Practical 1: Graphs of Elementary Functionsmathskthm.6te.net/F.Y.B.Sc. Calculus Practical.pdfDepartment of Mathematics, K.T.H.M. College, Nashik F.Y.B.Sc. Calculus Practical (Academic

Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)

Practical 1: Graphs of Elementary Functions

1. a) Graph of y1 = −f(x) ≡ mirror image of Graph of y = f(x) about X axisb) Graph of y2 = f(−x) ≡mirror image of Graph of y = f(x) about Y axisc) Graph of y3 = f(x− a) ≡ Graph of y = f(x) displaced along X axis by amount ad) Graph of y4 = f(x) + b ≡ Graph of y = f(x) displaced along Y axis by amount b

2. Graph of f−1(x) ≡ mirror image of Graph of f(x) about the line y = x

F.Y.B.Sc. Calculus Practical, Page: 1










Practical 2: Real Numbers

1. If 0 < a < b, show thata) a <

√ab < b b) 1

b< 1

a

Ans:

a) 0 < a < b=⇒ a.a < a.b and a.b < b.b=⇒ a2 < ab and ab < b2

=⇒ a <√ab and

√ab < b

=⇒ a <√ab < b

b) 0 < a < b=⇒ 1

aba < 1

abb (∵ ab > 0)

=⇒ 1b< 1

a

2. Find all x ∈ R satisfying the following inequalities.a) |x− 1| > |x+ 1| b) |x|+ |x+ 1| < 2

Ans:

a) |x− 1| > |x+ 1|⇐⇒ |x− 1| − |x+ 1| > 0 . . . (I)End points are −1 and 1.I Case: x ≤ −1In this case x− 1 ≤ −2 < 0 and x+ 1 ≤ 0. So from (I),−(x− 1)− (−(x+ 1)) > 0=⇒ 2 > 0This is true. So all x ≤ −1 satisfy (I).The solution set in I case is (−∞,−1]II Case: −1 < x ≤ 1In this case, x− 1 < 0 and x+ 1 > 0. So from (I),−(x− 1)− (x+ 1) > 0=⇒ −2x > 0=⇒ x < 0The solution set in II case is (−1, 0)III Case: x ≥ 1In this case x− 1 > 0 and x+ 1 > 2 > 0. So from (I),(x− 1)− (x+ 1) > 0=⇒ −2 > 0. This is impossible.The solution set in III case is φ

Solution Set = (−∞,−1] ∪ (−1, 0) ∪ φ = (−∞, 0)

b) |x|+ |x+ 1| < 2 . . . (II)End points are 0 and −1.I Case: x < −1



In this case x < 0 and x+ 1 ≤ 0. So from (II),−x− (x+ 1) < 2=⇒ −2x− 1 < 2=⇒ −2x < 1=⇒ x > −1

2

This is impossible as x < −1The solution set in I case is φ.II Case: −1 ≤ x < 0In this case x < 0 and x+ 1 ≥ 0. So from (II),−x+ (x+ 1) < 2=⇒ 1 < 2This is true.The solution set in II case is [−1, 0).III Case: x ≥ 0In this case x ≥ 0 and x+ 1 > 0. So from (II),x+ (x+ 1) < 2=⇒ 2x+ 1 < 2=⇒ x < 1

2

The solution set in III case is [0, 12).

Solution Set = φ ∪ [−1, 0) ∪ [0, 12) = [−1, 1

2)

3. Prove that the following are not rational numbers.a)√

3 b)√

3 +√

5

Ans:

a) Suppose on contrary that√

3 is rational. Take√

3 = pq, where p and q are integers

and p, q do not have any common factor. Consider√3 = p

q

=⇒ 3 = p2

q2

=⇒ p2 = 3q2

=⇒ 3 divides p (∵ 3 is prime)=⇒ p = 3k (k is some integer)=⇒ q2 = 3k2

=⇒ 3 divides q (∵ 3 is prime)Thus 3 is a common factor of p and q. This contradicts the fact that p, q do not haveany common factor. Hence

√3 must be irrational.

b) Consider√3 +√

5 is rational√3−√

5 is rational (∵√

3−√

5 = −2√3+√5)

=⇒ (√

3 +√

5) + (√

3−√

5) is rational=⇒ 2

√3 is rational

=⇒√

3 is rationalThis is a contradiction. Hence

√3 +√

5 is irrational.

4. Let K = {s+ t√

2 : s, t ∈ Q}. Show that K satisfies the following conditions.



a) If x1, x2 ∈ K, then x1 + x2 ∈ Kb) If x 6= 0 is in K, then 1

x∈ K

Ans:

a) Take x1 = s1 + t1√

2, x2 = s2 + t2√

2 in K where, s1, s2, t1, t2 ∈ Q. Thenx1 + x2 = (s1 + s2) + (t1 + t2)

√2 = s+ t

√2, where s = s1 + s2, t = t1 + t2 ∈ Q.

