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..........

P(x) = x³ - 2x³ - 3x = - x³ - 3x

Put P(x) = 0,

So, - x³ - 3x = 0

or,  x³ + 3x = 0

or, x (x² + 3) = 0

So, the only real value of x is 0................... (Ans)...............

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If there might be any typing mistake and the question is

P(x) = x³ - 2x² - 3x

So, again put P(x) = 0

So, x³ - 2x² - 3x = 0

or, x ( x² - 2x - 3) = 0

So, x = 0 .............. or  x² - 2x - 3 = 0

x = 0 ..................... or  (x - 3) (x + 1) = 0 implies x = 3 or x = -1.

So, real values of x are 0 , -1 and 3................. (Ans)....................

f(x) = x² - 6x + 25

f(x) = (x² - 6x + 9) + 16

f(x) = (x - 3)² +  16

This is the standard form of the given quadratic equation.

a sin(x) + b cos( x + y ) + b cos( x - y ) = d

or, a sin(x) + b [ 2 cos(x) cos(y) ] = d

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

or, cos(y) = [ d - a sin(x) ] / 2 b (1 - sin²x)^1/2

Now first term is minimum when sin(x) = 0 and second term is minimum when sin(x) = 1

So, min (cos(y)) = (d / 2b) - 0 = d / 2b

min( |cos(y)| = | d / 2b |

The number of equivalence relations on any set A having n number of elements is the number of partitions of A. Let T(n), called the bell mumber, be the number of partition on A. Clearly T(0) = 1. In general, one can fix a particular element x belonging to A, and calculate the number of partitions by increasing the number of elements in [x] from 1 to n by

Applying l-hospital's rule

lt x-> infinity 5x⁴ / 5^x log(5)

= (5/log(5)) ( lt x-> infinity x⁴ / 5^x )

Apply l-hospital's rule again and again and finally you will get

= 0

Hope its clear now.

Let us first integrate without limit,

I = int (2 sec²x + x³ + 2). dx

I = 2 tanx + (1/4)x⁴ + 2x

Now put the limits,

So, I = 2 (tan(pi/4) - tan0) + (1/4) ((pi/4)⁴ - 0) + 2(pi/4 - 0)

I = 2 + (1/4) (pi/4)⁴ + pi/2 (Ans)..........

As the three points are collinear,

So, (a) (2a - 1) - (2) (a + 1) = 0

or, 2a² - 3a - 2 = 0

or, (2a - 1) (a + 2) = 0

or, a = 1/2 , or a = -2 (Ans)......

Let point b = (6,k).

As, ab < 4

So, (ab)² < 16

or, (6 - 3)² + (k - 4)² < 16

or, (k - 4)² < 7

So,  - (7)^1/2 < k - 4 <  (7)^1/2

or, 4 - (7)^1/2 < k < 4 + (7)^1/2

or, 1.35 < k < 6.64

Since k is an integral value,

So, K = 2,3,4,5,6

So, point b has 5 positions namely (2,6) , (3,6), (4,6), (5,6), (6,6).

Just see the limit is approaching infinity.

So, alwyas remember 1/infinity approches to 0.

So, we try to form such an equation in which such a term comes.

For eg. In questions like lt x --> infinity x² + x / x² + 2x + 2

If highest power of both N and D are same, divide the N and D both by that power.

lt x--> infinity 1 + (1/x) / 1 + (2/x) + 2/x² = 1 / 1 = 1

In the given question asked by you, divide both N and D by n^3/2.

Ans = 0

I = int x⁵ . dx / x² + 1

I = int (x²)² x . dx / x² + 1

Put y = x² + 1

so, dy = 2x dx

or, x.dx = dy / 2

So, I = int (y - 1)² . dy / 2y = int (y - 2 + y^-1) . dy / 2

I = (1/2) ( (y²/2) - 2y + logy ) + C

I = (1/2) ( (x² + 1)²/2) - 2(x² + 1) + log(x² + 1) ) + C

I = ∫(e^1/x+1) / x²  dx

Put y = 1/x

So, - dy = dx / x²

So, I = - ∫ (e^y + 1)dy

I = - (e^y + y) + C

I = - e^1/x - (1/x) + C

@||Mr.Insomaniac|| and siDDhESwaR MuKhErjEE

You say that cos(x) = -1/2 has two solutions.

But just for a minute check the domain of x for which i have taken the second case. x is in 2nd and 4th quadrant. cos(x) is positive in 4th quadrant. so, there is only one solution i.e. 2pi/3 in 2nd quadrant.

The second solution you are talking about is x = 4pi/3 which lies in third quadrant. So, it is not the solution.

Anyhow i have made the graph there you can clearly see thta there's only one solution.

Hope you agree with me this time.................

Let u = f(cos(x))

So, du/dx = f ' (cos(x)) (-sin(x))

|du/dx|_x=pi/3 = f ' (1/2) (-(3)^1/2 / 2) = (3)^1/2 [ because f ' (1/2) = -2 is given] ------- (1)

Let v = g(sin(x))

So, dv/dx = g ' (sin(x)) (cos(x))

|dv/dx|_x=pi/3 = g ' ((3)^1/2/2) (1 / 2) = -1 [ because g ' ((3)^1/2/2) = -2 is given] ------- (2)

Dividing (1) by (2),

du/dv|_x=pi/3 = - (3)^1/2 (Ans)

So, correct option is (d).

DIFFERENTIAL COEFFICIENT

Differential coefficient of a function f(x) is the derivative d(f(x))/dx, the (not necessarily constant) multiplicative factor or coefficient of the differential dx in the differential d(f(x)).

I = int dx / x log(x)

Put y = log(x)

so, dy = dx / x

So, I = int dy / y

I = log|y| + C

I = log(log(y)) + C