Since, both the sphere and the plane are smooth, so there is no change in velocity parallel to the horizontal plane.
∴ v sinθ = u sinα …..(i) N
u
And by Newton’s law, along the normal CN, velocity of separation v
= e.(velocity of approach) ….(ii) αθ
∴ v cosθ − 0 = eu cosα om
v cosθ = eu cosα
Dividing (i) by (ii), we get cotθ = e cotα c
Particular case : If α = 0 then from (i), v sinθ = 0 ⇒ sinθ = 0
∴θ = 0 ; v ≠ 0 and from (ii) v = eu
Thus if a smooth sphere strikes a smooth horizontal plane normally, then it will rebound along the normal
with velocity, e times the velocity of impact i.e. velocity of rebound = e.(velocity before impact).
Rebounds of a particle on a smooth plane : If a smooth ball falls from a height h upon a fixed smooth
horizontal plane, and if e is the coefficient of restitution, then whole time before the rebounding ends = 2h . 1 + e
g 1 − e
And the total distance described before finishing rebounding = 1 + e2 h .
1 − e2
Example: 30 A ball is dropped from a height h on a horizontal plane and the coefficient of restitution for the impact is e, the velocity
Solution: (d)
Example: 31 with which the ball rebounds from the floor is [Roorkee 1994]
Solution: (c)
(a) eh (b) egh (c) e gh (d) e 2gh
Example: 32
Let u be the velocity with which the ball strikes the ground. Then, u2 = 2gh ⇒ u = 2gh .....(i)
Suppose the ball rebounds with velocity v, then V = −e(−u − 0) ⇒ v = ue ⇒ v = e 2gh , [from (i)].
A sphere impinges directly on a similar sphere at rest. If the coefficient of restitution is 1 , the velocities after impact are in
2
the ratio [MNR 1987; UPSEAT 2000]
(a) 1 : 2 (b) 2 : 3 (c) 1 : 3 (d) 3 : 4
From principle of conservation of momentum ,
mv1 + m′v1′ = mu1 + m′u1′ ......(i)
From Newton's rule for relative velocities before and after impact.
v1 − v1′ = −e(u1 − u1′ ) ......(ii)
Here m = m′ , e = 1 and let u1′ = 0
2
Then from (i) mv1 + mv1′ = mu1 or v1 + v1′ = u1 ......(iii)
From (ii), v1 − v1′ = −1 u1 .......(iv)
2
Adding (iii) and (iv) , 2v1 = 1 u1 or v1 = 1 u1 ......(v)
2 4
From (iii) and (v), v1′ = u1 − 1 u1 = 3 u1 . Hence v1 : v1′ = 1: 3.
4 4
A ball impinges directly upon another ball at rest and is itself reduced to rest by the impact. If half of the K.E. is destroyed
in the collision, the coefficient of restitution, is [MNR 1984]
(a) 1 (b) 1 (c) 3 (d) 1
4 3 4 2
Solution: (d) Let masses of the two balls be m1 and m2 .
By the given question, we have
m1 m2
u1 u2 = 0
v1 = 0 v2
∴ m1u1 + m2.0 = m1.0 + m2v2 or m1u1 = m2v2 .....(i) and 0 − v2 = e(0 − u1) i.e., v2 = eu1 .....(ii)
Again for the given condition, K.E. (before impact) = 2 × K.E. after impact
∴ 1 m1u12 + 0 = 2 0 + 1 m2v22 .....(iii) or 1 m1u12 = 2. 1 m2v22
2 2 2 2
But from (i), m1u1 = m2v2 , ∴ 1 u1 = v2 i.e., u1 = 2v2 . Putting in (ii), we get v2 = e.2v2 , ∴ e = 1 .
2 2
Projectile Motion.
When a body is thrown in the air, not vertically upwards but making an acute angle α with the horizontal,
then it describes a curved path and this path is a Parabola.
The body so projected is called a projectile. The curved path described by the body is called its trajectory.
The path of a projectile is a parabola whose equation is
y = x tan α − gx 2 Y v sin θ v
2u 2 cos 2 θ v cosθ
α u
Its vertex is A u 2 sin α cos α , u 2 sin 2 α .
g 2g
Focus is S u 2 sin 2α , − u 2 cos 2α α BX
2g 2g O
L
(1) Some important deductions about projectile motion: Let a particle is projected with velocity u in a
direction making angle α with OX . Let the particle be at a point P(x, y) after time t and v be its velocity making an
angle θ with OX .
(i) Time of flight = 2u sinα
g
(ii) Range of flight i.e., Horizontal Range R = u2 sin 2α
g
Maximum horizontal Range = u2 and this happens, when α = π
g 4
(iii) Greatest height u2 sin 2 α
2g
=
Time taken to reach the greatest height = u sinα
g
(iv) Let P(x, y) be the position of the particle at time t. Then x = (u cosα)t , y = (u sin α ) t − 1 gt 2
2
(v) Horizontal velocity at time t= dx = u cosα
dt
Vertical velocity at time t= dy = u sinα − gt
dt
(vi) Resultant velocity at time t = u2 − 2ugt sinα + g 2t 2 . And its direction is θ = tan −1 u sinα − gt
u cosα
(vii) Velocity of the projectile at the height h = u2 − 2gh u2 sin 2 α − 2gh
and its direction isθ = tan−1 u cosα
(2) Range and time of flight on an inclined plane
Yu
α RA
O β (R cos β,R sin β)
X
Let OX and OY be the coordinate axes and OA be the inclined plane at an angle β to the horizon OX. Let a
particle be projected from O with initial velocity u inclined at an angle α with the horizontal OX. The equation of
the trajectory is y = x tanα − gx 2 .
2u2 cos 2 α
(i) Range and time of flight on an inclined plane with angle of inclination β are given by
R = 2u 2 cosα sin(α − β) and maximum range up the plane = u 2 , where 2α − β = π
g cos 2 β + 2
g(1 sin β )
(ii) Time of flight T = 2u sin(α − β)
g cos β
(iii) Range down the plane = 2u 2 cosα sin(α + β)
g cos 2 β
(iv) Maximum range down the plane = u2 , where 2α + β = π
g(1 − sin β ) 2
(v) Time of flight = 2u sin(α + β) , down the plane
g cos β
(vi) Condition that the particle may strike the plane at right angles is cot β = 2 tan(α − β ) .
Example: 33 Two stones are projected from the top of a cliff h metres high, with the same speed u so as to hit the ground at the same
Solution: (d)
spot. If one of the stones is projected horizontally and the other is projected at an angle θ to the horizontal then tanθ
equals [AIEEE 2003]
(a) 2u (b) 2g u (c) 2h u (d) u 2
gh h g gh
R=u 2h = (u cosθ )× t ; t = 1 2h .....(i) Y
g cosθ g θ
Now h = (−u sinθ )t + 1 gt 2 , ‘t’ from (i) u
2
h u sinθ 2h 1 g 2h , h = −u 2h tanθ h sec2 θ h
cosθ g 2 cos 2 g
= − + g θ +
h = −u 2h tanθ + h tan2 θ +h, tan2 θ −u 2 tanθ =0; ∴ tan θ =u 2 . OR X
g hg hg
Example: 34 A gun projects a ball at an angle of 45o with the horizontal. If the horizontal range is 39.2 m, then the ball will rise to
Solution: (a)
Example: 35 [AMU 1999]
Solution: (b)
(a) 9.8 m (b) 4.9 m (c) 2.45 m (d) 19.6 m
Example: 36
Solution: (c) Horizontal range = u2 sin 2α = 39.2 ⇒ u2 sin 90o = 39.2 ⇒ u2 = 39.2
g g g
Example: 37
Solution: (b) Greatest height attained = u2 sin2 α = 39.2 sin2 45o = (19.6) 1 = 9.8m .
2g 2 2
If the range of any projectile is the distance equal to the height from which a particle attains the velocity equal to the
velocity of projection, then the angle of projection will be [UPSEAT 2000]
(a) 60o (b) 75o (c) 36o (d) 30o
Range = u2 sin 2α , where α is the angle of projection.
g
Also, u2 = 2gh ⇒ u2 = h
2g
By the given condition, u2 = u2 sin 2α ⇒ 1 = sin 2α , ∴ 2α = 30o or 150o
2g g 2
∴ α = 15o or 75o . Hence, required angle of projection = 75o .
A particle is thrown with the velocity v such that its range on horizontal plane is twice the maximum height obtained. Its
range will be [UPSEAT 2000]
(a) 2v2 (b) 4v2 (c) 4v2 (d) v2
3g 3g 5g 7g
Range = 2 × (maximum height)
∴ v2 sin 2α = 2 v2 sin 2 α ⇒ sin 2α = sin2 α
g 2g
⇒ 2 sinα cosα − sin2 α = 0 ⇒ sinα(2cosα − sinα) = 0 ⇒ sinα = 0 or tanα = 2
But α ≠ 0 , ∴sinα ≠ 0 , ∴tanα = 2 . Now R = v2 sin 2α = v2 . 2 tanα ⇒ v2 . 2.2 = 4v2 .
g g + tan2 α g 1+ 4 5g
1
Let u be the velocity of projection and v1 be the velocity of striking the plane when projected so that the range up the
plane is maximum, and v2 be the velocity of striking the plane when projected so that the range down the plane is
maximum, then u is the
(a) A.M. of v1,v2 (b) G.M. of v1,v2 (c) H.M. of v1,v2 (d) None of these
Let u be the velocity of projection and β be the inclination of the plane with the horizon. Let R be the maximum range up
the plane. Then, R = u2
g(1 + sin β )
The coordinates of the point where the projectile strikes the plane are ( R cos β, R sin β ) . It is given that v1 is the velocity
of striking the plane when projected so that the range up the plane is maximum.
∴ v12 = u2 − 2gR sin β
⇒ v12 = u2 − 2g sin β . u2 ⇒ v12 = u 2 1 − sin β .....(i)
+ sin 1 + sin β
g(1 β )
Replacing β by −β in (i) , we get v22 = u2 1 + sin β .....(ii)
1 − sin β
From (i) and (ii), we get v12.v22 = u4 ⇒ v1v2 = u2 i.e., u is the G.M. of v1, v2 .
Work, Power and Energy.
(1) Work : Work is said to be done by a force when its point of application undergoes a displacement. In
other words, when a body is displaced due to the action of a force, then the force is said to do work. Work is a
scalar quantity.
WWo=rkFd.odne = force × displacement of body in the direction of force d
W = F d cos θ , where θ is the angle between F and d.
θ
AF
(2) Power : The rate of doing work is called power. It is the amount of work that an agent is capable of doing
in a unit time. 1 watt = 107 ergs per sec = 1 joule per sec., 1 H.P. = 746 watt.
(3) Energy : Energy of a body is its capacity for doing work and is of two kinds :
(i) Kinetic energy : Kinetic energy is the capacity for doing work which a moving body possesses by virtue of
its motion and is measured by the work which the body can do against any force applied to stop it, before the
velocity is destroyed. K.E. = 1 mv 2
2
(ii) Potential energy : The potential energy of a body is the capacity for doing work which it possesses by
virtue of its position of configuration. P.E. = mgh.
Example: 38 Two bodies of masses m and 4m are moving with equal momentum. The ratio of their K.E. is [MNR 1990; UPSEAT 2000]
Solution: (b)
(a) 1 : 4 (b) 4 : 1 (c) 1 : 1 (d) 1 : 2
Example: 39
Solution: (b) Momentum of first body = mu, where u is the velocity of the first body.
Example: 40 ∴ Momentum of second body = 4mv, where v is the velocity of the second body.
Solution: (c)
By the given condition, mu = 4mv ⇒ u = 4v
1 mu 2 (4v)2
2 4v2
∴ Ratio of K.E. = 1 = =4. Hence, the ratio of K.E. is 4 : 1.
2 .4mv2
A bowler throws a bumper with a speed of 24m / sec . The moment the ball touches the ground, it losses its energy by 1.5
kgm. If the weight of the ball is 225 gm, then speed of the ball at which it hits the bat is [UPSEAT 2000]
(a) 2.22 m/sec. (b) 22.2 m/sec. (c) 4.44 m/sec. (d) 44.4 m/sec.
Weight of the ball = w = 225 gm = 0.225 kg
Velocity of the ball = v = 25 m/sec.
Loss of energy = 1.5 kg.m
∴ K.E. of the ball before it touches the ground = 1 wv2 = 1 . 0.225 × (25)2 = 7.17kgm .
2g 2 9.81
K.E. of the ball after bumping = K.E. before touching the ground – loss of K.E. = 7.17 – 1.5 = 5.67 kg-m.
Let v1 be the velocity of the ball at which it hits the bat.
Now K.E. of the ball at the time of hitting the bat = 1 w v12 ⇒ 5.67 = 1 × 0.225 v12 ⇒ v12 = 494 ⇒ v1 = 22.2 m / sec .
2 g 2 9.81
A particular starts from rest and move with constant acceleration. Then the ratio of the increase in the K.E. in mth and
(m + 1)th second is [UPSEAT 2000]
(a) m : m (b) m + 1 : m + 1 (c) 2m − 1 : 2m + 1 (d) None of these
Required ratio = Increase in K.E. in mthsecond = Work done against the force in mthsecond
Increase in K.E. in (m + 1)th second Work done against the force in (m + 1)th second
P × 0 + 1 f (2m − 1) 2m −1 2m −1
2 2m + 2−1 2m +1
= = = .
0 1 f(2(m + 1) − 1)
P × + 2
Exponential and logarithmic series
Exponential Series
Definition (The number e) .
The limiting value of 1 + 1 n when n tends to infinity is denoted by e
n
lim1 + 1 n 1 1 1 1
n→∞ n 1! 2! 3! 4!
i.e., e = e = =1+ + + + + .........∞ = 2.71 (Nearly)
Properties of e .
(1) e lies between 2.7 and 2.8. i.e., 2.7 < e < 2.8 (since 1 1 for n ≥ 2 )
n! ≤ 2n−1
(2) The value of e correct to 10 places of decimals is 2.7182818284
(3) e is an irrational (incommensurable) number
(4) e is the base of natural logarithm (Napier logarithm) i.e. ln x = loge x and log10 e is known as Napierian
constant. log10 e = 0.43429448 , ln x = 2.303 log10 x
since ln x = log10 x. loge 10 and loge10 = 1 e = 2.30258509
log10
Exponential Series .
x∈R, ex x x2 x3 xr or ∞ xn
∑For 1 1! 2! 3! r! n!
= + + + + ...... + + ......∞ ex =
n=0
The above series known as exponential series and ex is called exponential function. Exponential function is
also denoted by exp. i.e. exp A = e A ; ∴ exp x = e x
Exponential Function ax, where a > 0 .
a x = e loge ax = e x loge a
∴ax = eα x ....(i) , where α = log e a. We have, ex =1+ x + x2 + x3 ........ + xr + ........∞
1! 2! 3! r!
Replacing x by αx in this series, eα x 1 αx α 2x 2 α 3x 3 ..... α rx r + .....∞
1! 2! 3! r!
= + + + + +
Hence from (i), ax =1+ log e a x + (loge a)2 x2 + ....... + (loge a)r xr + ........∞
1! 2! r!
Some Important Results from Exponential Series .
We have the exponential series
∑(1) ex 1 x x2 ∞ xn …..(i)
= + 1! + 2! + ........∞ = n=0 n! …..(ii)
e−x x x2 x3 ∞ xn
∑(2) Replacing x by –x in (i), we obtain 1! 2! 3! n!
