Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Continuity and Differentiability

34. Kindly Consider the following

$s i n (x^{2} + 5)$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

34. Let f (x) = sin (x2 + 5)

Differentiating w. r t. x we get,

f'¢ (x) = $\frac{d}{d x}$ sin (x² + 5)

= cos (x² + 5) $\frac{d}{d x}$ = cos (x² + 5)

= cos (x² + 5) [2x].

= 2x cos (x² + 5).

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New answer posted•a year ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

20. Find the shortest distance between the lines $\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} a n d \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}$ .

14 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} \frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1} \\ \frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1} \end{array}$

Shortest distance between two lines is given by,

given equation we have

$\begin{array}{l} x_{1} = - 1, y_{1} = - 1, z_{1} = - 1 \\ a_{1} = 7, b_{1} = - 6, c_{1} = 1 \\ x_{2} = 3, y_{2} = 5, z_{2} = 7 \\ a_{2} = 1, b_{2} = - 2, c_{2} = 1 \end{array}$

Then,

$\begin{array}{l} = 4 (- 6 + 2) - 6 (7 - 1) + 8 (- 14 + 6) \\ = - 16 - 36 - 64 \\ = - 116 \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

19. Find the shortest distance between the $\vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} + \hat{j} + \hat{k}) a n d \vec{r} = 2 \hat{i} + \hat{j} + \hat{k} + μ (2 \hat{i} + \hat{j} + 2 \hat{k})$ lines

22 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} \vec{r} = (\hat{i} + 2 \hat{j} + \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k}) a n d - - - - - (1) \\ \vec{r} = 2 \hat{i} - \hat{j} - \hat{k} + μ (2 \hat{i} + \hat{j} + 2 \hat{k}) - - - - - (2) \end{array}$

Solution. Comparing (1) and (2) with $\vec{r} = {\vec{a}}_{1} + λ {\vec{b}}_{1}$ and $\vec{r} = {\vec{a}}_{2} + μ {\vec{b}}_{2}$ respectively.

We get,

$\begin{array}{l} a_{1} = \hat{i} + 2 \hat{j} + \hat{k}, b_{1} = \hat{i} - \hat{j} + \hat{k} \\ a_{2} = 2 \hat{i} - \hat{j} - \hat{k}, b_{2} = 2 \hat{i} + \hat{j} + 2 \hat{k} \end{array}$

Therefore,

$\begin{array}{l} {\vec{a}}_{2} - {\vec{a}}_{1} = \hat{i} - 3 \hat{j} - 2 \hat{k} \\ {\vec{b}}_{1} \times {\vec{b}}_{2} = (\hat{i} - \hat{j} + \hat{k}) \times (2 \hat{i} + \hat{j} + 2 \hat{k}) \end{array}$

$\begin{array}{l} = (2 - 1) \hat{i} - (2 - 2) \hat{j} + (1 + 2) \hat{k} \\ = - 3 \hat{i} + 3 \hat{k} \\ | {\vec{b}}_{1} \times {\vec{b}}_{2} | = = = = 3 \\ ({\vec{b}}_{1} \times {\vec{b}}_{2}) . ({\vec{a}}_{2} - {\vec{a}}_{1}) = (- 3 \hat{i} + 3 \hat{k}) (\hat{i} - 3 \hat{j} - 2 \hat{k}) \\ = - 3 - 6 \\ = - 9 \end{array}$

Hence, the shortest distance between the given line is given by

$\begin{array}{l} d = | \frac{({\vec{b}}_{1} \times {\vec{b}}_{2}) . ({\vec{a}}_{2} - {\vec{a}}_{1})}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |} | \\ = | \frac{- 9}{3} | \\ = \frac{3}{} = \frac{3}{2} \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

18. Show that the lines $\frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} a n d \frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular to each other.

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\begin{array}{l} \frac{x - 5}{7} = \frac{y + 2}{- 5} = \frac{z}{1} \\ \frac{x}{1} = \frac{y}{2} = \frac{z}{3} \end{array}$

Direction ratios of given lines are (7, -5,1) and (1,2,3).

i.e., $\begin{array}{l} a_{1} = 7, b_{1} = - 5, c_{1} = 1 \\ a_{2} = 1, b_{2} = 2, c_{2} = 3 \end{array}$

Now,

$\begin{array}{l} = a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} \\ = 7 \times 1 + (- 5) \times 2 + 1 \times 3 \\ = 7 - 10 + 3 \\ = 10 - 10 \\ = 0 \end{array}$

$∴$ These two lines are perpendicular to each other.

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New answer posted•a year ago

Ncert Solutions Maths class 12th Continuity and Differentiability

33. Find all the points of discontinuity of $f$ defined by $f x x x$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

33. Let g (x) = $x$ is continuous being a modules f x and h (x) = $x$ is also continuous being a modules $\forall x \in ?$

Then, f (x) = g (x) h (x).is also continuous for all x. E. R.

Hence, there is no point of discontinuous for f (x).

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New answer posted•a year ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

17. Find the values of p so that the lines $\frac{1 - x}{3} = \frac{7 y - 14}{2} p = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 p} = \frac{y - 5}{1} = \frac{6 - z}{5}$ are at right angles.

