Q

Q 15)
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$

G Gautam harsolia

Given function is
$f(x)\left\{\begin{matrix} 2x & if &x<0 \\ 0& if &0\leq x\leq 1 \\ 4x&if & x>1 \end{matrix}\right.$
Given function is satisfies for the all real values of x
case (i)  k < 0
$f(k) = 2k$
$\lim_{k \rightarrow 0^-}f(x)= 2k = f(k)$
Hence, function is continuous for all values of x < 0

case (ii)  x = 0
$f(0 )= 0$
L.H.L at x= 0
$\lim_{x\rightarrow 0^-}f(x)= 2(0)= 0$
R.H.L. at x = 0
$\lim_{x\rightarrow 0^+}f(x)= 0$
L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii)  k > 0
$f(k)=0$
$\lim_{k\rightarrow 0^+}f(x)= 0= f(k)$
Hence , function is continuous for all values of x > 0

case (iv) k < 1
$f(k) = 0$
$\lim_{x\rightarrow 1^-}f(x)= 0 = f(k)$
Hence , function is continuous for all values of x < 1

case (v)  k > 1
$f(k) = 4k$
$\lim_{x\rightarrow 1^+}f(x)= 4k = f(k)$
Hence , function is continuous for all values of x > 1

case (vi)  x = 1
$f(1)= 0$
$\lim_{x\rightarrow 1^-}f(1)= 0$
$\lim_{x\rightarrow 1^+}f(1)= 4(1) = 4$
Hence, function is not continuous at x = 1

Q.6 Write the following as intervals :

(i) {x : x $\in$R, – 4 $<$ x $\leq$ 6}

(ii) {x : x $\in$ R, – 12 $<$x $<$–10}

(iii) {x : x $\in$ R, 0 $\leq$ x $<$ 7}

(iv) {x : x $\in$ R, 3 $\leq$  x $\leq$  4}

S safeer

The following can be written in interval as :

(i) {x : x $\in$R, – 4 $<$ x $\leq$ 6} $= \left ( -4 ,6 \right ]$

(ii) {x : x $\in$ R, – 12 $<$x $<$–10} $= \left ( -12,-10 \right )$

(iii) {x : x $\in$ R, 0 $\leq$ x $<$ 7} $= \left \left [ 0,7\right )$

(iv) (iv) {x : x $\in$ R, 3 $\leq$  x $\leq$  4} $=\left [ 3,4 \right ]$

Q.5 How many elements has P(A), if A = $\phi$?

S safeer

Let the elements in set A be m,then n$\left ( A \right ) =$ m

then, number of elements in power set of A         n$\left ( p\left ( A \right )\right ) =$ $2^{m}$

Here,  A = $\phi$   so    n$\left ( A \right ) =$ 0

n$\left [ P\left ( A \right ) \right ] =$ $2^{0}$ $=$1

Hence,we conclude P(A) has 1 element.

Q.4 Write down all the subsets of the following sets:

(iv) $\phi$

S safeer

Subset  of  $\phi$ is  $\phi$ only.

The subset of a null set is null set itself

Q.4 Write down all the subsets of the following sets:

(iii) {1,2,3}

S safeer

Subsets of

$\left \{ 1,2,3 \right \} = \left \{ 1 \right \},\left \{ 2 \right \},\left \{ 3 \right \},\phi ,\left \{ 1,2 \right \},\left \{ 2,3 \right \},\left \{ 3,1 \right \},\left \{ 1,2,3 \right \}$

Q.4 Write down all the subsets of the following sets:

(ii) {a, b}

S safeer

Subsets of  $\left \{ a,b \right \}\ are \ \phi , \left \{ a \right \},\left \{ b \right \} and \left \{ a,b \right \}$. Thus the given set has 4 subsets

Q.4 Write down all the subsets of the following sets

(i) {a}

S safeer

Subsets of $\left \{ a \right \} = \phi \, and \left \{ a \right \}$.