= 1 − + − + ......∞ = (−1)n
n=0
in (i) and (ii), 1 1 1 ∞ 1
∑(3) we get, 1! 2! 3! n=0 n!
Putting x =1 e = 1 + + + + ......∞ =
∑e −1 1 1 1 1 ∞ (−1)n
= − 1! + 2! − 3! + ......∞ = n=0 n!
ex + e−x x2 x4 x6 ∞ x 2n
2 2! + 4! + 6! n=0 (2n)!
∑(4) From (i) and (ii), we obtain 1+
= + .......∞ =
ex − e−x x x3 x5 ∞ x 2n+1
3! + 5! n=0 (2n + 1)!
∑2
= + + .......∞ =
e + e−1 1 1 ∞ 1 ∑e − e−1 1 1 ∞1
2 2! 4! n=0 (2n)! 2 3! 5! n=0 (2n + 1)!
∑(5) 1+ , =1+ .......∞
= + + .......∞ = + + =
∑Note −1 1 1 1 1 ∞ 1 e 1 1 1 ∞ 1
: e 1! 2! 3! 4! .........∞ r =1 r! −2 2! 3! 4! + .........∞ r=2 r!
= + + + + = ∑ = + + =
Some Standard results.
∞1 ∞1 ∞1 ∞1 1 1 1
∑ ∑ ∑(1)n=0 n! = n=0 (n − k)! = e ∑(2)n=1 n! 1! + 2! + 3! +..........∞ e 1
= (n − 1)! = = −
n=0
∑(3) ∞1 = 1 + 1 + 1 + .........∞ = e−2 ∑(4)∞1 = 1 + 1 + 1 + .........∞ = e −1
n=2 n! 2! 3! 4! n=0 (n + 1)! 1! 2! 3!
∑(5) ∞1 = 1 1 1 + .........∞ =e−2 ∑(6)∞1 1 + 1 + 1 + .........∞ = e − 2
n=0 (n + 2)! 2! + 3! + 4! n=1 (n + 1)! = 2! 3! 4!
∞ 1 1 1 1 e + e −1 ∞1
n=0 (2n)! 2! 4! 6! 2 n=1 (2n − 2)!
∑ ∑(7) 1
= + + + + .........∞ = =
∞ 1 1 1 1 e − e −1 ∞1
n=1 (2n − 1! 3! 5! 2 = n=0 (2n + 1)!
∑ ∑(8)
1)! = + + + .........∞ =
(9) ex = 1 + x + x2 + x3 + ......... + xn + .........∞
1! 2! 3! n!
∴ Tn+1 = General term in the expansion of ex = xn and coefficient of xn in ex = 1
n! n!
(10) e−x = 1 − x + x2 − x3 + ........ + (−1)n xn + ........∞
1! 2! 3! n!
∴ Tn+1 = General term in the expansion of e−x = (−1)n xn and coefficient of xn in e−x = (−1)n
n! n!
(11) e ax =1+ (ax) (ax)2 + (ax)3 + ......... + (ax)n + .........∞
1! + 2! 3! n!
∴ Tn+1 = General term in the expansion of e ax (ax)n and coefficient of xn in e ax an
= n! = n!
∞n ∞n ∞ n2 ∞ n2
n=0 n! n=1 n! n=0 n! n=1 n!
∑ ∑(12) ∑ ∑(13)
=e = = 2e =
∞ n3 ∞ n3 ∞ n4 ∞ n4
n=0 n! n=1 n! n=0 n! n=1 n!
∑ ∑(14) ∑ ∑(15)
= 5e = = 15e =
Example: 1 2 + 4 + 6 + 8 + .........∞ = [JMI CET 2000]
Solution: (b) 1! 3! 5! 7!
Example: 2
Solution: (d) (a) 1 / e (b) e (c) 2e (d) 3e
Example: 3 2 4 6 8 + .........∞ = (1 + 1) + (1 + 3) + (1 + 5) + (1 + 7) + ........∞
Solution: (d) 1! + 3! + 5! + 7! 1! 3! 5! 7!
Example: 4 = 1 + 1 + 1 + 1 + ........∞ + 1 + 1 + 1 + 1 + ........∞ = e − e−1 + e + e−1 =e
1! 3! 5! 7! 2! 4! 6! 2 2
Solution: (d)
Example: 5 2 + 4 + 6 + .........∞ = [MNR 1979; MP PET 1995, 2002]
Solution: (d) 3! 5! 7!
(a) e (b) 2 e (c) e2 (d) 1/e
(2n + 1) − 1 1 1 ∞ 1 1 1 1 1 1 + .......∞
(2n + 1)! (2n)! (2n + 1)! n=1 2! 4! 6! 3! 5! 7!
∑Here .......∞
Tn = = − ⇒S= Tn = + + + − + +
S e +e −1 1 e −e −1 − 1 ⇒ e−1 1
2 2 e
⇒ = − − =
1+ 23 + 33 + 43 + .........∞ = [MNR 1976; MP PET 1997]
2! 3! 4!
(d) 5 e
(a) 2e (b) 3 e (c) 4 e
S = 13 + 23 + 33 + ........ + n3 + ........
1! 2! 3! n!
∑Here Tn = n3 ⇒ Sn = ∞ n3 = 5e
n! n=1 n!
The coefficient of xn in the expansion of e7x + e x is [MP PET 1999]
e 3x
(a) 4n−1 + (−2)n (b) 4n−1 + 2n (c) 4n−1 + (−2)n−1 (d) 4n + (−2)n
n! n! n! n!
e7x + e x ∞ (4 x)n ∞ (−2x)n e7x + e x 4n + (−2)n
e 3x n=0 n! n! e 3x n!
∑ ∑We have + xn
= e4x + e−2x = ∴coefficient of in =
n=0
1 + 1 +2 + 1 + 2+ 3 + 1 + 2 +3 + 4 + ........∞ = [Roorkee 1999; MP PET 2003]
2! 3! 4!
(a) e (b) 3 e (c) e/2 (d) 3e/2
∑Tn = n = n(n + 1) = 1 (n + 1) = 1 n−1 + 2 = 1 1 + 2
n! 2.n! 2 (n − 1)! 2 (n − 1)! 2
(n − 1)! (n − 2)! (n − 1)!
∑ ∑ ∑Sn = ∞ Tn = 1 ∞ (n 1 + ∞ 1 = e +e= 3e
n=1 2 n=1 − 2)! n=1 (n − 1)! 2 2
Logarithmic Series
Logarithmic Series.
An expansion for log e (1 + x)as a series of powers of x which is valid only when, | x| < 1 ,
Expansion of loge(1 + x); if | x |< 1, then log e (1 + x) = x − x2 + x3 − x4 + ........∞
2 3 4
Some Important Results from the Logarithmic Series .
(1) Replacing x by − x in the logarithmic series, we get
log e (1 − x) = −x − x2 − x3 − x4 − ........∞ or − log e (1 − x) = x + x2 + x3 + x4 + ........∞
2 3 4 2 3 4
(2) (i) log e (1 + x) + log e (1 − x) = log e (1 − x 2 ) = −2 x2 + x4 + ........∞,(−1 < x < 1)
2 4
(ii) log e (1 + x) − log e (1 − x) = 2 x + x3 + x5 + ........∞ or log e 1 + x = 2 x + x3 + x5 + ........∞
3 5 1 − x 3 5
(3) The series expansion of loge(1 + x) may fail to be valid if |x| is not less than 1. It can be proved that the
logarithmic series is valid for x=1. Putting x=1 in the logarithmic series.
We get, log e 2 =1− 1 + 1 − 1 + 1 − 1 + ........∞ = 1 + 1 + 1 + ........∞
2 3 4 5 6 1.2 3.4 5.6
(4) When x = −1 , the logarithmic series does not have a sum. This is in conformity with the fact that log(1-1) is
not a finite quantity.
Difference between the Exponential and Logarithmic Series.
(1) In the exponential series ex = 1+ x + x2 x3 + ..........∞ all the terms carry positive signs whereas in the
1! 2! + 3!
logarithmic series log e (1 + x) = x− x2 + x3 − x4 + ........∞ the terms are alternatively positive and negative.
2 3 4
(2) In the exponential series the denominator of the terms involve factorial of natural numbers. But in the
logarithmic series the terms do not contain factorials.
(3) The exponential series is valid for all the values of x. The logarithmic series is valid when |x|< 1.
Example: 6 0.5 − (0.5)2 (0.5)3 (0.5)4 + ........ [MP PET 1995]
2 +3 −4
(a) log e 3 (b) log 10 1 (c) log e (n!) (d) log e 1
2 2 2
Solution: (a) We know that, x− x2 + x3 x4 ........∞ = loge(1 + x)
2 3 −4
Putting x = 0.5, we get, 0.5 − (0.5)2 (0.5)3 (0.5)4 + ........∞ = log e (1 + 0.5) = log e 3
2 +3 −4 2
Example: 7 1 − 1 + 1 − 1 + ........∞ = [Roorkee 1992; MP PET 1999; AIEEE 2003]
1.2 2.3 3.4 4.5
(a) log e 4 (b) log e e (c) loge 4 (d) loge 2
e 4
Solution: (a) We know that, log e 2= 1 + 1 + 1 + .......∞ ………(i)
1.2 3.4 5.6
Also log e 2 = 1 − 1 − 1 − 1 − .......∞ ………(ii)
2.3 4.5 6.7
By adding (i) and (ii), we get, 2 loge 2 = 1 + 1 − 1 + 1 − 1 + ........
1.2 2.3 3.4 4.5
⇒ 2 loge 2−1 = 1 − 1 + 1 − 1 + ........ ⇒ 1 − 1 + 1 − 1 + ........∞ = log e 4 − log e e = log e 4
1.2 2.3 3.4 4.5 1.2 2.3 3.4 4.5 e
Example: 8 The coefficient of xn in the expansion of loge(1 + 3x + 2x2) is [UPSEAT 2001]
Solution: (b)
(a) (−1)n 2n + 1 (b) (−1)n+1 [2n + 1] (c) 2n + 1 (d) None of these
Example: 9 n n n
Solution: (c)
Example: 10
We have, loge(1 + 3x + 2x2) = loge(1 + x) + loge(1 + 2x)
Solution: (c)
∞ xn ∞ (2x)n ∞ (−1)n−1 1 2n x n ∞ 1 + 2n x n
n n=1 n n=1 n n n=1 n
(−1)n−1
n=1
∑ ∑ ∑ ∑= (−1)n−1
+ (−1)n−1 = + =
So coefficient of xn = (−1)n−1 2n + 1 = (−1)n+1(2n + 1) [ ](−1)n = (−1)n+2 = ........
n n
The equation x log x (2+x)2 = 25 holds for [MP PET 1992]
(a) x = 6 (b) x = −3 (c) x = 3 (d) x =7
Given equation x log x (2+x)2 = 25 ⇒ (2 + x)2 = 25 hold for x = 3
If y = − x3 + x6 x9 + ........ , then x= [MNR 1975]
2 +3
(a) 1+ ey (b) 1− ey 1 (d) (1 − ey)3
3 3
(c) (1 − e y ) 3
y = x 3 + (x 3 )2 (x 3 )3 + ........ = loge(1 − x3) ⇒ ey = 1− x3 ⇒ x 1
− 2 +3
= (1 − ey)3
Hyperbolic-functions
Definition.
We know that parametric co-ordinates of any point on the unit circle x2 + y2 = 1 is (cosθ ,sinθ ) ; so that these
functions are called circular functions and co-ordinates of any point on unit hyperbola x2 − y2 = 1 is
eθ + e −θ , eθ − e −θ i.e., (coshθ , sinhθ ) . It means that the relation which exists amongst cosθ ,sinθ and unit
2 2
circle, that relation also exist amongst coshθ , sinhθ and unit hyperbola. Because of this reason these functions are
called as Hyperbolic functions.
For any (real or complex) variable quantity x,
(1) sinh x = ex − e−x [Read as 'hyperbolic sine x'] (2) cosh x = ex + e−x [Read as 'hyperbolic cosine x']
2 2
(3) tanh x = sinh x = ex − e−x (4) coth x = cosh x = ex + e−x
cosh x ex + e−x sinh x ex − e−x
(5) cosechx = 1 x = ex 2 (6) sec hx = 1 = ex 2
sinh − e−x coshx + e−x
Note : sinh 0 = 0, cosh 0 = 1, tanh 0 = 0
Domain and Range of Hyperbolic Functions.
Let x is any real number
Function Domain Range
sinh x R R
cosh x R
tanh x R [1, ∞)
coth x R0 (−1,1)
sech x R R − [−1,1]
cosech x (0,1]
R0
R0
Graph of Real Hyperbolic Functions.
(1) sinh x Y (2) cosh x
(4) coth x
X (6) sech x (0, 1) X
O O
(3) tanh x Y Y
(5) cosech x Y=1
y =1
OX OX
Y = –1
y = –1
Y
Y
OX (0, 1)
OX
Formulae for Hyperbolic Functions.
The following formulae can easily be established directly from above definitions
(1) Reciprocal formulae
(i) cosech x = 1 x (ii) sech x = 1 x (iii) coth x = 1 x
sinh cosh tanh
(iv) tanh x = sinh x (v) coth x = cosh x
cosh x sinh x
(2) Square formulae
(i) cosh2 x − sinh2 x = 1 (ii) sec h 2 x + tanh 2 x = 1
(iii) coth2 x − cosech2x = 1 (iv) cosh 2 x + sinh 2 x = cosh 2x
(3) Expansion or Sum and difference formulae
(i) sinh(x ± y) = sinh x cosh y ± cosh x sinh y (ii) cosh (x ± y) = cosh x cosh y ± sinh x sinh y
(iii) tanh (x ± y) = tanh x ± tanh y
1 ± tanh x tanh y
(4) Formulae to transform the product into sum or difference
(i) sinh x + sinh y = 2 sinh x + y cosh x − y (ii) sinh x − sinh y = 2 cos h x + y sin h x − y
2 2 2 2
(iii) cosh x + cosh y = 2 cosh x + y cosh x − y (iv) cosh x − cosh y = 2 sinh x + y sinh x − y
2 2 2 2
(v) 2 sinh x cosh y = sinh (x + y) + sinh (x − y) (vi) 2 cosh x sinh y = sinh (x + y) − sinh (x − y)
(vii) 2 cosh x cosh y = cosh (x + y) + cosh (x − y) (viii) 2 sinh x sinh y = cosh (x + y) − cosh (x − y)
(ix) cosh x + sinh x = ex (x) cosh x − sinh x = e−x (xi) (cosh x + sinh x)n = cosh nx + sinh nx
(5) Trigonometric ratio of multiple of an angle
2 tanh x
(i) sinh 2x = 2 sinh x cosh x = 1 − tanh2 x
(ii) cosh 2x = cosh2 x + sinh2 x = 2 cosh2 x −1 = 1+ 2 sinh2 x = 1 + tanh 2 x
1 − tanh 2 x
(iii) 2 cosh2 x = cosh 2x + 1 (iv) 2 sinh2 x = cosh 2x − 1 (v) tanh 2x = 2 tanh x
1 + tanh2 x
(vi) sinh 3x = 3 sinh x + 4 sinh3 x (vii) cosh 3x = 4 cosh3 x − 3 cosh x (viii) tanh 3x = 3 tanh x + tanh 3x
1 + 3 tanh2 x
(6) (i) cosh x + sinh x = ex (ii) cosh x − sinh x = e−x (iii) (cosh x + sinh x)n = cosh nx + sinh nx
Example: 1 sinh x − sinh y is equal to
Solution: (c) cosh x − cosh y
Example: 2
Solution: (b) (a) 2coth(x + y) (b) tanh x + y (c) coth x + y (d) coth x − y
Example: 3 2 2 2
Solution: (a)
sinh x − sinh y 2 cosh x + y sinh x − y x y
Example: 4 cosh x − cosh y 2 2
= x y x y = coth + .
