Messages posted by devilguy_deepak@yahoo.com

f(x) = (x - 3) / (3 - x)

f(x) = (-1)(3 - x) / (3 - x)

f(x) = -1

So, range of the function is -1.

Analysis: It is a constant function i.e for all x belongs to R, f(x) = -1.

|cot(x)| = cot(x) + 1/sin(x)

|cot(x)| = cot(x) + cosec(x)

Now, clearly note that |cot(x)| = cot(x) , when x belongs to 0 to pi/2 or x belongs from pi to 3pi/2

= - cot(x), when x belongs to pi/2 to pi or x belongs from 3pi/2 to 2pi

CASE1. When x belongs to 0 to pi/2 or x belongs from pi to 3pi/2,

cot(x) = cot(x) + cosec(x)

or, cosec(x) = 0 which is not possible.

CASE2. When x belongs to pi/2 to pi or x belongs from 3pi/2 to 2pi

-cot(x) = cot(x) + cosec(x)

(cosec(x))(1 + 2 cos(x)) = 0

cosec(x) # 0

so, 1 + 2cos(x) = 0

or, cos(x) = -1/2

so, x = 2pi/3 (because x belongs to pi/2 to pi or x belongs from 3pi/2 to 2pi)

Hence there is only one solution.

You can see it from the graph even---

clearly there is only one solution, in 0<x<2pi.

( 5 + 2(6)^1/2 ) ( 5 - 2(6)^1/2 ) = 25 - 24 = 1

So, ( 5 + 2(6)^1/2 ) = 1 / ( 5 - 2(6)^1/2 )

According to equation now,

Let ( 5 + 2(6)^1/2 )^{(x^2 - 3)} = y

So, y - (1/y) = 10

or, y² - 10y - 1 = 0

So, y = (10 + (104)^1/2) / 2 = 5 + (26)^1/2

So, x² - 3 = log (5 + (26)^1/2) / log (5 + 2(6)^1/2)

Therefore, x = [ { log (5 + (26)^1/2) / log (5 + 2(6)^1/2) } + 3 ]^1/2 (Ans)...........

(4/9)^x + (6/9)^x- 1 = 0

or (2/3)^2x + (2/3)^x - 1 = 0

Put (2/3)^x = y

So, y² + y - 1 = 0

y = (-1 + (5)^1/2) / 2

or, (2/3)^x = (-1 + (5)^1/2) / 2

Taking log on both sides,

x log(2/3) = log (-1 + (5)^1/2) - log(2)

x = [ log (-1 + (5)^1/2) - log(2) ] / [ log(2/3) ] (Ans)............

The slope of tangent at any point of parabola is equal to the slope of parabola ta that point. So, the angle between parabola and tangent is 0.

A function is differentiable at a point if it has a tangent at every point. That is, a function is differentiable at $x$ if the limit

$\lim_{a \to 0} {{f(x+a)-f(x)} \over\ a}$ exists.

It fails to be differentiable if:

$f(x)$ is not continuous at $a$
The graph has a sharp corner at $a$
The graph has a vertical tangent line

Note: While all differentiable functions are continuous, all continuous functions may not be differentiable.

very gud shiva.........salute from me !!!!!!!!

integration of log(cos(x)) is not in JEE syllabus.

Anyhow i am giving u the method to solve it,

cos(x) = (e^ix + e^-ix) / 2

So, I = int log((e^ix + e^-ix) / 2) . dx

Put (e^ix + e^-ix) = y

So, ((i)e^ix + (-i)e^-ix) dx = dy

or, (ie^ix - ie^-ix) dx = dy

or, (i) (e^ix - e^-ix) dx = dy

or, dx = dy / i (4 - y²)

So, I = (1/i) int log( y ) / (4 - y²)^1/2 dy + ((log2) / (i)) int dy / (4 - y²)^1/2

Now to solve first part of integral use polylogarithmic functions.

Finally,

Polylogarithm

The polylogarithm , also known as the Jonquière's function, is the function

defined in the complex plane over the open unit disk. Its definition on the whole complex plane then follows uniquely via analytic continuation.

Note that the similar notation is used for the logarithmic integral.

Consider a triangle with coordinates of 3 vertices A(x₁,y₁), B(x₂,y₂) and C(x₃,y₃).

So, ATQ,

AB = a, BC = b and AC = c.

The given determinant is equal to twice the area of triangle ABC.

also area of triangle = (s(s-a)(s-b)(s-c))^1/2, where s = semi-perimeter

So, given determinant = (1/2) ((a + b + c)(a + b - c)(a - b + c)(b + c - a) (Ans)..........

Linear Combination of Vectors

A linear combination of vectors is a sum of scalar multiples of those vectors. That is, given a set of vectors $underline{x}_i$ of the same type, such as ${f R}^N$ (they must have the same number of elements so they can be added), a linear combination is formed by multiplying each vector by a scalar and summing to produce a new vector of the same type:

$displaystyle underline{y}= alpha_1 underline{x}_1 + alpha_2underline{x}_2 + cdots + alpha_M underline{x}_M$

For example, let $underline{x}_1=(1,2,3)$ , $underline{x}_2=(4,5,6)$ , , and . Then the linear combination of $underline{x}_1$ and $underline{x}_2$ is given by

$displaystyle underline{y}= alpha_1underline{x}_1 +alpha_2underline{x}_2 = 2cdot(1,2,3) + 3cdot(4,5,6)= (2,4,6)+(12,15,18) = (14,19,24).$

In signal processing, we think of a linear combination as a signal mix. Thus, the output of a mixing console may be regarded as a linear combination of the input signal tracks.