This proves that x1 + x2 ∈ K.

b) Take x = s+ t√

2 ∈ K, where s, t ∈ Q. Then1x

= 1s+t√2

= 1s+t√2× (s−t

√2)

(s−t√2)

= ss2−2t2 −

ts2−2t2

√2 = s′ + t′

√2

where s′ = ss2−2t2 and t′ = −t

s2−2t2 are in Q.

This proves that 1x∈ K.

5. Find supremum and infimum of the following sets.a) { 1

n: n ∈ N} b) {−1

n: n ∈ N} c) {3 + (2

3)n : n ∈ N} d) {(1

2)n : n ∈ N}

Ans:a) infimum of { 1

n: n ∈ N} = 0 and supremum of { 1

n: n ∈ N} = 1

b) infimum of {−1n

: n ∈ N} = −1 and supremum of {−1n

: n ∈ N} = 0c) infimum of {3+(2

3)n : n ∈ N} = 3 and supremum of {3+(2

3)n : n ∈ N} = 3+ 2

3= 11

3

d)infimum of {(12)n : n ∈ N} = 0 and supremum of {(1

2)n : n ∈ N} = 1

2

6. If L ∈ R, L < M + ε for every ε > 0, prove L ≤M .

Ans: Suppose on contrary that L > M . Take ε = L−M > 0. By hypothesis,

L < M + ε

=⇒ L < M + (L−M)

=⇒ L < L

This is a contradiction. Hence L ≤M .

7. If L ∈ R, L ≤M + ε for every ε > 0, prove L ≤M .

Ans: Suppose on contrary that L > M . Take ε = L−M2

> 0. By hypothesis,

L < M + ε

=⇒ L < M + L−M2

=⇒ L−M < L−M2

=⇒ 1 < 12

This is a contradiction. Hence L ≤M .



Practical 3: Limits and Continuity

1. a) Using ε− δ definition of limit, show that limx→1

(2x+ 2) = 4.

b) Show that limx→0

sin( 1x) does not exist.

Ans:

a) Let ε > 0. Take δ = ε2. Consider

0 < |x− 1| < δ

=⇒ |x− 1| < ε2

=⇒ 2|x− 1| < ε

=⇒ |2x− 2| < ε

=⇒ |(2x+ 2)− 4| < ε

=⇒ |f(x)− 4| < ε

Thus

0 < |x− 1| < δ =⇒ |f(x)− 4| < ε.

This proves that limx→1

(2x+ 2) = 4.

b) Take ε = 1 and δ be any positive real number. Select a positive integer n such that

1(2nπ−π

2)< δ. Take y = 1

(2nπ−π2)

and y′ = 1(2nπ+π

2). Then

−δ < y′ < y < δ. i.e. 0 < |y − 0| < δ, 0 < |y′ − 0| < δ

Also

|f(y)− f(y′)| = |1− (−1)| = 2 > 1 = ε

So

0 < |x− 0| < δ =⇒ |f(x)− f(0)| < ε is not possible for any δ > 0 .

This proves that limx→0

sin( 1x) does not exist.

2. If exists, evaluate the following limits.

a) limx→−1

x+52x+3

b) limx→0

|x|+2x3x−5|x| c) lim

x→0

sin( 1x)

1x

d) limx→1

√x−1x−1

Ans:

a) limx→−1

x+52x+3

= −1+52(−1)+3

= 41

= 4



b) limx→0+

|x|+2x3x−5|x| = lim

x→0+

x+2x3x−5x = lim

x→0+

3x−2x = −3

2

and

limx→0−

|x|+2x3x−5|x| = lim

x→0−−x+2x3x+5x

= limx→0+

x8x

= 18

So limx→0+

|x|+2x3x−5|x| 6= lim

x→0−|x|+2x3x−5|x| and lim

x→0

|x|+2x3x−5|x| does not exist.

c) Consider

−1 ≤ sin 1x≤ 1

=⇒ −x ≤ x sin 1x≤ x if x ≥ 0 and −x ≥ x sin 1

x≥ x if x ≤ 0

=⇒ limx→0+

−x ≤ limx→0+

x sin 1x≤ lim

x→0+x and lim

x→0−−x ≥ lim

x→0−x sin 1

x≥ lim

x→0−x

=⇒ 0 ≤ limx→0+

x sin 1x≤ 0 and 0 ≥ lim

x→0−x sin 1

x≥ 0

=⇒ limx→0+

x sin 1x

= limx→0−

x sin 1x

= 0

=⇒ limx→0

x sin 1x

= 0

=⇒ limx→0

sin 1x

1x

= 0

d) Consider

limx→1

√x−1x−1 = lim

x→1

√x−1x−1 ×

√x+1√x+1

= limx→1

x−1(x−1)(

√x+1)

= limx→1

1√x+1

= 1√1+1

= 12

3. a) Show that limx→0

e1x−1e1x+1

does not exist.

b) Discuss the continuity of the function χ : R→ R defined as

χ(x) =

{1 if x ∈ Q0 if x 6∈ Q

Ans:

a) limx→0+

e1x−1e1x+1

= limx→0+

e1x

(1− 1

e1x

)e1x

(1+ 1

e1x

) = limx→0+

(1− 1

e1x

)(1+ 1

e1x

) = limx→0+

1−01+0

= 1

limx→0−

e1x−1e1x+1

= 0−10+1

= −1

So limx→0+

e1x−1e1x+16= lim

x→0−e1x−1e1x+1

and limx→0

e1x−1e1x+1

does not exist.

b) Let c ∈ R. Take ε = 12> 0 and δ be any positive real number. Take a rational r

and irrational s in the interval (c− δ, c+ δ)− {c} (i.e. in deleted δ neighborhood ofc. Then|χ(r)− χ(s)| = |1− 0| > 1

2= ε.

