Problem based on General Terms

Type : 1_ Q4 Find the 7th term in the expansion of

[4x – (1 / 2vx)]13

Sol : T7 = T6+1 = 13C6(4x)13-6 - (1/2vx) 6

= 13C6.47x7. 1 /(26.x3)

= 13C6. 28.x4

= 13!/ (6!x7!) . 28. x4

= T7 = 439296x4

Type II : Find the coefficient of x-7 in the expansion of (ax – 1/ bx2) 11

Sol.: General Term , Tr+1 = 11Cr(ax)11-r - (1/ bx2 ) r

Tr+1 = (-1)r 11Cr. (a11-r / br) x11-3r --------------> (i)

Putting 11 – 3r = -7

Or 3r = 18

r = 6

From (i) to T7 = (-1)6 11C6. ( a5 / b6) x-7 --------------> (i)

Hence, the coefficient of x-7 in ax- (1 / b x2) 11 is 11C6a5b-6

Type III : Find the term independent of ‘x’ in [(3 x2 / 2) – (1/ 3x) ] 9

Sol.: General Term, Tr+1 = 9Cr (3 x2 / 2) 9-r – (1/3x) r

= (-1)r 9Cr ( 3/2) 9-r x18-2r (1 / 3r. xr )

Tr+1 = (-1)r9 Cr (39-2r / 29-r). x18-3r -------> (i)

Putting 18- 3 r = o

r = 6

So, from (i), 7th term is independent of ‘x’, and its value is:

T7 = (-1)6 . 9C6. (3-3 / 23) xo

= 9 ! /(6! X 3!) . 1/ (33 x 23)

= T7 = (7/18)

Pth term from end:

‘P’th term from end in the expansion of (x+y)n is (n-P+2)th term from beginning.

Ex.: Find the 4th term from the end in the expansion of [ (x3/2) - (2/x2) ] 7

Sol.: 4th term from end = (7-4+2)th or 5th term from beginning.

T5 = T4+1 = 7C4 (x3/2)7-4 . (-2/x4) 4

= 7C4 (x3 /2) 3 ( -2/ x2) 4

= 7! / (4! X 3!) . (x9/8) . (16/ x8)

= (7.6.5 / 3.2.1) .2x

T5 = 70x

Hence ‘4’ term, from the end = 70x.

Middle Terms: It depends upon the value of ‘n’.

Case -1 : When ‘n’ is even, then total number of terms in (x+y)n is odd. So there is only one middle term i.e. [(n/2) + 1] th them is the middle term.

So we find (Tn+1/2). th term in this case, if ‘n’ is even.

Case II : When ‘n’ is odd, then total number of terms in (x+y)n is even. So there are two middle terms i.e. (n+1) /2 th and (n+3) /2 th are true middle terms.

so we find T(n+1)/2 th and T(n+3)/2 th in this case if ‘n’ is odd.



Ex.: Find the middle – term in the expansion of [ 3x – (x3 / 6)]9

Sol.: Here total no. of terms are 10 (even). So there are true middle-terms

i.e (9+1) / 2 th and (9+3) / 2 th. So we have to find – out ‘T5’ and ‘T6’.

T5 = T4+1 = 9C4(3x)9-4 (-x3 / 6) 4

= 9! / (4! X 5!) .35 x5 ( x12 / 64)

= (9.8.7.6 / 4.3.2.1) 35 / (24 x 34) x17

T5 = (189 / 8) x17

T6 = T5+1 = 9C5(3x)9-5 (-x3 / 6) 5

= 9! / (5! X 4!) .34 x4 (x15 / 65)

= -(9.8.7.6 / 4.3.2.1) 34(25 x 35) x19

T6 = - ( 21 / 16) x19

Greatest – term in (1+x)n : If ‘Tr’ and ‘Tr+1’ be the ‘r’ th and (r+1)th terms in the

Expansion of (1+x)n, then :

Tr+1 = nCr(1)n-r xr = nCr xr

And Tr = nCr-1. xr-1

So: Tr+1 / Tr = (nCr xr / nCr-1 xr-1) = (n-r+1)/r x

If ‘Tr+1 be the greatest term, then Tr+1 ³ Tr

Or Tr+1 / Tr ³ 1



since (n-r+1) / r. x >=1, where ‘r’ is a ‘+’ ve integer.

