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let y=2cos^z+cosz-1

comp. it with ax^2+bx+c

here a=2
therefore it is having minima @ -b/2a= -1/4

this imp cosz=-1/4

2*(-1/16)-1/4-1

=-9/8

ans= -9/8

we want more..

no only fr bengal

for more detaile check jee 2008 brochure

lol
this info is abt bengal jee not IIT

ans2:
let E be mid pt of AB
& F be mid pt of AC
B(x1,y1)
C(x2,y2)
E=((x1+1)/2,(y1-2)/2)
F=((x2+1)/2,(y2-2)/2)
putting in perp eq.
x1-y1=-13
x2+2y2-3=0
slope of AB=-1 AC= 2
so x1+y1+1=0
2x2-y2-4=0
so
B=(-7/6)
C(11/5,2/5)
eq of BC 14x+23y-40=0

ans 3:
3:4 internally
firsty the lines r on opposite side of the orgin
op=9/2root5
oq=6/root5
so op / oq=3/4

ans 1:
int. pt. of lines 7x-2y+10=0 & 7x+2y-10=0 is(0,5)
n for y=2 n 7x-2y+10=0 is (-6/7,2) n for y=2 & 7x+2y-10=0 is (6/7,2)
now find the area

www.graphcalc.com

www.grapgcalc.com

they r only same if m1=m2

anyone can be used if not specified
but half rxn should be preffered if u r comfortable with it

see in 2nd u r multiplying n divding -1 to -1/0 so u r getting infinity

for sure the answer is -infinity

in feb
ur school has to decide the dates

physical-p bahadur or rc mukherjee

inorganic-ncert is enough but jdlee is really much if u want