+ − 2
2 sinh 2 sinh 2
If tanh2 x = tan2 θ , then cosh 2x is equal to [EAMCET 1998]
(a) − sin 2θ (b) sec 2θ (c) cos 3θ (d) cos 2θ
cosh 2x = 1 + tanh2 x = 1 + tan2 θ = 1 − 1 = 1 = sec 2θ .
1 − tanh2 x 1 − tan2 θ tan2 θ cos 2θ
1 + tan2 θ
If u= log tan π + x , then cosh u is equal to [EAMCET 1991; Rajasthan PET 1999]
4 2
(a) sec x (b) cosec x (c) tan x (d) sin x
u = log tan π + x ⇒ ∴ eu = 1+ tan x
4 2 1 1− tan 2
x
2
1+ tan x 2 2 1 + tan2 x π
1− tan 2 2
1+ x +1 2
1− π
cosh u = eu + e−u = e2u + 1 = 2. 2 = = 1 + tan2 = 1 = 1 = sec x .
2 2eu 1 − tan2 2 cos
tan x 21 − tan x 1 + tan x 1 − tan2 π x
2 2 2 2
x π
tan 2 1 + tan2 2
cosh 2 + sinh 2 = [EAMCET 2000]
(a) 1 (b) e (c) 1 (d) e2
e e2
Solution: (d) cosh 2 + sinh 2 = e 2 + e−2 + e 2 − e−2 = e2.
Example: 5 2 2
Solution: (d) If coshα = sec x, then tan 2 x = [Rajasthan PET 1998]
2
(a) cos 2 α (b) sin 2 α (c) cot 2 α (d) tanh2 α
2 2 2 2
coshα = sec x = 1 = 1 x
cos x 1 − tan2 2
1+ tan 2 x
2
coshα 1+ tan 2 x coshα − 1 2 tan2 x x 2 sinh 2 α x x
1 1− tan 2 2 coshα + 1 2 2 2 2 2 2
= x ⇒ = = tan2 ⇒ = tan2 ⇒ tanh2 α = tan 2 .
2 2 α 2
2 cosh 2
Transformation of a Hyperbolic Functions.
Since, cosh 2 x − sinh 2 x = 1
⇒ sinh x = cosh 2 x − 1 ⇒ sinh x = 1 − sech 2 x
sec h x
⇒ sinh x = tanh x ⇒ sinh x = 1
1 − tanh 2 x
coth 2 x − 1
Also, sinh x = 1 x
cosech
In a similar manner we can express cosh x , tanh x, coth x ,................ in terms of other hyperbolic functions.
Expansion of Hyperbolic Functions.
(1) sinh x = ex − e−x = x + x3 + x5 + x7 + ....
2 3! 5! 7!
(2) cosh x = ex + e−x =1+ x2 + x4 + x6 + ....
2 2! 4! 6!
(3) tanh x = ex − e−x = x − x3 + 2x5 − 17 x7 + .....
ex + e−x 3 315
The expansion of coth x, cosech x does not exist because coth(0) = ∞, cosech(0) = ∞ .
Relation between Hyperbolic and Circular Functions.
We have from Euler formulae,
eix = cos x + i sin x ........(i) and e−ix = cos x − i sin x ........(ii)
Adding (i) and (ii) ⇒ cos x = e ix + e −ix
2
Subtracting (ii) from (i) ⇒ sin x = e ix − e −ix
2i
Replacing x by ix in these values, we get cos(ix) = e−x + e x = cosh x
2
∴ cos(ix) = cosh x
sin(ix) = e−x − ex =i e x − e−x
2i 2
∴ sin(ix) = i sinh x
Also tan(ix) = sin(ix) = i sinh x
cos(ix) cosh x
tan(ix) = i tanh x
Similarly replacing x by ix in the definitions of sinh x and cosh x , we get cosh (ix) = e ix + e −ix = cos x
2
Also, tanh (ix) = sinh (ix) = i sin x = i tan x
cosh (ix) cos x
Thus, we obtain the following relations between hyperbolic and trigonometrical functions.
(1) sin(ix) = i sinh x (2) cos(ix) = cosh x
sinh (ix) = i sin x cosh(ix) = cos x
sinh x = −i sin(ix) cosh x = cos (ix)
sin x = −i sin h (ix) cos x = cos h (ix)
(3) tan(ix) = i tanh x (4) cot(ix) = −i coth x
tanh(ix) = i tan x coth(ix) = −i cot x
tanh x = −i tan(ix) coth x = i cot(ix)
tan x = −i tanh(ix) cot x = i coth(ix)
(5) sec(ix) = sechx (6) cosec(ix) = −i cosech x
sec h(ix) = sec x cosech (ix) = i cosecx
sec hx = sec(ix) cosech x = i cosec (ix)
sec x = sech(ix) cosecx = i cosech(ix)
Important Tips
• For obtaining any formula given in (5)th article, use the following substitutions in the corresponding formula for trigonometric functions.
sin x →i sinh x cos x → cosh x tan x → i tanh x
sin2 x → − sinh2 x cos2 x → cosh2 x tan2 x → − tanh2 x
For example,
For finding out the formula for cosh 2x in terms of tanh x , replace tan x by i tanh x and tan 2 x by tan 2 x by − tanh 2 x in the
following formula of trigonometric function of cos 2x :
cos 2x = 1 − tan 2 x we get, cosh 2x = 1 + tanh 2 x
1 + tan 2 x 1 − tanh 2 x
Period of Hyperbolic Functions.
If for any function f(x), f(x + T) = f(x), then f(x) is called the Periodic function and least positive value of T is
called the Period of the function.
sinh x = sinh(2πi + x)
cosh x = cosh(2πi + x)
and tanh x = tanh(πi + x)
Therefore the period of these functions are respectively 2πi, 2πi and πi . Also period of cosech x, sech x and
coth x are respectively 2πi , 2πi and πi .
Note : Remember that if the period of f(x) is T, then period of f (nx) will be T .
n
Hyperbolic function are neither periodic functions nor their curves are periodic but they show the algebraic
properties of periodic functions and having imaginary period.
Example: 6 If cos(x + iy) = A + i B , then A equals [Rajasthan PET 1994]
Solution: (a)
Example: 7 (a) cos x cosh y (b) sin x sinh y (c) − sin x sinh y (d) cos x sinh y
Solution: (c)
cos(x + iy) = A + iB [Rajasthan PET 1999]
Example: 8
⇒ cos x cos(iy) − sin x sin(iy) = A + iB ⇒ cos x cosh y − i sin x sinh y = A + iB (d) sin2 u + sinh2 v
∴ A = cos x cosh y [Rajasthan PET 1999]
If cos(u + iv) = x + iy, then x 2 + y2 + 1 is equal to
(a) cos2 u + sinh2 v (b) sin2 u + cosh2 v (c) cos2 u + cosh2 v
cos(u + iv) = x + iy
⇒ cos u cos(iv) − sin u sin(iv) = x + iy ⇒ cos u cosh v − i sin u sinh v = x + iy
∴ x = cos u cosh v
y = − sin u sinh v
x 2 + y2 = cos2 u. cosh2 v + sin2 u. sinh2 v
= (1 − sin2 u) cosh2 v + sin2 v[cosh2 v − 1] = cosh2 v − sin2 v
∴ x2 + y2 + 1 = cosh2 v + 1 − sin2 u = cosh2 v + cos2 u .
The value of sec h(iπ ) is
(a) – 1 (b) i (c) 0 (d) 1
(d) e −x
Solution: (a) sec h(iπ ) = eπi 2 = sec π = −1 .
Example: 9 + e −πi
Solution: (d)
cos ix + i sin ix equals
Example: 10
(a) eix (b) e −ix (c) e x
Solution: (b)
Example: 11 cos ix + i sin ix
Solution: (c) = cosh x + i.i sinh x = cosh x − sinh x = ex + e−x − ex + e−x = e−x .
Example: 12 2
Solution: (a)
Example: 13 sin π i is equal to
6
Solution: (c)
(a) − i (b) i (c) i 3 (d) −i 3
2 2 2 2
sinh π i = i. sin π = i. 1 = i .
6 6 2 2
sec h(πi) + cosech π i equals
2
(a) 1 − i (b) −1 + i (c) −1 − i (d) 1 + i
(d) 9πi
sec h(πi) + cosech π i = sec π − icosec π = −1 − i .
2 2
The period of cosh θ is
3
(a) 6πi (b) 2πi (c) πi
Since the period of coshθ is 2πi, so the period of cosh θ is 3.2πi = 6πi .
3
The period of sinh x is
2
(a) 2πi (b) 2π (c) 4πi (d) 4π
Since period of sinh x is 2πi , therefore period of sinh x will be 4πi .
2
Inverse Hyperbolic Functions.
If sinh y = x, then y is called the inverse hyperbolic sine of x and it is written as y = sinh−1 x . Similarly
cosech−1x, cosh−1 x, tanh−1 x etc. can be defined.
(1) Domain and range of Inverse hyperbolic function
Function Domain Range
sinh−1 x R R
cosh−1 x R
tanh−1 x [1, ∞) R
coth−1 x (−1,1) R0
sech−1 x R – [–1, 1] R
(0, 1] R0
cosech−1 x
R0
(2) Relation between inverse hyperbolic function and inverse circular function
Method : Let sinh−1 x = y
⇒ x = sinh y = − i sin(iy) ⇒ ix = sin(iy) ⇒ iy = sin−1(ix)
⇒ y = −i sin −1(ix) ⇒ sinh−1 x = −i sin−1(ix)
Therefore we get the following relations
(i) sinh−1 x = −i sin−1(ix) (ii) cosh−1 x = −i cos−1 x (iii) tanh−1 x = −i tan−1(ix)
(iv) sec h−1x = −i sec−1 x (v) cosech –1 x = i cosec −1(ix)
(3) To express any one inverse hyperbolic function in terms of the other inverse hyperbolic
functions
To express sinh−1 x in terms of the others
(i) Let sinh−1 x = y ⇒ x = sinh y ⇒ cosech y = 1 ⇒ y = cosec −1 1
x x
(ii) cosh y = 1 + sinh2 y = 1 + x2
y = cosh−1 1 + x2 ⇒ sinh−1 x = cosh−1 1 + x2
(iii) tanh y = sinh y = sinh y = x
cosh y 1 + sinh2 y 1+ x2
∴ y = tanh−1 x ⇒ sinh−1 x = tanh−1 x
1+ x2 1+ x2
(iv) coth y = 1 + sinh2 y = 1+ x2
sinh y x
∴ y = coth−1 1+ x2 ⇒ sinh−1 x = coth−1 1+ x2
x 11 x
=
(v) sec hy = 1 =
cosh y 1 + sinh2 y 1 + x 2
y = sec h−1 1 ⇒ sinh−1 x = sec h−1 1
1+ x2 1+ x2
(vi) Also, sinh−1 x = cosech −1 1
x
From the above, it is clear that
coth−1 x = tanh −1 1
x
sec h −1 x = cosh −1 1
x
cosech−1 = sinh −1 1
x
Note : If x is real then all the above six inverse functions are single valued.
(4) Relation between inverse hyperbolic functions and logarithmic functions
Method : Let sinh−1 x = y
⇒ x = sinh y = ey − e−y ⇒ e2y − 2xey − 1 = 0 ⇒ ey = 2x ± 4x2 +4 = x± x2 +1
2 2
But ey > 0,∀y and x < x 2 + 1
∴ ey = x + x2 + 1 ⇒ y = log(x + x 2 + 1)
∴ sinh−1 x = log(x + x 2 + 1)
By the above method we can obtain the following relations between inverse hyperbolic functions and principal
values of logarithmic functions.
(i) sinh−1 x = log(x + x 2 + 1) (−∞ < x < ∞) (ii) cosh−1 x = log(x + x 2 − 1) (x ≥ 1)
(iii) tanh−1 x = 1 log 1 + x | x|< 1 (iv) coth−1 x = 1 log x + 1 | x|> 1
2 1 − x 2 x − 1
(v) sec h−1x = log 1 + 1 − x 2 0< x ≤1 (vi) cosech−1x = log 1 + 1 + x 2 (x ≠ 0)
x x
Note : Formulae for values of cosech –1 x, sec h –1 x and coth−1 x may be obtained by replacing x by 1 in
x
the values of sinh−1 x, cosh−1 x and tanh−1 x respectively.
Separation of Inverse Trigonometric and Inverse Hyperbolic Functions.
If sin(α + iβ ) = x + iy then (α + iβ ) , is called the inverse sine of (x + iy). We can write it as,
sin−1(x + iy) = α + iβ
Here the following results for inverse functions may be easily established.
(1) cos −1(x + iy) = 1 cos −1 (x 2 + y2)− (1 − x2 + y 2 )2 + 4x 2y 2 + i cosh −1 (x 2 + y2)+ (1 − x2 + y 2 )2 + 4x 2y 2
2 2
(2) sin −1 (x + iy) = π − cos −1 (x + iy)
2
= π − 1 cos −1 (x 2 + y2)− (1 − x2 + y 2 )2 + 4x 2y 2 − i cosh −1 (x 2 + y2)+ (1 − x2 + y 2 )2 + 4x 2y 2
2 2 2
(3) tan−1(x + iy) = 1 tan−1 1 − 2x y2 + i tanh−1 1 + 2y y2 = 1 tan −1 1 − 2x y2 + i log x 2 + (1 + y)2
2 x2 − 2 x2 + 2 x2 − 4 x 2 + (1 − y)2
(4) sin−1(cosθ + i sinθ ) = cos−1( sinθ ) + i sinh−1( sinθ ) or cos−1( sinθ ) + i log( sinθ + 1 + sinθ )
(5) cos−1(cosθ + i sinθ ) = sin−1( sinθ ) − i sinh−1( sinθ ) or sin−1( sinθ ) − i log( sinθ + 1 + sinθ )
(6) tan −1(cosθ + i sinθ ) = π + i log 1 + sinθ , (cosθ ) > 0
4 4 1 − sinθ
and tan −1(cosθ + i sinθ ) = − π + 1 log 1 + sinθ , (cosθ ) < 0
4 4 1 − sinθ
Since each inverse hyperbolic function can be expressed in terms of logarithmic function, therefore for
separation into real and imaginary parts of inverse hyperbolic function of complex quantities use the appropriate
method.
Note : Both inverse circular and inverse hyperbolic functions are many valued.
Example: 14 If x = log(y + y 2 + 1), then y = [EAMCET 1995]
Solution: (c)
Example: 15 (a) tanh x (b) coth x (c) sinh x (d) cosh x
Solution: (a) (c) 3
Example: 16 x = sinh−1 y ⇒ y = sinh x .
