Binomial Theorem
BINOMIAL THEOREM FOR POSITIVE INTEGRAL INDEX Synopsis:1. f x and a are real numbers, then for all n∈N,
1 1 2 2 1 2 ( )n n n o n n n n .... o x+a = C x a + C x −a + C x − a + 1 1 1 .... nC xr n r− ar ... nCn−x an− nC x an o n + + + + + i.e Formula: 0 ( ) . n n n n r r r r x a C x − a = + =
∑
2. MIDDLE TERM IN A BINOMIAL EXPANSION Synopsis:
If n is an even natural number, then in the binomial expansion of ( ) , 1 2
n n
x+a + th
term is the middle term.
Formula: If is odd natural number , then 1 2 n th + and 3 2 n th +
are the middle terms
In the binomial expansion of (x+a)n. 3. GREATEST TERM
Synopsis:
Let Tr+1and Trbe(r+1)th and rth terms respectively in the expansion of (x+a)n. Then, 1
n n r r r r T+ = C x − a and Tr=nCr-1 xn-r+1ar-1. ∴ 1 1 1 1 n n r r r r n n r r r r T C x a T C x a − + − + − − = = ! ( 1)!( 1)!. ( )! ! ! n r n r a n r r n x − − + × − = 1 . n r a r x − + CASE1 When 1 1 n x a + + is an integer Let 1 . 1 n m x a + = +
Then, from (i), we have
Tr+1> Trfor r=1,2,3,…..(m-1) ……(ii)
Tr+1= Tr for r = m ….(iii)
and, Tr+1< Trfor r = m +1, ….n …(iv)
∴T2> T1, T3> T2, T4> T3, ……, Tm>Tm-1 [From (ii)]
Tm+1= Tm [From (iii)]
⇒ T1< T2< ….<Tm-1 < Tm=Tm+1> Tm+2> …> Tn
This shows that mth and (m+1)th terms are greatest terms.
Case II: When 1 1 n x a + + is not an integer.
Let m be the integral part of 1 1 n x a + +
. Then, from (i), we have
1 1, 2...,
r r
T+ >T for= m …(v)
and, Tr+1<T for rr = +m 1,m+2,....n ….(vi) ∴ T2 >T T1, 3>T2,....,Tm+1>Tm [From (v)]
and, Tm+2 <Tm+1,Tm+3<Tm+2,...,Tn+1<Tn [From (vi)]
⇒ T1< < <T2 T3 ....<Tm<Tm+1>Tm+2>Tm+3...>Tn+1
⇒ (m+1)th term is the greatest term. 4. MULTINOMIAL THEOREM
Using binomial theorem, we have
0 ) , n n n n r r r r x a C x − a n N = + =
∑
∈ = 0 ! ( )! ! n n r r r n x a n r r − = −∑
= ! , ! ! s r r s n n x a r s + =∑
where s=n-r.This result can be generalized in the following form :
1 2 ( .... )n k x + + +x x = 1 2 1 2 1 2 .... 1 2 ! .... ! !.... ! k k r r r k r r r n k n x x x r r r + + + =
∑
The general term in the above expansion is
3 1 2 1 2 3 1 2 3 ! .... ! ! !... ! k r r r r k k n x x x x r r r r
The number of terms in the above expansion is equal to the number of non-negative integral solution of the equation.
r1+r2+…..+rk= n, because each solution of this equation gives a term in the above
5. PARTICULAR CASES (i) ( ) ! ! ! ! n r s t r s t n n x y z x y z r s t + + = + + =
∑
The above expansion has 3 1 2
3 1 2 n+ −C n+ C − = terms. (ii) ( ) ! ! ! ! ! n p q r s p q r s n n x y z u x y z u p q r s + + + = + + + =
∑
There are n+ −4 1C4 1− =n+3C3 terms in the above expansion.
REMARK The greatest coefficient in the expansion of ( 1 2 .... )n m
x + + +x x is ! , ( !)m r[( 1)!]r
n
q − q+ where q and r are the quotient and remainder respectively when n is divided by m.
PROPERTIES OF THE BINOMIAL COEFFICEINT
PROPERTY I In the expansion of (1+x)nthe coefficients of terms equidistant from the beginning and the end are equal.
PROPERTY IIThe sum of the binomial coefficients in the expansion of (1+x)nis 2n.
i.e, Co+ +C1 C2+ +... Cn=2n or, 0 2 n n n r r C = =
∑
.PROPERTY IIIThe sum of the coefficients of the odd terms in the expansion of (1+x)nis equal to the sum of the coefficients of the even terms and each is equal to 2n-1.
i.e, Co+C2+C4+….=C1+C3+C5+……=2n-1.
PROPERTY IVProve that:
1 2 1 2 1 . . . 1 n n n r r r n n n C C C r r r − − − − − = = − and so on. PROPERTY VCo-C1+C2-C3+C4-…+(-1)nCn=0
1. Find the coefficient of xmin the expression (1+x)n+2(1+x)n-1+3(1+x)n-2+….+(n-m+1) (1+x)m, where 0≤n.
2. Find the sum of the series : 2 3 4
0 1 3 7 15 ( 1) .... 2 2 2 2 r r r n r n r r r r r r C upto m terms = − + + + +
∑
\3. Find the sum of the series 2 3 4
0 1 3 7 15 ( 1) .... 2 2 2 2 r r r n r n r r r r r r C to = − + + + + + ∞
∑
4. If k and n be positive integers and sk=1k+ …+ nk, then show that 1 1 1 ( 1) ( 1) m m m r r r C s n n + + = = + − +
∑
5. Prove that : 2 2 1 1 2 2 2 0 1 1 2 2 k n n k n n k n n k k k − − − − − + − − +…..+ +(-1)k-1 .... ( 1) 0 n n k n k k − + + − = , where n k n C k =
6 In the expansion of the binomial expression (x+a)15, if the eleventh term is the geometric mean of the eighth and twelfth terms, which term in the expansion is the greatest.
7. Prove that the greatest term in the expansion of (1+x)2n has also the greatest coefficient, then 1 , 1 n n x n n + ∈ + . 8. 1 1 1 1 2 2 2 1 3 3 3 .... 1 (1 ) (1 ) x x x C C C nx nx nx + + + − + − + + + + 1 ( 1) 0 (1 ) n n n nx C nx + + − = +
9. If nCo, nC1, nC1, nC2, …..,nCn denote the binomial coefficients in the expansion of (1+x)n and p+q=1,
then prove that
(i) 0 n n r n r r r r C p q − np = =
∑
(ii) 2 2 2 0 n n r n r r r r C p q − n p npq = = +∑
10. Evaluate 3 2 0 3 2 6 11 6 n n r r r C r r r = + + + +∑
, Where nC0, nC1,...nCn are the binomial coefficients in the expansion of (1+x)n.11. When
(32) (32)
32 is divided by 7, prove that the remainder is 4.
12. A is a set containing n elements. A subset P of A is chosen at random. The set A is reconstructed by replacing the elements of P. A subset Q is again chosen at random. Find the number of ways selecting P and Q so that.
