Two lines are perpendicular if the product of their slopes is -1 or one has a slope of 0 (a horizontal line) and the other has an undefined slope (a vertical line).

Refer: http://en.wikipedia.org/wiki/Slope

Discuss various components of laser and explain its working taking into account terms like stimulated emission, optical inversion, photon multiplication and cavity oscillations.

Diagonal relationship exists between certain pairs of diagonally adjacent elements in the second and third periods of the periodic table.

These pairs (Li & Mg, Be & Al, B & Si etc.) exhibit similar properties.

REASON -

Such a relationship occurs because crossing and descending the periodic table have opposing effects. On moving across a period of the periodic table, the size of the atoms decreases, and on moving down a group the size of the atoms increases. Similarly, on moving across the period, the elements become progressively more covalent, less basic and more electronegative, whereas on moving down the group the elements become more ionic, more basic and less electronegative. Thus, on both descending a group and crossing by one element the changes "cancel" each other out, and elements with similar properties which have similar chemistry are often found - the atomic size, electronegativity, properties of compounds (and so forth) of the diagonal members are similar

UNCERTAINTY PRINCIPLE

In quantum mechanics, the Heisenberg uncertainty principle states that certain pairs of physical properties, like position and momentum, cannot simultaneously be known to arbitrary precision.

(delta x) . (delta p) > h / 4 pi

or, (delta x) . (m) (delta v) > h / 4 pi

or, (delta x) . (delta v) > h / 4 m pi

So, for a massless particle, say photon, m = 0

So, (delta x) . (delta v) > infinity

@sahell, photon is massless.

@canorous i hope you have got the point.

Nudge me if you still have any doubt regarding this.

tan(3A) = tan(2A + A) =(  tan(2A) + tan(A) ) / ( 1 - tan(2A) tan(A) )

tan(3A) - tan(A) tan(2A) tan(3A) = tan(2A) + tan(A)

tan(3A) - tan(2A) - tan(A) = tan(A) tan(2A) tan(3A)

I think there's some problem in the question. Its not 4A but 2A.

f(x) = sin(sin^-1x)

or, f(x) = x

So, the graph is a straight line passing through origin and inclined at an angle of pi/4 from X-axis.

Ans is 1/100..........

The ball will reach maximum height (h) when velocity of ball is zero.

Using, ( 0 )² - u² = 2 ( -g ) h

So,  h  = u² / 2 g = ( 100 )² / 2 g

So, velocity of ball at half the max. height ( h / 2 ) will be given by,

v² - u² = 2 ( - g ) ( h / 2 )

or, v² = ( 100 )² - ( 100 )² / 2

or, v = 100 ( 1 / 2 )^1/2 = 50 ( 2 )^1/2   (Ans)............

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

      = 2x - 3 , x > 2

               1   , 1 < x < 2

        - 2x + 3 , x < 1

Clearly, range of f(x) is [1,5].

f(x) = |x| is the modulus function.

or, f(x) = x , x > 0

               0 , x = 0

              -x . x < 0

Clearly now x can take any real value, so domain of function is all real values.

Now for any real x, f(x) is always positive so range of f(x) is [0,infinity).

It is quite clear from graph of modulus function also. ( See Below )

Using triangular inequality,

So, minimum ( x² + y² ) = 1

Solving the equaitons now, you'll get

x = 5 / 13 and y = - 12 / 13

If an equation f(x) = c has a root x = a of multiplicity 2 then f'(a) = 0. If the root has multiplicity greater than 2, then also f"(a) = 0.

As, x(x + 1)(x + 2) .............. (x + 2009) = c

Taking log on both sides,

Let .

or,

Also,

Since all the terms in the sum have the same sign, it's clear that their sum cannot be zero.

So has no roots of multiplicity greater than 2.

To find the number of roots of multiplicity 2, note that the derivative of a function has a root between each pair of consecutive roots of the function. Since x(x+1)(x+2)...(x+2009) = 0 has 2010 roots, its derivative will have 2009 roots.