Solution: (b)
Example: 17 If cosh−1 x = log(2 + 3), then x = [EAMCET 2000]
Solution: (c) (a) 2 (b) 1 (d) 5
Example: 18
Solution: (b) cosh−1 x = log(x + x 2 − 1) = log(2 + 3) (c) tanh−1 3 [Rajasthan PET 1990]
Example: 19 ∴ x=2. (b) cosh−1 3
log(3 + 2 2) = (d) cosh−1 3
Solution: (a) (a) sinh−1 3
Example: 20 log(3 + 2 2) = log(3 + 8 ) = log(3 + 9 − 1) = log(3 + 32 − 1)
Solution: (a) log(3 + 2 2) = cosh−1 3 .
sec h -1 1 is [Rajasthan PET 1998]
2
(d) None of these
(a) log( 3 + 2) (b) log( 3 + 1) (c) log(2 + 3)
[EAMCET 2003]
sec h −1 1 = cosh −1 (2) = log(2 + 22 − 2) = log(2 + 3) .
2
sinh−1(23 / 2 ) is
(a) log(2 + 18 ) (b) log(3 + 8 ) (c) log(3 − 8 ) (d) log( 8 + 27)
sinh−1(23 / 2) = log(23 / 2 + (23 / 2)2 + 1) = log(3 + 8 ) . [Rajasthan PET 2002]
If tan−1(α + iβ ) = x + iy, then x =
(a) 1 tan−1 1 2α β 2 (b) 1 tan−1 1 + 2α β 2 (c) tan−1 1 − 2α β 2 (d) None of these
2 −α2 − 2 α2 + α2 −
tan−1(α + iβ ) = x + iy
tan−1(α − iβ ) = x − iy
2x = x + iy + x − iy = tan−1(α + iβ ) + tan−1(α − iβ )
∴ x = 1 tan −1 2α = 1 tan −1 1 α+ iβ + α − iβ ) = 1 tan−1 2α β 2 .
2 1−α2 − β 2 2 − (α + iβ )(α − iβ 2 1−α2 −
If − π < x < π , then the value of log sec x is
2 2
(a) 2coth−1 cosec 2 x − 1 (b) 2coth−1 cosec 2 x + 1 (c) 2cosech-1 cot 2 x − 1 (d) 2cosech-1 cot 2 x + 1
2 2 2 2
Let log sec x =y; ∴ 1 x = ey / 2
cos e−y / 2
By componendo and Dividendo rule, 1 + cos x = ey / 2 + e−y / 2 ⇒ cot 2 x = coth y
1 − cos x ey / 2 − e−y / 2 2 2
⇒ y = 2coth−1 cosec 2 x − 1 .
2
Example: 21 The value of cosh−1(sec x) is
Solution: (a) (a) log 1 + sin x (b) log 1 − sin x (c) log 1 + cos x (d) log 1 − cos x
Example: 22 cos x cos x sin x sin x
Solution: (a)
Here cosh−1(sec x) = log(sec x + sec 2 x −1) = log(sec x + tan x) = log 1 + sin x .
Example: 23 cos x
Solution: (a)
2 sinh−1(θ ) is equal to
Example: 24
(a) sinh−1(2θ 1 + θ 2 ) (b) sinh−1(2θ 1 − θ 2) (c) sinh−1(θ 1 + θ 2 ) (d) None of these
Solution: (c)
Example: 25 We know that, 2 sin−1 x = sin−1(2x 1 − x 2 )
Solution: (b) Putting the value of x = iθ
Example: 26
Solution: (c) 2 sin−1(iθ ) = sin−1(2iθ 1 − i2θ 2 )
Example: 27
2i sinh−1(θ ) = sin−1(2iθ 1 + θ 2 ) or 2i sinh−1(θ ) = i sinh−1(2θ 1 + θ 2 ) ( sin−1(ix) = i sinh−1 x )
⇒ 2 sinh−1(θ ) = sinh−1(2θ 1 + θ 2 ) .
The value of sinh−1 x is
1 − x 2
(a) tanh−1 x (b) coth−1 x (c) sinh−1(2x) (d) cosh−1(2x)
Let x = tanh y , then x = tanh y = sinh y
1− x2 sec hy
∴ sinh−1 x = sinh −1 (sinh y) ⇒ y = tanh−1(x) .
1− x2
If cosα cosh β = 1, then β is equal to
(a) log sec α (b) log tanα (c) log(secα + tanα) (d) log sin α
2 2
cosα.cosh β = 1 ⇒ cosh β = sec α ⇒ β = cosh−1(secα) = log(sec α + sec2 α − 1) = log(secα + tanα)
sinh−1(sinh−1 θ ) is equal to
(a) iθ (b) θ (c) −iθ (d) π + iθ
sinh−1(sinh−1 θ ) = − i sin(i sinh−1 θ ) = − i sin[i(−i sin−1(iθ )] = − sin[sin−1(iθ )] = − i.iθ = −i2θ = θ .
If sinh−1 x = cosech–1 y ,then the correct statement is
(a) x = y (b) xy = −1 (c) xy = 1 (d) x + y = 0
Given that, sinh−1 x = cosech–1y or sinh −1 x = sinh−1 1 or x = sinh sinh−1 1 or x = 1 ⇒ xy = 1 .
y y y
Find real part of tan−1(1 + i)
(a) − 1 tan −1 (2) (b) 1 tan −1 (2) (c) − 1 tan−1 1 (d) 0
2 2 2 2
Solution: (a) Real part = 1 tan −1 2(1) = − 1 tan−1(2) .
Example: 28 2 1−1−1 2
Solution: (c)
Example: 29 Find real part of cosh−1(1)
Solution: (b)
Example: 30 (a) – 1 (b) 1 (c) 0 (d) None of these
Solution: (b) We know that cosh−1 x = log x + x 2 − 1
Example: 31 ∴ cosh−1(1) = log1 + 12 − 1 = log 1 = 0 .
Solution: (a)
Find imaginary part of sin −1 5 7 − 9i
16
(a) log 2 (b) − log 2 (c) 0 (d) None of these
−1 57 − 9i = − log 9 1 + 9 = − log(2) .
sin 16 16 16 + 16
Find real part of cos −1 3 + i
2 2
(a) π (b) π (c) log 3 − 1 (d) None of these
3 4 2
Expression cos−1(cosθ + i sinθ ) = sin−1 sinθ − i log( sinθ + 1 + sinθ )
Where θ = π
6
∴ cos −1 3 + i = sin−1 1 − i log 1 + 1+ 1 = π − i log 1 + 3 = π + i log 3 − 1
2 2 2 2 2 4 2 4 2
Real part = π , Imaginary part = log 3− 1 .
4 2
Find imaginary part of sin−1(cosecθ )
(a) log cot θ (b) π (c) 1 log cot θ (d) None of these
2 2 2 2
Let sin−1(cosecθ ) = x + iy
∴ cosecθ = sin(x + iy) = sin x cosh y + i cos x sinh y
By comparing we get, sin x cosh y = cosecθ ......(i) and cos x sinh y = 0 .........(ii)
From (ii), cos x = 0 ⇒ x = π
2
∴ from (i) sin π . cosh y = cosecθ or y = cosh−1(cosecθ ) = log[cosecθ ]
2
⇒ y= log[cosecθ + cot θ ] = log cot θ
2
∴ sin−1(cosecθ ) = π + i log cot θ
2 2
Real part = π , Imaginary part = log cot θ .
2 2
Logarithms indices and surds & partial fractions
Definition.
“The Logarithm of a given number to a given base is the index of the power to which the base must be raised
in order to equal the given number.”
If a > 0 and ≠ 1, then logarithm of a positive number N is defined as the index x of that power of 'a' which
equals N i.e., log a N = x iff a x = N ⇒ a logaN = N, a > 0, a ≠ 1 and N > 0
It is also known as fundamental logarithmic identity.
The function defined by f(x) = log a x, a > 0, a ≠ 1 is called logarithmic function.
Its domain is (0, ∞) and range is R. a is called the base of the logarithmic function.
When base is 'e' then the logarithmic function is called natural or Napierian logarithmic function and when
base is 10, then it is called common logarithmic function.
Note : The logarithm of a number is unique i.e. No number can have two different log to a given base.
log e a = log e 10.log10 a or log10 a = log e a = 0.434 log e a
log e 10
Characteristic and Mantissa.
(1) The integral part of a logarithm is called the characteristic and the fractional part is called mantissa.
log10 N = integer + fraction (+ve)
↓ Man↓tissa
Characterstics
(2) The mantissa part of log of a number is always kept positive.
(3) If the characteristics of log10 N be n, then the number of digits in N is (n+1)
(4) If the characteristics of log10 N be (– n) then there exists (n – 1) number of zeros after decimal part of N.
Example: 1 For y = loga x to be defined 'a' must be [IIT 1990]
Solution: (d) (a) Any positive real number
(c) ≥ e (b) Any number
It is obvious (Definition). (d) Any positive real number ≠ 1
Example: 2 Logarithm of 325 4 to the base 2 2 is
(a) 3.6 (b) 5 (c) 5.6 (d) None of these
Solution: (a) Let x be the required logarithm , then by definition (2 2)x = 325 4
(2.21/ 2 )x = 25.22 / 5 ; ∴ 3x = 25 + 2
5
22
Here, by equating the indices, 3 x = 27 , ∴ x = 18 = 3.6
2 5 5
Properties of Logarithms.
Let m and n be arbitrary positive numbers such that a > 0, a ≠ 1, b > 0, b ≠ 1then
(1) log a a = 1, log a 1 = 0 (2) log a b.log b a = 1= log a a = log b b ⇒ log a b = 1 a
log b
(3) log c a = log b a. log c b or log c a = log b a (4) log a (mn) = log a m + log a n
log b c
(5) log a m = log a m − log a n (6) log a mn = n log a m
n
(7) a loga m = m (8) log a 1 = − log a n
n
(9) log a β n= 1 loga n (10) log aβ nα = α log a n , (β ≠ 0)
β β
(11) a logc b = b logc a , (a,b, c > 0 and c ≠ 1)
Example: 3 The number log2 7 is [IIT 1990]
(a) An integer (b) A rational number (c) An irrational number (d) A prime number
Solution: (c) Suppose, if possible, log2 7 is rational, say p / q where p and q are integers, prime to each other.
Example: 4 Then, p = log 2 7 ⇒ 7 = 2p / q ⇒ 2p = 7q ,
Solution: (b) q
Example: 5
Which is false since L.H.S is even and R.H.S is odd. Obviously log2 7 is not an integer and hence not a prime number
Solution: (a)
Example: 6 If log 7 2 = m, then log 49 28 is equal to [Roorkee 1999]
Solution: (c)
Example: 7 (a) 2 (1 + 2m) (b) 1 + 2m (c) 2 (d) 1 + m
2 1 + 2m
log 49 28 = log 28 = log 7 + log 4 = log 7 + log 4 = 1 + 1 log7 4 = 1 + 1 .2 log7 2 = 1 + log7 2 = 1 +m = 1 + 2m
log 49 2 log 7 2 log 7 2 log 7 2 2 2 2 2 2 2
If loge a +b = 1 (log e a + log e b) , then relation between a and b will be [UPSEAT 2000]
2 2
(a) a = b (b) a = b (c) 2a = b (d) a = b
2 3
loge a +b = 1 (log e a + log e b) = 1 log e (ab) = log e ab
2 2 2
( )⇒a +b
2 = ab ⇒ a + b = 2 ab ⇒ a− b 2 =0 ⇒ a− b =0 ⇒ a=b
If log 10 3 = 0.477 , the number of digits in 340 is [IIT 1992]
(a) 18 (b) 19 (c) 20 (d) 21
Let y = 340 is
Taking log both the sides, log y = log 340 ⇒ log y = 40 log 3 ⇒ log y = 19.08
∴ Number of digits in y = 19 + 1 = 20 [Roorkee 2000]
Which is the correct order for a given number α in increasing order
(a) log2 α, log3 α, loge α, log10 α (b) log10 α, log3 α, loge α, log2 α
(c) log10 α, loge α, log2 α, log3 α (d) log3 α, loge α, log2 α, log10 α
Solution: (b) Since 10, 3, e, 2 are in decreasing order
Obviously, log10 α, log3 α, loge α, log2 α are in increasing order.
Logarithmic Inequalities. (2) If 0 < a < 1, p > 1 ⇒ log a p < 0
(1) If a > 1, p > 1 ⇒ log a p > 0 (4) If p > a > 1 ⇒ log a p > 1
(3) If a > 1, 0 < p < 1 ⇒ log a p < 0 (6) If 0 < a < p < 1 ⇒ 0 < log a p < 1
(5) If a > p > 1 ⇒ 0 < log a p < 1
(7) If 0 < p < a < 1 ⇒ log a p > 1 (8) If logm a > b ⇒ a > mb, if m > 1
a < mb, if 0 < m < 1
(9) logm a < b ⇒ a < mb, if m > 1
a > mb, if 0 < m < 1
(10) log p a > log p b ⇒ a ≥ b if base p is positive and >1 or a ≤ b if base p is positive and < 1 i.e.,
0< p<1
In other words, if base is greater than 1 then inequality remains same and if base is positive but less than 1
then the sign of inequality is reversed.
Example: 8 If x = log 3 5, y = log17 25, which one of the following is correct [W.B. JEE 1993]
Solution: (c)
(a) x < y (b) x = y (c) x > y (d) None of these
Example: 9
Solution: (a) y = log17 25 = 2 log17 5
∴ 1 = 1 log 5 17
y 2
1 = log 5 3 = 1 log 5 9
x 2
Clearly 1 > 1 ,∴x > y
y x
If log 0.3 (x − 1) < log 0.09 (x − 1), then x lies in the interval
(a) (2, ∞) (b) (– 2, –1) (c) (1, 2) (d) None of these
log 0.3 (x − 1) < log (0.3)2 (x − 1) = 1 log 0.3 (x − 1)
2
∴ 1 log 0.3 (x − 1) < 0
2
or log0.3(x − 1) < 0 = log 1 or (x − 1) > 1 or x > 2
As base is less than 1, therefore the inequality is reversed, now x>2 ⇒ x lies in (2, ∞) .
Logarithms, indices and surds & partial fractions
Definition of Indices.
If a is any non zero real or imaginary number and m is the positive integer, then am = a.a.a.a..........a
(m times). Here a is called the base and m the index, power or exponent.
Laws of Indices.
(1) a0 = 1 , (a ≠ 0)
(2) a −m = 1 , (a ≠ 0)
am
(3) a m + n = a m .a n , where m and n are rational numbers
(4) a m−n = am , where m and n are rational numbers, a ≠ 0
an (6) a p / q = q a p
(5) (am )n = amn
(7) If x = y , then a x = a y , but the converse may not be true.
For example: (1)6 = (1)8 , but 6 ≠ 8 (ii) If a = 1, then x, y may be any real number
(i) If a ≠ ±1, or 0, then x = y (iv) If a = 0, then x, y may be any non-zero real number
(iii) If a = −1, then x, y may be both even or both odd
But if we have to solve the equations like [ f(x)]φ(x) = [ f(x)]Ψ(x) then we have to solve :
(a) f(x) = 1 (b) f(x) = −1 (c) f(x) = 0 (d) φ(x) = Ψ(x)
Verification should be done in (b) and (c) cases
(8) am.bm = (ab)m is not always true
In real domain, a b = (ab) , only when a ≥ 0,b ≥ 0
In complex domain, a. b = (ab) , if at least one of a and b is positive.
(9) If a x = b x then consider the following cases :
(i) If a ≠ ±b, then x = 0 (ii) If a = b ≠ 0, then x may have any real value (iii) If a = −b , then x is even.