So0 < |x− c| < δ =⇒ |χ(x)− χ(c)| < ε is not possible for any δ > 0.



This proves that limx→c

χ(x) does not exist. Therefore χ is discontinuous at c. Since c

was arbitrary point of R, χ is discontinuous at all points of R.

4. Discuss the continuity of the following functions.

a) f(x) =

2x− 1 if x ≤ 1x2 if 1 < x < 23x− 4 if 2 ≤ x < 4

x32 if x ≥ 4

b) f(x) =

e1x2

1−e1x2

if x 6= 0

1 if x = 0

c) f(x) =√

x−1x+3

for all x ∈ R

Ans:

a) Possible points of discontinuity are 1,2 and 4.i) c = 1 :

limx→1−

f(x) = limx→1−

(2x− 1) = 2(1)− 1 = 1

limx→1+

f(x) = limx→1+

(x2) = 1

So limx→1−

f(x) = limx→1+

f(x) = limx→1

f(x) = 1 = f(1).

f is continuous at 1.

ii) c = 2 :

limx→2−

f(x) = limx→2−

(x2) = 22 = 4

limx→2+

f(x) = limx→2+

(3x− 4) = 3(2)− 4 = 2

So limx→2−

f(x) 6= limx→2+

f(x)

limx→2

f(x) does not exist.

f is discontinuous at 2.

iii) c = 4 :

limx→4−

f(x) = limx→4−

(3x− 4) = 3(4)− 4 = 8

limx→4+

f(x) = limx→4+

(x32 ) = 4

32 = 8

So limx→4−

f(x) = limx→4+

f(x) = limx→4

f(x) = 8 = f(4).

f is continuous at 4.

f is continuous at all points of R except at c = 2.

b) Possible point of discontinuity: 0.

limx→0

e1x2

1−e1x2

= 01−0 = 0 6= 1 = f(1)



So f is discontinuous at c =0.

f is continuous at all points of R except at c = 0.

c) By the algebra of continuous functions, f(x) will be continuous at all points atwhich it is defined. It is defined only when x+ 3 6= 0 and x−1

x+3≥ 0.

I Condition: x+ 3 6= 0In this case x 6= −3

II Conditiion x−1x+3≥ 0

In this case,(x− 1) ≥ 0 and x+ 3 ≥ 0 OR x− 1 <≤ 0 and x+ 3 ≤ 0=⇒ x ≥ 1 and x ≥ −3 OR x ≤ 1 and x ≤ −3=⇒ x ≥ 1 OR x ≤ −3

The function f(x) is continuous on (−∞,−3) ∪ [1,∞)

5. Determine the set of points of discontinuity of the following functions.a) f(x) = x2−2

x2+1, b) f(x) = x2+2x+3

x+2

Ans:

a) f(x) = x2−2x2+1

is a rational function (ratio of two polynomials). It is discontinuousonly where denominator vanishes. SoPoints of discontinuity of f(x) = {x : x2 + 1 = 0} = φ.The function f(x) is continuous at all points of R.

b) f(x) = x2+2x+3x+2

is a rational function (ratio of two polynomials). It is discontinuousonly where denominator vanishes. SoPoints of discontinuity of f(x) = {x : x+ 2 = 0} = {−2}.The function f(x) is continuous at all points of R except at x = −2.

6. If function f(x) is continuous on [−2, 2] where

f(x) =

sin axx

+ 2 if − 2 ≤ x < 03x+ 5 if 0 ≤ x ≤ 1√x2 + 8− b if 1 < x ≤ 2

Show that a+ b+ 2 = 0.

Ans: f(x) is continuous at x = 0

=⇒ limx→0−

f(x) = f(0)

=⇒ limx→0

(sin axx

+ 2)

= 3(0) + 5

=⇒ limx→0

(sin axax× a)

+ 2 = 5

=⇒ 1× a+ 2 = 5

=⇒ a+ 2 = 5



=⇒ a = 3

Also

f(x) is continuous at x = 1

=⇒ limx→1+

f(x) = f(1)

=⇒√

12 + 8− b = 3(1) + 5

=⇒√

9− b = 8

=⇒ 3− b = 8

=⇒ b = −5

Thus a = 3 and b = −5. Therefore a+ b+ 2 = 3 + (−5) + 2 = 0.