I = int ( (1 + cos4x) / (cotx + sinx) ) . dx

I = 2 int ( (cos²2x) ( - sinx) / (cos²x - cosx  - 1) ) . dx

Put y = cosx

so, dy = - sinx . dx

Now, I = 2 int ( (2y² -1)² / ( y² - y - 1 ) . dy

I = 2 int ( (4y²)(y² - -y - 1 + y) + 1) / ( y² - y -1 ) ) . dy

I = 8 int y² . dy + 8 int ( y³ / ( y² - y - 1 ) ) . dy + 2 int ( 1 / ( y² - y - 1 ) ) . dy

I = 8 int y² . dy + 8 int ( 2y + 1 / ( y² - y - 1 ) ) . dy + 8 int (y + 1) . dy+ 2 int ( 1 / ( y² - y - 1 ) ) . dy

I = 8 int y² . dy + 8 int ( 2y - 1 / ( y² - y - 1 ) ) . dy + 2 int ( 1 / ( y² - y - 1 ) ) . dy + 8 int (y + 1) . dy+ 2 int ( 1 / ( y² - y - 1 ) ) . dy

Now, you can integrate. If you still have any problem nudge me.

2. Acute angle between lines x² + 4xy + y² = 0 is tan^-1(3)^1/2 = pi / 3

Angle bisectors of x² + 4xy + y² = 0 are given by

x² - y² = xy

 1 - 1       2

or, x = + y

As, x + y = 0 is perpendicular to x - y = 4, given triangle is isosceles with vertical angle equal to pi / 3 and hence it is equilateral.

1. Let us assume that the common line is y=mx. So, it satisfies the two homogeneous equations.

So, am² + 2m + 1 = 0

and , m² + 2m + a = 0

Solving them,

    m²      =      m                =          1              

2(1 - a)        a² - 1                   2(1 - a)

or, m² = 1 and m = -( a + 1)/2

or, (a + 1)² = 4m² = 4

So, a = -3 ( bcoz when a = 1 the two pairs have both the lines common)

and therefore, m = 1

Now given pairs of equations become,

x² + 2xy - 3y² = 0

or, (x - y) (x + 3y) = 0

and -3x² + 2xy + y² = 0

or, (x - y) (-3x - y) = 0

So, required equation is - (x + 3y) (3x + y) = 0

or, 3x² + 10xy + 3y² = 0

Actually the given function is not defined when cos^-1(cos(x)) > sin^-1(sin(x)).

Here is a comparison of these two functions through graph.

Clearly see that from [0,pi/2] these functions are equal and from (pi/2,2pi) cos^-1(cos(x)) > sin^-1(sin(x)) and so our functions in not defined. Here i am only able to show the graph from (-pi,pi). 

Similarly on negative x-direction our function is not defined in (-3pi/2,0) and so on.

So, our function is defined in the interval ..........(-4pi,-7pi/2)U(-2pi,-3pi/2)U(0,pi/2)U(2pi,5pi/2)U(4pi,9pi/2)............

and it is the domain of given function.

Let us assume that the particle takes time ' t ' sec. after being dropped from rest from height ' h ' to reach the ground.

As it covers 9h/25 in last second,

so the particle covers (h - 9h/25) = 16h/25 in ' t - 1 ' sec.

Now, using equations of motion

16h/25 = (1/2)g( t - 1)²  ------ (1)

Also, h = (1/2)g( t )² ------- (2)

Solving (1) and (2),

(16/25)t² = (t - 1)²

or, 4t = + (5t - 5)

or, t = 5 sec. ( t = 5/9 is rejected because in question it is given that t>1 as the bosy covers 9h/25 in last second)

So, h = (1/2)g (5)² = (25)(9.8) / 2 = 122.5 m (Ans)..................

yeah thnx himsy..i was in a hurry nd didn't noticed this one.

I am continuing the question from second last step now,

cos(y) = ( sec(x) ) [ ( d / 2b ) - ( a / 2b ) ( sin(x) ) ]

This value will be minimum when sin(x) is maximum i.e. 1 but when sinx = 1, secx is not defined.

So, this expression has no global minima according to me.

Nudge me if you still have any problem..........