please solve the following questions frm ncert

part 1

class xii

q no 19,23,32,33,34........

one salute 4 each post....

please reply at the earliest with all the solutions in detail.......

Q19

LHL

(lim x tends to integer){x-h}=1-h

RHL

(lim x tends to integer){x+h}=h

LHL not equal to RHL hence discontinuous

LHL

(lim h tends to 0)sin(-h)/(-h)=sinh/h=1

RHL

(lim h tends to 0)(h+1)=1

f(0)=1

hence no discontinuity

point of discontinuity can be only (2n+1)(pi/2)

LHL

lim(x tends to pi/2)|cosx|

(lim h tends to 0)|cos(pi/2+h)|=0

RHL

(lim h tends to 0)|cos(pi/2-h)|=0

f(pi/2)=0

hence continuous

|x| changes its nature at 0

LHL

(lim h tends to 0)sin|h|=sin|0|=0

RHL

(lim h tends to 0)sin|-h|=sin|h|=sin|0|=0

f(0)=0

hecne continuous

Q34

for x>0

x-(x+1)=-1 so continuous everywhere

-1<x<0

-x-(x+1)=-2x-1

x<-1

-x+(x+1)=1

so we have to check continuoty on -1 and 0

for zero

LHL

(lim h tends to 0)-1=-1

RHL

(lim h tends to 0)2(h)-1=-1

f(0)=-1

hence continuous at x=0

for -1

LHL

(lim h tends to 0)1=1

RHL

(lim h tends to 0)2(-1+h)-1=-3+2g=-3

LHL is not equal to RHL hence not continuous at x=-1

for q 19

u can also draw the graph of {x}

u will find that it is discontinious at all integral points

for q 32

again draw graph of |cosx|, only all the branches of cos x will come above the x-axis

and since it does  not  break at any point it is continious

ankit NCERT mein tum graph draw karke reply nahin kar sakte warna ye saare question kuch nahin the

@ genius

 i was just providing the alternate method