(i) P and Q are disjoint sets. (ii) P∩Q contains just one element (iii) P∪Q contains just one element
(iv) Q is a subset of P (v) P∩ =Q φ
(vi) P and Q have equal number of elements (vii) Q contains just one element more than P (viii) P∪ =Q A (ix) P = Q
PASSAGE – 1 Numerically greatest Term in the Expansion of (x + a)n
(x+a)n. Then Tr=nCr-1 xn-r+1ar-1 and Tr+1=nCrxn-r ar ∴ r 1 1 r T n r a T r x + = − + Now, Tr+1 > , = , < Tr According as 1 , , 1, n r a r x − + > = < i.e. according as 1 1 , , , n x r+ − > = <a i.e. according as , , 1 1 n r x a + < = > + . So, if 1 1 n x a + +
is an integer, say p, then Tr+1> Trif r < p otherwise Tr+1 ≤ Tr
So, Tp= Tp+1(numerically) and these are greater than any other term in the expansion.
Next, if 1 1 n x a +
+ is a non-integer, suppose m be its integral part then Tr+1< Trif r≤m and
Tr+1< Trif r > m.
So, Tm+1is the numerically greatest term among the terms of the expansion.
Again we can also write that kthterm is numerically greatest if Tk> Tk+1and Tk-1.
1. The numerically greatest term in the expansion of (1-2x)8, when x=2 is
a) 8C646 b) 8C444 c) 21 7 d) None of these
2. Magnitude wise the greatest term in the expansion of (3, -2x)9when x=1 is
a) 9C23722 b)9C336 23 c)9C43524 d) both (b) and (c)
3. If x > 0 and the 4thterm in the expansion of
10 3 2 8x +
has maximum value, then
a) 2< x < 3 b) 3 < x < 10
3 c) 4 < x < 5 d) None of these
4. If n is even positive integer, then the condition that the numerically greatest term in the expansion of (1+x)nmay have the greatest coefficient also is
a) | | 2 2 n n x n n + < < + b) 1 | | 1 n n x n+ < < n+ c) | | 4 4 n n x n n + < < + d) None of these
5. The interval in which x must lies so that the numerically greatest term in the expansion of (1-x)21 has the greatest coefficient is, (x > 0).
a) 5 6, 6 5 b) 5 6 , 6 5 c) 4 5 , 5 4 d) 4 5 , 5 4
Definite Integrals
THE NEWTON – LEIBNITZ FORMULA OR THE FUNDAMENTAL THEOREM OF INTEGRAL CALCULUS
Ifφ(x) is one the primitives or anti derivatives of a function f(x)defined on [a, b], then the definite integral of f(x) over [a, b] is given byφ(b) -φ (a) and is denoted by ( )
b a f x dx
∫
. Thus, if d ( ( ))x f x( ), dx φ = then ( ) ( ) ( ) b a f x dx=φ b −φ a∫
The numbers a and b are called the limits of integration ‘a’ is called the lower limit and ‘b’ is the upper limit. The interval [a, b] is called the interval of integration.
INTEGRAL FUNCTION AND ITS PROPERTIES
PROPERTY I :The integral function of an integrable function is always continuous PROPERTY II :If φ (x) is the integral function of a continuous function f(x) defined on [a, b], thenφ(x) is differentiable on (a, b) and φ1(x) = f(x) for all x∈ (a, b). PROPERTY III :The integral function of an odd function is an even function If f(x) is an odd function, thenφ (x) = ( )
x
a
f t dt
∫
is an even function PROPERTY IV :If f(t) is an even function, then φ (x) =0 ( ) x f t dt
∫
an odd function (1) Prove that / 2 2 2 2 2 2 0 sin cos , ( cos sin ) 4 ( ) x x x dx a x b x ab a b π π = + +∫
where a, b > 0. (2) Prove that 2 2 0 2 , 1 1 1 2 cos 1 , 0 1 1 if a a I dx a a x if a a π π π > − = = − + < < − ∫
(3) Prove that 2 2 2 2 0 2 , 0 sin 2 2 cos , 0 2 if a b x a dx a ab x b if b a b π π π > > = − + > > ∫
(4) Show that 2 0 1 , 1. cos dx 1 a a x a π π = > − −∫
Also, deduce that 20 1 2 (2 cos )x dx 3 3 π π = −
∫
(5) If 1 2 2 1 2 2 0 0 x x x e dx e dxβ+ ∫ − = ∫ − , then find the value of β. (6) For x > 0, let f(x) = 1 log . 1 x et dt t +
∫
Find the function f(x) + f 1 x
and show that
1 1 ( ) 2 f e f e + =
(7) If in is positive integer, prove that ! 0 a x n e− x dx=n ∫ 2 1 1 .... 2! ! n a a a e a n − − + + + +
Also, deduce the value
of 0
x n e x dx ∞ −∫
PROPERTIES OF DEFINITE INTEGRALS
(12) ( ) ( )
b b
a a
f x dx= f t dt
∫
∫
i.e., integration is independent of the change of variable.(13) ( ) ( ) ( ) b c b a a c f x dx= f x dx+ f x dx
∫
∫
∫
, where a < c < b. (14) Property ( ) ( ) . b b a a f x dx= f a b+ −x dx∫
∫
(15) Property 0 ( ) 2 ( ) , ( ) a a a f x dx f x dx if f x is an even function − = ∫
∫
0 2 ( ) , ( ) ( ) 0 , ( ) a a a f x dx if f x is an even function f x dx if f x is an odd function − = ∫
∫
(16) Prove that 2 0sin 2 sin cos
8 2 2 x x x dx x π π π π = −
∫
(17) Evaluate 2 (1 sin )2 1 cos x x dx x π π − + +∫
(19) 0 ( ) { ( ) ( )} a a a f x dx f x f x dt − = + −∫
∫
(18) Property 2 0 0 2 ( ) , (2 ) ( ) ( ) 0 , (2 ) ( ) a a f x dx if f a x f x f x dx if f a x f x − = = − = − ∫
∫
(19) Property 2 0 0 ( ) { ( ) (2 )} a a f x dx= f x = f a−x dx∫
∫
(20) Property 1 ( ) ( ) {( ) } b a a f x dx= −b a f b a x− +a dx∫
∫
(21) Property f(x) is a periodic function with period T, then (i) 0 0 ( ) ( ) , nT T f x dx=n f x dx n∈Z
∫
∫
(ii) ( ) ( ) , , 0 a nT T f x dx n f x dx n Z a R a + = ∈ ∈ ∫ ∫ (iii) 0 ( ) ( ) ( ) , , T nT f x dx n m f x dx m n Z mT = − ∈ ∫∫
(iv) 0 ( ) ( ) , , a a nT f x dx f x dx n Z a R nT + = ∈ ∈ ∫∫
(v) ( ) ( ) , , b a b nT f x dx f x dx n Z a R a nT + = ∈ ∈ ∫ +∫
INDEFINITE INTEGRALS
(7) Evaluate the following integral sin 2
sin( / 3) sin( / 3) x dx x π x π ∫ − +
(11) Evaluate the following integral 3 1 sin sin( ) dx x x+a
∫
(12) Evaluate the following integral 2 sin sin 23 ( cos ) a x b x dx b a x + +
∫
(13) Evaluate the following integral3/ 2 3/ 2
3 3
sin cos
sin cos sin( )
x x dx x x x θ + +
∫
(14) Evaluate the following integral ∫x13 / 2(1+x5 / 2 1/ 2) dx (15) Evaluate 4 1 1 x x − ∫ + dx (16) Evaluate 2 1 3 (x 2x 10) dx ∫ + +