Source: https://ccrma.stanford.edu/~jos/st/Linear_Combination_Vectors.html

The question is totally correct. Here's the solution.

Let E_i denote the evet that the bag contains i white balls and so (4 - i) balls of any other colour(s). Let A denote the event that the 2 balls drawn at random are white.

P(E_i) = 1/5 (i =0,1,2,3,4)

P(A/E_i) =0 for i = 0,1

P(A/E_i) = ⁱC₂/⁴C₂ for i>2

Now by total probability rule,

P(A) = summation i from 0 to 4 P(E_i) P(A/E_i) = (1/5)(1/⁴C₂)(²C₂ + ³C₂ + ⁴C₂)

P(A) = (1/5)(1/6) (1 + 3 + 6) = 1/3

By the Bayes' rule

P(E₄/A) = P(E₄) P(A/E₄) / P(A) = (1/5) (1) / (1/3) = 3/5 = 0.6 (Ans)............

very nice.............salute to u :)

This is the graph of |x| + |y| = 1.

So, area within the curve = area of square ABCD = 2 sq. units.

or, by integration,

area within the curve = 4 times area of triangle AOB = 4 int x from 0 to 1 (1 - x).dx = 4(x - (x²/2)) with limit of x from 0 to 1

= 4(1 - (1/2)) = 4(1/2) = 2 sq. units.

How to draw the graph

|x| + |y| = 1

For x,y>0; eq. is x + y = 1

For x>0 and y<0; eq. is x - y =1

For x<0 and y>0; eq. is - x + y =1

For x,y<0; eq. is - x - y = 1

Let p be the prob. of getting an odd no. in a single throw of a die. Then the prob. of getting an even no. is 2p.

We have,

P(1) + P(2) + P(3) + P(4) +P(5) + P(6) = 1

or, p + 2p + p + 2p + p + 2p = 1

or, p = 1/9

So, prob. of getting an even no. = 2/9 and prob. of getting an odd no. = 1/9.

Since an even no. is considered as success,

Let X = no. of even no. on 2 throws.

When X = 0, P(X) = (1/9)(1/9)(1/9) = 1/729

When X = 1, P(X) = (1/9)(1/9)(2/9)(3) = 6/729 = 2/243

When X = 2, P(X) = (1/9)(2/9)(2/9)(3) = 4/243

When X = 3, P(X) = (2/9)(2/9)(2/9) = 8/729

This is the graph of greatest integer function (or the floor function).

Now break the limit for the integral first from 1 to 2 and then from 2 to 2.5.

I = int x from 1 to 2.5 [x] . dx

I = int x from 1 to 2 [x] . dx + int x from 2 to 2.5 [x] . dx

I = int x from 1 to 2 (1) . dx + int x from 2 to 2.5 (2) . dx

I = limit from 1 to 2 (x) + limit from 2 to 2.5 (2x)

I = (2 - 1) + 2(2.5 - 2) = 2 (Ans)

Find the intersection of parabola 2y = x² and line y = mx + 2. (U may find only the x coordinate).

U will get x = m + (m² + 4)^1/2

Since x₂> x₁,

So, x₂ = m + (m² + 4)^1/2

and x₁ = m - (m² + 4)^1/2

I = (mx²/2) + 2x + (x³/6) limit of x from x₁ to x₂

Put the limits and then find dI/dm and put it equal to 0 to get a value of m.

Check thru double derivative that d²I/dm² > 0.

So, value of m corresponds to minimum value of I.

Solve it by differentiating both sides.

I = int x from 0 to pi/4 (sinx + cosx)dx / (9 + 16sin2x)

I = int x from 0 to pi/4(sinx + cosx)dx / (25 - 16(sinx - cosx)²)

Put y = sinx - cosx

or, dy = (cosx + sinx).dx

When x = 0, y = -1 and when x = pi/4 , y = 0

I = int y from -1 to 0 (dy / 25 - 16y²)

I = (1/16) int y from -1 to 0 ( dy/ (5/4)² - y² )

I = (1/16) (2/5) |log(5 + 4y / 5 - 4y)| with limit of y from -1 to 0

I = (1/40) ( 0 - log(1/9) )

I = (1/40) (log9)

(1) Probability of getting 1 = P(1) = 1/6

Probability of getting 2 = P(2) = 2/6

and, Probability of getting 3 = P(3) = 3/6

(i) For the sum to be 4, it can occur as (1,1,2)

Further these three can arrange in (3!/2!) ways (as 1 comes twice).

So, P(sum 4) = (1/6) (1/6) (2/6) (3!/2!) = 1 / 36

(ii) For the sum to be 6, it can occur in two ways as (2,2,2) and (1,2,3).

So, P(sum 6) = (2/6)³ + (1/6) (2/6) (3/6) (3!) = (1/27) + (1/6) = 11/54

So, req. Prob. = P(sum 4) + P(sum 6) = (1/36) + (11/54) = 25/108 (Ans).............