If we have to solve the equation of the form [ f(x)]φ(x) = [g(x)]φ(x) i.e., same index, different bases, then we
have to solve
(a) f(x) = g(x) , (b) f(x) = −g(x) , (c) φ(x) = 0
Verification should be done in (b) and (c) cases.
Example: 1 For x ≠ 0, xl (l 2 +lm+m2) xm (m2 +nm+n2) xn (n2 +nl+l 2) =
Solution: (a) xm xn xl
(a) 1 (b) x (c) Does not exist (d) None of these
xl l 2 +lm+m2 xm m2 +nm+n2 xn n2 +nl +l 2
xm xn xl
= (x l−m )(l 2 +lm+m2 ) (x m−n )m2 +nm+n2 (x n−l )n2 +nl+l 2 = x l3 −m3 .x m3 −n3 .x n3 −l3 = x l 3 −m3 +m3 −n3 +n3 −l 3 = x 0 =1
Example: 2 If 2x = 4y = 8 z and xyz = 288, then 1 + 1 + 1 =
2x 4y 8z
(a) 11/48 (b) 11/24 (c) 11/8 (d) 11/96
Solution: (d) 2x = 22y = 23z i.e., x = 2y = 3z = k (say). Then xyz = k3 = 288 , So k = 12
6
∴ x = 12, y = 6, z = 4 . Therefore, 1 + 1 + 1 = 11
2x 4y 8z 96
Example: 3 2.3n+1 + 7.3n−1 =
3n+2 − 2(1 / 3)1−n
(a) 1 (b) 3 (c) –1 (d) 0
Solution: (a) 2.3n+1 + 7.3n−1 = 2.3n−1.3 2 + 7.3n−1 = 3n−1[18 + 7] = 1
Example: 4 3n−1.3 3 − 2.3n−1 3n−1[27 − 2]
Solution: (c) 3n+2 2 1 1−n
3
−
2 x +2 3 2−2 x
3 2
If = , then x = [UPSEAT 1999]
(a) 1 (b) 3 (c) 4 (d) 0
2 x + 2 3 2− 2 x 2 x + 2 2 2x − 2
3 2 3 3
= ⇒ = . Clearly x + 2 = 2x − 2 ⇒ x =4
Example: 5 The equation 4(x2 +2) − 9.2(x2 +2) + 8 = 0 has the solution
(a) x = 1 (b) x = −1 (c) x = 2 (d) x = − 2
Solution: (a, b) 4(x2 +2) − 9.2(x2 +2) + 8 = 0 ⇒ 2(x2 +2) 2 − 9.2(x2 +2) + 8 = 0
Put 2(x2 +2)2 = y . Then y2 − 9y + 8 = 0 , which gives y = 8, y = 1
When y = 8 ⇒ 2x2 +2 = 8 ⇒ 2x2 +2 = 23 ⇒ x2 + 2 = 3 ⇒ x2 = 1 ⇒ x = 1,−1
When y = 1 ⇒ 2x2 +2 = 1 ⇒ 2x2 +2 = 2o ⇒ x2 + 2 = 0 ⇒ x2 = −2 , which is not possible.
Definition of Surds .
Any root of a number which can not be exactly found is called a surd.
Let a be a rational number and n is a positive integer. If the nth root of x i.e., x1/ n is irrational, then it is called
surd of order n.
Order of a surd is indicated by the number denoting the root.
For example 7, 3 9,(11)3 / 5 , n 3 are surds of second, third, fifth and nth order respectively.
A second order surd is often called a quadratic surd, a surd of third order is called a cubic surd.
Note : If a is not rational, n a is not a surd.
For example, (5 + 7) is not a surd as 5 + 7 is not a rational number.
Types of Surds.
(1) Simple surd : A surd consisting of a single term. For example 2 3,6 5, 5 etc.
(2) Pure and mixed surds : A surd consisting of wholly of an irrational number is called pure surd.
Example : 5, 3 7
A surd consisting of the product of a rational number and an irrational number is called a mixed surd.
Example : 5 3 .
(3) Compound surds : An expression consisting of the sum or difference of two or more surds.
Example : 5 + 2, 2 − 3 + 3 5 etc.
(4) Similar surds : If the surds are different multiples of the same surd, they are called similar surds.
Example : 45, 80 are similar surds because they are equal to 3 5 and 4 5 respectively.
(5) Binomial surds : A compound surd consisting of two surds is called a binomial surd.
Example : 5 − 2 , 3 + 3 2 etc.
(6) Binomial quadratic surds: Binomial surds consisting of pure (or simple) surds of order two i.e., the
surds of the form a b ± c d or a ± b c are called binomial quadratic surds.
Two binomial quadratic surds which differ only in the sign which connects their terms are said to be conjugate
or complementary to each other. The product of a binomial quadratic surd and its conjugate is always rational.
For example: The conjugate of the surd 2 7 + 5 3 is the surd 2 7 − 5 3 .
Properties of Quadratic Surds.
(1) The square root of a rational number cannot be expressed as the sum or difference of a rational number
and a quadratic surd.
(2) If two quadratic surds cannot be reduced to others, which have not the same irrational part, their product
is irrational.
(3) One quadratic surd cannot be equal to the sum or difference of two others, not having the same irrational part.
(4) If a + b = c + d, where a and c are rational, and b, d are irrational, then a = c and b= d .
Example: 6 The greatest number among 3 9, 4 11, 6 17 is
Solution: (a)
(a) 3 9 (b) 4 11 (c) 6 17 (d) Can not be determined
Example: 7
Solution: (c) 3 9, 4 11, 6 17
Example: 8 L.C.M of 3, 4, 6 is 12
Solution: (a)
∴ 3 9 = 91/ 3 = (9 4 )1/12 = (6561)1/12 , 4 11 = (11)1 / 4 (113)1 / 12 = (1331)1 / 12 , 6 17 = (17)1/ 6 = (17 2 )1/ 2 = (289)1/12
Hence 3 9 is the greatest number.
The value of 15 is
10 + 20 + 40 − 5 − 80
(a) 5(5 + 2) (b) 5(2 + 2) (c) 5(1 + 2) (d) 5(3 + 2)
Given fraction = 10 + 15 5− 80 = 15 5 −4 5
20 + 40 − 10 + 2 5 + 2 10 −
= 15 = 5 . 10 + 5 = 10 + 5 = 5( 2 + 1)
3 10 − 3 5 10 − 5 10 + 5
If x = 3 ( 2 + 1) − 3 ( 2 − 1); then x3 + 3x =
(a) 2 (b) 6 (c) 6x (d) None of these
x = ( 2 + 1)1 / 3 − ( 2 − 1)1 / 3
x3 = ( 2 + 1) − ( 2 − 1) − 3( 2 + 1)1 / 3 ( 2 − 1)1 / 3 3 ( 2 + 1) − 3 2 − 1
x3 = 2 − 3 (2 − 1)1 / 3 x ⇒ x3 + 3x = 2 .
Rationalisation Factors .
If two surds be such that their product is rational, then each one of them is called rationalising factor of the other.
Thus each of 2 3 and 3 is a rationalising factor of each other. Similarly 3 + 2 and 3 − 2 are
rationalising factors of each other, as ( 3 + 2)( 3 − 2) = 1 , which is rational.
To find the factor which will rationalize any given binomial surd :
Case I: Suppose the given surd is P a − q b
suppose a1 / P = x,b1 / q = y and let n be the L.C.M. of p and q. Then x n and yn are both rational.
Now x n − yn is divisible by x − y for all values of n, and x n − yn = (x − y)(x n−1 + x n−2y + x n−3y 2 + ..... + yn−1) .
Thus the rationalizing factor is x n−1 + x n−2y + x n−3y 2 + ..... + yn−1 and the rational product is x n − yn .
Case II: Let the given surd be p a + q b .
Let x, y,n have the same meaning as in Case I.
(1) If n is even, then x n − yn is divisible by x+ y and x n − yn = (x + y)(x n−1 − x n−2y + x n−3y 3 − ..... − yn−1)
Thus the rationalizing factor is x n−1 − x n−2y + x n−3y 2 − ..... − yn−1 and the rational product is x n − yn .
(2) If n is odd, x n + yn is divisible by x + y , and x n + yn = (x + y)(x n−1 − x n−2y + .... + yn−1)
Thus the rationalizing factor is x n−1 − x n−2y + ..... − xyn−2 + yn−1 and the rational product is x n + yn .
Example: 9 The rationalising factor of a1 / 3 + a−1 / 3 is
Solution: (d)
(a) a1 / 3 − a −1 / 3 (b) a 2 / 3 + a −2 / 3 (c) a 2 / 3 − a −2 / 3 (d) a 2 / 3 + a −2 / 3 − 1
Let x = a1/ 3 , y = a −1/ 3 then a = x 3 , a −1 = y 3
x3 + y3 = (x + y)(x 2 − xy + y2)
So rationlising factor is (x2 − xy + y2) . Put the value of x and y
Thus the required rationlising factor is a 2 / 3 + a −2 / 3 − 1 .
Square Roots of a +√b and a + √b + √c + √d Where √b , √c , √d are Surds .
Let (a + b) = x + y, where x, y > 0 are rational numbers.
Then squaring both sides we have, a + b = x + y + 2 x y
⇒ a = x + y, b = 2 xy ⇒ b = 4 xy
So, (x − y)2 = (x + y)2 − 4 xy = a 2 − b
After solving we can find x and y.
Similarly square root of a − b can be found by taking (a − b) = x − y, x > y
To find square root of a + √b + √c + √d : Let (a + b + c + d ) = x + y + z,(x, y, z > 0) and take
(a + b − c − d ) = x + y − z . Then by squaring and equating, we get equations in x, y, z. On solving these
equations, we can find the required square roots.
Note : If a 2 − b is not a perfect square, the square root of a + b is complicated i.e., we can't find the
value of (a + b) in the form of a compound surd.
If (a + b) = x + y, x > y then (a − b) = x − y
a + (b) = a + a2 − b + a − a 2 − b
2
2
a − (b) = a + a2 − b − a − a 2 − b
2 2
If a is a rational number, b, c, d , are surds then
(i) a+ b+ c+ d= bd + bc + cd (ii) a− b− c+ d= bd + cd + bc ,
4c 4d 4b 4c 4b 4d
(iii) a− b− c+ d= bc bd cd
4d − 4c − 4b
Example: 10 (3 + 5) is equal to
(a) 5 + 1
(b) 3 + 2 (c) ( 5 + 1) / 2 (d) 1 ( 5 + 1)
2
Solution: (c) Let 3 + 5 = x + y
3 + 5 = x + y + 2 xy . Obviously x + y = 3 and 4 xy = 5 . So (x − y)2 = 9 − 5 = 4 or (x − y) = 2
After solving x = 5 , y = 1 . Hence 3+ 5 = 5 1 5 +1
2 2 2+ 2= 2
Example: 11 [10 − (24) − (40) + (60)] =
(a) 5 + 3 + 2 (b) 5 + 3 − 2 (c) 5 − 3 + 2 (d) 2 + 3 − 5
Solution: (b) Let 10 − 24 − 40 + 60 = ( a − b + c )2
10 − 24 − 40 + 60 = a + b + c − 2 ab − 2 bc + 2 ca, a, b, c > 0 . Then a + b + c = 10, ab = 6 , bc = 10, ca = 15
a2b2c2 = 900 ⇒ abc = 30 (≠ ±30) . So a = 3, b = 2, c = 5
Therefore, (10 − 24 − 40 + 60) = ±( 3 + 5 − 2)
Example: 12 4 (17 + 12 2) =
(a) 2 + 1
(b) 21 / 4( 2 + 1) (c) 2 2 + 1 (d) None of these
Solution: (a) (17 + 12 2) = [32 + (2 2)2 + 2.3.2 2] = 3 + 2 2
∴ 4 (17 + 12 2) = (3 + 2 2) = 2 + 1 . y, where a is a rational number and b is a surd.
Cube Root of a Binomial Quadratic Surd.
If (a + b)1/ 3 = x + y then (a − b)2 / 3 = x −
Procedure of finding (a + b)1/ 3 is illustrated with the help of an example :
Taking (37 − 30 3)1/ 3 = x + y we get on cubing both sides, 37 − 30 3 = x 3 + 3xy − (3x 2 + y) y
∴ x 3 + 3xy = 37
(3x 2 + y) y = 30 3 = 15 12
As 3 can not be reduced, let us assume y = 3 we get 3x 2 + y = 3x 2 + 3 = 30 ∴ x = 3
Which doesn't satisfy x 3 + 3xy = 37
Again taking y = 12, we get
3x2 +12 = 15 , ∴ x = 1
x = 1, y = 12 satisfy x 3 + 3xy = 37
∴ 3 37 − 30 3 = 1 − 12 = 1 − 2 3
Example: 13 3 (61 − 46 5) =
(a) 1 − 2 5 (b) 1 − 5 (c) 2 − 5 (d) None of these
Solution: (a) 3 61 − 46 5 = a − b ⇒ 61 − 46 5 = (a − b)3 = a3 + 3ab − (3a2 + b) b
⇒ 61 = a3 + 3ab, 46 5 = (3a2 + b) b ⇒ 61 = (a2 + 3b)a , 23 20 = (3a2 + b) b
So a = 1, b = 20 . Therefore 3 61 − 46 5 = 1 − 20 = 1 − 2 5 .
Equations Involving Surds.
While solving equations involving surds, usually we have to square, on squaring the domain of the equation
extends and we may get some extraneous solutions, and so we must verify the solutions and neglect those which do
not satisfy the equation.
Note that from ax = bx , to conclude a = b is not correct. The correct procedure is x(a − b) =0 i.e. x = 0 or
a = b . Here, necessity of verification is required.
Example: 14 The equation (x + 1) − (x − 1) = (4 x − 1) , x ∈ R has
Solution: (d)
(a) One solution (b) Two solution (c) Four solution (d) No solution
Given (x + 1) − (x − 1) = (4 x − 1) .....(i)
Squaring both sides, we get, − 2 (x 2 − 1) = 2x − 1
Squaring again, we get, x = 5 , which does not satisfy equation (i)
4
Hence, there is no solution of the given equation.
Logarithms indices and surds & partial fractions
Definition.
An expression of the form f(x) , where f(x) and g(x) are polynomial in x, is called a rational fraction.
g(x)
(1) Proper rational functions: Functions of the form f(x) ,where f(x) and g(x) are polynomials and
g(x)
g(x) ≠ 0 ,are called rational functions of x.
If degree of f(x) is less than degree of g(x) ,then f(x) is called a proper rational function.
g(x)
Example: x+2 is a proper rational function.
x2 + 2x + 4
(2) Improper rational functions : If degree of f(x) is greater than or equal to degree of g(x), then f(x) is
g(x)
called an improper rational function.
For example: x3 is an improper rational function.
(x − 1)(x − 2)
(3) Partial fractions : Any proper rational function can be broken up into a group of different rational
fractions, each having a simple factor of the denominator of the original rational function. Each such fraction is
called a partial fraction.
If by some process, we can break a given rational function f(x) into different fractions, whose denominators
g(x)
are the factors of g(x),then the process of obtaining them is called the resolution or decomposition of f(x) into its
g(x)
partial fractions.
Different Cases of Partial Fractions.
(1) When the denominator consists of non-repeated linear factors: To each linear factor (x − a)
occurring once in the denominator of a proper fraction, there corresponds a single partial fraction of the form x A a ,
−
where A is a constant to be determined.
