)}^{2} \times x^{2} \times 4}{{(1)}^{2} \times x^{2}}$

= 4

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths Integrals

Kindly consider the following

$\int e^{- 3 x} c o s^{3} x d x$

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t ? ? ? ? I = ? \underset{I I}{e^{? 3 x}} \underset{I}{c o s^{3} x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . ? e^{? 3 x} ? d x ? ? (D (c o s^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = c o s^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 c o s^{2} x (? s i n x) . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? c o s^{2} x s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? (1 ? s i n^{2} x) s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + ? \underset{I}{s i n^{3} x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x ? e^{? 3 x} d x ? ? (D (s i n^{3} x) . ? e^{? 3 x} ? d x) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x + s i n^{3} x . \frac{e^{? 3 x}}{? 3} ? ? (3 s i n^{2} x . c o s x . \frac{e^{? 3 x}}{? 3}) d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? s i n^{2} x c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? ? s i n x . e^{? 3 x} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? (1 ? c o s^{2} x) c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{3} e^{? 3 x} c o s^{3} x ? [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? ? c o s^{3} x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{3} e^{? 3 x} c o s^{3} x + s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x ? \frac{1}{3} e^{? 3 x} s i n^{3} x + ? c o s x . e^{? 3 x} d x ? I \\ ? ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] ? [s i n x . \frac{e^{? 3 x}}{3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{3} ? c o s x . e^{? 3 x} d x + ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? 2 I = \frac{e^{? 3 x}}{? 3} [c o s^{3} x + s i n^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} + \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ N o w, ? I_{1} = \frac{2}{3} ? \underset{I}{c o s x} . \underset{I I}{e^{? 3 x}} d x \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . ? e^{? 3 x} d x ? ? (D (c o s x) . ? e^{? 3 x} ? d x) d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? ? ? s i n x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{2}{3} [c o s x . \frac{e^{? 3 x}}{? 3} ? \frac{1}{3} ? s i n x . e^{? 3 x} d x] \end{array}$

$\begin{array}{l} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} ? s i n x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = \frac{2}{? 9} c o s x . e^{? 3 x} ? \frac{2}{9} [s i n x . \frac{e^{? 3 x}}{? 3} ? ? c o s x . \frac{e^{? 3 x}}{? 3} d x] \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{2}{27} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . \frac{2}{3} ? c o s x . e^{? 3 x} d x \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} ? \frac{1}{9} . I_{1} \\ I_{1} + \frac{1}{9} I_{1} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? \frac{10 I_{1}}{9} = ? \frac{2}{9} c o s x . e^{? 3 x} + \frac{2}{27} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? I_{1} = ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ S o, ? ? ? ? ? 2 I = ? \frac{1}{3} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{3} s i n x . e^{? 3 x} ? \frac{1}{10} c o s x . e^{? 3 x} + \frac{1}{15} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? I = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{6} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} + \frac{1}{30} s i n x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = ? \frac{1}{6} e^{? 3 x} [s i n^{3} x + c o s^{3} x] + \frac{1}{5} s i n x . e^{? 3 x} ? \frac{1}{20} c o s x . e^{? 3 x} \\ ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? = \frac{e^{? 3 x}}{24} [s i n 3 x ? c o s 3 x] + \frac{3 e^{? 3 x}}{40} [s i n x ...Read more \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths Integrals

Kindly consider the following

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Maths Integrals

Kindly consider the following

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Maths Integrals

$\int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} L e t I = \int e^{t a n^{- 1} x} (\frac{1 + x + x^{2}}{1 + x^{2}}) d x \\ P u t t a n^{- 1} x = t \Rightarrow \frac{1}{1 + x^{2}} . d x = d t \\ = \int e^{t} (1 + t a n t + t a n^{2} t) d t = \int e^{t} (s e c^{2} t + t a n t) d t \\ H e r e, f (t) = t a n t \\ ∴ f^{'} (t) = s e c^{2} t \\ = e^{t} . f (t) = e^{t} t a n t = e^{t a n^{- 1} x} . x + C \\ [? \int e^{t} [f (x) + f^{'} (x)] d x = e^{x} f (x) + C] \\ H e n c e, I = e^{t a n^{- 1} x} . x + C \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Maths Integrals

$\int \frac{2 x - 1}{(x - 1) (x - 2) (x - 3)} d x$

1 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Maths Integrals

Kindly consider the following

$\int^{} \frac{x^{2} d x}{x^{4} - x^{2} - 12}$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Long Answer Type Questions as classified in NCERT Exemplar

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The solution of $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{1 + x^{2}}$ is:

(A) $y (1 + x^{2}) = c +tan^{- 1} x$

(B) $\frac{y}{1 + x^{2}} = c +tan^{- 1} x$

(C) $ylog (1 + x^{2}) = c +tan^{- 1} x$

(D) $y (1 + x^{2}) = c +sin^{- 1} x$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ \frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{{(1 + x^{2})}^{2}} \\ since i t i s l i n e a r d i f f e r e n t i a l e q u a t i o n w h e r e P = \frac{2 x y}{1 + x^{2}} a n d Q = \frac{1}{{(1 + x^{2})}^{2}} \\ ∴ I . F . = e^{\int P . d x} = e^{\int \frac{2 x y}{1 + x^{2}} . d x} = e^{l o g (1 + x^{2})} = (1 + x^{2}) \\ S o, t h e s o l u t i o n i s \\ y \times I . F . = \int Q \times I . F . d x + c \\ \Rightarrow y (1 + x^{2}) = \int \frac{1}{{(1 + x^{2})}^{2}} \times (1 + x^{2}) d x + c \\ \Rightarrow y (1 + x^{2}) = \int \frac{1}{(1 + x^{2})} . d x + c \\ \Rightarrow y (1 + x^{2}) = t a n^{- 1} x + c \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

Upvote

New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The solution of the differential equation $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ is:

(A) $y = e^{y - y} - x^{2} e^{- y} + c$

(B) $e^{y} - e^{x} = x^{3} / 3 + c$

(C) $e^{x} + e^{y} = x^{3} / 3 + c$

(D) $e^{x} - e^{y} = x^{3} / 3 + c$

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ \frac{d y}{d x} = e^{x - y} + x^{2} e^{- y} \\ \Rightarrow \frac{d y}{d x} = e^{x} . e^{- y} + x^{2} e^{- y} \Rightarrow \frac{d y}{d x} = e^{- y} (e^{x} + x^{2}) \\ \Rightarrow \frac{d y}{e^{- y}} = (e^{x} + x^{2}) d x \Rightarrow e^{y} . d y = (e^{x} + x^{2}) d x \\ I n t e g r a t i n g b o t h s i d e s, w e h a v e \\ \Rightarrow \int e^{y} . d y = \int (e^{x} + x^{2}) d x \\ \Rightarrow e^{y} = e^{x} + \frac{x^{3}}{3} + c \\ \Rightarrow e^{y} - e^{x} = \frac{x^{3}}{3} + c \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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Ncert Solutions Maths class 12th

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