7. Show that the equation 2xx− 1 = 0 has at least one root in (0, 1).

Ans: Take f(x) = 2xx− 1 = 0 on [0, 1]. Then

f(0) = 20(0)− 1 = −1 < 0 and

f(1) = 21(1)− 1 = 2− 1 = 1 > 0

Therefore by Bolzano’s Theorem, there is at least one c ∈ (0, 1) such that f(c) = 0.This implies that

2cc− 1 = 0

=⇒ c is root of 2xx− 1 = 0 in (0, 1).

8. Show that the equation cos x = x has a solution in the interval [0, π2].

Ans: Take f(x) = cos x− x on [0, π2]. Then

f(0) = cos 0− 0 = 1− 0 = 1 > 0 and

f(π2) = cos(π

2)− π

2= 0− π

2= −π

2< 0

Therefore by Bolzano’s Theorem, there is at least one c ∈ (0, π2) such that f(c) = 0.

This implies that

cos c− c = 0

=⇒ c is solution of cosx = x in [0, π2].



Practical 4: Differentiation

1. Use the definition of derivative, to find derivative of the following functions.a) f(x) = x3 b) f(x) = 1

x

Ans: a) f ′(x) = limh→0

f(x+h)−f(x)h

= limh→0

(x+h)3−x3h

= limh→0

(x3+3x2h+3xh2+h3)−x3h

= limh→0

3x2h+3xh2+h3

h

= limh→0

(3x2 + 3xh+ h2)

= 3x2

f ′(x) = 3x2

Ans: b) f ′(x) = limh→0

f(x+h)−f(x)h

= limh→0

1x+h− 1x

h

= limh→0

x−(x+h)x(x+h)h

= limh→0

−hx2h+h2

= limh→0

−1x2+h

= −1x2

f ′(x) = −1x2

2. a) If f(x) = |x|, show that f ′(x) = 1 when x > 0 and f(x) = −1 when x < 0. Alsoshow that the function f(x) is not differentiable at x = 0 .b) If f(x) = x|x| (∀x ∈ R), show that f ′′(x) = 2 when x > 0 and f”(x) = −2 whenx < 0. Also show that the function f(x) is differentiable at x = 0.

c) If f(x) =

{1− x if x ≤ 1(1− x)2 if x > 1

,

show that f ′+(1) = 0 and f ′−(1) = −1

Ans:

a) f(x) = |x| = x if x > 0 implying that f ′(x) = 1 when x > 0f(x) = |x| = −x if x < 0 implying that f ′(x) = −1 when x < 0

limh→0+

f(0+h)−f(0)h

= limh→0+

h−0h

= limh→0+

1 = 1

andlimh→0−

f(0+h)−f(0)h

= limh→0−

−h−0h

= limh→0−

−1 = −1

So limh→0−

f(0+h)−f(0)h

6= limh→0+

f(0+h)−f(0)h

and limh→0

f(0+h)−f(0)h

does not exist. This proves

that f(x) is not differentiable at x = 0.

b) f(x) = x|x| = xx = x2 if x > 0 implying that f ′(x) = 2x and f”(x) = 2 whenx > 0f(x) = x|x| = x(−x) = −x2 if x > 0 implying that f ′(x) = −2x and f”(x) = −2when x < 0

limh→0+

f(0+h)−f(0)h

= limh→0+

h2−0h

= limh→0+

h = 0

limh→0−

f(0+h)−f(0)h

= limh→0−

−h2−0h

= limh→0+

−h = 0

So limh→0−

f(0+h)−f(0)h

= limh→0+

f(0+h)−f(0)h

= 0 and limh→0

f(0+h)−f(0)h

exists and equals to

0. This proves that f(x) is differentiable at x = 0 and f ′(0) = 0

c) f ′+(1) = limh→0+

f(1+h)−f(1)h

= limh→0+

[1−(1+h)]2−0h

= limh→0+

h2

h= lim

h→0+h = 0



andf ′−(1) = lim

h→0−f(1+h)−f(1)

h= lim

h→0−[1−(1+h)]−0

h= lim

h→0−−hh

= limh→0−

−1 = −1

3. Test whether the conditions of Rolles Mean Value Theorem hold for given functionson the interval and if so find the value of c.a) f(x) = log

(x2+ab(a+b)x

)on [a, b] b) f(x) = (x− a)m(x− b)n on [a, b]

Ans:

a) f(x) is continuous on [a, b] and f(x) is differentiable in (a, b). Also

f(a) = log(a2+ab(a+b)a

)= log

(a2+ab

(a2+ab)x

)= log 1 = 0

and

f(a) = log(b2+ab(a+b)b

)= log

(b2+ab(a+b2)x

)= log 1 = 0

The hypothesis of Rolle’s Mean Value theorem is satisfied. To find c, consider

f ′(c) = 0

=⇒ 1(c2+ab)(a+b)c

(2c) = 0

=⇒ c2 + ab = 0

=⇒ c =√ab

b) f(x) is continuous on [a, b] and f(x) is differentiable in (a, b). Also

f(a) = (a− a)m(a− b)n = 0

and

f(a) = (b− a)m(b− b)n = 0

The hypothesis of Rolle’s Mean Value theorem is satisfied. To find c, consider

f ′(c) = 0

=⇒ m(c− a)m−1(c− b)n + n(c− a)m(c− b)n−1 = 0

=⇒ (c− a)m−1(c− b)n−1(mc−mb+ nc− na) = 0

=⇒ mc−mb+ nc− na = 0

=⇒ c(m+ n)− (mb+ na) = 0



=⇒ c = mb+nam+n

4. Test whether the conditions of Lagranges Mean Value Theorem hold for given func-tions on the interval and if so find the value of c.