(17) For any natural number m, evaluate ∫(x3m+x2m+xm)(2x2m+3x+6)1/m dx x, >0
(18) Evaluate ∫sec25 /13xcosec27 /13x dx (19) ∫sin3xcos 22 x dx (20) 7 2 5 (1 ) x dx x ∫ − (21) 3 5 1 sin xcs x dx ∫ (22) Evaluate 6 1 6 cos x sin x dx ∫
+ (23) Evaluate ( tan∫ x+ cot )x dx (24) Evaluate 3 5 cos cos 2 x x dx + ∫ (25) Evaluate 3 2 1 ( 1) x dx x x x x − ∫ + + + (28) Prove that 2 2 2 2 ( ) x dx x a x b ∫ + + 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 tan , 1 log , 2 x b if a b a b a b x b b a if a b b a x b b a − + − > − + − − < − + + −
(29) Evaluate ∫ cosecx−1dx (30) 1 x dx x + ∫ (31) Evaluate 2 cos sin sin cos x x dx x x ∫ − (32) Evaluate 2 2 cos sin 2 (2 cos sin ) x x dx x x + ∫ − (33) 1 2 3/ 2 sin (1 ) x dx x − ∫ − (34) ∫sin x dx (35) ∫(sin−1x)2dx (36) 1 1 1 1 sin cos sin cos x x x x − − − − − ∫ + (37) ∫cot (1−1 − +x x2)dx (38) 2 1 2 cos 2 tan sec 2 sin 2 d θ θ θ θ − ∫ − (39) sin 1 x dx a x − ∫ + (40) 2 2 cos sin 2 (2 cos sin ) x x dx x x + ∫ − (41) 3 sin 2 cos sin cos x x x x e dx x − ∫
(42 ∫cos 2 log(1 tan )x + x dx
(43) If f(x) is a polynomial function of the degree prove that
( ) { ( ) '( ) ''( ) '''( ) .... ( 1) ( )} x x n n e f x dx e f x f x f x f x f x ∫ = − + − + + − , where fn(x) = ( ) n n d f x dx
(44) Evaluate
4cos3 sin cos sin cos . 2cos2 x x x x x x x x e x x − + + ∫ dx (45) 1 2 2 tan (1 ) x dx x x − ∫ + (46) 2 2
sec xlog(1 sin x dx)
∫ + (47) 2 2 ( 1){ 1 1} 1 x x x e dx x + + + ∫ + (48) 2 1 sin (2 cosx x 1) dx ∫ − (49) 6 6 21 2 2 2 ( ) dx x a a x x a ∫ − + − (50) 2 1 (1 x) x dx x xe + ∫ +
Limits
FORMAL APPROACH TO LIMITBefore we proceed to formulate a definition of the limit of a function at a point on the basis of the discussion made in the previous article, we intend to discuss some basic concepts which will be used in further discussions.
NEIGHBOURHOOD (NBD) OF A POINT
Let a be a real number and let δ be a positive real number. Then the set of all real numbers lying between a-δ and a+δ is called the neighbourhood of a of radius ‘δ ’ and is denoted by Nδ (a).
Thus Nδ (a) = (a-δ , a+δ ) = {x. R|a-δ <x<a+δ }
The set Nδ (a) – {a} is called deleted nbd of a of radiusδ . The set (a-δ , a) is called the left nbd of a and the set (a, a+δ ) is known as the right nbd of a.
If δ is very small and x lies in the interval (a-δ , a), then x is said to approach to a from the left and we write x→a. If x. (a, a+δ ), then x is said to approach to a from the right which is denoted by x→a+.
Consider the statement |x-a|<δ . We have |x-a|<δ ⇔-δ <x-a<δ ⇔a-δ ⇔x. Nδ (a).
Thus, |x – a| <δ means that x lies in the nbd of ‘a’ of radiusδ as shown in Fig.
• • •
Let f(x) be a function with domain D and let ‘a’ be point such that every nbd of a contains infinitely many points of D.A real number l is limit of f(x) as x tends to a, if for every nbd of l, there exists a nbd of ‘a’, such that images of all points in the deleted nbd of a are in the nbd of l.
In other words, a real number l is lmit of f(x) x tends to a, if for every. > 0 there exists aδ >0 such that 0 < |x – a| <δ ⇒ |f(x) – l | < ε ⇔x. (a - δ , a+δ ), x≠a ⇒ f(x). (l+ε)
If l is the limit of f(x) as x tends to a, then we write ( ) 1
x a
Lim f x
→ =
THE ALGEBRA OF LIMITS
Let f and g be two real functions with domain D. We define four new functions f ±g fg f g, , / on domain D by setting.
(f±g) (x) = f(x) ± g(x), (fg) (x) =f(x) g(x) (f/g) (x) = f(x)/g(x), if g(x)≠0 for any x∈D.
Following are some results concerning the limits of these functions. Let lim
x→a f(x) =land lim ( )x→ag x =m. If l and m exist, then
(i) lim( )( ) lim ( ) lim ( )
x→a f ±g x = x→a f x ±x→ag x = ±l m
(ii) lim( )( ) lim ( ).lim ( )
x→a fg x = x→a f x x→ag x =lm (iii) lim ( ) lim ( ) , lim ( ) x a x a x a f x f l x g g x m → → → = = providedm≠0
(iv) lim ( ) .lim ( ),
x→aKf x =K x→a f x where K is constant
(v) lim | ( ) lim ( ) | | |
x→a f x =x→a f x = l
(vi) lim{ ( )} ( )x m
x→a f x g =l
(vii) If f(x) ≤ g(x) for every x in the deleted nbd of a, then lim
x→a f(x) ≤ limx→ag(x
(viii) If f(x) ≤ g(x) ≤ h (x) for every x in the deleted nbd of a and, lim ( ) lim ( )
x→a f x = =l x→ah x then
lim ( ) .
x→ag x =l
(ix) lim ( )
(
lim ( ))
( )x→a fog x = f x→ag x = f m
In particular
(i) lim
x→a log f(x) =log
(
lim ( )x→a f x)
=logl(ii) lim f x( ) limx af x( ) 1.
x ae e e
→
→ = =
(iii) If lim ( ) , lim 1 0 ( )
x→a f x = +∞or− ∞ then x→a f x =
(iv) Evaluate
2 2 2
3
[1 (sin ) ] [2 (sin ) ] .... [ 9 sin ) ] lim lim 0 x x x n x x n x n x →∞ + + + → (v) Evaluate 3 3 3 8 lim 8 n n r r r →∞ = − +
∏
Here Π stands for the product. (ix) 2 2 1/ 1 cos( ) lim (1 ) x cx bx a x α α → − + + −(x) If f(x+y)=f(x)+f(y) for all x, y∈ R and f(1) = 1, then evaluate (tan ) (sin ) 2 2 lim 2 0 (sin ) f x f x x x f x − →
Maxima and Minima
MAXIMUM Let f(x) be a function with domain D ⊂ R. Then, f(x) is said to attaining the maximum value at a point a∈ D if f(x) ≤ f (a0 for all x∈ D.
MINIMUM Let f(x) be a function with domain D ⊂ R. Then, f(x) is said to attain the minimum value at a point a∈ D if f(x) ≥f(a0 for all x∈D.
LOCAL MAXIMUM A function f(x) is said to attain a local maximum at x = 0 a if there exists a neighbourhood (a-δ m a+δ ) of a such that
f(x) < f(a) for all x∈ (a -δ , a + δ ), x ≠ a (or)
f(x) – f(a) < 0 for all x∈(a - δ , a+δ ), x ≠ a. In such a case f(a) is called the local maximum value of f(x) at x = a.