This inequality, changes either to the form r<=m+f pr r <= m, where ‘m’ is a ‘+’ ve integer and ‘f’ is a fraction. So we get: r <= m + f ---------------> (i)

or r <= m ------------------> (ii)

In case (i), ‘T’m+1 is the greatest term, and in case (i) ‘T’m and ‘T’m+1 are the greatest terms, and both are equal.

Short-cut: First calculate m = x (n+1) / (x + 1)



Case (1) If ‘m’ is an integer, then ‘T’m and ‘T’m+1 are the greatest terms and both are equal.

Case (2) If ‘m’ is not an integer, then T[m]+1 will be the greatest term, where [.] denotes greatest integer function.

Binomial Theorem

Binomial expression: An algebraic expression consisting of two terms with a positive or negative sign between them is called a binomial expression.

Example: (a+b), ( P / x2) – (Q / x4) etc.

Binomial Theorem: When a binomial expression is raised to a power ‘n’ we would like to be able to expand it. The binomial theorem assists us in doing this. It converts such an expression into a series.

Binomial Theorem for positive integral index:

(x+y)n = xn + nC1xn-1y+nC2xn-2y2+-----+nCrxn-ryr+ -------+---------+nCn-1xyn-1 + ncnyn.

It can be represented as:

(x+y)n = nCrxn-ryr

Particular – Cases :

(i) Replacing ‘y’ by ‘-y’, we have :

(x-y)n = nCoxyo-nC1xn-1y+nC2xn-2y2-------+(-1)r nCrxn-ryr+------+(-1)n nCnxoyn.

It can be represented as :

(x+y)n = (-1)r nCrxn-ryr

(ii) Replacing ‘x’ by ‘1’ and ‘y’ by ‘x’, we have :

(1+x)n = nCoxo+nC1x+nC2x2+---------+nCrxr+------+nCn-1xn-1+nCnxn.



or = nCrxr

(ii) Replacing ‘x’ by ‘-x’, we have :

(1+x)n = nCoxo-nC1x1+nC2x2 - ---------+(-1)r nCrxr+------+nCn-1(-1) n-1 +(-1)n nCnxn.

or = (-1)rnCrxr

Properties of Binomial – Expansion (x+y)n :

(i) There are (n+1) terms in the expansion.

(ii) In each term, sum of the indices of ‘x’ and ‘y’ is equal to ‘n’.

(iii) In any term, the lower suffix of ‘c’ is equal to the index of ‘y’, and the index of x = n-(lower suffix of c).

(iv) Because nCr = nCn-r,

so we have :

nCo = nCn

nC1=nCn-1

nC2=nCn-2 etc.

It follows that the coefficients of terms equidistant from the beginning and the ends are equal

ASYMPTOTE

A straight line is called an asymptote to the curve y=f(x) if
the distance from the variable point M of the curve to the
straight line approaches zero as the point M receds to infinity
along some branch of the curve.

Asymptote are of three kinds:-
1.Horizontal asymptote

2.Vertical asymptote

3.Inclined asymptote

Continuity

The derivative does not exist (the function is not differentiable) at any point where the function is not continuous. In other words, wherever the function is not continuous it’s also not differentiable.

But even if the function is continuous at a point, it still might not be differentiable there. These three conditions must be fulfilled for the function to be differentiable at a given point:

1. The function must be continuous there.
2. The graph must be smooth there, with no sharp bend.
3. The tangent line must be oblique (slanted) or horizontal, not vertical.