a) f(x) = (x− 1)(x− 2)(x− 3) on [0, 4] b) f(x) = tan−1 x on [0, 1]

Ans:

a) f(x) is continuous on [0, 4] and f(x) is differentiable in (0, 4). So the hypothesisof Lagrange’s Mean Value Theorem is satisfied. To find value of c, consider

f ′(c) = f(4)−f(0)4−0

=⇒ (c− 2)(c− 3) + (c− 1)(c− 3) + (c− 1)(c− 2) = −6−64

=⇒ 3c2 − 12c+ 11 = −3

=⇒ 3c2 − 12c+ 8 = 0

=⇒ c = 2 + 2√3

3or c = 2− 2

√3

3

=⇒ c = 3.1547 or c = 0.8453

b) f(x) is continuous on [0, 1] and f(x) is differentiable in (0, 1). So the hypothesisof Lagrange’s Mean Value Theorem is satisfied. To find value of c, consider

f ′(c) = f(1)−f(0)1−0

=⇒ 11+c2

=π4−01

=⇒ 11+c2

= π4

=⇒ 1 + c2 = 4π

=⇒ c = ±√

4π− 1

=⇒ c =√

4π− 1 (∵ c ∈ (0, 1))

5. a) Show that the value of c when Cauchy’s Mean Value Theorem is applied to thefunctions f(x) = sinx and g(x) = cos x on [a, b] is the arithmetic mean of a and b.b) Show that the value of c when Cauchy’s Mean Value Theorem is applied to the

functions f(x) =√x and g(x) =

√1x

on [a, b] is the geometric mean of a and b.

Ans:

a) The functions f(x) = sin x and g(x) = cosx are continuous on [a, b] and differen-tiable in (a, b). So the hypothesis of Cauchy’s Mean Value Theorem is satisfied. To



find c, consider

f ′(c)g′(c)

= f(b)−f(a)g(b)−g(a)

=⇒ cos c− sin c

= sin b−sin acos b−cos a

=⇒ − cot c =2 sin( b−a

2) cos( b+a

2)

−2 sin( b−a2

) sin( b+a2

)

=⇒ − cot c = − cot (b+a)2

=⇒ c = b+a2

=⇒ c is arithmetic mean of a and b.

b) The functions f(x) =√x and g(x) =

√1x

are continuous on [a, b] and differentiable

in (a, b). So the hypothesis of Cauchy’s Mean Value Theorem is satisfied. To find c,consider

f ′(c)g′(c)

= f(b)−f(a)g(b)−g(a)

=⇒12c−

12

− 12c−

32

=√b−√a√

1b−√

1a

=⇒ −c = −√ba(√

b−√a√

a−√b

)=⇒ c =

√ab

=⇒ c is geometric mean of a and b

6. Show that the function f(x) = x3 − 3x2 + 3x− 100 is increasing on R.

Ans: f(x) = x3 − 3x2 + 3x− 100

=⇒ f ′(x) = 3x2 − 6x+ 3

=⇒ f ′(x) = 3(x− 1)2

=⇒ f ′(x) > 0 for all x ∈ R except for x = 1.

=⇒ f(x) is increasing on R.

7. Find values of x for which the function f(x) = 2x3 − 9x2 + 12x+ 2 is decreasing.

Ans: f(x) = 2x3 − 9x2 + 12x+ 2

=⇒ f ′(x) = 6x2 − 18x+ 12



=⇒ f ′(x) = 6(x− 1)(x− 2)

Therefore f ′(x) < 0

=⇒ 6(x− 1)(x− 2) < 0

=⇒ x− 1 < 0 and x− 2 > 0 OR x− 1 > 0 and x− 2 < 0

=⇒ x < 1 and x > 2 OR x > 1 and x < 2

=⇒ x > 1 and x < 2

=⇒ 1 < x < 2

=⇒ x ∈ (1, 2)

So f(x) is decreasing for x ∈ (1, 2).