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{1 - x}{3} = \frac{7 y - 14}{2 ρ} = \frac{z - 3}{2} a n d \frac{7 - 7 x}{3 ρ} = \frac{y - 5}{1} = \frac{6 - z}{5}$

The standard form of a pair of Cartesian lines is;

$\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}} a n d \frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}} - - - - - (1)$

So,

$\begin{array}{l} \frac{- (x - 1)}{3} = \frac{7 (y - 5)}{2 ρ} = \frac{z - 3}{2} a n d \frac{- 7 (x - 1)}{3 ρ} = \frac{y - 5}{1} = \frac{- (z - 6)}{5} \\ \frac{x - 1}{- 3} = \frac{y - 2}{2 ρ / 7} = \frac{z - 3}{2} a n d \frac{x - 1}{- 3 ρ / 7} = \frac{y - 5}{1} = \frac{z - 6}{- 5} - - - - - (2) \end{array}$

Comparing (1) and (2) we get

$\begin{array}{l} a_{1} = - 3, b_{1} = \frac{2 ρ}{7}, c_{1} = 2 \\ a_{2} = \frac{- 3 ρ}{7}, b_{2} = 1, c_{2} = - 5 \end{array}$

Now, both the lines are at right angles

So, $a_{1} a_{2} + b_{1} b_{2} + c_{1} c_{2} = 0$

$\begin{array}{l} \Rightarrow (- 3) \times \frac{(- 3 ρ)}{7} + \frac{2 ρ}{7} \times 1 + 2 \times (- 5) = 0 \\ \Rightarrow \frac{9 ρ}{7} + \frac{2 ρ}{7} + (- 10) = 0 \\ \Rightarrow \frac{9 ρ + 2 ρ}{7} = 10 \\ \Rightarrow 11 ρ = 70 \\ ρ = \frac{70}{11} \end{array}$

$∴$ The value of $ρ$ is $\frac{70}{11}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

16. Find the angle between the following pair of lines

$\begin{array}{l} (i) \frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{- 3} & \frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4} \\ (i i) \frac{x}{y} = \frac{y}{2} = \frac{z}{1} & \frac{x - 5}{4} = \frac{y - 2}{1} = \frac{z - 3}{8} \end{array}$

9 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(i) Let ${\vec{b}}_{1}$ and ${\vec{b}}_{2}$ be the vectors parallel to the pair of lines, $\frac{x - 2}{2} = \frac{y - 1}{5} = \frac{z + 3}{- 3} & \frac{x + 2}{- 1} = \frac{y - 4}{8} = \frac{z - 5}{4}$ , respectively.

$\begin{array}{l} {\vec{b}}_{1} = 2 \hat{i} + 5 \hat{j} - 3 \hat{k} & {\vec{b}}_{2} = - \hat{i} + 8 \hat{j} + 4 \hat{k} \\ ∴ | {\vec{b}}_{1} | = = \\ | {\vec{b}}_{2} | = = = 9 \\ {\vec{b}}_{1} . {\vec{b}}_{2} = (2 \hat{i} + 5 \hat{j} - 3 \hat{k}) . (- \hat{i} + 8 \hat{j} + 4 \hat{k}) \\ = 2 (- 1) + 5 \times 8 + (- 3) . 4 \\ = - 2 + 40 - 12 \\ = 26 \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Continuity and Differentiability

32. Examine that sin $| x |$ is a continuous function.

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

32. Given, f (x) = sin $x$

Let g (x) = sin x and h (x) = $x$ then as sine f x and modulus f x are continuous in x e R

g and h are continuous.

So, (goh) (x) = g (h (x) = g (|x|) = sin |x| = f (x)

Is a continuous f x being a competitive f x of two continuous f x.

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New answer posted•a year ago

Ncert Solutions Maths class 12th Three Dimensional Geometry

15. Find the angle between the following pairs of lines:

$\begin{array}{l} (i) \vec{r} = 2 \hat{i} - 5 \hat{j} + \hat{k} + λ (3 \hat{i} - 2 \hat{j} + 6 \hat{k}) & \vec{r} = 7 \hat{i} - 6 \hat{k} + μ (\hat{i} + 2 \hat{j} + 2 \hat{k}) \\ (i i) \vec{r} = 3 \hat{i} + \hat{j} - 2 \hat{k} + λ (\hat{i} - \hat{j} - 2 \hat{k}) & \vec{r} = 2 \hat{i} - \hat{j} - 56 \hat{k} + μ (\hat{3 i} - 5 \hat{j} - 4 \hat{k}) \end{array}$

8 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(i) Let Q be the angle between the given lines.

The angle between the given pairs of lines is given by, $c o s Q = | \frac{{\vec{b}}_{1} . {\vec{b}}_{2}}{| {\vec{b}}_{1} | | {\vec{b}}_{2} |} |$

The given lines are parallel to the vectors, ${\vec{b}}_{1} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k} & {\vec{b}}_{2} = \hat{i} + 2 \hat{j} + 2 \hat{k}$ , respectively.

$\begin{array}{l} ∴ | {\vec{b}}_{1} | = = 7 \\ | {\vec{b}}_{2} | = = 3 \\ {\vec{b}}_{1} . {\vec{b}}_{2} = (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) . (\hat{i} + 2 \hat{j} + 2 \hat{k}) \\ = 3 \times 1 + 2 \times 2 + 6 \times 2 \\ = 3 + 4 + 12 \\ = 19 \\ \Rightarrow c o s Q = \frac{19}{7 \times 3} \\ \Rightarrow Q = c o s^{- 1} (\frac{19}{21}) \end{array}$