Wherever any of those conditions is violated, the derivative does not exist at that point. Turning it around, if the derivative does exist at a given point, then you know that all three of the above are true at that point.

Slope of a Curve

One of the two main problems of calculus is to find the slope of a curve at a given point. Today we’ll solve this problem!

f(x) = –x²+9x–14 is graphed at right. We will solve the classic tangent line problem by finding the slope of this curve at some point, such as (6,4).

How to find the slope of the curve? It’s the same as the slope of a line that is tangent to the curve at that point. (A tangent line touches the curve at just one point.)

How to find the slope of the tangent line? Draw a secant line, which crosses through the curve at two points, (6,4) and another point nearby. Find the slope of that secant line. Then use a limit process to find the slope as you bring the second point closer to (6,4).

How to find the slope of the secant line? Use the normal formula for slope, Δy/Δx. The first point is at (6,4) and the second point is at (6+Δx, 4+Δy). But since y = f(x), you can also say that the y coordinate of the second point is f(6+Δx). Therefore the slope of the secant line is the difference quotient, as described on pages 95–97

Rational Roots

Assuming you’ve already factored out the easy monomial factors and special products, what do you do if you’ve still got a polynomial of degree 3 or higher?

The answer is the Rational Root Test. It can show you some candidate roots when you don’t see how to factor the polynomial, as follows.

Consider a polynomial in standard form, written from highest degree to lowest and with only integer coefficients:

f(x) = a_nxⁿ + ... + a_o

The Rational Root Theorem tells you that if the polynomial has a rational zero then it must be a fraction p/q, where p is a factor of the trailing constant a_o and q is a factor of the leading coefficient a_n.

Example:

p(x) = 2x⁴ − 11x³ − 6x² + 64x + 32

The factors of the leading coefficient (2) are 2 and 1. The factors of the constant term (32) are 1, 2, 4, 8, 16, and 32. Therefore the possible rational zeroes are ±1, 2, 4, 8, 16, or 32 divided by 2 or 1:

±1/2, 1/1, 2/2, 2/1, 4/2, 4/1, 8/2, 8/1, 16/2, 16/1, 32/2, 32/1

reduced: ± ½, 1, 2, 4, 8, 16, 32

What do we mean by saying this is a list of all the possible rational roots? We mean that no other rational number, like ¼ or 32/7, can be a root of p(x) = 0.

Caution: Don’t make the Rational Root Test out to be more than it is. It doesn’t say those rational numbers are roots, just that no other rational numbers can be roots. And it doesn’t tell you anything about whether some irrational or even complex roots exist. The Rational Root Test is only a starting point.

Suppose you have a polynomial with non-integer coefficients. Are you stuck? No, you can factor out the least common denominator (LCD) and get a polynomial with integer coefficients that way. Example:

(1/2)x³ − (3/2)x² + (2/3)x − 1/2

The LCD is 1/6. Factoring out 1/6 gives the polynomial

(1/6)(3x³ − 9x² + 4x − 3)

The two forms are equivalent, and therefore they have the same roots. But you can’t apply the Rational Root Test to the first form, only to the second. The test tells you that the only possible rational roots are ±1/3, 1, 3.

Once you’ve identified the possible rational zeroes, how can you screen them? The brute-force method would be to take each possible value and substitute it for x in the polynomial: if the result is zero then that number is a root. But there’s a better way.

Use Synthetic Division to see if each candidate makes the polynomial equal zero. This is better for three reasons. First, it’s computationally easier, because you don’t have to compute higher powers of numbers. Second, at the same time it tells you whether a given number is a root, it produces the reduced polynomial that you’ll use to find the remaining roots. Finally, the results of synthetic division may give you an upper or lower bound even if the number you’re testing turns out not to be a root.