If g(x) = (x − a1)(x − a2)(x − a3).......(x − an), then we assume that, f(x) = A1 + A2 + ...... + An
g(x) x − a1 x − a2 x − an
Where A1, A2, A3 ....... An are constants, can be determined by equating the numerator of L.H.S. to the
numerator of R.H.S. (after L.C.M.) and substituting x = a1, a2......an .
Note : Remainder of polynomial f(x) , when divided by (x − a) is f(a) .
e.g., Remainder of x 2 + 3x − 7 ,when divided by x − 2 is (2)2 + 3(2) − 7 = 3 .
(x px + q b) = (x pa + q + (b pb + q
− a)(x − − a)(a − b) − a)(x − b)
Example: 1 The remainder obtained when the polynomial x 64 + x 27 + 1 is divided by (x + 1) is [EAMCET 1992]
(a) 1 (b) – 1 (c) 2 (d) – 2
Solution: (a) Remainder of x 64 + x 27 + 1 , when divided by x + 1 is (−1)64 + (−1)27 + 1 = 1 − 1 + 1 = 1 .
Example: 2 If (x 2x + 3 3) = x a 1 + (x b 3) , then a+b [MNR 1993]
+ 1)(x − + −
(a) 1 (b) 2 (c) 9 (d) −1
4 4
Solution: (b) 2x + 3 = a(x − 3) + b(x + 1)
Put x = −1 ; 2(−1) + 3 = a(−1 − 3) ⇒ 1 = −4a ⇒ a = −1
4
Now put x = 3 ; 2(3) + 3 = b(3 + 1) ⇒ 9 = 4b ⇒ b = 9
4
Therefore, a+b = −1 + 9 = 2.
4 4
Example: 3 If 3x + a 2 = (x A − 10 , then
x2 − 3x + − 2) x −1
(a) a = 7 (b) a = −7 (c) A = −13 (d) A = 13
Solution: (a, d) 3x + a 2 = (x A − 10
x2 − 3x + − 2) (x − 1)
⇒ (3x + a) = A(x − 1) − 10(x − 2) ⇒ 3 = A − 10 , a = − A + 20 (On equating coefficients of x and constant term)
⇒ A = 13, a = 7.
(2) When the denominator consists of linear factors, some repeated: To each linear factor (x – a)
occurring r times in the denominator of a proper rational function, there corresponds a sum of r partial fractions.
Let g(x) = (x − a)k(x − a1)(x − a2).......(x − ar ) .Then we assume that
f(x) = A1 + A2 + ...... + Ak + B1 + ...... + Br
g(x) x−a (x − a)2 (x − a)k (x − a1) (x − ar )
Where A1, A2,......, Ak are constants. To determined the value of constants adopt the procedure as above.
Example: 4 If (x 3x + 4 1) = (x A 1) + (x B 1) + (x C , then A= [EAMCET 1994]
Solution: (c) + 1)2(x − − + + 1)2 [IIT 1992]
Example: 5
Solution: (c) (a) −1 (b) 15 (c) 7 (d) −1
2 4 4 4
We have, (x 3x + 4 = (x A + (x B 1) + (x C
+ 1)2(x − 1) − 1) + + 1)2
⇒ 3x + 4 = A(x + 1)2 + B(x + 1)(x − 1) + C(x − 1)
Putting x = 1 , we get 7 = A(2)2 ⇒ A = 7 .
4
The partial fraction of x2 are
(x − 1)3(x − 2)
(a) −1 + 3 − 4 + 4 (b) −1 − (x 3 + 4 1) + (x 4 2)
(x − 1)3 (x − 1)2 (x − 1) (x − 2) (x − 1)3 − 1)2 (x − −
(c) −1 + −3 + −4 + 4 (d) None of these
(x − 1)3 (x − 1)2 (x − 1) (x − 2)
Put the repeated factor (x − 1) = y ⇒ x = y + 1
∴ (x x2 − 2) = (1 + y)2 = 1 + 2y + y 2
− 1)3 (x y3 (y − 1) y3 (−1 + y)
Dividing the numerator, 1 + 2y + y2 by −1 + y till y3 appears as factor, we get
1 + 2y + y 2 = (−1 − 3y − 4y2) + 4y3
−1+ y −1+ y
Given expression = −1 − 3 − 4 + 4 y = −1 + −3 + −4 + 4 .
y3 y2 y −1+ (x − 1)3 (x − 1)2 (x − 1) (x − 2)
(3) When the denominator consists of non-repeated quadratic factors: To each irreducible non
repeated quadratic factor ax 2 + bx + c , there corresponds a partial fraction of the form Ax +B c , where A and
ax 2 + bx +
B are constants to be determined.
Example : (x 2 4x2 + 2x + 3 = Ax + B + C + D
+ 4 x + 9)(x − 2)(x + 3) x2 + 4x +9 x−2 x+3
Note : px + q = −q − pa + q + pa + q
x 2 (x − a) ax 2 a2x a 2 (x − a)
px + q = q − q a) + pa + q
x(x − a)2 a2x a2(x − a(x − a)2
px +q = q + pa 2 − qx
x(x 2 + a2) a2x a2(x 2 + a2)
Example: 6 The partial fractions of 3x − 1 are [MNR 1995]
Solution: (c) (1 − x + x 2 )(2 + x)
Example: 7 (a) (x 2 x + 1) + x 1 2 (b) 1x (c) x1 (d) x2 −1 +1 + x x 2
Solution: (a) −x + x2 − x +1 + x + 2 x2 − x +1 − x + 2 −x +
(1 − x 3x − 1 x) = Ax + B + C
+ x 2 )(2 + x2 − x +1 x+2
⇒ (3x − 1) = (Ax + B)(x + 2) + C(x 2 − x + 1)
Comparing the coefficient of like terms, we get A + C = 0 , 2A + B − C = 3 , 2B + C = −1 ⇒ A = 1 , B = 0 , C = −1
∴ (1 − x 3x − 1 + x) = x2 x +1 − x 1 2 .
+ x 2 )(2 −x +
If (x + 1)2 = A + Bx + C , then sin −1 A = [EAMCET 1997, 98]
x3 + x x x2 + 1 C
(a) π (b) π (c) π (d) π
6 4 3 2
(x + 1)2 = A + Bx + C
x3 + x x x2 +1
⇒ (x + 1)2 = A(x 2 + 1) + (Bx + C) x ⇒ A + B = 1 , C = 2 , A = 1 ⇒ B = 0
Therefore sin −1 A = sin −1 1 = 30 o = π .
C 2 6
(4) When the denominator consists of repeated quadratic factors: To each irreducible quadratic factor
ax 2 + bx + c occurring r times in the denominator of a proper rational fraction there corresponds a sum of r partial
fractions of the form.
A1 x + B1 + A2 x + B2 + ......... + Ar x + Br
ax 2 + bx + c (ax 2 + bx + c)2 (ax 2 + bx + c)r
Where, A’s and B’s are constants to be determined.
Example: 8 If x = 1 1 1) − x +1 + y then y =
Solution: (a) (x − 1)(x2 + 1)2 4 (x − x2 +1
(a) (1 − x) (b) (1 − x) (c) 1+ x (d) None of these
2(x 2 + 1)2 3(x 2 + 1) 2(x 2 − 1)2
x + 1)2 = 1 1 − x +1 + y
(x − 1)(x 2 4 (x − 1) x2 +1
⇒ x + 1)2 = 1 1 − x +1 + Ax + B ⇒ 4 x = (x2 + 1)2 − (x + 1)(x − 1)(x2 + 1) + 4(Ax + B)(x − 1)
(x − 1)(x 2 4 (x − 1) x 2 + 1 (x 2 + 1)2
⇒ 4A + 2 = 0 , 4B − 4A = 4 ⇒ A= −1 , B = 1
2 2
∴ y = Ax + B = 1 (1 − x)
(x 2 + 1)2 2 (x 2 + 1)2
Partial Fractions of Improper Rational Functions.
If degree of f(x) is greater than or equal to degree of g(x), then f(x) is called an improper rational function
g(x)
and every rational function can be transformed to a proper rational function by dividing the numerator by the
denominator.
We divide the numerator by denominator until a remainder is obtained which is of lower degree than the
denominator.
i.e., f(x) = Q(x) + R(x) , where degree of R(x) < degree of g (x) .
g(x) g (x)
For example, x2 x3 +6 is an improper rational function and can be expressed as (x + 5) + 19x − 30
− 5x x2 − 5x + 6
which is the sum of a polynomial (x + 5) and a proper rational function 19x − 30 .
x2 − 5x + 6
Example: 9 If x3 − 6x 2 + 10x − 2 = f(x) + A + B , then f(x) =
x2 − 5x + 6 − 2) −
(x (x 3)
(a) x − 1 (b) x + 1 (c) x (d) None of these
Solution: (a) x 2 − 5x + 6 x 3 − 6x 2 + 10x − 2 x − 1
x3 5x 2 6x
− +− +−
−x2 +4x −2
−x2+5x −6
+− +
−x+4
∴ f(x) = x − 1 .
General Method of Finding out the Constants.
(1) Express the given fraction into its partial fractions in accordance with the rules written above.
(2) Then multiply both sides by the denominator of the given fraction and you will get an identity which will
hold for all values of x.
(3) Equate the coefficients of like powers of x in the resulting identity and solve the equations so obtained
simultaneously to find the various constant is short method. Sometimes, we substitute particular values of the variable
x in the identity obtained after clearing of fractions to find some or all the constants. For non-repeated linear factors,
the values of x used as those for which the denominator of the corresponding partial fractions become zero.
Note : If the given fraction is improper, then before finding partial fractions, the given fraction must be
expressed as sum of a polynomial and a proper fraction by division.
Important Tips
Some times a suitable substitution transforms the given function to a rational fraction which can be integrated by breaking it into partial
fractions.
Example: 10 The coefficient of xn in the expression 5x + 6 when expanded in ascending order is [MNR 1993]
Solution: (a) (2 + x)(1 − x)
(a) − 2 (−1)n + 11 (b) 2 + (−1)n − 11 (c) − 2 + (−1)n − 11 (d) None of these
3 2n 3 3 2n 3 3 3 2n
5x + 6 −4 11
(2 + x)(1 +
x) = 3 + 3
2+ x 1− x
Rewriting the denominators for expressions, we get
−4 11 −1
= 3 x + 3 = −2 1 + x + 11 (1 − x)−1
21 + 2 1− x 3 2 3
= −2 1 − x + x2 − x3 + ...... + (−1)n xn + ...... + 11 [1 + x + x2 + ....... + xn + .....]
3 2 4 8 2n 3
The coefficient of x n in the given expression is −2 (−1)n 1 + 11 .
3 2n 3
Numerical-method
Introduction.
The limitations of analytical methods have led the engineers and scientists to evolve graphical and numerical
methods. The graphical methods, though simple, give results to a low degree of accuracy. Numerical methods can,
however, be derived which are more accurate.
Significant digits and Rounding off of Numbers.
(1) Significant digits : The significant digits in a number are determined by the following rules :
(i) All non-zero digits in a number are significant.
(ii) All zeros between two non-zero digits are significant.
(iii) If a number having embedded decimal point ends with a non-zero or a sequences of zeros, then all these
zeros are significant digits.
(iv) All zeros preceding a non-zero digit are non-significant.
Number Number of significant digits
3.0450 5
0.0025 2
102.030070 9
35.9200 6
0.0002050 4
20.00 4
2000 1
(2) Rounding off of numbers : If a number is to be rounded off to n significant digits, then we follow the
following rules :
(i) Discard all digits to the right of the nth digit.
(ii) If the (n+1)th digit is greater then 5 or it is 5 followed by a nonzero digit, then nth digit is increased by 1. If
the (n+1)th digit is less then 5, then digit remains unchanged.
(iii) If the (n+1)th digit is 5 and is followed by zero or zeros, then nth digit is increased by 1 if it is odd and it
remains unchanged if it is even.
Error due to Rounding off of Numbers.
If a number is rounded off according to the rules, the maximum error due to rounding does not exceed the
one half of the place value of the last retained digit in the number.
The difference between a numerical value X and its rounded value X1 is called round off error is given by
E = X − X1.
Truncation and Error due to Truncation of Numbers.
Leaving out the extra digits that are not required in a number without rounding off, is called truncation or
chopping off.
The difference between a numerical value X and its truncated value X1 is called truncation error and is given
by E = X − X1 .
The maximum error due to truncation of a number cannot exceed the place value of the last retained digit in
the number.
Remark 1 : In truncation the numerical value of a positive number is decreased and that of a negative
number is increased.
Remark 2 : If we round off a large number of positive numbers to the same number of decimal places, then
the average error due to rounding off is zero.
Remark 3 : In case of truncation of a large number of positive numbers to the same number of decimal
places the average truncation error is one half of the place value of the last retained digit.
Remark 4 : If the number is rounded off and truncated to the same number of decimal places, then
truncation error is greater than the round off error.
Remark 5 : Round of error may be positive or negative but truncation error is always positive in case of
positive numbers and negative in case of negative numbers.
Number Approximated number obtained by
0.335217... Chopping off Rounding off
0.666666...
0.123451... 0.3352 0.3352
0.213450...
0.213950... 0.6666 0.6667
0.335750... 0.1234 0.1235
0.999999... 0.2134 0.2134
0.555555... 0.2139 0.2140
0.3357 0.3358
0.9999 1.0000
0.5555 0.5556
Relative and Percentage errors of Numbers .
The difference between the exact value of a number X and its approximate value X1, obtained by rounding off
or truncation, is known as absolute error.
The quantity X − X1 is called the relative error and is denoted by ER .
X
Thus ER = X − X1 = ∆X . This is a dimensionless quantity.
X X
The quantity ∆X × 100 is known as percentage error and is denoted by Ep , i.e. Ep = ∆X × 100 .
X X
Remark 1 : If a number is rounded off to n decimal digits, then | ER |< 0.5 ×10−n+1
Remark 2 : If a number is truncated to n decimal places, then | ER |< 10−n+1
Example: 1 The number of significant digits in 0.0001 is
Solution: (c)
Example: 2 (a) 5 (b) 4 (c) 1 (d) None of these
Solution: (c) 0.0001 has only one significant digit 1.
Example: 3 When a number is rounded off to n decimal places, then the magnitude of relative error does not exceed [DCE 1998]
(a) 10 −n (b) 10 −n+1 (c) 0.5 × 10 −n+1 (d) None of these
When a number is rounded off to n decimal places, then the magnitude of relative error i.e. | ER | does not exceed
0.5 × 10 −n+1 .
When the number 2.089 is rounded off to three significant digits, then the absolute error is
(a) 0.01 (b) –0.01 (c) 0.001 (d) –0.001
Solution: (d) When the number 2.089 is rounded off to three significant digits, it becomes 2.09. Hence, the absolute error that occurs is
2.089 − 2.09 = −0.001
Algebraic and Transcendental Equation.
An equation of the form f(x)=0, is said to an algebraic or a transcendental equation according as f(x) is a
polynomial or a transcendental function respectively.
e.g. ax 2 + bx + c = 0 , ax 3 + bx 2 + cx + d = 0 etc., where a, b, c, d∈Q, are algebraic equations whereas
ae x + b sin x = 0 ; a log x + bx = 3 etc. are transcendental equations.
Location of real Roots of an Equation.
By location of a real root of an equation, we mean finding an approximate value of the root graphically or otherwise.
(1) Graphical Method : It is often possible to write f(x) = 0 in the form f1(x) = f2(x) and then plot the
graphs of the functions y = f1(x) and y = f2(x) .