8. Differentiate log(1 + x2) with respect to tan−1 x

Ans: Take u = log(1 + x2) and v = tan−1 x. Then

dudx

= 11+x2× 2x = 2x

1+x2

and

dvdx

= 11+x2

So,

dudv

= dudx× dx

dv= 2x

1+x2× 1+x2

1= 2x

9. Differentiate tan−1(cosx+sinxcosx−sinx

)w.r.t x

Ans: Take y = tan−1(cosx+sinxcosx−sinx

). Then

y = tan−1(1+tanx1−tanx

)=⇒ y = tan−1

(tan(π

4+ x)

)=⇒ y = π

4+ x

=⇒ dydx

= 1



Practical 5: Integration

1. Evaluate the following integrals.a)∫

x2+1(x2−1)(2x+1)

dx b)∫x3+4x2−7x+5

x+2dx

Ans: a) x2+1(x2−1)(2x+1)

= x2+1(x+1)(x−1)(2x+1)

Partial fractions:

x2+1(x2−1)(2x+1)

= A(x+1)

+ B(x−1) + C

(2x+1). . . (I)

x2 + 1 = A(x− 1)(2x+ 1) +B(x+ 1)(2x+ 1) + C(x− 1)(2x+ 1) . . . (II)

Put x = −1 in (II), 2 = A(−1− 1)(2(−1) + 1) =⇒ A = 1

Put x = 1 in (II), 2 = B(1 + 1)(2(1) + 1) =⇒ B = 13

Put x = −12

in (II), 54

= C(−12

+ 1)(−12− 1) =⇒ C = −5

3

Substituting these values of A,B and C in (I), we get

x2+1(x2−1)(2x+1)

= 1(x+1)

+ 13

1(x−1) −

53

1(2x+1)

Therefore∫x2+1

(x2−1)(2x+1)dx = log |x+ 1|+ 1

3log |x− 1| − 5

6log |2x+ 1|+ c

b) x2 + 2x− 11

x+ 2)

x3 + 4x2 − 7x + 5− x3 − 2x2

2x2 − 7x− 2x2 − 4x

− 11x + 511x + 22

27

Therefore

x3 + 4x2 − 7x+ 5 = (x+ 2)(x2 + 2x− 11) + 27

=⇒ x3+4x2−7x+5x+2

= (x2 + 2x− 11) + 27x+2∫

x3+4x2−7x+5x+2

dx = x3

3+ x2 − 11x+ 27 log |x+ 2|+ c


1x3−1dx b)

∫x2

x4+5x2+6dx



Ans: a) 1(x3−1) = 1

(x−1)(x2+x+1)

Partial Fractions are:

1(x−1)(x2+x+1)

= A(x−1) + Bx+C

x2+x+1. . . (I)

=⇒ 1 = A(x2 + x+ 1) + (Bx+ c)(x− 1)

=⇒ 1 = A(x2 + x+ 1) + (Bx2 −Bx+ Cx− C)

=⇒ 1 = (A+B)x2 + (A−B + C)x+ (A− C)

Equating coefficients of equal powers of x

A+B = 0

A−B + C = 0

A− C = 1

From first and last equation, B = −A and C = A−1. Using these in middle equation,A+A+A−1 = 0. Therefore A = 1

3. So B = −A = −1

3and C = A−1 = 1

3−1 = −2

3.

Substituting these values of A,B and C in (I), we get

1(x−1)(x2+x+1)

= 13

1(x−1) −

13

x(x2+x+1)

− 23

1(x2+x+1)

=⇒ 1(x−1)(x2+x+1)

= 13

1(x−1) −

16(2x+1)−1(x2+x+1)

− 23

1(x2+x+1)

=⇒ 1(x−1)(x2+x+1)

= 13

1(x−1) −

16

2x+1(x2+x+1)

− 12

1(x2+x+1)

=⇒ 1(x3−1) = 1

31

(x−1) −16

2x+1(x2+x+1)

− 12

1

(x+ 12)2+(

√32)2

=⇒∫

1x3−1dx = 1

3log |x− 1| − 1

6log |x2 + x+ 1| − 1√

3tan−1(2x+1√

3) + c

b) Put x2 = t. Then x2

x4+5x2+6= t

(t+2)(t+3)

The partial fractions are

t(t+2)(t+3)

= A(t+2)

+ B(t+3)

. . . (I)

t = A(t+ 3) +B(t+ 2) . . . (II)

Put t = −2 in (II). −2 = A(−2 + 3) =⇒ A = −2

Put t = −3 in (II). −3 = B(−3 + 2) =⇒ B = 3

Substituting these values in (I)



t(t+2)(t+3)

= −2(t+2)

+ 3(t+3)

=⇒ x2

x4+5x2+6= −2

x2+2+ 3

x2+3

=⇒∫

x2

x4+5x2+6dx = −2√

2tan−1( x√

2) + 3√

3tan−1( x√

3) + c

=⇒∫

x2

x4+5x2+6dx =

√3 tan−1( x√

3)−√

2 tan−1( x√2) + c


1sin(x−a) cos(x−b)dx b)

∫4ex+6e−x

9ex−4e−xdx

Ans: a) Put k =cos(a− b). Consider

1sin(x−a) cos(x−b)

= 1cos(a−b) [

cos(a−b)sin(x−a) cos(x−b) ]

= 1k[ cos((x−b)−(x−a))sin(x−a) cos(x−b) ]

= 1k[ cos(x−b) cos(x−a)+sin(x−b) sin(x−a)

sin(x−a) cos(x−b) ]

= 1k(cot(x− a) + tan(x− b))

Therefore∫1

sin(x−a) cos(x−b)dx

= 1k

∫(cot(x− a) + tan(x− b))dx

= 1k[log | sin(x− a)|+ log | sec(x− b)|] + c

= 1cos(a−b) log | sin(x−a)

cos(x−b) |+ c

b)∫

4ex+6e−x

9ex−4e−xdx =∫ 4(ex)2+6

9(ex)2−4exdx.