Find    $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$     in

Q (10)

$\small \sin x = \frac{1}{4}$   ,x in quadrant II

S seema garhwal

$\frac{\pi}{2} < x < \pi\\ \\ \frac{\pi}{4} < \frac{x}{2} < \frac{\pi}{2}$ all functions are positive in this range
We know that
$\cos^{2}x = 1 - \sin^{2}x$
= 1 - $\left ( \frac{1}{4} \right )^{2}$     =   $1 - \frac{1}{16}$ =   $\frac{15}{16}$

cos x = $\sqrt\frac{15}{16} = \pm \frac{\sqrt15}{4} = - \frac{\sqrt15}{4}$                 (cos x is -ve in II quadrant)

We know that
cosx = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} = \cos x + 1 = -\frac{\sqrt15}{4} + 1 = \frac{-\sqrt15+4}{4}$

$\cos^{2}\frac{x}{2} = \frac{-\sqrt15+4}{8}$
$\cos\frac{x}{2} = \pm \sqrt\frac{-\sqrt15+4}{8} = \frac{\sqrt{-\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8-2\sqrt15}}{4}$             (because all functions are posititve in given range)

similarly,
cos x = $1-2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x\\ \\ 2\sin^{2}\frac{x}{2} = 1 -\left (\frac{-\sqrt15}{4} \right ) = \frac{4+\sqrt15}{4}$
$\sin\frac{x}{2} = \pm \sqrt\frac{\sqrt15+4}{8} = \frac{\sqrt{\sqrt15+4}}{2\sqrt2} = \frac{\sqrt{8+2\sqrt15}}{4}$     (because all functions are posititve in given range)
$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt{8+2\sqrt15}}{4}}{\frac{\sqrt{8-2\sqrt15}}{4}} = \frac{{8+2\sqrt15}}{\sqrt{64 - 15\times4}} = \frac{{8+2\sqrt15}}{\sqrt{4}} = 4 + \sqrt15$

Find     $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$   in

Q (9)

$\small \cos x = -\frac{1}{3}$   , x in quadrant III

S seema garhwal

$\pi < x < \frac{3\pi}{2}\\ \\ \frac{\pi}{2} < \frac{x}{2} < \frac{3\pi}{4}$

We know that
cos x  = $2\cos^{2}\frac{x}{2} - 1$
$2\cos^{2}\frac{x}{2} =$  cos x + 1
=  $\left ( -\frac{1}{3} \right )$   + 1   =  $\left ( \frac{-1+3}{3} \right )$   =   $\frac{2}{3}$

$\cos\frac{x}{2} = \sqrt{ \frac{1}{3}} = \pm \frac{1}{\sqrt3}$

$\cos\frac{x}{2} = - \frac{1}{\sqrt3} = - \frac{\sqrt3}{3}$
Now,
we know that
cos x = $1 - 2\sin^{2}\frac{x}{2}$
$2\sin^{2}\frac{x}{2} = 1 - \cos x$
=  1 - $\left ( -\frac{1}{3} \right )$   =  $\frac{3+1}{3}$  = $\frac{4}{3}$

$2\sin^{2}\frac{x}{2} = \frac{4}{3} \\ \\ \sin^{2}\frac{x}{2} = \frac{2}{3}\\ \sin\frac{x}{2} = \sqrt{ \frac{2}{3}} = \pm \sqrt{ \frac{2}{3}} = \frac{\sqrt6}{3}$
Because  $\sin\frac{x}{2}$  is +ve in given quadrant

$\tan\frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{\sqrt6}{3}}{\frac{-\sqrt3}{3}} = - \sqrt2$

Find    $\small \sin\frac{x}{2} , \cos\frac{x}{2} , and \tan\frac{x}{2}$    in

Q (8)

$\small \tan x = - \frac{4}{3}$        , x in quadrant  II

S seema garhwal

tan x = $-\frac{4}{3}$
We know that ,
$\sec^{2}x = 1 + \tan^{2}x$
$= 1 +\left ( -\frac{4}{3} \right )^{2}$
$= 1 + \frac{16}{9}$  =  $\frac{25}{9}$
$sec x = \sqrt{\frac{25}{9}}$ = $\pm\frac{5}{3}$
x lies in II quadrant  thats why sec x is -ve
So,