Y y=f2(x) y=f1(x)
P
X′ O A real root of X
Y′ f1(x) = f2(x)
The abscissae of the points of intersection of these two graphs are the real roots of f(x) = 0 .
(2) Location Theorem : Let y = f(x) be a real-valued, continuous function defined on [a, b]. If f(a) and f(b)
Y
y = fx
A real root of
f(x) = 0
f(a)
X′ O a bX
Y′ f(b)
have opposite signs i.e. f(a).f(b) < 0, then the equation f(x)=0 has at least one real root between a and b.
Position of Real Roots
If f(x) = 0 be a polynomial equation and x1, x2............xk are the consecutive real roots of f(x) = 0 , then
positive or negative sign of the values of f(−∞), f(x).............. f(xk ), f(∞) will determine the intervals in which the root
of f(x) = 0 will lie whenever there is a change of sign from f(xr ) to f(xr+1) the root lies in the interval [xr , xr+1] .
Example: 4 If all roots of equation x 3 − 3x + k = 0 are real, then range of value of k
(a) (–2, 2) (b) [–2, 2] (c) Both (d) None of these
Solution: (a) Let f(x) = x 3 − 3x + k , then f(x) = 3x 2 − 3 and so f(x) = 0 ⇒ x = ±1 . The values of f(x) at x = −∞, − 1,1, ∞ are :
x : −∞ −1 1 ∞
f(x) : − ∞ k + 2 k − 2 ∞
If all roots of given equation are real, then k + 2 > 0 and k − 2 < 0 ⇒ −2 < k < 2 . Hence the range of k is (–2, 2)
Example: 5 For the smallest positive root of transcendental equation x − e −x = 0 , interval is [MP PET 1996]
(a) (0, 1) (b) (–1, 0) (c) (1, 2) (d) (2, 3)
Solution: (a) Let f(x) = x − e −x = 0 ⇒ xe x − 1 = 0 but f(0) = −ive and f(1) = +ive . Therefore root lie in (0,1).
Example: 6 The maximum number of real roots of the equation x 2n − 1 = 0 is [MP PET 2001]
Solution: (a)
(a) 2 (b) 3 (c) n (d) 2n
Let f(x) = x 2n − 1 , then f ' (x) = x 2n−1 = 0 ⇒ x = 0
Sign of f(x) at x = −∞, 0, + ∞ are x: −∞ 0 +∞
f(x) : + ive − ive + ive
This show that there are two real roots of f(x) = 0 which lie in the interval (−∞,0) and (0,+∞) . Hence maximum number
of real roots are 2.
Solution of Algebraic and Transcendental Equations
There are many numerical methods for solving algebraic and transcendental equations. Some of these methods
are given below. After locating root of an equation, we successively approximate it to any desired degree of accuracy.
(1) Iterative method: If the equation f(x) = 0 can be expressed as x = g(x) (certainly g(x) is non-constant),
the value g(x0) of g(x) at x = x0 is the next approximation to the root α . Let g(x0 ) = x1 , then x2 = g(x1) is a third
approximation to α . This process is repeated until a number, whose absolute difference from α is as small as we
please, is obtained. This number is the required root of f(x) = 0 , calculated upto a desired accuracy.
Thus, if xi is an approximation to α , then the next approximation xi+1 = g(xi ) ........(i)
The relation (i) is known as Iterative formula or recursion formula and this method of approximating a real
root of an equation f(x) = 0 is called iterative method. Y
(2) Successive bisection method : This method consists in locating
y=f(x)
the root of the equation f(x) = 0 between a and b. If f(x) is continuous between
a and b, and f(a) and f(b) are of opposite signs i.e. f(a).f(b)<0, then there is O a x2 X
(at least one) root between a and b. For definiteness, let f(a) be negative and x3 x1 b
f(b) be positive. Then the first approximation to the root x1 = 1 (a + b) .
2
Working Rule : (i) Find f(a) by the above formula.
(ii) Let f (a) be negative and f (b) be positive, then take x1 = a + b .
2
(iii) If f(x1) = 0 , then c is the required root or otherwise if f(x1) is negative then root will be
in (x1,b) and if f(x1) is positive then root will be in (a, x1) .
(iv) Repeat it until you get the root nearest to the actual root.
Note : This method of approximation is very slow but it is reliable and can be applied to any type of
algebraic or transcendental equations.
This method may give a false root if f(x) is discontinuous on [a, b].
Example: 7 By bisection method, the real root of the equation x 3 − 9x + 1 = 0 lying between x = 2 and x = 4 is nearer to
Solution: (b)
[MP PET 1997]
(a) 2.2 (b) 2.75 (c) 3.5 (d) 4.0
Since f(2) = 23 − 9(2) + 1 < 0 and f(4) = 4 3 − 9(4) + 1 > 0
∴ Root will lie between 2 and 4.
At x = 2 + 4 = 3 , f(3) = 33 − 9(3) + 1 > 0
2
∴ Root lie between 2 and 3.
At x = 2+3 = 2.5 , f(2.5) is –ive,∴ root lies between 2.5 and 3
2
At x = 2.5 + 3 = 2.75 and f(2.75) = (2.75)3 − 9(2.75) + 1 < 0 . ∴Root is near to 2.75.
2
(3) Method of false position or Regula-Falsi method : This is the oldest method of finding the real root
of an equation f(x) = 0 and closely resembles the bisection method. Here Y A[x0, f(x0)]
we choose two points x0 and x1 such that f(x0 ) and f(x1) are of opposite
signs i.e. the graph of y = f(x) crosses the x-axis between these points. This
indicates that a root lies between x0 and x1 consequently f(x0 )f(x1) < 0 . O x0 x3 x2 x1 X
p(x)
B[x1, f(x1)]
Equation of the chord joining the points A[x0 , f(x0 )] and B[x1, f(x1)] is
y− f(x0) = f(x1) − f(x0 )(x − x0) ........(i)
x1 − x0
The method consists in replacing the curve AB by means of the chord AB and taking the point of intersection
of the chord with the x-axis as an approximation to the root. So the abscissa of the point where the chord cuts the
x-axis (y = 0) is given by x2 = x0 − x1 − x0 f(x0 ) .........(ii)
f(x1) − f (x 2 )
which is an approximation to the root.
If now f(x0 ) and f(x2 ) are of opposite signs, then the root lies between x0 and x2 . So replacing x1 by x2
in (ii), we obtain the next approximation x3 . (The root could as well lie between x1 and x2 and we would obtain
x3 accordingly). This procedure is repeated till the root is found to desired accuracy. The iteration process based on
(i) is known as the method of false position.
Working rule
(i) Calculate f(x0 ) and f(x1), if these are of opposite sign then the root lies between x0 and x1 .
(ii) Calculate x2 by the above formula.
(iii) Now if f(x2 ) = 0 , then x2 is the required root.
(iv) If f(x2 ) is negative, then the root lies in (x2, x1) .
(v) If f(x2 ) is positive, then the root lies in (x0 , x2 ).
(vi) Repeat it until you get the root nearest to the real root.
Note : This method is also known as the method of false position.
The method may give a false root or may not converge if either a and b are not sufficiently close to
each other or f(x) is discontinuous on [a, b].
Geometrically speaking, in this method, part of the curve between the points P(a, f(a)) and Q (b,
f(b)) is replaced by the secant PQ and the point of intersection of this secant with x-axis gives an
approximate value of the root.
It converges more rapidly than bisection.
Example: 8 The root of the equation x 3 + x − 3 = 0 lies in interval (1, 2) after second iteration by false position method, it will be in
(a) (1.178, 2.00) (b) (1.25, 1.75) (c) (1.125, 1.375) [MP PET 2003]
(d) (1.875, 2.00)
Solution: (a) f(x) = x 3 + x − 3
f(1) = −1 and f(2) = 7
Therefore, root lie in (1, 2).
Now, take x0 = 1 , x1 = 2
⇒ x2 = x0 − x1 − x0 ) .f(x0 ) ⇒ x2 = 1 − + 2−1 .(−1) =1.125 and so f(x 2 ) = −0.451
f (x1 ) − f(x0 7 − (−1)
Hence, roots lie in (1.125, 2)
⇒ x3 = 1.125 − 2 − 1.125 (−0.451) = 1.178 . So required root lie in (1.178, 2)
7 − (−0.451)
(4) Newton-Raphson method : Let x0 be an approximate root of the equation f(x) = 0 . If x1 = x0 + h be
the exact root, then f(x1) = 0
∴ Expanding f(x0 + h) by Taylor's series
f(x0 ) + hf '(x0 ) + h2 f ''(x0 ) + ....... =0
2!
Since h is small, neglecting h2 and higher powers of h, we get
f(x0 ) + hf '(x0 ) = 0
or h = − f(x0 ) ........(i)
f '(x0 )
∴ A closer approximation to the root is given by
x1 = x0 − f(x0 )
f '(x0 )
Similarly, starting with x1 , a still better approximation x2 is given by
x2 = x1 − f(x1)
f '(x1 )
In general, x n+1 = xn − f (xn ) ........(ii)
f '(xn )
Which is known as the Newton-Raphson formula or Newton's iteration formula.
Working rule : (i) Find | f(a)| and | f(b)|. If | f(a)|<| f(b)|, then let a = x0 , otherwise b = x0 .
(ii) Calculate x1 = x0 − f(x0 )
f '(x0 )
(iii) x1 is the required root if f(x1) = 0 .
(iv) To find nearest to the real root, repeat it.
Note : Geometrically speaking, in Newton-Raphson method, the part of the graph of the function y = f(x)
between the point P(a, f(a)) and the x-axis is replaced by a tangent to the curve at the point at each
step in the approximation process.
This method is very useful for approximating isolated roots.
The Newton-Raphson method fails if f '(x) is difficult to compute or vanishes in a neighbourhood of
the desired root. In such cases, the Regula-Falsi method should be used.
The Newton-Raphson method is widely used since in a neighbourhood of the desired root, it
converges more rapidly than the bisection method or the Regula-Falsi method.
If the starting value a is not close enough to the desired root, the method may give a false root or
may not converge.
If f(x0 ) / f '(x0 ) is not sufficiently small, this method does not work. Also if it work, it works faster.
Geometrical Interpretation
Let x0 be a point near the root α of the equation f(x) = 0 . Then the equation of the tangent at
A0[x0 , f(x0 )] is y − f(x0 ) = f '(x0 )(x − x0 ) . Y
It cuts the x-axis at x1 = x0 − f(x0 ) y= f(x) A[x0, f(x0)]
f '(x0 )
A1 X
A2
O x2 x1
α
Which is a first approximation to the root α . If A1 is the point corresponding to x1 on the curve, then the
tangent at A1 will cut the x-axis of x2 which is nearer to α and is, therefore, a second approximation to the root.
Repeating this process, we approach to the root α quite rapidly. Hence the method consists in replacing the part of
the curve between the point A0 and the x-axis by means of the tangent to the curve at A0 .
Example: 9 The value of the root nearest to the 2, after first iteration of the equation x 4 − x − 10 = 0 by Newton-Raphson method is
Solution: (d)
[MP PET 2003]
Example: 10
Solution: (c) (a) 2.321 (b) 2.125 (c) 1.983 (d) 1.871
Let f(x) = x 4 − x − 10 , then f(1) = −10 and f(2) = 4
Thus roots lie in (1, 2). Also | f(2)|<| f(1)|, So take x0 = 2
Also f ' (x) = 4 x 3 − 1
f ' (2) = 4(8) − 1 = 31
∴ By Newton's rule, the first iteration,
x1 = x0 − f(x0 ) =2− 4 = 1.871
f'(x0 ) 31
For finding real roots of the equation x 2 − x = 2 by Newton-Raphson method, choose x0 = 1 , then the value of x 2 is
(a) –1 (b) 3 (c) 11 [MP PET 2000, 02]
5
(d) None of these
Let f(x) = x 2 − x − 2 . Given x0 = 1 .
By Newton-Raphson method, x1 = x0 − f(x0 )
f'(x0 )
f ' (x) = 2x − 1 = 2(1) − 1 = 1 and f(1) = −2 . Now x1 =1− −2 = 3
1
f(x1 ) = f(3) = 9 − 3 − 2 = 4
f ' (x1 ) = f ' (3) = 2(3) − 1 = 5 ; x2 = 3 − 4 = 11
5 5
Example: 11 The value of 12 correct to 3 decimal places by Newton-Raphson method is given by [DCE 1999, 2000]
(a) 3.463 (b) 3.462 (c) 3.467 (d) None of these
Solution: (a) Let x = 12 ⇒ x 2 − 12 = 0 ⇒ f(x) = x 2 − 12, f ' (x) = 2x
∴ f(3) = −3 i.e. –ive and f(4) = +4 i.e. +ive
Hence roots lie between 3 and 4. | f(3)|<| f(4)|, ∴ x0 = 3
First iteration , x1 = x0 − f(x0 ) = 3 − −3 = 3.5
f'(x0 ) 6
Now second iteration, f(3.5) = 0.25, f ' (3.5) = 7 , x2 = 3.5 − 0.25 = 3.463
7
Numerical Integration
It is the process of computing the value of a definite integral when we are given a set of numerical values of
the integrand f(x) corresponding to some values of the independent variable x.
∫If I = by.dx . Then I represents the area of the region R under the curve y = f(x) between the ordinates
a
x = a, x = b and the x-axis.
(1) Trapezoidal rule
Let y = f(x) be a function defined on [a, b] which is divided into n equal sub-intervals each of width h so that
b − a = nh.
Let the values of f(x) for (n + 1) equidistant arguments x0 = a, x1 = x0 + h, x2 = x0 + 2h,......., xn =
x0 + nh = b be y0 , y1, y2,.........yn respectively.
∫ ∫ [ ]bf(x)dx = 1 h
2 2
a
Then x0 +nh y dx = h (y 0 + yn ) + (y1 + y2 + ...... + yn−1) = (y0 + yn ) + 2(y1 + y2 + ...... + yn−1)
x0
This rule is known as Trapezoidal rule .
The geometrical significance of this rule is that the curve y = f(x) is replaced by n straight lines joining the
points (x0 , y0 ) and (x1, y1) ; (x1, y1) and (x2, y2 ) ; (xn−1, yn−1) and (xn, yn ) . The area bounded by the curve
y = f(x) . The ordinate x = x0 and x = xn and the x-axis, is then approximately equivalent to the sum of the areas
of the n trapeziums obtained.
