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Thank u atul :)

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I guess, the question is already solved.

Nudge me, if its not

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What has to be done with this ?
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|x-3| should not be 0 => x should not = 3

Also, 3x2 - 10x + 3 = 0

=>  3x2 - 9x - x + 3 = 0
=>  (3x-1)(x-3) = 0
=>   x = 1/3

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
beautifully solved Jagdish :)

Thanks for sharing the work

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
practice as much as u can, and u'll develop ur own tricks and strategies to solve the question
 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
y2+ 2xy + 40|x| = 400

Case 1: x > 0
y2+ 2xy + 40x = 400
=>  (y2+ 2xy + x2) - (x2- 40x + 400) = 0
=> (y+x)2 - (x-20)2 = 0
=>   (y+2x-20) (y+20) = 0

The two lines are shown on the graph, in green. Since x> 0 , the area bounded is that by the triangle PQC.

Similarly, when x< 0 (Case 2) ,
The two lines are those shown in red, and the area bounded is that by triangle APQ

therefore, the area bounded = 20X 40 = 800 sq units

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

x2+y2-6x-10y+c=0
=>  (x-3)2+ (y-5)2 = 34-c

Now(1,4) is an interior point
=>  (1-3)2+ (4-5)2 -(34-c) < 0
=>  c < 29

The circle doesnt touch the co-ordinate axes
=>the distance of centre from each of the axis > radius of circle
=> sqrt (34-c) < 3    and  sqrt (34-c) < 5
=> c> 25

combining, 25< c < 29

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teachers are not really necesaary for solving the doubts, u can ask ur friends or some othr batch mates, to discuss the problems with u, n solve them thru discussion only.

the choice between the two, depends on u totally... u can post ur doubts here at goiit.com ...we'll try to solve it for u :)

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

whats the question???

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

are u sure, thr is no other condition specified??

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

well done hemang, beautifully solved :)

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]

for board exams, go thru the NCERT book, and for othr exams, try arihant  or TMH!

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Let z be a complex number satisfying z4 + z3 + 2z2 + z + 1 = 0, then find the value of |z|

z4 + z3 + 2z2 + z + 1 = 0
=> (z4 + 2z2 + 1) + (z3+z) = 0
=>   (z4 + 2z2 + 1) + z(z2+1) = 0
=> (z2+1)2 + (z2+1) = 0
=> (z2+1)(z2+z+1) = 0

Case 1:
(z2+1)= 0
=>  z = + i or -i

Case 2:
z2+z+1 = 0
=> (z2+z+ 1/4) + 3/4 = 0
=> (z+1/2)2 = -3/4
=>  (z+1/2) =(+/-) i (sqrt3)/2
=>   z =(-1/2) (+/-) i (sqrt3)/2

 Discussion Forums -> This Post 0 points    (0    in 0 votes )   [?]
from the figure,  tan-1x  = @ (assume)
then sin(tan-1x) = sin @ = x/(1+x2)

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go thru any text book, NCERT is the most basic one...arihant, TMH ne will do!

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whr do u want to store the formulas, prasanth?

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