Sometimes Descartes’ Rule of Signs can help you screen the possible rational roots further. For example, the Rational Root Test tells you that if

q(x) = 2x⁴ + 13x³ + 20x² + 28x + 8

has any rational roots, they must come from the list ±½, 1, 2, 4, 8. But don’t just start off substituting or synthetic dividing. Since there are no sign changes, there are no positive roots. Are there any negative roots?

q(−x) = 2x⁴ − 13x³ + 20x² − 28x + 8

has four sign changes. Therefore there could be as many as four negative roots. (There could also be two negative roots, or none.) There’s no guarantee that any of the roots are rational, but any root that is rational must come from the list −½, −1, −2, −4, −8.

(If you have a graphing calculator, you can pre-screen the rational roots by graphing the polynomial and seeing where it seems to cross the x axis. But you still need to verify the root algebraically, to see that f(x) is exactly 0 there, not just nearly 0.)

Remember, the Rational Root Test guarantees to find all rational roots. But it will completely miss real roots that are not rational, like the roots of x²−2=0, which are ±√2, or the roots of x²+4=0, which are ±2i.

Finally, remember that the Rational Root Test works only if all coefficients are integers. Look again at this function, which is graphed at right:

p(x) = 2x⁴ − 11x³ − 6x² + 64x + 32

The Rational Root Theorem tells you that the only possible rational zeroes are ±½, 1, 2, 4, 8, 16, 32. But suppose you factor out the 2 (as I once did in class), writing the equivalent function

p(x) = 2(x⁴ − (11/2)x³ − 3x² + 32x + 16)

This function is the same as the earlier one, but you can no longer apply the Rational Root Test because the coefficients are not integers. In fact −½ is a zero of p(x), but it did not show up when I (illegally) applied the Rational Root Test to the second form. My mistake was forgetting that the Rational Root Theorem applies only when all coefficients of the polynomial are integers

Significant digits and Rounding

What strikes you about the difference between a measurement of 1.615 in and 1.6 in? The first measurement is more precise than the second. Specifically, the first measurement is accurate to the nearest thousandth (0.001) of an inch, while the second measurement is accurate to only the nearest tenth (0.1) of an inch.

No measurement in the real world is exact; all have some degree of “slop”. When we present a numerical result of 1.615 in, it is understood to be accurate to the nearest thousandth: we know that the true measurement is between 1.6145 and 1.6155 in. By contrast, if we present a value of 1.6 in, we are really saying that it is 1.6 to the nearest tenth: the true measurement could be anywhere between 1.55 and 1.65 in.

You can see that every real-world number carries information both about its magnitude and about its precision. 1.615 in and 1.6 in have about the same magnitude, but the first one is more precise. We talk about that level of precision as how many significant digits the number has.

The significant digits in a number start at the first non-zero digit and end at the last digit. Examples: 1417 has four significant digits, and so do 1.417 and 0.00001417. What about 14.1700? It has six significant digits, not four, because only zeroes at the start of a number are non-significant. Finally, 14.07 has four significant digits.

There can be some ambiguity with trailing zeroes in a large whole number. For instance, we quote the average distance from earth to sun as 93 million miles. In that form, the number has two significant digits. (Remember what significant digits mean: they mark the non-“slop” part of the measurement. All we’re saying is that the average distance is 92½ to 93½ million. But suppose we write the number as 93,000,000? Does it now have eight significant digits? Are we saying the average distance is between 92,999,999.5 and 93,000,000.5 miles? Surely not!

When you see a large round whole number, the zeroes may represent “slop”. To get around this problem, numbers are often expressed in scientific notation. For instance, the figure of 93 million miles is 9.3×10⁷ miles (“nine point three times ten to the seventh”). On your calculator it appears as 9.3E7. You’ll find some more about scientific notation

How to round up

Once you’ve computed an unrounded answer, how do you round it correctly? Decide how many significant digits (or decimal places) you’ll need, and then round all at once. Example: if the TI-83 computes a mean of numbers with one decimal place as 3.876, that needs to be rounded to one decimal place. Draw a line at the spot where the rounding must happen: 3.8|76. Because what’s to the right of the line is bigger than 5, you round up to 3.9.