Put t = ex. Then exdx = dt. Therefore∫4ex+6e−x

9ex−4e−xdx =∫

4t2+69t2−4dt . . . (I)

49

9t2 − 4)

4t2 + 6− 4t2 + 16

9709

4t2 + 6 = 49(9t2 − 4) + 70

9



4t2+69t2−4 = 4

9+ 70

91

9t2−4

=⇒ 4t2+6t(9t2−4) = 4

9t+ 70

91

t(3t+2)(3t−2) . . . (II)

The partial fractions are

1t(3t+2)(3t−2) = A

t+ B

(3t+2)+ C

(3t−2) . . . (III)

=⇒ 1 = A(3t+ 2)(3t− 2) +Bt(3t− 2) + Ct(3t+ 2) . . . (IV )

Taking t = 0 in (IV ), we get

1 = A(−2− 2) =⇒ A = −14

Taking t = −23

in (IV ), we get

1 = B(−23

(−2− 2)) =⇒ B = 38

Taking t = 23

in (IV ), we get

1 = C(23(2 + 2)) =⇒ C = 3

8

Substituting these values in (III)

1t(3t+2)(3t−2) = −1

4t+ 3

8(3t+2)+ 3

8(3t−2)

So from (II),

=⇒ 4t2+6t(9t2−4) = 4

9t+ 70

9(−14t

+ 38(3t+2)

+ 38(3t−2))

=⇒ 4t2+6t(9t2−4) = −3

2t+ 35

12(3t+2)+ 35

12(3t−2)

Using this in (I),∫4ex+6e−x

9ex−4e−xdx

= 32

log |t|+ 3536

log |3t+ 2|+ 3536

log |3t− 2|+ c

= 32

log |t|+ 3536

log |(3t+ 2)(3t− 2)|+ c

= 32

log |ex|+ 3536

log |9t2 − 4|+ c

= 32x+ 35

36log |9e2x − 4|+ c


x2

x4+1dx b)

∫√3x2 − 4x+ 1dx



Ans:a)x2

x4+1

= 12(x

2+1x4+1

+ x2−1x4+1

)

= 12(

1+ 1x2

x2+ 1x2

) + 12(

1− 1x2

x2+ 1x2

)

= 12(

1−(−1

x2)

(x− 1x)2+2

) + 12(

1+(−1

x2)

(x+ 1x)2−2)

Therefore∫x2

x4+1dx = 1

2

∫ 1−(−1

x2)

(x− 1x)2+2

dx+ 12

∫ 1+(−1

x2)

(x+ 1x)2−2dx . . . (I)

Put u = x − 1x

in First integral on R.H.S. and v = x + 1x

in Second integral onR.H.S. Then

1− (−1x2

)dx = du and 1 + (−1x2

)dx = dv. So∫x2

x4+1dx = 1

2

∫1

u2+2du+ 1

2

∫1

v2−2dv

=⇒∫

x2

x4+1= 1

2√2

tan−1(u√

2) + 14√2

log |v−√2

v+√2|+ c

=⇒∫

x2

x4+1= 1

2√2

tan−1(x2−1x√2) + 1

4√2

log |x2−√2x+1

x2+√2x+1|+ c

b)

√3x2 − 4x+ 1

=√

3√x2 − 4x

3+ 1

3

=√

3√

(x− 23)2 − 4

9+ 1

3

=√

3√

(x− 23)2 − (1

3)2

We use the standard formula

√x2 − a2 = x

2

√x2 − a2 − a2

2log(x +

√x2−a2

log a) + c∫√

3x2 − 4x+ 1dx

=√

3∫√

(x− 23)2 − (1

3)2dx

=√

3(x− 2

3)

2

√(x− 2

3)2 − (1

3)2 −

√3

18log(

(x− 23) +√

(x− 23)2−( 1

3)2

13

) + c

= (3x−2)6√3

√(3x− 2)2 − 1−

√3

18log((3x− 2) +

√(3x− 2)2 − 1) + c

= (3x−2)6√3

√9x2 − 12x+ 3−

√3

18log((3x− 2) +

√(3x− 2)2 − 1) + c



= (3x−2)6

√3x2 − 4x+ 1−

√3

18log((3x− 2) +

√(3x− 2)2 − 1) + c


cosx√1+sinx+sin2 x

dx b)∫(1− x)

√4x2 − 5x+ 1dx

Ans: a)Put sin x = t. Then cos xdx = dt. Therefore∫

cosx√1+sinx+sin2 x

dx =∫√1+t+t2

dt =∫√t2+t+1

dt . . . (I)