LOCAL MINIMUM A function f(x) is said to attain a local minimum at x = a if there exists a neighbourhood (a -δ , a+δ ) of a such that
f(x) > f(a) for all x∈(a - δ , a + δ ), x ≠ a (or) f(x) – f(a0 > 0 for all x∈ (a - δ , a +δ ), x ≠ a
The value of the function at x=a i.e., f(a0 is called the local minimum value of f(x) at x = a.
NECESSARY CONDITION FOR EXTREME VALUES: We have the following theorem which we state without proof.
THEOREM A necessary condition for f (a) to be an extreme value of a function f(x) is that f’(a) = 0, in case it exists. ILLUSTRATION Let f(x) = 3 2 10 , 0 3sin , 0 x x x x x x + + < − ≥
Investigate x = 0 for local maximum/minimum. PROPERTIES OF MAXIMA AND MINIMA
(I) If f(x) is continuous function in its domain, then at least one maxima and one minima must lie between two equal values of x.
(II) Maxima and Minima occur alternately, that is, between two maxima there is one minimum and vice-versa.
(III) If f(x)→ ∞as x →a or b and f’(x) = 0 only for one value of x (say c) between a and b, then f© is necessarily the minimum and the least value.
If f(x) → ∞as x →a or b, f(c) is necessarily the maximum the greatest value.
(1) The circle x2+y2=1 cuts the x-axis at P and Q. Another circle with center at Q and variable radius intersects the first circle at R above the x-axis and the line segment PQ at S. Find the maximum area of ∆QSR.
(2) P is a point on the ellipse
2 2
2 2 1
x y
a +b = whose center is O and N is the foot of the perpendicular from O upon the tangent at P. Find the maximum area of ∆OPN and the coordinates of P.
(3) Let A(p2, -p), B(q2, q) and C(r2, -r) be the vertices of the triangle ABC. A parallelogram AFDE is drawn with D, E and F on the lines segments BC, CA and AB respectively. Using calculus show that the maximum area of such a parallelogram is 1
(4) (at12, 2ati); i=1, 2, 3 are the vertices of a triangle inscribed in the parabola y2= 4ax. A parallelogram
AFDE is drawn with D, E, F on the line segments BC, CA and AB respectively. Show that the maximum area of such a parallelogram is
2 1 2 2 3 1 3 ( ) ( )( ). 2 a t −t t −t t −t
(8) From a fixed point P on the circumference of a circle of radius a, the perpendicular PM is let fall on the tangent at point Q. Prove that the maximum area of ∆PQM is
2
3 3 8 a
.
(9) Find the values of p for which f(x) = x3+ 6(p-3)x2+3(p2-4)x+10 has positive point of maximum. (10) Find the condition that f(x) = x3+ ax2+ bx + c has
(i) a local minimum at a certain x∈ R+ (ii) a local maximum at a certain x∈ R
-(iii) a local maximum at certain x∈ R-and minimum at certain x∈ R+.
(11) If f(x) = cos3x + λ cos2x, x ∈(0, π). Find the range of λ so that f(x) has exactly one maximum and exactly one minimum.
Monotonic Functions
STRICTLY INCREASING FUNCTION A function f(x) is said to be a strictly increasing function on (a, b) if
1 2 ( )1 ( 2)
x <x ⇒ f x < f x all x x1, 2∈( , )a b
STRICTLY INCREASING FUNCTIONA function f(x) is said to be a strictly decreasing function on (a, b) if
1 2 ( )1 ( 2)
x <x ⇒ f x > f x for all x1, x2∈ (a, b)
NECESSARY CONDITION Let f(x) be a differentiable function defined on (a, b). Then f’(x) > 0 or < 0 according as f(x) is increasing or decreasing on (a, b).
SUFFICIENT CONDITION
THEOREM Let f be a differentiable real function defined on an open interval (a, b). COROLLARY Let f(x) be a function defined on (a, b)
(a) If f’(x) > 0 for all x∈ (a, b) except for a finite number of points, where f’(x)=0, then f(x) is increasing on (a, b).
(b) If f’(x) < 0 for all x∈(a, b0 except for a finite number of points, where f’(x)=0, then f(x) id decreasing on (a, b).
SOME USEFUL PROPERTIES OF MONOTONIC FUNCTIONS
(1) If f (x) is strictly increasing function on an interval [a, b], then f-1 exists and it is also positive. (2) If f(x) is strictly increasing function on an interval [a, b] such that it is continuous, then f-1 I
continuous on [f(a), f(b)].
(3) If f(x) is continuous on [a, b] such that f’(c ) ≥0 (f’©>0) for each c∈(a, b), then f(x) is monotonically (strictly) increasing function on [a, b].
(4) If f(x) is continuous on [a, b] such that f’ (c ) ≤0 (f’© < 0) for each∈ (a, b), then f(x) is monotonically (strictly) decreasing function on [a, b]
(5) If f(x) and g(x) are monotonically (or strictly) increasing (or decreasing) functions on [a, b], then gof(x) is a monotonically (or strictly) increasing function on [a, b]
(6) If one of the two functions f(x) and g(x) is strictly (or monotonically) increasing and other a strictly (monotonically) decreasing, then g of (x) is strictly (monotonically) decreasing on [a, b].
(1) Let f(x) = 2 3 , 0 , 0 ax xe x x ax x x ≤ + − >
, where a is a positive constant.
Find the intervals in which f’ (x) is increasing.
(2) If φ (x) =f(x) + f(1-x) and f’’ (x) < 0 for all x∈[0, 1]. Prove that φ(x) is increasing in [0, ½] and decreasing in (1/2, 1].
(3) Let g(x) = 2f(x/2) + f(2-x) and f’’(x) < 0 for all x∈(0, 2). Find the intervals of increases and decrease of g(x).
Mean Values Theorems and Some Other Applications of Derivatives
ROLE’S THEOREMSTATEMENT Let f be a real valued function defined on the closed interval [a, b] such that (i) It is continuous on the closed interval [a, b],
(ii) it is differentiable on the open interval (a, b), and (iii) f(a) = f(b).
LAGRANGE’S MEAN VALUE THEOREM
STATEMENT Let f(x) be a function defined on [a, b] such that it is continuous on [a, b] (ii) it is differentiable on (a, b).
Then there exists a real number c∈ (a, b) such that f c'( ) f b( ) f a( ) b a
− =
− (1) Prove that (b-a)sec2a<tanb-tana<(b-a)sec2b, where 0 < a < b <
2 π .
(2) If f(x) and g(x) are continuous functions in [a, b] and they are differentiable in (a, b), then prove that there exists c∈ (a, b) such that ( ) ( ) ( ) ( ) '( )
( ) ( ) ( ) '( )
f a f b f a f b
b a
g a f b = − g a g c (3) If the function of f: [0, 4] → R is differentiable, then show that
(1) (f(4))2– (f(0))2= 8f’(a) f(b) for a, b∈ (0, 4) (2) 4 2 2 0 ( ) 2[ ( ) ( ) )] f t dt= α αf +β βf
∫
for all 0 < α β, < 2.(4) If 2a+3b+6c=0, then prove that the equation ax2+bx+c=0 has at least one real root in (0, 1)
(5) If aoxn+ a1xn-1 +….+an-1x=0 has a real positive root α, then prove that the equation
1 2
0 ( 1) 1 ... 1 0
n n
n
Permutations and Combinations
FACTORIALThe continued product of first n natural numbers is called the “n factorial” and is denoted by n or n! i.e. ! 1 2 3 4 ... ( 1)
n = × × × × × − ×n n EXPONENT OF PRIME p in n!