If the first digit after the line is 5 to 9, round up; if the first digit after the line is 0 to 4, round down. Example: to round 2.884 and 2.885 to the nearest hundredth, write 2.88|4 → 2.88 and 2.88|5 → 2.89.

It’s important to round all at once, not digit by digit. Example: Round 30.4746 to the nearest hundredth.

Incorrect solution: 30.4746 → 30.475 → 30.48.

Correct solution: 30.47|46 → 30.47.

What’s to the right of the line is less than 5, and you round down to 30.47.

Scientific Notation

Scientific notation was developed to express very large and very small numbers.

To write a large number in scientific notation, move the decimal point to the left until it is between the first and second significant digits; the number of places moved is the exponent. For example, 167 becomes 1.67, but the decimal point moved two places left. Therefore 167 = 1.67×10² or 1.67E2.

Scientific notation removes the guesswork about how significant a large number is. 9.3E7 has two significant digits; it is accurate to the nearest million miles. 9.30000E7 has five significant digits, and it is accurate to the nearest hundred miles (0.00001×10⁷ = 100 miles).

To write a small number in scientific notation, move the decimal point right until it has just passed the first non-zero digit. Write the number of places moved as a negative number in the exponent. Example: 0.0000894 must move the decimal point five places right to become 8.94E–5 or 8.94×10^–5.

Notice that big numbers end up with a positive exponent after the E and small numbers end up with a negative exponent after the E.

To convert a number from scientific notation to ordinary decimals, reverse the process. A positive exponent indicates a big number: move the decimal point to the right. A negative exponent indicates a small number: move the decimal point to the left. Example: if the probability of an event is 6.014E–4 or 6.014×10^–4, you must move the decimal point four places left to convert it to 0.0006014. If the population of the earth is about 6.1×10⁹ people, you move the decimal point right nine places to convert to 6,100,000,000.

To enter scientific notation in your calculator, find the EE symbol in gold just over the comma key (which is above the 7). Example: to enter 6.1×10⁹, press [6] [.] [1] [2nd] [EE] 9. If you then press the [ENTER] key, you will probably see 6100000000. In normal mode, the calculator tries to present results in ordinary numbers if they fit on the screen.

Solving Trigonometric Equation

Step 1. Get one function of one angle.

Trig equations, like any equations, are really about numbers, not angles. You are looking for all possible numbers that could be substituted for the variable in the equation to make it true. But it simplifies things to think about the angles first and worry about the variables later.

Example A:

cos(4A) − sin(2A) = 0

Here the “angles”, the arguments to the trig functions, are 4A and 2A. True, you want to solve for A ultimately. But if you can solve for the angle 4A or 2A, it is then quite easy to.

As you see, that equation involves two functions (sine and cosine) of two angles (4A and 2A). You need to get it in terms of one function of one angle. Note well: a function of one angle, not necessarily a function of just the variable A.

This is where it is essential to have a nodding acquaintance with all the trig identities. If you do, you’ll remember that cos(2u) can be expressed in terms of sin(u). Specifically, cos(2u) = 1 − 2sin²(u).

How does that help? Well, 4A is 2×2A, isn’t it?

cos(2u) = 1 − 2sin²(u)

cos(2×2A) = 1 − 2sin²(2A)

cos(4A) = 1 − 2sin²(2A)

That transforms the original equation to

1 − 2sin²(2A) − sin(2A) = 0

which can be rewritten in standard form as

2sin²(2A) + sin(2A) − 1 = 0

Now you have the equation in terms of only one function (sine) and only one angle (2A)

Step 2. Solve for the value(s) of a trig function.

Now that the equation involves only a single function of a single angle, your next task is to solve for that function value.