Example: 12 If e 0 = 1, e1 = 2.72, e 2 = 7.39, e 3 = 20.09, e 4 = 54.60 , then by Trapezoidal rule ∫0 4 x dx = [MP PET 1995, 2001]
Solution: (c)
e
(a) 53.87 (b) 53.60 (c) 58.00 (d) None of these
h = b −a = 4 − 0 =1
n 4
x: 0 1 2 3 4
y: 1 2.72 7.39 20.09 54.60
[ ]4 4 x h
2
f(x) dx = e
0
∫ ∫By Trapezoidal rule, 0 dx = (y 0 + y4 ) + 2(y1 + y2 + y3)
= 1 [(1 + 54.6) + 2(2.72 + 7.39 + 20.09)] = 1 [55.6 + 60.4] = 58.00
2 2
Example: 13 ∫By trapezoidal rule the value of the integral 5 2 dx on dividing the interval into four equal parts is [MP PET 2004]
Solution: (a)
x
Example: 14 1
Solution: (c)
(a) 42 (b) 41.3 (c) 41 (d) 40
h = 5− 1 =1
4
x: 1 2 3 4 5
y : 1 4 9 16 25
∫5 2 dx = h [(y0 + y4 ) + 2(y1 + y2 + y3 )] = 1 [(1 + 25) + 2(4 + 9 + 16)] = 1 [26 + 58] = 84 = 42
x 2 2 2 2
1
∫If for n = 4 , the approximate value of integral 9 2 dx by Trapezoidal rule is 2 1 (1 + 92) + α 2 + β 2 + 7 2 , then
2
x
1
[MP PET 2002]
(a) α = 1, β = 3 (b) α = 2, β = 4 (c) α = 3, β = 5 (d) α = 4, β = 6
h = b − a = 9 − 1 = 2
n 4
x0 = 1 , x1 = x0 + nh = 1 + 1.2 = 3 , x 2 = x0 + 2.2 = 5, x 3 = x0 + 3.2 = 7, x 4 = x0 + 4.2 = 9
y0 = 1, y1 = 9, y2 = 25, y3 = 49, y4 = 81
By trapezoidal rule,
∫b = h [(y0 ]+ y4 ) + 2(y1 + y2 + y3 ) = 2 [(1 + 81) + 2(9 + 25 + 49)] = 2 1 (1 + 92) + (3 2 + 52 + 7 2 )
f(x) dx 2 2 2
a
Obvious from above equation, α = 3, β = 5 .
(2) Simpson's one third rule : Let y = f(x) be a function defined on [a, b] which is divided into n (an even
number) equal parts each of width h so that b − a = nh.
Suppose the function y = f(x) attains values y0 , y1, y2,.......... yn at n + 1 equidistant points
x0 = a, x1 = x0 + h , x2 = x0 + 2h,..........., xn = x0 + nh = b respectively. Then
∫ ∫b f(x)dx = x0 +nh ydx = h [(y0 + yn) + 4(y1 + y3 + y5 + .... + yn−1) + 2(y2 + y4 + ... + yn−2 )]
3
a x0
= (one-third of the distance between two consecutive ordinates)
× [(sum of the extreme ordinates)+4(sum of odd ordinates)+2(sum of even ordinates)]
This formula is known as Simpson's one-third rule. Its geometric significance is that we replace the graph of
the given function by n arcs of second degree polynomials, or parabolas with vertical axes. It is to note here that
2
the interval [a, b] is divided into an even number of subinterval of equal width.
Simpson's rule yield more accurate results than the trapezoidal rule. Small size of interval gives more accuracy.
Example: 15 6 [MP PET 2004]
∫By Simpson's rule the value of the interval x dx on dividing, the interval into four equal parts is
1
(a) 16 (b) 16.5 (c) 17 (d) 17.5
Solution: (d) h = 6 −1 = 5 = 1.25
4 4
x 0 = a = 1, x1 = x 0 + h = 1 + 1.25 = 2.25
x 2 = x0 + 2h = 1 + 2(1.25) = 3.50
x 3 = x0 + 3h = 1 + 3(1.25) = 4.75
x 4 = x0 + 4h = 1 + 4(1.25) = 6.0
[ ]b 6 1.25
3
f(x)dx = xdx
a 1
∫ ∫By Simpson's rule, = (y0 + y4 ) + 4(y1 + y3 ) + 2(y2 )
= 1.25 [(1 + 6) + 4(2.25 + 4.75) + 2(3.5)] = 1.25 [7 + 28 + 7] = 17.5
3 3
Example: 16 ∫Considering six sub-interval, the value of 6 dx by Simpson's rule is
Solution: (b) 0 1+ x2
Example: 17 (a) 1.3562 (b) 1.3662 (c) 1.3456 (d) 1.2662
Solution: (a)
h = 6 − 0 = 1
6
x0 = 0, x1 = 0 + 1.1 = 1
x 2 = 0 + 2.1 = 2, x 3 = 3, x 4 = 4, x 5 = 5, x 6 = 6
and y0 = 1 = 1, y1 = 1 , y2 = 1 , y3 = 1 , y4 = 1 , y5 = 1 , y6 = 1
1+0 2 5 10 17 26 37
By Simpson's rule ,
∫b f (x)dx = h [(y 0 + y6) + 4(y1 + y3 + y5) + 2(y 2 + y4 )] = 1 1 + 1 + 4 1 + 1 + 1 + 2 1 + 1 = 1.3662
a 3 3 37 2 10 26 5 17
∫By Simpson's rule taking n = 4 , the value of the integral 11 dx is equal to [Kurukshetra CEE 1998; DCE 2001]
0 1+ x2
(d) None of these
(a) 0.785 (b) 0.788 (c) 0.781
h = 1−0 = 0.25
4
x0 = 0 x1 = 0 + 0.25 = 0.25
x 2 = 0 + 2(0.25) = 0.50, x 3 = 0.75, x 4 = 1
and y0 = 1 1 =1, y1 1 = 0.941 , y2 1 = 0.8 , y3 = 0.64 and y4 = 0.5
= 1+ 0 = 1 + (0.25)2 = 1 + (0.50)2
1 + x 2
0
∫ [ ]1 dx 0.25
3
0 1+ x2
= (y 0 + y4 ) + 4(y1 + y3 ) + 2(y2 )
= 0.25 [(1 + 0.5) + 4(0.941 + 0.64) + 2(0.8)] = 0.785 .
3
Progression
Introduction.
(1) Sequence : A sequence is a function whose domain is the set of natural numbers, N.
If f : N → C is a sequence, we usually denote it by < f(n) > = < f(1), f(2), f(3),.... >
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. Terms of a sequence are connected by commas. Example : 1, 1, 2, 3, 5,
8, …………. is a sequence.
(2) Series : By adding or subtracting the terms of a sequence, we get a series.
If t1, t2, t3 ,..... tn,..... is a sequence, then the expression t1 + t2 + t3 + ..... + tn.... is a series.
A series is finite or infinite as the number of terms in the corresponding sequence is finite or infinite.
Example : 1+ 1 + 1 + 1 + 1 + .... is a series.
2 3 4 5
(3) Progression : A progression is a sequence whose terms follow a certain pattern i.e. the terms are
arranged under a definite rule.
Example : 1, 3, 5, 7, 9, …….. is a progression whose terms are obtained by the rule : Tn = 2n − 1, where Tn
denotes the nth term of the progression.
Progression is mainly of three types : Arithmetic progression, Geometric progression and Harmonic
progression.
However, here we have classified the study of progression into five parts as :
• Arithmetic progression
• Geometric progression
• Arithmetico-geometric progression
• Harmonic progression
• Miscellaneous progressions
Arithmetic progression(A.P)
Definition.
A sequence of numbers < tn > is said to be in arithmetic progression (A.P.) when the difference tn − tn−1 is a
constant for all n ∈ N. This constant is called the common difference of the A.P., and is usually denoted by the letter d.
If ‘a’ is the first term and ‘d’ the common difference, then an A.P. can be represented as
a, a + d, a + 2d, a + 3d,........
Example : 2, 7, 12, 17, 22, …… is an A.P. whose first term is 2 and common difference 5.
Algorithm to determine whether a sequence is an A.P. or not.
Step I: Obtain an (the nth term of the sequence).
Step II: Replace n by n – 1 in an to get an−1 .
Step III: Calculate an − an−1 .
If an − an−1 is independent of n, the given sequence is an A.P. otherwise it is not an A.P. An arithmetic
progression is a linear function with domain as the set of natural numbers N.
∴ tn = An + B represents the nth term of an A.P. with common difference A.
General Term of an A.P..
(1) Let ‘a’ be the first term and ‘d’ be the common difference of an A.P. Then its nth term is a + (n − 1)d .
Tn = a + (n − 1)d
(2) pth term of an A.P. from the end : Let ‘a’ be the first term and ‘d’ be the common difference of an A.P.
having n terms. Then pth term from the end is (n − p + 1)th term from the beginning.
p th term from the end = T(n−p+1) = a + (n − p)d
Important Tips
• General term (Tn) is also denoted by l (last term).
• Common difference can be zero, +ve or –ve.
• n (number of terms) always belongs to set of natural numbers.
• If Tk and Tp of any A.P. are given, then formula for obtaining Tn is Tn − Tk = Tp − Tk .
n−k p−k
• If pTp = qTq of an A.P., then Tp + q = 0.
• If pth term of an A.P. is q and the qth term is p, then Tp + q = 0 and Tn = p + q – n.
• If the pth term of an A.P. is 1 and the qth term is 1 , then its pqth term is 1.
q p
• If Tn =pn + q, then it will form an A.P. of common difference p and first term p + q.
Example: 1 Let Tr be rth term of an A.P. whose first term is a and common difference is d. If for some positive integers m, n, m ≠ n,
Solution: (d) Tm = 1 and Tn = 1 , then a – d equals [AIEEE 2004]
n m
Example: 2
Solution: (c) (a) 11 (b) 1 (c) 1 (d) 0
Example: 3 m+n mn
Solution: (a)
Tm = 1 ⇒ a + (m − 1) d = 1 …..(i)
n n
and Tn = 1 ⇒ a + (n − 1) d = 1 …..(ii)
m m
Subtract (ii) from (i), we get (m − n) d = 11 ⇒ (m − n) d = (m − n) ⇒ d= 1 , as m – n ≠ 0
n−m mn mn
a = 1 − (n − 1) d = 1 − n−1 = 1 = d . Therefore a – d = 0
m m mn mn
The 19th term from the end of the series 2 + 6 + 10 + …. + 86 is
(a) 6 (b) 18 (c) 14 (d) 10
86 = 2 + (n − 1) 4 ⇒ n = 22
19th term from end = tn−19+1 = t22−19+1 = t4 = 2 + (4 − 1) 4 = 14 [AMU 1991]
In a certain A.P., 5 times the 5th term is equal to 8 times the 8th term, then its 13th term is
(a) 0 (b) – 1 (c) – 12 (d) – 13
We have 5 T5 = 8 T8
Let a and d be the first term and common difference respectively
∴ 5{a + (5 − 1)d} = 8{a + (8 − 1) d}
⇒ 3a + 36d = 0 ⇒ a + 12d = 0 , i.e. a + (13 − 1) d = 0 . Hence 13th term = 0
Example: 4 If 7th and 13th term of an A.P. be 34 and 64 respectively, then its 18th term is
Solution: (c)
(a) 87 (b) 88 (c) 89 (d) 90
Example: 5
Solution: (c) Let a be the first term and d be the common difference of the given A.P., then
T7 = 34 ⇒ a + 6d = 34 …..(i)
T13 = 64 ⇒ a + 12d = 64 …..(ii)
From (i) and (ii), d = 5, a = 4
∴ T18 = a + 17d = 4 + 17 × 5 = 89
Trick: Tn − Tk = Tp − Tk ⇒ T18 − T7 = T13 − T7 ⇒ T18 − 34 = 64 − 34 ⇒ T18 = 89
n−k p−k 18 − 7 13 − 7 11 6
am an ap
If < an > is an arithmetic sequence, then ∆ = m n p equals
111
(a) 1 (b) –1 (c) 0 (d) None of these
Let a be the first term and d the common difference. Then ar = a + (r − 1) d
a + (m − 1) d a + (n − 1)d a + (p − 1) d a a a m − 1 n − 1 p − 1
∆= m n p = m n p +d m n p
1 1 1 1 11 1 11
1 11 mn p
= a m n p +d m n p = a.0 + d.0 = 0
1 11 1 11
Example: 6 The nth term of the series 3 + 10 + 17 + ….. and 63 + 65 + 67 + …… are equal, then the value of n is
Solution: (c)
[Kerala (Engg.) 2002]
(a) 11 (b) 12 (c) 13 (d) 15
nth term of 1st series = 3 + (n − 1)7 = 7n − 4
nth term of 2nd series = 63 + (n − 1) = 2n + 61
∴ we have, 7n − 4 = 2n + 61 ⇒ n = 13
Selection of Terms in an A.P..
When the sum is given, the following way is adopted in selecting certain number of terms :
Number of terms Terms to be taken
3 a – d, a, a + d
4 a – 3d, a – d, a + d, a + 3d
5 a – 2d, a – d, a, a + d, a + 2d
In general, we take a – rd, a – (r – 1)d, ……., a – d, a, a + d, ……, a + (r – 1)d, a + rd, in case we have to
take (2r + 1) terms (i.e. odd number of terms) in an A.P.
And, a − (2r − 1)d,a − (2r − 3)d,......., a − d,a + d,......., a + (2r − 1)d , in case we have to take 2r terms in an A.P.
When the sum is not given, then the following way is adopted in selection of terms.
Number of terms Terms to be taken
3 a, a + d, a + 2d
4 a, a + d, a + 2d, a + 3d
5 a, a + d, a + 2d, a + 3d, a + 4d
Sum of n terms of an A.P. : The sum of n terms of the series a + (a + d) + (a + 2d) + ....... + {a + (n − 1)d} is
given by Sn = n [2a + (n − 1)d]
2
Also, Sn = n (a + l) , where l = last term = a + (n − 1)d
2
Important Tips
• The common difference of an A.P is given by d = S2 − 2S1 where S2 is the sum of first two terms and S1 is the sum of first term or
the first term.
• The sum of infinite terms = ∞, when d > 0 .
− ∞, when d < 0
• If sum of n terms Sn is given then general term Tn = Sn − Sn−1 , where Sn−1 is sum of (n – 1) terms of A.P.
• Sum of n terms of an A.P. is of the form An2 + Bn i.e. a quadratic expression in n, in such case, common difference is twice the
coefficient of n2 i.e. 2A.
•• If for the different A.P’s Sn = fn , then Tn = f (2n − 1)
Sn′ φn Tn′ φ(2n − 1)
A n + 1 + B
2
• If for two A.P.’s Tn = An + B then Sn =
Tn′ Cn + D Sn′ C n + 1
2 + D
• Some standard results
∑• n n (n + 1)
Sum of first n natural numbers = 1 + 2 + 3 + ........ + n = r =1 r = 2
n
∑• Sum of first n odd natural numbers = 1 + 3 + 5 + ..... + (2n − 1) = (2r − 1) = n2
r =1
n
∑• Sum of first n even natural numbers = 2 + 4 + 6 + ...... + 2n = 2r = n (n + 1)
r =1
• If for an A.P. sum of p terms is q and sum of q terms is p, then sum of (p + q) terms is {–(p + q)}.
• If for an A.P., sum of p terms is equal to sum of q terms, then sum of (p + q) terms is zero.
• If the pth term of an A.P. is 1 and qth term is 1 , then sum of pq terms is given by Spq = 1 (pq + 1)
q p 2
Example: 7 7th term of an A.P. is 40, then the sum of first 13 terms is [Karnataka CET 2003]
Solution: (b)
Example: 8 (a) 53 (b) 520 (c) 1040 (d) 2080
Solution: (a)
Example: 9 S13 = 13 {2a + 12d} = 13{a + 6d} = 13 × T7 = 13 × 40 = 520
2
Solution: (c)
The first term of an A.P. is 2 and common difference is 4. The sum of its 40 terms will be [MNR 1978; MP PET 2002]
(a) 3200 (b) 1600 (c) 200 (d) 2800
S = n [2a + (n − 1) d] = 40 [2 × 2 + (40 − 1)4] = 3200
2 2
The sum of the first and third term of an A.P. is 12 and the product of first and second term is 24, the first term is
[MP PET 2003]
(a) 1 (b) 8 (c) 4 (d) 6
Let a − d, a, a + d, ........ be an A.P.
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