Let p be a prime number and n be a positive integer. Then, the last integer amongst 1, 2, 3…., (n-1), n which is divisible by p is n p, p where n p
denotes the greatest integer
less than or equal to n p.
(1) Find the number of zeros at the end of 100!.
(2) Find the largest positive integer n for which 35! Is divisible by 3n.
SYMBOLnPr OR, P(n, r) :
If n is a natural number and r is a positive integer satisfying 0≤ ≤r n, then the natural number !
( )! n
n r− is denoted by the symbol
nP
ror, P(n, r)
PROPERTIES OFnCror, C(n, r)
We shall now discuss some important properties ofnCr.
PROPERTY I nCr=nCn-rfor 0≤ ≤r n
PROPERTY II Let n and r be non-negative such that r ≤ n. Then, nCr+nCr−1=n+1Cr PROPERTY III Let n and r be non-negative integers such that 1≤ ≤r n. Then,
1 1 . n n r r n C C r − − = PROPERTY IV If 1≤ ≤r n, then 1 1 1 .n ( 1)n . r r n − C− = − +n r C − PROPERTY V IfnCx=nCy ⇒ x = y or, x+y=n.
PROPERTY VIIf If n is even, then the greatest value of nCr(0≤ ≤r n) is nCn/ 2 PROPERTY VII If n is odd then the greatest value ofnCr (0≤ ≤r n) is
1 2 n n C + or, 1 2 n n C −
FUNDAMENTAL PRINCIPLES OF MULTIPLICATION
If there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then the two jobs in succession can be completed in m × n wasy.
FUNDAMETNAL PRINCIPLE OF ADDTION
If there are two jobs such that they can be performed independently in m and n ways respectively, then either of the two jobs can be performed in (m+n) ways.
(1) Five persons entered the lift cabin on the ground floor of an 8 floor house. Suppose each of them can leave the cabin independently at any floor beginning with the first. Find the total number of ways in which each of the five persons can leave the cabin (i) at any one of the 7 floors (ii) at different floors. COMBINATIONSEach of the different selections made by taking some or all of a
number of distinct objects or items, irrespective of their arrangements or order in which they are placed, is called a combination.
PERMUTATIONS Each of the different arrangements which can be made by taking some or all of a number of distinct objects is called a permutation.
THEOREM 1: Let r and n be positive integers such that 1≤ ≤r n. Then prove that the number of all permutations of n distinct items or objects taken r at a time is.
THEOREM 2:The number of all permutations (arrangements) of n distinct objects taken all at a time is n!. THEOREM 3:The number of ways of selecting items or objects from a group of n distinct items or objects is
PERMUTATIONS UNDER CERTAIN CONDITIONS
THEOREM:The number of all permutations (arrangements) of n different objects taken r at a time, (i) When a particular object is to be always included in each arrangement is 1
1 ! n r C r − − ×
(ii) When a particular object is never taken in each arrangement is n−1Cr−1×r! (iii) When two specified objects always occur together is 2
1 ! 2! n r C r − − × ×
(1) Find the number of ways of arranging m white and n black balls in a row (m > n) so that no two black balls are placed
PERMUTATIONS OF OBJECTS NOT ALL DISTINCT
THEOREM:The number of mutually distinguishable permutations of n things, take all at a time, of which p are alike of one kind, q alike of second such that p+q=n, is
! ! ! n p q
(1) Find the number of ways of selecting two integers a and b from the set {1, 2, …, 5n}, n∈N so that a4– b4is divisible by 5.
Progressions SEQUENCE
A sequence is a function whose domain is the set N of natural numbers. REAL SEQUENCE
A Sequence whose range is a subset of R is called a real sequence.
In other words, a real sequence is a function with domain N and the range a subset of the set R of real numbers.
PROGRESSIONS: It is not necessary that the terms of a sequence always follow a Certain pattern or they are described by some explicit formula for the nth term. Those sequences whose terms follow certain patterns are called progressions.
ILLUSTRATION 1 11, 7, 3, -1, …is an A.P. whose first term is 11 and the common difference 7-11=-4.
ILLUSTRATION 2 Sow that the sequence <an> defined by an= 2n2+1 is not an A.P.
PROPERTIES OF AN ARITHMETIC
PROPERTY I: If a is the first term and d the common difference of an A.P., then its nth terms anis given by an= a+(n-1)d
PROPERTY II: A sequence is an A.P iff its nth term is of the form An+B i.e. a linear expression in n. The common difference in such a case is A i.e. the coefficient of n. PROPERTY III: If a constant is added to or subtracted from each term of an A.P., then the resulting sequence is also A.P. with the same common difference.
PROPERTY IV: If each term of a given A.P. is multiplied or divided by a non-zero constant k, then the resulting sequence is also an A.P. with common difference kd or d/k, where d is the common difference of the given A.P.
PROPERTY V: In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of first and last term i.e.
ak+ an-(k-1)= a1+ anfor all k=1, 2, 3, …, n-1.
PROPERTY VI: Three numbers a, b, c are in A.P. iff 2b=a+c.
PROPERTY VII: If the terms of an A.P. are chosen at regulr intervals then they from an A.P.
PROPERTY VIII: If an, an+1and an+2are three consecutive terms of an A.P., then
2an+1= an+an+2.
INSERTION OF ARITHMETIC MEANS
If between two given quantities a and b we have to insert n quantities A1, A2, …,Ansuch that a, A1, A2,
…..An, b form an A.P., then we say that A1, A2, ….,Anare arithmetic means between a and b.
INSERTION OF n ARITHMETIC MEANS BETWEEN a AND b.
Let A1, A2, ….,Anbe n arithmetic means between two quantities a and b. Then,
a, A1, A2,…., An,b is an A.P.
Let d be the common difference of this A.P. Clearly, it contains (n+2) terms. ∴ b = (n+2)th term
⇒ b = a+(n+1) d ⇒ 1 b a d n − = + Now, A1= a + d ⇒ A1= 1 1 b a A a n − = + + A2= a + 2d ⇒ 2 2( ) 1 b a A a n − = + + An= a + nd ⇒ An= ( ) 1 n n b a A a n − = + + ILLUSTRATIVE EXAMPLES
(1) Between 1 and 31 are inserted m arithmetic means so that the ratio of the 7thand (m-1)th means is 5 : 9. Find the value of m.
GEOMETRIC PROGRESSION
A sequence of non-zero numbers is called a geometric progression (abbreviate as G.P.) if the ratio of a term and the term preceding to it is always a constant quantity.
The constant ratio is called the common ratio of the G.P.
In other words a sequence, a1, a2, a3,. …,an…. is called a geometric progression if
1 n n a a +
= constant for all n∈N.
PROPERTIES OF GEOMETIC PROGRESSIONS
In this section, we shall discuss some properties of geometric progressions and geometric series PROPERTY I: If all the terms of a G.P. be multiplied or divided by the same non-zero
constant, then it remains G.P. with the same common ratio.
PROPERTY II: The reciprocals of the terms of a given G.P. form a G.P.
PROPERTY III: If each terms of a G.P. be raised to the same power, the resulting sequence also forms a G.P.
PROPERTY IV: In a finite G.P the product of the terms equidistant from the beginning and the end is always same and is equal to the product of the first and the last term.