Example A continues. Recapping what was done so far,

cos(4A) − sin(2A) = 0 ⇒

2sin²(2A) + sin(2A) − 1 = 0

You want to solve for sin(2A). You should recognize that the equation is really a quadratic,

2y² + y − 1 = 0, where y = sin(2A)

It can be factored in a straightforward way:

(sin(2A) + 1) (2sin(2A) − 1) = 0

From algebra you know that if a product is 0 then you solve by setting each factor to 0:

sin(2A) + 1 = 0 or 2sin(2A) − 1 = 0

sin(2A) = −1 or sin(2A) = 1/2

Example B:

This one is a bit simpler. Solve:

3tan²(B/2) − 1 = 0

To solve it, add 1 to both sides and divide by 3:

tan²(B/2) = 1/3

and then take square root of both sides:

tan(B/2) = ±√(1/3) or ±√(3)/3

It’s important to remember to use the plus-or-minus sign ± when taking the square root of both sides; otherwise you could overlook some solutions.

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Step 3. Solve for the angle.

After solving for a function value, now you solve for the angle. If it’s a multiple of π/6 (30°) or π/4 (45°), you can easily solve it exactly. Otherwise you must write the solution as an arcfunction.

Trig equations have one important difference from other types of equations. Trig functions are periodic, meaning that they repeat their values over and over. Therefore a trig equation has an infinite number of solutions if it has any.

Think about an equation like sin u = 1. π/2 is a solution, but the sine function repeats its values every 2π. Therefore π/2±2π, π/2±4π, and so on are equally good solutions. To show this, write the solution as u = π/2 + 2πn, where n is understood to be any integer, positive, negative, or zero. (The tangent and cotangent functions repeat all their values every π radians, so the solution to tan v = 1 is v = π/4 + πn, not +2πn.)

Example A continues. Recapping what was done so far,

cos(4A) − sin(2A) = 0 ⇒

2sin²(2A) + sin(2A) − 1 = 0 ⇒

sin(2A) = −1 or sin(2A) = 1/2

The sine of 3π/2 is −1, so the first possibility reduces to 2A = 3π/2. But remember that the sine function is periodic, so write

2A = 3π/2 + 2πn.

For the second possibility, sin(2A) = 1/2, there are two solutions. sin(π/6) = sin(5π/6) = 1/2, and again we add 2πn to the angle to account for all solutions:

2A = 3π/2 + 2πn or π/6 + 2πn or 5π/6 + 2πn

Example B continues. Recapping what was done so far,

3tan²(B/2) − 1 = 0 ⇒

tan(B/2) = ±√(3)/3

What angle has a tangent value of √(3)/3? the angle π/6. And where does the tangent have a value of −√(3)/3? at the angle 5π/6. This gives the solutions

B/2 = π/6 + πn or 5π/6 + πn

Remember that the tangent and cotangent have period π and not 2π.

Example C:

Of course, you don’t always luck out with nice angles. Take a look at this equation:

sec(3C) = 2.5

What are the possible values of the angle 3C? It’s hard to work with the secant function, but 1/sec(3C) = cos(3C) so rewrite the equation as

1/sec(3C) = 1/2.5

cos(3C) = 0.4

For what angles is that true? We write arccos(0.4) to mean the angle in quadrant I that has a cosine equal to 0.4. (Some books write cos⁻¹(0.4) instead of arccos(0.4). I prefer the arccos notation because the superscript −1 makes many students think of 1/cos(0.4), which has a different meaning entirely.)

So initially we would write 3C = arccos(0.4) + 2πn. But that’s not the whole story: any angle in quadrant I has a reflection in quadrant IV with the same cosine value, so we need to account for both angles:

3C = arccos(0.4) + 2πn or 2π−arccos(0.4) + 2πn

Note that the base angle is always nonnegative and less than 2π: 2π−arccos(0.4) + 2πn, not simply −arccos(0.4) + 2πn. This is necessary to make step 5 come out right.

