Maths

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.

41. L.H.S. = cos 4x.

= cos 2 (2x)

= 1 – 2 Sin² (2x) [ cos 2x = 1 – 2 Sin²x]

= 1 – 2 [2 sin xcosx]² [ sin 2x = 2 sin xcos x]

= 1 – 2 [4 sin²xcos²x]

= 1 – 8 sin²xcos²x

= R.H.S.

40. $$L.H.S. = tan 4x

We know that,

$tan2=\frac{2tan}{1-ta{n}^2}.$ , we can write

$L.H.S=tan2(2x)=\frac{2tan2x}{1-ta{n}^{2}2x.}$

$=\frac{2\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}{1-{\left(\frac{2tanx}{1-ta{n}^{2}x}\right)}^2}$

$=\frac{\frac{4tanx}{\left(1-ta{n}^{2}x\right)}}{\frac{{\left(1-ta{n}^{2}x\right)}^2-4ta{n}^{2}x}{{\left(1-ta{n}^{2}x\right)}^2}}$

$=\frac{4tanx*\left(1-ta{n}^{2}x\right)}{1+ta{n}^{4}x-2ta{n}^{2}x-4ta{n}^{2}x}$

$=\frac{4tanx\left(1-ta{n}^{2}x\right)}{1-6ta{n}^{2}x+ta{n}^{4}x}$

= R.H.S.

39. L.H.S. = cot x cot 2x – cot 2x cot 3x – cot 3x cot x.

= cot x cot 2x – cot 3x (cot 2x + cot x)

= cot x cot 2x – (cot 2x + cot x) [cot (2x + x)]

We know that,

$cot(A+B)=\frac{co\mathrm{tA}co\mathrm{tB}-1}{co\mathrm{tA}+co\mathrm{tB}}$ we can write

$L.H.S=cotxcot2x-(cot2x+cotx)\left[\frac{cot2x·cotx-1}{cot2x+cotx}\right]$

= cot x cot 2x – cot 2x cot x + 1

= R.H.S.

38. L.H.S$=\frac{cos4x+cos3x+cos2x}{sin4x+sin3x+sin2x.}$

$=\frac{(cos4x+cos2x)+cos3x}{(sin4x+sin2x)+sin3x}.$

$=\frac{2cos\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)+cos3x}{2sin\left(\frac{4x+2x}{2}\right)cos\left(\frac{4x-2x}{2}\right)sin3x}.$

$=\frac{2cos\frac{6x}{2}cos\frac{2x}{2}+cos3x}{2sin\frac{6x}{2}cos\frac{2x}{2}+sin3x}$

$=\frac{2cos3xcosx+cos3x}{2sin3xcosx+3x}$

$=\frac{cos3x}{sin3x}*\frac{(2cosx+1)}{(2cosx+1)}$

$=cot3x=R.H.S$

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.

37. $=\frac{sinx-sin3x}{si{n}^{2}x-co{s}^{2}x}$

$=\frac{2cos\left(\frac{x+3x}{2}\right)sin\frac{x-3x}{2}}{-\left(co{s}^{2}x-si{n}^{2}x\right)}$

$=\frac{2cos\frac{4x}{2}sin\left(-\frac{2x}{2}\right)}{-cos2x}\left[?cos2x=co{s}^{2}x-si{n}^{2}x\right]$

$=-\frac{2cos2xsinx}{-cos2x}[?sin(-x)=-sinx]$

= 2 sin x

= R.H.S.

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,

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