PROPERTY V Threenon-zero numbers a, b, c are in G.P. iff b2= ac
PROPERTY VI Ifthe terms of a given G.P. are chosen at regular intervals, then the new sequence so formed also forms a G.P.
PROPERTY VII Ifa1, a2, a3….,an, …be a G.P. of nonzero non negative terms, then
log a1, loga2, …, logan,…. is an A.P and vice – versa.
ILLUSTRATIVE EXAMPLES
(1) Find all the complex numbers x and y such that x, x+2y, 2x+y are in A.P. and (y+1)2, xy + 5, (x+1)2 are in G.P. Also, find the progression.
(2) If mth term of a G.P.is m and nth term is n, then prove that its rth term is 1/ m n r n r m m n − − − SUM OF n TERMS OF A G.P.
THEOREM To prove that the sum of n terms of a G.P. With first term ‘a’ and common ratio ‘r’ is
given by 1 1 n n r S a r − = − or, 1 , 1 1 n n r S a r r − = ≠ − . ILLUSTRATIVE EXAMPLES
(1) Let S denote the sum of the terms of an infinite G.P. and σ2 denote the sum of the squares of the terms. Show that the sum of the first n terms of this geometric progression is given by
2 2 2 2 1 n S S S σ σ − − +
(2) Find the geometric progression of real number such that the sum of its first four terms is equal to 30 and the sum of the squares of the first four terms is 340.
(3) Prove that
91
111...1
times is not a prime number.
INSERTION OF n GEOMETRIC MEANS BETWEEN TWO GIVEN NUMBERS A AND b. Let G1, G2, … Gn be n geometric means between two given numbers a and b. Then,
A, G1, G2, …., Gn, b is a G.P. consisting of (n+2) terms. Let r be the common ratio of this G.P.
Then, B = (n+2)th term = arn+1 ⇒ rn+1= b a ⇒ r=(b/a)1/n+1 ∴ 1/( 1) 1 , n b G ar a a + = = ∴ 2 /( 1) 2 2 ..., n b G ar a a + = = ( 1) 1 n n b G ar a a + = =
AN IMPORTANT PROPERTY OF GEOMETRIC GEANS THEOREM:
If geometric means are inserted between two quantities, then the product of n geometric means is the nth power of the single geometric mean between the two quantities.
SOME IMPORTANT PROPERTIES OF ARITHMETIC AND GEOMETRIC MEANS BETWEEN TWO GIVEN QUANTITIES.
PROPERTY IIf A and G are respectively arithmetic and geometric means between two positive number a and b, then A > G.
PROPERTY II If A and G are respectively arithmetic and geometric means between two positive quantities a and b, then the quadratic equation having a, b as its roots is x2– 2Ax + G2= 0
PROPERTY III If A and G be the A.M. and G.M. between two positive numbers, then the numbers are A± A2−G2
(1) If one geometric mean G and two arithmetic means A1and A2be inserted between two given
quantities, prove that G2= (2A1– A2) (2A2-A1).
(2) The A.M. between m and n and the G.M. between a and b are each equal to ma nb. m n
+
+ Find m and n in terms of a and b.
ARITHMETICO-GEOMETRIC SERIES Let a, (a + d)r, (a+2d)r2, (a+3d)r3, … be an arithmetico –geometric sequence. Then, a+(a+d)r+(a+2d)r2+(a+3d)r3+… is an arithmetico geometric series.
ILLUSTRATION Find the nth term of the series1 2 32 43 ... 3 3 3 + + + + SUM OF n TERMS OF AN ARITHMETICO-GEOMETRIC SEQUENCE THEOREMThe sum of n terms of an arthmetico-geometric sequence a, (a + d) r, (a+2d)r2, (a+3d)r3, … is given by
1 (1 ) { ( 1) } , 1 1 1 1 [2 ( 1) ], 1 2 n n n a r a n d r dr when r r r r S n a n d when r − + − − + − ≠ − − − = + − =
EXAMPLE Show that the sum of the series
2 2 1 2 1 1 5 .... 2 1 2 1 n n to n n + + + + + − −
n terms is an even or an odd
number according as n is even or odd. HARMONIC PROGRESSION
DEFINITIONA sequence a1, a2,…., an, …. Of non-zero number is called a Harmonic
progression, if the sequence
1 2 3 1 1 1 1 , , ,..., ,... n a a a a is an Arithmetic progression . nth TERM OF A HP
The nth term of a H.P is the reciprocal of the nth term of the corresponding A.P. Thus, if a1, a2, a3, …, an, … is a HP and the common difference of the corresponding AP is d
i.e. 1 1 1 n n d a + a = − , then 1 1 1 ( 1) n a n d a = + − If a, b, c are in HP, then 1 1 1, ,
a b c are in AP. Therefore
2 1 1 b = +a c ⇒ 2ac b a c = + .
(1) If S1, S2and S3denote the sum up to n(>1) terms of three non-constant sequence in A.P., whose first
terms are unity and common differences are in H.P., prove that 1 3 1 2 2 3
1 2 3 2 2 S S S S S S n S S S − − = − +
(2) If a, b, c are in HP and a > c, show that 1 1 4 b c− +a b− > a c−
(3) Let a, b, c be positive real numbers. If a, A1, A2, b are in A.P., a, G1, G2, b are in G.P. and a, H1, H2, b
are in H.P., show that 1 2 1 2
1 2 1 2 (2 ) ( 2 ) 9 G G A A a b a b H H H H ab + + + = = +
PROPERTIES OF ARITHMETIC, GEOMETRIC AND HARMONIC MEANS BETWEEN TWO GIVEN NUMBERS
Let A, G and H be arithmetic, geometric and harmonic means of two positive numbers a and b. Then. , 2 a b A= + G ab and H 2ab a b =
PROPERTY1 H 2ab a b =
+ A ≥ G ≥ H.
PROPERTY2 A, G, H form a GP i.e, G2– AH.
PROPERTY3 The equation having a and b as its roots x2– 2Ax + G2= 0
PROPERTY4 If A, G, H are arithmetic, geometric and harmonic means between three given numbers a, b and c, then the equation having a, b, c as its roots is
3 3 2 3 3 3 G 0 x Ax x G H − + − =
(1) For what value of n,
1 1 n n n n a b a b + + +
+ is the harmonic mean of a and b?
(2) If A1, A2; G1, G2; H1, H2be two A.M.’s, G.M’s and H.M.s between two number a and b, then prove
that :
(3) If H1, H2,….,Hnbe n harmonic means between a and b and n is a root of the equation x2
(1-ab)-x(a2+b2)-(1+ab)=0, then prove that H1= ab(a-b) 1
1 n H n r H nr + = +
METHOD OF DIFFERENCES: Sometimes the nth term of a sequence or a series cannot be
determined by the methods discussed in the earlier sections. In such cases, we use the following steps to find the nth term Tnof the given sequence.
STEP 1 Obtain the terms of the sequence and compute the differences between the successive terms of the given sequence. If these differences are in A.P, then take Tn= an2+ bc +c, where a, b, c are constants.
Determine a, b, c by putting n=1, 2, 3 and putting the values of T1, T2, T3.
STEP 2 If the successive differences computed in step 1 are in G.P. with common ratio r, then take Tn
= arn-1 + bn+c.
STEP 3 If the differences of the differences computed in step 1 are in A.P., then take Tn= an3+ bn2+
cn + d and find the values of constants a, b, c, d.
