Ncert Solutions Maths class 12th

46. Given, ax + by2 = cos y.

Differentiating w r t 'x' we get,

$\frac{d}{dx}\left(ax+b{y}^2\right)=\frac{dx}{dx}cosy$

= a + b 2y = - sin y $\frac{dy}{dx}$ + sin y $\frac{dy}{dx}$ = -a

= $\frac{dy}{dx}=\frac{-9}{2by+siny}.$

= 2by $\frac{dy}{dx}$

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

Differentiating w r t x. we get,

$\frac{d}{dx}(2x+3y)=\frac{d}{dx}siny$

$\Rightarrow 2+3\frac{dy}{dx}=cosy\frac{dy}{dx}$

=cos y $\frac{dy}{dx}-3\frac{dy}{dx}=2$

$\Rightarrow \frac{dy}{dx}(cosy-3)=2$

= $\frac{dy}{dx}=\frac{2}{cosy-3}$

44. Kindly go through the solution

43. The given f x ⁿ is

f(x) = 0 < $\left|x\right|$ < 3

At x = 1

L*H*L* = $\underset{h\to 0^{-}}{lim}\frac{f(1+h)-f\left(0\right)}{h}$

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

$\underset{h\to \sigma }{lim}\frac{0-1}{h}$ $\left\{\begin{array}{ccc}\hfill ?h<0\hfill & \hfill ,1+h<1-1\hfill & \hfill So, [1+h]=0\hfill \end{array}\right\}$

$=\underset{h\to 0}{lim}\frac{-1}{h}=\infty$

Hence lines does not exist

Qf is not differentiable at x = 1

At x = 2

L*H*L = $\underset{h\to 0^{-}}{lim}\frac{f(2+h)-f(2)}{h}$ $\left\{\begin{array}{ccc}\hfill ?h<0\hfill & \hfill 2+h<2\hfill & \hfill 30,[2+h]=1\hfill \end{array}\right\}$

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

$=\underset{h\to 0^{-}}{lim}\frac{1-2}{h}=\underset{h\to 0^{-}}{lim}\frac{-1}{h}$

Hence, limit does not exist.

Qf is not differentiable at x = 2

42. The given f x v is

f(x) = |x- 1|, x ε R

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

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

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

Now,

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

= $\underset{h\to 0^{-}}{lim}\frac{|1+h-1|-0.}{h}=\underset{h\to 0^{-}}{lim}-\frac{h}{h}$ $\left\{\begin{array}{l}\therefore h<0\\ \left|h\right|=-h\end{array}\right\}$

$=\underset{h\to {\sigma }^{-}}{lim}(-1)$

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

$=\underset{h\to 0^{+}}{lim}\frac{(1+h-1)-0}{h}=\underset{h\to 0^{+}}{lim}\frac{h}{h}=\underset{h\to 0^{+}}{lim}1$ $\left\{\begin{array}{cc}\hfill ?f_{n}h>0\hfill & \hfill \left|h\right|=h\hfill \end{array}\right\}$

= 1

Hence, L*H*S ¹ R*H*L*

So, f is not differentiable at x = 2.

The equations of the planes are

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

It can be seen that,

$\begin{array}{l}\frac{a_1}{a_2}=\frac{2}{5}\\ \frac{b_1}{b_2}=\frac{-1}{-2.5}=\frac{2}{5}\\ \frac{c_1}{c_2}=\frac{4}{10}=\frac{2}{5}\\ \therefore \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\end{array}$

Therefore, the given planes are parallel.

Hence, the correct answer is B.

The equations of the planes are

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

It can be seen that the given planes are parallel.

It is known that the distance between two parallel planes,   $ax + by + cz = d_1 and ax + by + cz = d_2,$ is given by,

$\begin{array}{l}D=\left|\frac{d_2-d_1}{}\right|\\ \Rightarrow D=\left|\frac{6-4}{}\right|\\ D=\frac{2}{}\end{array}$

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

Hence, the correct answer is D.

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,

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

Let the required line be parallel to the vector  $\overrightarrow{b}$  given by,  $\overrightarrow{b}=b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}$

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

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

$\begin{array}{l}\overrightarrow{r}=\overrightarrow{a}+\lambda \overrightarrow{b}\\ \Rightarrow \overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-4\widehat{k}\right)+\lambda \left(b_1\widehat{i}+b_2\widehat{j}+b_3\widehat{k}\right).......(1)\end{array}$

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.

$\therefore 3b_1-16b_2+7b_3=0........(4)$

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

$\therefore 3b_1+8b_2-5b_3=0........(5)$

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

$\begin{array}{l}\frac{b_1}{\left(-16\right)\left(-5\right)-8*7}=\frac{b_2}{7*3-3\left(-5\right)}=\frac{b_3}{3*8-3\left(-16\right)}\\ \Rightarrow \frac{b_1}{24}=\frac{b_2}{36}=\frac{b_3}{72}\\ \Rightarrow \frac{b_1}{2}=\frac{b_2}{3}=\frac{b_3}{6}\end{array}$

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

$\therefore \overrightarrow{b}=2\widehat{i}+3\widehat{j}+6\widehat{k}$

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

$\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-4\widehat{k}\right)+\lambda \left(2\widehat{i}+3\widehat{j}+6\widehat{k}\right)$

This is the equation of the required line.