$sec x =-\frac{5}{3}$
Now,  $cos x = \frac{1}{\sec x}$  =  $-\frac{3}{5}$
We know that,
$cos x = 2\cos^{2}\frac{x}{2}- 1$                                                                               ($\because \cos2x = 2\cos^{2}x - 1 \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1$  )
$-\frac{3}{5}+ 1 = 2$   $\cos^{2}\frac{x}{2}$

=  $\frac{-3+5}{5}$    =    $2\cos^{2}\frac{x}{2}$

$\frac{2}{5}$   =    $2\cos^{2}\frac{x}{2}$
$\cos^{2}\frac{x}{2}$ =  $\frac{1}{5}$
$\cos\frac{x}{2}$   = $\sqrt{\frac{1}{5}}$  = $\pm\frac{1}{\sqrt5}$
x lies in II quadrant so value of           $\cos\frac{x}{2}$     is +ve

$\cos\frac{x}{2}$   =  $\frac{1}{\sqrt5} = \frac{\sqrt5}{5}$
we know that
$cos x =1 - 2\sin^{2}\frac{x}{2}$

$2\sin^{2}\frac{x}{2}$  =  1 -  $(-\frac{3}{5})$   =  $\frac{8}{5}$

$\sin^{2}\frac{x}{2} = \frac{4}{5}\\ \\=\sin\frac{x}{2} = \sqrt{ \frac{4}{5}} = \pm \frac{2}{\sqrt{5}}$
x lies in II quadrant So value of sin x is +ve

$\sin\frac{x}{2} = \frac{2}{\sqrt{}5} = \frac{2\sqrt5}{5}$

$\tan \frac{x}{2} = \frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} = \frac{\frac{2\sqrt5}{5}}{\left ( \frac{\sqrt5}{5} \right )} = 2$

Q(7)   Prove that

$\small \sin3x + \sin2x - \sin x = 4\sin x \cos\frac{x}{2}\cos\frac{3x}{2}$

S seema garhwal

We know that
$cosA + cosB = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$sinA - sinB = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

we use these identities
$sin3x - sinx = 2\cos\frac{3x+x}{2}\sin\frac{3x-x}{2}$

$= 2\cos2x\sin x$

sin2x + $2\cos2x\sin x$  =   2sinx cosx  + $2\cos2x\sin x$
take 2 sinx common
$2sinx ( cosx + cos2x) = 2sinx(2\cos\frac{2x+x}{2}\cos\frac{2x-x}{2})$

$= 2sinx(2\cos\frac{3x}{2}\cos\frac{x}{2})$
$= 4sinx\cos\frac{3x}{2}\cos\frac{x}{2}$

=  R.H.S.

Q (6)  Prove that

$\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)} = \tan6x$

S seema garhwal

We know that

$sinA + sinB = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
and
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$

We use these two identities in our problem

sin7x + sin5x  =  $2\sin\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$    =       $2\sin6x\cos x$

sin 9x + sin 3x = $2\sin\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$  =    $2\sin6x\cos 3x$

cos 7x + cos5x = $2\cos\frac{7x+5x}{2}\cos\frac{7x-5x}{2}$   =   $2\cos6x\cos x$

cos 9x + cos3x = $2\cos\frac{9x+3x}{2}\cos\frac{9x-3x}{2}$  =$2\cos6x\cos 3x$

$\small \frac{(\sin 7x + \sin 5x) + (\sin9x + \sin 3x)}{(\cos7x + \cos5x) + (\cos9x + \cos3x)}$      =      $\small \frac{(2\sin 6x\cos x) + (2\sin6x \cos3x)}{(2\cos6x cos x) + (2\cos6x cos3x)}$

=     $\small \frac{2\sin6x(\cos x + \cos3x)}{2\cos6x (cos x + cos3x)} = \tan6x$       = R.H.S.                       $\small \left ( \because \frac{\sin x}{\cos x} = \tan x\right )$