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Step 4. Solve for the variable.

Now it’s time to abandon angular thinking and go for the variable. As you will see, it is very important to do this step after step 3. 2πn or πn must be added to the angle, not the variable, to reflect the period of the trig function.

Example A continues. Recapping what was done so far,

cos(4A) − sin(2A) = 0 ⇒

2sin²(2A) + sin(2A) − 1 = 0 ⇒

sin(2A) = −1 or sin(2A) = 1/2 ⇒

2A = 3π/2 + 2πn or π/6 + 2πn or 5π/6 + 2πn

Now divide both sides by 2:

A = 3π/4 + πn or π/12 + πn or 5π/12 + πn

Be sure to divide the entire equation, so that the 2πn becomes πn.

Why is the order of steps so important? The 2πn came in because the sine function has a period of 2π: if you take an angle and add 2π to it, it looks like the same angle and all six of its function values are unchanged. But now we’re no longer dealing with the angle 2A, we’re dealing with the variable A. In this equation, we say that adding π to any solution for A will give another solution for A.

For instance, set n=1 and obtain solutions A = 7π/4, 13π/12, or 17π/12. True, sin(7π/4) doesn’t equal −1. But the equation was sin(2A) = −1, not sin(A) = −1. If you substitute A = 7π/4 in sin(2A), you get sin(2*7π/4) = sin(7π/2) which does equal −1. Always pay attention to whether you’re dealing with the angle or the variable.

Example B continues. Recapping what was done so far,

3tan²(B/2) − 1 = 0 ⇒

tan(B/2) = ±√(3)/3 ⇒

B/2 = π/6 + πn or 5π/6 + πn

Multiplying both sides by 2 gives

B = π/3 + 2πn or 5π/3 + 2πn

Once again, the angle was B/2 and had a period of π; the variable is B and has a period of 2π.

Example C continues. Recapping what was done so far,

sec(3C) = 2.5 ⇒

3C = arccos(0.4) + 2πn or 2π−arccos(0.4) + 2πn

Divide both sides by 3:

C = (1/3)arccos(0.4) + (2π/3)*n or (2π/3)−(1/3)arccos(0.4) + (2π/3)*n

It’s a matter of taste whether to combine terms in that second solution:

C = (1/3)arccos(0.4) + (2π/3)*n or −(1/3)arccos(0.4) + (2n+1)*π/3

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Step 5. Apply any restrictions.

Does the problem specify a solution interval for the variable? Sometimes this is done in interval notation, like [0,2π); other times it’s done as an inequality, 0 <= x <>

But if solutions are restricted to a particular interval, you have a bit more work to do after solving for the variable.

Example A continues. Recapping what was done so far,

cos(4A) − sin(2A) = 0 ⇒

2sin²(2A) + sin(2A) − 1 = 0 ⇒

sin(2A) = −1 or sin(2A) = 1/2 ⇒

2A = 3π/2 + 2πn or π/6 + 2πn or 5π/6 + 2πn ⇒

A = 3π/4 + πn or π/12 + πn or 5π/12 + πn

Now suppose that only solutions on the interval [0,2π) were wanted.

The general solutions have a period of π (from +πn), and therefore there will be two cycles between 0 and 2π:

Probability of Shared dates

In a group of 30 people, would you be surprised if two of them have the same birthday? As it turns out, you should be more surprised if they don’t.

There are 365 possible birthdays. (To keep the numbers simpler, we’ll ignore leap years.) The key to assigning the probability is to think in terms of complements: “Two (or more) people share a birthday” is the complement of “All people in the group have different birthdays.” Each probability is 1 minus the other.

(a) What is the probability that any two people have different birthdays? The first person could have any birthday (p = 365÷365 = 1), and the second person could then have any of the other 364 birthdays (p = 364÷365). Multiply those two and you have about 0.9973 as the probability that any two people have different birthdays, or 1−0.9973 = 0.0027 as the probability that they have the same birthday.