STEP 4 If the differences of the differences computed in step 1 are in G.P. with common ratio r, then take Tn= arn-1 + bn2+ cn + d The following examples will illustrate the above procedure.
(1) Sum the following series to n terms : 5 + 7+ 13 + 31 + 85 +…. (2) If 1 1 1 .... 1 2 3 n H n = + + + + and, 1 ' 2 n n H = + - 1 2 3 ... 2 ( 1) ( 1)( 2) ( 2)( 3 2.3 n n n n n n n + + + + − − − − − − show that Hn= Hn1
(3) Find the sum of first n terms of the series whose nth term is 1 ( 4 2 3 2 2 1)
( 1) n n n n n n+ + + + − (4) If 1 ( 1)( 2)( 3) , 12 n r r n n n n T = + + + =
∑
where Trdenotes the rth term of the series. Find, 1 1 lim . n n r Tr →∞ =∑
(5) Sum the series to n terms: 1 1 1 ....
Scalar or Dot Product of Vectors
DEFINITIONLet a and b be two non-zero vectors inclined at an angleθ. Then, the scalar product of a with b is denoted by a b . and is defined as the scalar | | | |cosa b θ
(1) Show that the angle between two diagonals of a cube iscos 1 1 3
−
.
(2) The length of the sides a, b, c of a triangle ABC are related as a2+b2= 5c2. Prove, using vector methods, that the medians drawn to the sides a and b are perpendicular.
(3) Determine the lengths of the diagonals of a parallelogram constructed on the vectors a=2α β − and
2
b= −α β, where α and β are unit vectors forming an angle of 60o.
(4) Two points A and B are given on the curve y=x2such that OA. i=1 and OB. i =-2. Find |2OA−3OB|
VECTOR (CROSS) PRODUCT OF VECTORS
DEFINITIONVECTOR (CROSS) PRODUCTLet a b , be two non –zero non-parallel vectors. Then the vector product a b × , in that order, is defined as a vector whose magnitude is | || |a b
sinθ whereθ is the angle between a and b and whose direction is perpendicular to the plane of a and b in such a way that a b , and this direction constitute a right handed system.
PROPERTIES OF VECTOR PRODUCT The vector product has the following properties:
(1) Vector product I not commutative i.e. if a and b are any two vectors, then a × b ≠ b ×a, however - b a × .
(2) If a and b are two vectors and m is a scalar, then
( )
ma b × =m a b × = ×a mb
(3) If a b , are two vectors and m, n are scalars, then ma nb× =mn a b( × =) m a nb(× )=n ma b( × ) (4) Vector product is distributive over vector addition i.e.
( )
a × + = × + ×b c a b a c and, (b c + × = × + ×) a b a c a
(5) For any three vectors , ,a b c , we havea× − = × − ×(b c ) a b a c .
(6) The vector product of two non-zero vectors is she the null vector if they are collinear or parallel i.e. 0 ||
a b × = ⇔a b , where a b , are non-null vectors
(7) i i × = × = × = 0,j j k k i× = j k j i, × = − k, j k× =i k, × = j i k i, × = j i k,× = − j (8) If a=a i1+a k2, and b=b i b j1+ 2+b k3, then 1 2 3 1 2 3 i j k a b a a a b b b × =
SOME USEFUL RESULTS
RESULT I If a and b are two non-zero, non-parallel vectors, then unit vectors normal to the plane of a and b are | | a b a b × ± ×
RESULT II Vectors of magnitude λ normal to the plane of a and b are | | a b a b λ × ± ×
RESULT III The area of the parallelogram with adjacent sides a and b is |a × b| RESULT IV The area of a triangle with adjacent sides a and b is 1| |
2 a b× RESULT V Area of 1| | 2 ABC AB AC ∆ = × = 1| | 2 BC BA× = 1| | 2 CA CB×
RESULT VI The area of a plane quadrilateral ABCD is 1| | 2 AC BD×
, where AC and BD are its diagonals.
(1) If a b c , , are the position vectors of the vertices A, B, C of a triangle ABC, show that the area of triangle ABC is 1| |
2 a× + + ×b c c a
.
(2) Show that the perpendicular distance of the point c from the line joining a and b is | | | | b c c a a b b a × + × + × −
(3) If A, B, C, D be any four points in space, prove that | AB CD× +BC×AD CA BD+ × |= 4 (Area of triangle ABC).
(4) Let OA =a OB, =10a+2 ,b and OC =b where O is origin. Let p denote the area of the quadrilateral OABC and q denote the area of the parallelogram with OA and OC as adjacent sides. Prove that p = 6q. (5) ABCD is a quadrilateral such that AB=b AD, =d, AC=mb+ pd. Show that the area of the quadrilateral ABCD is 1| || |
2 m+ p b d×
PRODUCT OF THREE VECTORS
SCALAR TRIPLE PRODUCTDEFINITIONLet a b c , , be three vectors. Then the scalar (a ×b c). is called the scalar product of a, b and c and is denoted by [a b c ].
Thus, [a b c ] = (a× b). c.
PROPERTIES OF SCALAR TRIPEL PRODUCT
PROPERTY-I If a b c , , are cyclically permuted the value of scalar triple product remains same, i.e (b ×c a). = ×(c a b ). (or), [a b c ] [= b c a ] [= c a b ]
PROPERTY-IIThe change of cyclic order of vectors in scalar triple product changes the sign of the scalar triple product but not the magnitude.
i.e.[a b c ]= −[b a c ]= −[c b a ]= −[a c b ]
PROPERTY-IIIIn scalar triple product the positions of dot and cross can be interchanged provided that the cyclic order of the vectors remains same i.e.
PROPERTY-IVThe scalar triple product of three vectors is zero if any two of them are equal. [λa b c ]=λ[a b c ]
PROPERTY-VIThe scalar triple product of three vectors is zero if any two of them are parallel or collinear.
PROPERTY-VIIIf a b c d , are four vectors, then [a +b c d]= [a c d ] + [b c d ]
PROPERTY-VIIIThe necessary and sufficient condition for three non-zero, non-collinear vectors a b c , , to be coplanar is that [a b c ]=0. i.e. a b c , , are coplanar ⇔ [a b c ]=0
PROPERTY-IXFor points with position vectors a b c , , and d will be coplanar if [ , , ] [ , , ] [ , , ] [ , , ]d b c + d c a + d a b = a b c
PROPERTY-X If a=a i,+a j2+a k3, b=b i b j1+ 2+b k3 and c=c i1+c j2+c k3 are three vectors, then
1 2 3 1 2 3 1 2 3 [ ] a a a a b c b b b c c c =
(1) If the vectors α= =ai a j+ck,β= +i k and γ= +ci c j+bk are coplanar, then prove that c is the geometric mean of a and b.
(2) Let a b c , , three non-zero vectors such that c is a unit vector perpendicular to both a and b. If the angle between a and b is π/ 6, prove that [ ]2 1
4
a b c = | | | | .a 2 b 2 (3) Consider the vectors: A= +i cos(β α− )j+cos(γ α− )k
cos( ) cos( )
B= α β− i+ +j γ β− k
and, C=cos(α γ− )i+cos(β γ− )j+ak
where α β, are γ are different angles in (0,π/ 2). If A B C, , are coplanar vectors, show that a is independent ofα β, and γ.
(4) If a b c , , be three on –coplanar unit vectors equally inclined to one another at an angleθ such that ,
a × + × =b b c pa+qb+rc find p, q, r in terms ofθ. Also, prove that
2 2 2 2 cos q p r θ + + = .
