Class 12th

49. Given,  x³ + x²y + xy² + y³ = 81.

Differentiating w r t 'x' we get,

$\frac{d}{dx}\left(x^3+x^{2}y+x{y}^2+y^3\right)$ = $\frac{d(81)}{dx}$

$\Rightarrow \frac{d{x}^3}{dx}+\frac{d}{dx}{x}^{2}y+\frac{d}{dx}x{y}^2+\frac{d}{dx}{y}^3=0$

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+y\frac{d{x}^2}{dx}+\frac{xd{y}^2}{dx}+y^2\frac{dx}{dx}+3y^2\frac{dy}{dx}=0$

$\Rightarrow 3x^2+x^2\frac{dy}{dx}+2xy+2xy\frac{dy}{dx}+y^2+3y^2\frac{dy}{dx}=0.$

$\Rightarrow \left(x^2+2xy+3y^2\right)\frac{dy}{dx}$= - (3x² + 2xy + y²)

$\frac{dy}{dx}=\frac{-\left(3x^2+2xy+y^2\right)}{\left(x^2+2xy+3y^2\right).}$

48. Given,  x² + xy + y² = 100.

Differentiating w r t 'x' we get,

$\frac{d}{dx}\left(x^2+xy+y^2\right)=\frac{d}{dx}(100)$

$\Rightarrow 2x+x\frac{dy}{dx}+y\frac{dx}{dx}+2y\frac{dy}{dx}=0.$

$\Rightarrow x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-y$

$\Rightarrow \frac{dy}{dx}=\frac{-(2x+y)}{(x+2y)}$

47. Given, xy + y² = tan x + y  Differentiating w r t x we get,

$\frac{d}{dx}\left(xy+y^2\right)=\frac{d}{dx}(tanx+y)$

$\Rightarrow x\frac{dy}{dx}+y\frac{dx}{dx}+\frac{d{y}^2}{dx}=d{x}^{2}x+\frac{dy}{dx}$

$\Rightarrow x\frac{dy}{dx}+2y\frac{dy}{dx}-\frac{dy}{dx}=se{n}^{2}x-y$

$\Rightarrow (x+2y-1)\frac{dy}{dx}=se{c}^{2}x-y$

$\Rightarrow \frac{dy}{dx}=\frac{si{n}^{2}x-y}{x+2y-1.}$

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.