(b) Now add a third person. What is the probability that her birthday is different from the other two? Since there are 363 days still “unused” out of 365, we have p = 363÷365 = about 0.9945. Multiply that by the 0.9973 for two people and you have about 0.9918, the probability that three randomly selected people will have different birthdays.

(c) Now add a fourth person, and a fifth, and so on until you have 22 people with different birthdays (p ≈ 52.4%). When you add the 23rd person, you should have p ≈ 49.3%.

(d) If the probability that 23 randomly selected people have different birthdays is 49.3%, what is the probability that two or more of them have the same birthday? 1−0.493 = 0.507 or 50.7%. In a randomly selected group of 23 people, it is slightly more likely than not that two or more of them share a birthday.

For n people (n ≤ 365), your chain of n fractions would be

and therefore

On your TI-83, to get ₃₆₅P_n you first enter the 365, then press [MATH] [◄] to get the PRB menu and [2] for nPr, then enter the second number. In Excel, it’s PERMUT(365,n).

What if n > 365? In this case there is no need for any calculations (and in fact the above formula won’t work). If there are 366 or more people, but only 365 possible birthdays disregarding leap year, then two or more of them must share a birthday.

You can see that the dividing line is between 22 and 23 people. In a group of 22 people, the odds are less than 50–50 that two share a birthday; in a group of 23, the odds are better than 50–50. In a bar with even a small crowd, if you can get someone to take your bet that two people share a birthday, you’ll win more often than you lose.

10 minutes Trigonometry

Degrees and Radians

circumference of circle: 2πr
angle around circle (like clock hand): 2π radians or 360°
T

herefore 2π = 360°,

or π = 180°, or 1 radian = 180°/π

sin A = cos(π/2−A)
cos A = sin(π/2−A)
tan A = cot(π/2−A)
cot A = tan(π/2−A)
sec A = csc(π/2−A)
csc A = sec(π/2−A)

Inverse Functions

arcsin 0.65 or sin^-10.65 means the angle whose sine is 0.65. That’s not the same as 1/sin 0.65
Function ranges: −π/2 ≤ arcsin x ≤ +π/2, 0 ≤ arccos x ≤ +π, −π/2 <>

Function composition (see diagram at right):
What is e.g. cos( arctan x ) ?
Solution: arctan x is the angle whose tangent is x; call it θ. Then you must find cos θ. Use Pythagoras to find the third side, √(x²+1), then read off function value: cos θ = 1 / √(x²+1)

Questions on permutation and combination

1.A man has 5 friends.In how many ways can he invite one or more of them
to a party?

2.Find the number of ways in which one or more letters can be selected from the
letters A A A A B B B C D E.

3.In how many ways can 12 books be equally distributed among 3 students?

4.Three tourist want to stay in five different hotels.In how many ways can they do
so if:
(a) each hotel cannot accomodate more than one tourist.
(b) each hotel can accomodate any number of tourist.

5. How many 3 lettered word can be formed by letters of the word SOCIETY,so that
(a) S is included in each word.
(b) S is not included in any word.

Permutation and Combination(M-1)

Hey frendz let me welcome you all to one of the most interesting topics of math..
I will guide you step by step to achieve mastry in this chapter lets start today
BEGIN
In this chapter we mainly desire to do counting without performing actual counting.
FUNDAMENTAL PRINCIPLE OF COUNTING
If an event can occur in m different ways following which another event can occur in
n different ways, the total number of ways in which all the event can occur is m.n

PERMUTATION:-means arrangement in a definite order of things which may be different
some alike and some different taken some or all at a time.

NOTE:In case of permutation order of occurence is important.

FACTORIAL:-Continued product of first n natural number is n factorial n!
n!=n.(n-1)!, 0!=1