(5) If a b c , , are three non-coplanar vectors, prove that any vector r is expressible as
[ [ [ . [ ] [ ] [ ] r b c r c a r b c r a b c a b c a b c a b c = + +
RECIPROCAL SYSTEM OF VECTORS
Let a b c , , be three non-coplanar vectors, vectors, so that [a b c ]≠0.We define another set of three vectors a b c , , as given below . ,
[ ] [ ] b c c a a b a b c a b c × × = = [ ] a b c a b c × =
(1) If a b c , , are three non-coplanar vectors and a b c , , form a reciprocal system of vectors, then prove that
(ii) a b . =a c. =0; .b c =b a. =0;c a . =c b. =0 (iii) [ ] 1 [ ] a b c a b c =
(2) If a b c and a b c ', ', ' be the reciprocal system of vectors, prove that (i) a a b b c c . + . + . =3 (ii) a a b b c c × + × + × =0
(3) If a and b are two vectors such that .a b ≠0, then solve the vector equations . 0, . 1,
r a= r b=
[r a b ] 1.=
(4) If r a × +( . )r b c =d, then prove that ( 2 , ( . )| | a d c r a a a c a λ × × = + × where λ is a scalar VOLUME OF A TETRAHEDRON
THEOREM (i) If two pairs of opposite edges of a tetrahedron are perpendicular, then the opposite edges of the third pair are also perpendicular to each other.
(ii) The sum of the squares of two opposite edges is the same for each pair of opposite edges (iii) Any two opposite edges in a regular tetrahedron
CENTROID OF A TETRAHEDRON
THEOREM The volume V of a tetrahedron whose three coterminous edges in the right-handed system are ,
a b c is given by 1[ ] 6
V = a b c
(1) A tetrahedron has three of its vertices of its vertices at A, B and C whre
3 2 , 3 ; 2
OA= +i j OB= +i j−k OC= j. Find the unit vector perpendicular to the face ABC. The fourth vertex D is such that DA AB . = =0 DA AC. . Find the vector equation of AD. If the volume of the tetrahedron is 3 2 cubic units and D is on the same side as the origin, find the coordinates of D.
(2) The position vectors of the vertices A, B and C of a tetrahedron ABCD are i+ + ,j k i and 3i respectively. The altitude from vertex D to the opposite face ABC meets the median line through A of the triangle ABC at a point E. If the length of the side AD is 4 and the volume of the tetrahedron is 2 2
3 cubic units, find the position vector of the point E for all its possible positions.
(3) OABC is a regular tetrahedron. D is the circumcentre of ∆OAB and E is the middle point of the edge AC. Use vector method to find distance DE. .
(4) A pyramid with vertex at the point P whose position vector is 4i+2j+2 3k has a regular hexagonal base ABCDEF. The points A and B have position vectors i and i+2j respectively. The centre of hexagon has position vector i+ +j 3k. Given that the volume of the pyramid is
6 3 and the perpendicular from the vertex meets the diagonal AD, locate the position vectors of the foot of this perpendicular.
(5) Let a=a i1+a j2+a k b3,=b i b j1+ 2+b k3 and c=c i1+c j2+c k3 be three non-zero vectors such that c is a unit vector perpendicular to both the vectors a and b. If the angle between a and b is , 6 π prove that 1 2 3 1 2 3 1 2 3 1 4 a a a b b b c c c = (a12+ a22+ a32) (b12+ b22+ b32)
(6) If a b c , , are three on-coplanar vectors and r is any vector in space, then prove that
. . . ( ) ( ) ( ) [ ] [ ] [ ] r a r a r c r b c c a a b a b c a b c a b c = × + × + ×
(7) If a b c , , and d are four vectors, then prove that (i) (a b × ).(c d× + ×) (b c ).(a d× + ×) (c a ).(b d× )=0 (ii) d a .[ × × ×{b c d)}] [ . ][= b d a c d ]
Tangents and Normals
ILLUSTRATIVE EXAMPLES:(i) The curve y=ax3+bx2+cx+5 touches the x-axis at P(-2, 0) and cuts the y-axis at the point Q where its gradient is 3. Find the equation of the curve completely.
(ii) Find the equation of the normal to the curve y= (1=x)y+ sin-1(sin2x)atx=0
(iii) Determine the constant c such that the straight line joining the points (0, 3) and 95, -2) is tangent to the curve 1 c y x = +
(iv) Prove that all normal to the curve x=a cost + at sint, y = a sint - at cost (v) Find the points at which the tangents to the curves y=x3– x – 1 and
y=3x2– 4x + 1 are parallel. Also, find the equations of tangents.
(vi) Find the equation of the tangent to x3= ay2at the point A (at2, at3). Find also the point where this tangent meets the curve again.
(vii) Tangent at point P1(other than (0, 0) on the curve y=x3meets the curve again at P2. The tangent at P2
meets the curve at P3and so on. Show that the abscissae of P1, P2, P3. ….., Pnform a GP. Also, find
the ratio 1 2 3 2 3 4 area P P P area P P P ∆ ∆ (viii) For the function F(x) =
0
2 | |
x
t
∫
dt, find the tangent lines which are parallel to the bisector of the angle in the first quadrant.(ix) If α β, are the intercepts made on the axes by the tangent at any point of the curve x=a cos3θ, y=bsin3θ, prove that
2 2
2 2 1
a b
α +β =
.
(x) If x1 and y1 be the intercepts on the axes of X an Y cut off by the tangent to the curve
1, n n x y a b + =
then prove that
/ 1 / 1 1 1 1. n n n n a b x y − − + =
(xi) Show that the normal to the rectangular hyperbola xy=c2 at the point P 1
1 , c ct t
meets the curve
again at the point 2
2 , c Q ct t , if t1 3 t2= -1
ANGLE OF INTERSECTION OF TWO CURVES
The angle of intersection of two curves is defined to be the angle between the tangents to the two curves at their point of intersection.
ORTHOGONAL CURVES
If the angle of intersection of two curves is a right angle, the two curves are said to intersect orthogonally and the curves are called orthogonal curves.
If the curves C1and C2are orthogonal, thenφ =π / 2
∴ m1m2= -1 ⇒ 1 2 1 C C dy dy dx dx = − EXAMPLES:
(i) Find the acute angle between the curves y = |x2–1| and y=|x2– 3| at their points of intersection (ii) Show that the curves x3– 3xy2= -2 and 3x2y-y3=2 cut orthogonally.
(iii) Find the acute angles between the curves y = |2x2– 4| and y=|x2– 5|.
(iv) Show that the curves y2=4ax and ay2=4x3intersect each other at an angle of tan-11
2 and also if PG1 and PG2be the normals to two curves at common point of intersection (other than the origin)
meeting the axis of X in G1and G2, then
G1G2= 4a.
LENGTHS OF TANGENT, NORMAL, SUBTANGENT AND SUBNORMAL
Let the tangent and normal at a point P(x, y) on the curve y=f(x), meet the x-axis at T and N respectively. If G is the foot of the ordinate at P, then TG and GN are called the Cartesian subtangent and subnormal, while the lengths PT and PN are called the lengths of the tangent and normal respectively.
If PT makes angle Ψ with x-axis, then tanΨ = dy
dx. From Fig we find that Subtangent = TG = y cot y dy dx Ψ = Subnormal = GN = y tan ydy dx Ψ =
Length of the tangent = PT = y cosecΨ
