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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.

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G Gautam harsolia
Answered Jun 06, 2019

Given function is

Given function is satisfies for the all real values of x
case (i)  k < 0


Hence, function is continuous for all values of x < 0

case (ii)  x = 0

L.H.L at x= 0

R.H.L. at x = 0

L.H.L. = R.H.L. = f(0)
Hence, function is continuous at x = 0

case (iii)  k > 0


Hence , function is continuous for all values of x > 0

case (iv) k < 1


 Hence , function is continuous for all values of x < 1

case (v)  k > 1


 Hence , function is continuous for all values of x > 1

case (vi)  x = 1



Hence, function is not continuous at x = 1

Q.6 Write the following as intervals :

(i) {x : x \inR, – 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}

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S safeer
Answered May 05, 2019

The following can be written in interval as : 

(i) {x : x \inR, – 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?

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S safeer
Answered May 05, 2019

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

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S safeer
Answered May 05, 2019

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}

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S safeer
Answered May 05, 2019

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}

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S safeer
Answered May 05, 2019

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}

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S safeer
Answered May 05, 2019

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

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S seema garhwal
Answered May 05, 2019

\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

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S seema garhwal
Answered May 05, 2019

\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

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S seema garhwal
Answered May 05, 2019

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}

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S seema garhwal
Answered May 05, 2019

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

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S seema garhwal
Answered May 05, 2019

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

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S seema garhwal
Answered May 05, 2019


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 )

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S seema garhwal
Answered May 05, 2019

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 )

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S seema garhwal
Answered May 05, 2019

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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