Messages posted by Manasi

Thank u atul :)

I guess, the question is already solved.

Nudge me, if its not

What has to be done with this

?

|x-3| should not be 0 => x should not = 3

Also, 3x² - 10x + 3 = 0

beautifully solved Jagdish :)

Thanks for sharing the work

practice as much as u can, and u'll develop ur own tricks and strategies to solve the question

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

x²+y²-6x-10y+c=0

=> (x-3)²+ (y-5)² = 34-c

=> radius = sqrt (43-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

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 :)

whats the question???

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

well done hemang, beautifully solved :)

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

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

from the figure, tan^-1x = @ (assume)

then sin(tan^-1x) = sin @ = x/(1+x²)

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

whr do u want to store the formulas, prasanth?

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