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Since its an bijective function it should be one-one and onto.
For one-one it should either be increasing or decreasing.
f(x) = ax + cosx
f'(x) = a - sinx
If f(x) is an increasing function f'(x) > 0 implies a > sinx implies a > 1
If f(x) is a decreasing function f'(x) < 0 implies a < sinx implies a < -1
Hence a R-(-1,1)
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I would suggest you to just go with your school studies and make your basics clear. Also you can buy fundamental books of 11th and 12th and can learn the topics which are in your 9th and 10th. But at this stage you should concentrate on the basics and also can solve some mind teasing problems as it would help to improve your analyticity.
All the best and do study well.
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d(tan-1x)/dx = [ h] [ 0] [tan-1(x+h) - tan-1x]/h
= [ h][ 0] sin2y - hsin4y/2 + h2sin6y/3 + ............
Neglecting terms which have higher powers of h as h tends to 0.
d(tan-1x)/dx = sin2y
1/(1+x2) = sin2y
siny = (1+x2)-1/2
coty = x
Hence y = cot-1x (c) option is correct.
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A+B+C =  tan(A/2 + B/2) = tan( /2 - C/2) = cotC/2 Arranging it we get tanA/2.tanB/2 + tanB/2.tanC/2 + tanC/2.tanA/2 = 1 Now tan2A/2 + tan2B/2 + tan2C/2 - 1 = tan2A/2 + tan2B/2 + tan2C/2 - (tanA/2.tanB/2 + tanB/2.tanC/2 + tanC/2.tanA/2) = 1/2[(tanA/2-tanB/2)2 + (tanB/2-tanC/2)2 + (tanC/2-tanA/2)2 >= 0 Hence tan2A/2 + tan2B/2 + tan2C/2 - 1 >= 0 tan2A/2 + tan2B/2 + tan2C/2 >= 1
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See actually no such marks are alloted in JEE to the chapters specifically. But you can flick through the previous JEE papers and can analyse those papers that certain topics are important and have greater weightage. My suggestion is that you should go for each topic and grasp its concepts.
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Such type of problems are analytic and there is no specific method, so practicing such problems and analyzing them is the best way to cope with them.
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Lets put x = 10.4
It satisfies [x+ 1/2 ] = [2x] - [x]
[ x + 1/3 ] + [x + 1/2 ] + [x] = 30
and [3x] = [31.2] = 31
So I dont think that option a) is correct.
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Differentiate x wrt to t and put t=0 to get the initial velocity.It comes out c.
Double differentiate x wrt to t and put t=0 to get the initial acceleration.It comes out 2b.
Hence the required ratio = c/2b
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If the forces form the sides of a triangle their resultant would be zero and hence the body will not accelerate.
Use triangle property a+b>c for all the sides.
Only (5N,10N,12N) forms the sides of a triangle.
Hence correct option is a).
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No doubt that DON is one of the good students in this forum and he has done it perfectly.
But I liked the way in which Arvind solved.
Definitely answer is 6.
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The conductors which connect the two resistors offer no resistance to the current and hence the voltage drop across each resistor is same.
Now current is divided because the higher resistance would allow less current to flow through and the remaining current would flow through the other lower resistance(this current is greater than that of flowing through higher resistor).
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(a + b + c) = 0
(a + b + c)2 = a.a + b.b + c.c + 2(a.b + b.c + c.a)
2(a.b + b.c + c.a) + 1+1+1 = 0
(a.b + b.c + c.a) = -3/2
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For algebra : Arihant, A Dasgupta
For Trigonometry : S. L. Loney
For organic chemistry : NCERT book, Morrison & Boyd, Sanyal.
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The possible combinations and no. of ways for product of three integers to be 30 are given as :
(1,1,30) no. of ways = 3!/2! = 3
(1,2,15) no. of ways = 3! = 6
(1,3,10) no. of ways = 3! = 6
(1,5,6) no. of ways = 3! = 6
(2,3,5) no. of ways = 3! = 6
Hence total no. of ways = 27
Please post your other queries on a new page.
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Yaar see.......obviously IIT is the best option and you should dedicate yourself towards JEE preparation but still you should handle class 12th side by side. Actually you should manage your time in such a way that you can prepare for both the exams. If your JEE preparations are good then obviously it will ad to your confidence which you can use for 12th preparations.
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Single slit diffraction :
When light from such a slit is focussed on a distant screen, an interference pattern like the one plotted at right appears. The angular width of the whole pattern increases as the slit width decreases.
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For physics H C Verma is the best, build your concepts and then go for Arihant.
For maths solve some elementary problems from R D Sharma and then go for arihant. A Dasgupta is another good book.
For chemistry go through NCERT books and then P Bahadur for numericals and J D Lee for inorganic(for details) otherwise O P Agrawal is also good.
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r(t) = (sint)i + (t2 - cost)j + (et)k
r(0) = -j + k
r'(t) = (cost)i + (2t+sint)j + et)k
r'(0) = i + k
Hence it passes through r(0) and its direction is in r'(0)
Equation of line :
r = r(0) +  r'(0)
r =-j + k + (i + k)
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KAB is right. Its computation involves polylograthmic function which is out of scope of JEE.
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Charged conductors which have reached electrostatic equilibrium share a variety of unusual characteristics. One characteristic of a conductor at electrostatic equilibrium is that the electric field anywhere beneath the surface of a charged conductor is zero. If an electric field did exist beneath the surface of a conductor (and inside of it), then the electric field would exert a force on all electrons that were present there. This net force would begin to accelerate and move these electrons. But objects at electrostatic equilibrium have no further motion of charge about the surface. So if this were to occur, then the original claim that the object was at electrostatic equilibrium would be a false claim. If the electrons within a conductor have assumed an equilibrium state, then the net force upon those electrons is zero. The electric field lines either begin or end upon a charge and in the case of a conductor, the charge exists solely upon its outer surface. The lines extend from this surface outward, not inward. (This of course presumes that our conductor does not surround a region of space where there was another charge.)
To illustrate this characteristic, let's consider the space between and inside of two concentric, conducting cylinders of different radii as shown in the diagram at the right. The outer cylinder is charged  positively. The inner cylinder is charged negatively. The electric field about the inner cylinder is directed towards the negatively-charged cylinder. Since this cylinder does not surround a region of space where there is another charge, it can be concluded that the excess charge resides solely upon the outer surface of this inner cylinder. The electric field inside the inner cylinder would be zero. When drawing electric field lines, the lines would be drawn from the outer surface of the inner cylinder to the inner surface of the outer cylinder. For the excess charge on the outer cylinder, there is more to consider than merely the repulsive forces between charges on its surface. While the excess charge on the outer cylinder seeks to reduce repulsive forces between its excess charge, it must balance this with the tendency to be attracted to the negative charges on the inner cylinder. Since the outer cylinder surrounds a region which is charged, the characteristic of charge residing on the outer surface of the conductor does not apply.
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