Class 12th

45. Kindly consider the following

$2 x + 3 y = s i n y$

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45. Given, 2x + 3y = sin y.

Differentiating w r t x. we get,

$\frac{d}{d x} (2 x + 3 y) = \frac{d}{d x} s i n y$

$\Rightarrow 2 + 3 \frac{d y}{d x} = c o s y \frac{d y}{d x}$

=cos y $\frac{d y}{d x} - 3 \frac{d y}{d x} = 2$

$\Rightarrow \frac{d y}{d x} (c o s y - 3) = 2$

= $\frac{d y}{d x} = \frac{2}{c o s y - 3}$

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44. Find $\frac{d y}{d x}$ in the following:

$2 x + 3 y = s i n x$

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44. Kindly go through the solution

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43. Prove that the greatest integer function defined by

$f (x) = [x], 0 < x < 3$ is not differentiable at $x$ = 1 and $x$ = 2.

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43. The given f x ⁿ is

f(x) = 0 < $| x |$ < 3

At x = 1

L*H*L* = $\underset{h \to 0^{-}}{l i m} \frac{f (1 + h) - f (0)}{h}$

$= \underset{h \to 0^{-}}{l i m} \frac{[1 + h] - [1]}{h}$

$\underset{h \to σ}{l i m} \frac{0 - 1}{h}$ ${\begin{matrix} ? h < 0 & , 1 + h < 1 - 1 & So, [1 + h] = 0 \end{matrix}}$

$= \underset{h \to 0}{l i m} \frac{- 1}{h} = \infty$

Hence lines does not exist

Qf is not differentiable at x = 1

At x = 2

L*H*L = $\underset{h \to 0^{-}}{l i m} \frac{f (2 + h) - f (2)}{h}$ ${\begin{matrix} ? h < 0 & 2 + h < 2 & 30, [2 + h] = 1 \end{matrix}}$

$= \underset{h \to 0^{-}}{l i m} \frac{[2 + h] - [2] .}{h}$

$= \underset{h \to 0^{-}}{l i m} \frac{1 - 2}{h} = \underset{h \to 0^{-}}{l i m} \frac{- 1}{h}$

Hence, limit does not exist.

Qf is not differentiable at x = 2

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42. Prove that the function $f$ given by

$f (x) = | x - 1 |, x \in$ is not differentiable at $x$ = 1.

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42. The given f x v is

f(x) = |x- 1|, x ε R

For a differentiable f x v f at x = c,

$\underset{h \to 0^{-}}{l i m} \frac{f (c + h) - f (c)}{h}$ and $\underset{h \to 0^{+}}{l i m} \frac{f (c + h) - f (c)}{h}$ are finite & equal.

So, at x = 1. f(1) = |1 - 1| = 0.

Now,

L*H*L* = $\underset{h \to 0^{-}}{l i m} \frac{f (1 + h - f (1)}{h}$

= $\underset{h \to 0^{-}}{l i m} \frac{| 1 + h - 1 | - 0 .}{h} = \underset{h \to 0^{-}}{l i m} - \frac{h}{h}$ ${\begin{cases} ∴ h < 0 \\ | h | = - h \end{cases}}$

$= \underset{h \to σ^{-}}{l i m} (- 1)$

R*H*L = $\underset{h \to 0^{+}}{l i m} \frac{f (1 + h) - f (1)}{h}$ = - 1.

$= \underset{h \to 0^{+}}{l i m} \frac{(1 + h - 1) - 0}{h} = \underset{h \to 0^{+}}{l i m} \frac{h}{h} = \underset{h \to 0^{+}}{l i m} 1$ ${\begin{matrix} ? f_{n} h > 0 & | h | = h \end{matrix}}$

= 1

Hence, L*H*S ¹ R*H*L*

So, f is not differentiable at x = 2.

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59. The planes: $2 x - y + 4 z = 5 a n d 5 x - 2.5 y + 10 z = 6$ are

(A) Perpendicular (B) Parallel (C) intersect y-axis

(D) Passes through (0,0,5/4)

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The equations of the planes are

$\begin{array}{l} 2 x - y + 4 z = 5 \dots (1) & 5 x - 2.5 y + 10 z = 6 \dots (2) \end{array}$

It can be seen that,

$\begin{array}{l} \frac{a_{1}}{a_{2}} = \frac{2}{5} \\ \frac{b_{1}}{b_{2}} = \frac{- 1}{- 2.5} = \frac{2}{5} \\ \frac{c_{1}}{c_{2}} = \frac{4}{10} = \frac{2}{5} \\ ∴ \frac{a_{1}}{a_{2}} = \frac{b_{1}}{b_{2}} = \frac{c_{1}}{c_{2}} \end{array}$

Therefore, the given planes are parallel.

Hence, the correct answer is B.

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Choose the correct answer in Exercise Q. 22 and 23.

58. Distance between the two planes: $2 x + 3 y + 4 z = 4 a n d 4 x + 6 y + 8 z = 12$ is

(A) 2 units (B) 4 units (C) 8 units (D) 2/√29 units

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The equations of the planes are

$\begin{array}{l} 2 x + 3 y + 4 z = 4 & 4 x + 6 y + 8 z = 12 & \Rightarrow 2 x + 3 y + 4 z = 6 \end{array}$

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes, $a x + b y + c z = d_{1} a n d a x + b y + c z = d_{2},$ is given by,

$\begin{array}{l} D = | \frac{d_{2} - d_{1}}{} | \\ \Rightarrow D = | \frac{6 - 4}{} | \\ D = \frac{2}{} \end{array}$

Thus, the distance between the lines is 2/√29 units.