Q(5)  Prove that

$\small \sin x + \sin 3x + \sin 5x + \sin 7x = 4\cos x\cos2x \sin4x$

S seema garhwal

we know that
$sinA + sinB =2\sin\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this identity in our problem
If we notice we need sin4x in our final result so it is better if we made a combination of sin7x and sin x , sin3x and  sin5x tp get sin4x

$(sin7x + sinx) + (sin5x + sin3x) = 2\sin\frac{7x+x}{2}\cos\frac{7x-x}{2}$ $+2\sin\frac{5x+3x}{2}\cos\frac{5x-3x}{2}$
$=$$2\sin4x\cos3x + 2\sin4x\cos x$
take 2sin4x common
= 2sin4x(cos3x + cosx)
Now,
We know that
$cosA + cosB =2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
We use this
$cos3x + cosx =2\cos\frac{3x+x}{2}\cos\frac{3x-x}{2}$
= $2\cos2x\cos x$
= 2sin4x($2\cos2x\cos x$)
= 4cosxcos2xsin4x = R.H.S.

Q(4)  Prove that

$\small (\cos x-\cos y)^{2} + (\sin x - \sin y)^{2} = 4\sin^{2}\left ( \frac{x-y}{2} \right )$

S seema garhwal

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$
and
$(\cos x-\cos y)^{2} = \cos^{2}x - 2\cos x\cos y + \cos^{2}y$

$\small (\cos x - \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x - 2\cos x\cos y + \cos^{2}y$ +   $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 - 2cosxcosy + 1 - 2sinxsiny                           $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 - 2(cosxcosy + sinxsiny)
= 2 - 2cos(x - y)                                                   $\small (\because \cos(x-y) =\cos x \cos y + \sin x \sin y)$
=  2(1 - cos(x - y) )
Now we can write
$cos(x + y) = 1 -2sin^{2}\frac{(x + y)}{2}$                                             $\left ( \because \cos2x = 1 - 2\sin^{2}x \ \Rightarrow \cos x = 1 - 2\sin^{2}\frac{x}{2} \right )$

so

$2(1 - cos(x - y) ) = 2(1 - ( 1 -2sin^{2}\frac{(x + y)}{2}))$

$= 4sin^{2}\frac{(x - y)}{2}$   =  R.H.S.

Q (3) Prove that

$\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2} = 4 \cos^{2}\left ( \frac{x+y}{2} \right )$

S seema garhwal

We know that $(a+b)^{2} = a^{2} + 2ab + b^{2}$
and
$(a-b)^{2} = a^{2} - 2ab + b^{2}$
We use these two in our problem

$(\sin x-\sin y)^{2} = \sin^{2}x - 2\sin x\sin y + \sin^{2}y$
and
$(\cos x+\cos y)^{2} = \cos^{2}x + 2\cos x\cos y + \cos^{2}y$

$\small (\cos x + \cos y)^{2} + (\sin x - \sin y)^{2}$ = $\cos^{2}x + 2\cos x\cos y + \cos^{2}y$ +   $\sin^{2}x - 2\sin x\sin y + \sin^{2}y$
= 1 + 2cosxcosy + 1 - 2sinxsiny                           $\left ( \because \sin^{2}x + \cos^{2}x = 1\ and \ \sin^{2}y + \cos^{2}y = 1 \right )$
= 2 + 2(cosxcosy - sinxsiny)
= 2 + 2cos(x + y)
=  2(1 + cos(x + y) )
Now we can write
$cos(x + y) =2cos^{2}\frac{(x + y)}{2} - 1$                                             $\left ( \because \cos2x = 2cos^{2}x - 1 \ \Rightarrow \cos x = 2\cos^{2}\frac{x}{2} - 1\right )$

=   $2(1 + 2cos^{2}\frac{(x + y)}{2} - 1)$
$=4cos^{2}\frac{(x + y)}{2}$

=  R.H.S.

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