Hence, the correct answer is D.

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57. Prove that if a plane has the intercepts a,b,c and is at a distance of P units from the origin, then $\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} = \frac{1}{p^{2}}$

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The equation of a plane having intercepts a, b, c with x, y, and z axes respectively is given by,

$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$

The distance (p) of the plane from the origin is given by,

$\begin{array}{l} p = | \frac{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}{} | \\ \Rightarrow p = \frac{1}{} \\ \Rightarrow p^{2} = \frac{1}{\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}}} \\ \Rightarrow \frac{1}{p^{2}} = \frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{c^{2}} \end{array}$

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56. Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines:

$\frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5}$

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Let the required line be parallel to the vector $\vec{b}$ given by, $\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, − 4) is $\vec{a} = \hat{i} + 2 \hat{j} - 4 \hat{k}$

The equation of the line passing through (1, 2, −4) and parallel to vector $\vec{b}$ is

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . (1) \end{array}$

The equations of the lines are

$\begin{array}{l} \frac{x - 8}{3} = \frac{y + 19}{- 16} = \frac{z - 10}{7} . . . . . . . . (2) \\ \frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{- 5} . . . . . . . . (3) \end{array}$

Line (1) and line (2) are perpendicular to each other.

$∴ 3 b_{1} - 16 b_{2} + 7 b_{3} = 0 . . . . . . . . (4)$

Also, line (1) and line (3) are perpendicular to each other.

$∴ 3 b_{1} + 8 b_{2} - 5 b_{3} = 0 . . . . . . . . (5)$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 16) (- 5) - 8 * 7} = \frac{b_{2}}{7 * 3 - 3 (- 5)} = \frac{b_{3}}{3 * 8 - 3 (- 16)} \\ \Rightarrow \frac{b_{1}}{24} = \frac{b_{2}}{36} = \frac{b_{3}}{72} \\ \Rightarrow \frac{b_{1}}{2} = \frac{b_{2}}{3} = \frac{b_{3}}{6} \end{array}$

Direction ratios of $\vec{b}$ are 2, 3, and 6.

$∴ \vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$

Substituting $\vec{b} = 2 \hat{i} + 3 \hat{j} + 6 \hat{k}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} - 4 \hat{k}) + λ (2 \hat{i} + 3 \hat{j} + 6 \hat{k})$

This is the equation of the required line.

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55. Find the vector equation of the line passing through (1, 2, 3) and parallel to the plane $\vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5$ and $\vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6$

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Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

$\begin{array}{l} \Rightarrow (\hat{i} - \hat{j} + 2 \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0 \\ \Rightarrow λ (b_{1} - b_{2} + 2 b_{3}) = 0 \\ \Rightarrow b_{1} - b_{2} + 2 b_{3} = 0 . . . . . . . . . . (4) \end{array}$

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 1) * 1 - 1 * 2} = \frac{b_{2}}{2 * 3 - 1 * 1} = \frac{b_{3}}{1 * 1 - 3 * (- 1)} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{b_{2}}{5} = \frac{b_{3}}{4} \end{array}$

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

Substituting the value of $\vec{b}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the equat

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Let the required line be parallel to vector $\vec{b}$ given by,

$\vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}$

The position vector of the point (1, 2, 3) is $\vec{a} = \hat{i} + 2 \hat{j} + 3 \hat{k}$

The equation of line passing through (1, 2, 3) and parallel to $\vec{b}$ is given by,

$\begin{array}{l} \vec{r} = \vec{a} + λ \vec{b} \\ \Rightarrow \vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) . . . . . . . . (1) \end{array}$

The equations of the given planes are

$\begin{array}{l} \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) = 5 . . . . . . . . (2) \\ \vec{r} . (3 \hat{i} + \hat{j} + \hat{k}) = 6 . . . . . . . . (3) \end{array}$

The line in equation (1) and plane in equation (2) are parallel. Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

Similarly, $(3 \hat{i} + \hat{j} + \hat{k}) . λ (b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k}) = 0$

From equations (4) and (5), we obtain

$\begin{array}{l} \frac{b_{1}}{(- 1) * 1 - 1 * 2} = \frac{b_{2}}{2 * 3 - 1 * 1} = \frac{b_{3}}{1 * 1 - 3 * (- 1)} \\ \Rightarrow \frac{b_{1}}{- 3} = \frac{b_{2}}{5} = \frac{b_{3}}{4} \end{array}$

Therefore, the direction ratios of $\vec{b}$ are −3, 5, and 4.

$∴ \vec{b} = b_{1} \hat{i} + b_{2} \hat{j} + b_{3} \hat{k} = - 3 \hat{i} + 5 \hat{j} + 4 \hat{k}$

Substituting the value of $\vec{b}$ in equation (1), we obtain

$\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (- 3 \hat{i} + 5 \hat{j} + 4 \hat{k})$

This is the equation of the required line.

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53. Find the equation of the plane which contains the line of intersection of the planes $\vec{r} (\hat{i} + 2 \hat{j} + 3 \hat{k}) - 4 = 0, \vec{r} . (2 \hat{i} + h t \hat{j} - \hat{k}) + 5 = 0$ and which is perpendicular to the plane $\vec{r} . (5 \hat{i} + 3 \hat{j} - 6 \hat{k}) + 8 = 0$

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