Consider

√t2 + t+ 1

=√t2 + 21

2t+ 1

=√

(t+ 12)2 − 1

4+ 1

=√

(t+ 12)2 + (

√32

)2

We use the standard formula∫1√

x2+a2dx = log

(x +√x2+a2

a

)+ c∫

1√t2+t+1

dt

From (I),∫cosx√

1+sinx+sin2 xdx

=∫

1√t2+t+1

dt

=∫

1√(t+ 1

2)2+(

√32)2

dt

= log( (t+ 1

2)+

√(t+ 1

2)2+(

√32)2

√32

)+ c

= log( (2t+1) +

√(2t+1)2+3

√3

)+c

= log(2t+1+2

√t2+t+1√

3

)+ c

= log(2 sinx+1+2

√sin2 x+sinx+1√3

)+ c

b)(1− x)

√4x2 − 5x+ 1

=√

4x2 − 5x+ 1− x√

4x2 − 5x+ 1



=√

4x2 − 5x+ 1− 18[(8x− 5)

√4x2 − 5x+ 1]− 5

8

√4x2 − 5x+ 1

= −18[(8x− 5)

√4x2 − 5x+ 1] + 3

8

√4x2 − 5x+ 1

So∫√3x2 − 4x+ 1dx = −1

12(4x2 − 5x+ 1)

32 + 3

8

∫√4x2 − 5x+ 1dx . . . (I)

Consider

√4x2 − 5x+ 1

= 2√x2 − 5

4x+ 1

4

= 2√

(x− 58)2 − 25

64+ 1

4

= 2√

(x− 58)2 − 9

64

= 2√

(x− 58)2 − (3

8)2

We use the standard formula

√x2 − a2 = x

2

√x2 − a2 − a2

2log(x +

√x2−a2

log a) + c∫√

4x2 − 5x+ 1dx

= 2∫√

(x− 58)2 − (3

8)2dx

= 2(

(x− 58)

2

√(x− 5

8)2 − (3

8)2 − 9

128log( (x− 5

8) +√

(x− 58)2−( 3

8)2

58

))+ c

Using this in (I), we get∫√3x2 − 4x+ 1dx

= −112

(4x2 − 5x+ 1)32 + 3

4

((x− 5

8)

2

√(x− 5

8)2 − (3

8)2− 9

128log( (x− 5

8) +√

(x− 58)2−( 3

8)2

58

))+ c


11+x−x2 dx b)

∫1√

(x−2)(x−3)dx

Ans: a) 1 + x− x2

= 1 + 14− 1

4+ x− x2

= 54− (x2 − x+ 1

4)

= (√52

)2 − (x− 12)2



Therefore∫1

1+x−x2 dx

=∫

1(√5

2)2−(x− 1

2)2

dx

= 1

2√5

2

log

∣∣∣∣∣√52+(x− 1

2)

√52−(x− 1

2)

∣∣∣∣∣+ c

= 1√5

log∣∣∣√5−1+2x√

5+1−2x

∣∣∣+ c

b) (x− 2)(x− 3)

= x2 − 5x+ 6

= x2 − 5x+ 254− 25

4+ 6

=√

(x− 52)2 − (1

2)2

Therefore,∫1√

(x−2)(x−3)dx

=∫

1√(x− 5

2)2−( 1

2)2

dx

= log∣∣∣(x− 5

2) +√

(x− 52)2 − (1

2)2∣∣∣+ c

= log∣∣∣(x− 5

2) +√x2 − 5x+ 6

∣∣∣+ c

7. Evaluate∫x3+4x2+7x+2x2+3x+2

dx

Ans: x + 1

x2 + 3x+ 2)

x3 + 4x2 + 7x + 2− x3 − 3x2 − 2x

x2 + 5x + 2− x2 − 3x− 2

2x

Therefore

x3 + 4x2 + 7x+ 2 = (x2 + 3x+ 2)(x+ 1) + 2x

=⇒ x3+4x2+7x+2x2+3x+2

= x+ 1 + 2xx2+3x+2

= x+ 1 + 2x+3x2+3x+2

− 3x2+3x+2



= x+ 1 + 2x+3x2+3x+2

− 3x2+3x+ 9

4+2− 9

4

= x+ 1 + 2x+3x2+3x+2

− 3(x+ 3

2)2−( 1

2)2

Thus

x3+4x2+7x+2x2+3x+2

= x+ 1 + 2x+3x2+3x+2

− 3(x+ 3

2)2−( 1

2)2

This implies that∫x3+4x2+7x+2x2+3x+2

dx

= x2

2+ x+ log |x2 + 3x+ 2| − 3 1

2 12

log∣∣∣x+ 3

2− 1

2

x+ 32+ 1

2

∣∣∣+ c

= x2

2+ x+ log |x2 + 3x+ 2| − 3 log

∣∣∣x+1x+2

∣∣∣+ c



Practical 6: Differential Equations

Solve the following differential equations.

1. dydx

= 4x−3y3x−2y

Ans: Put y = vx. Then

dydx

= v + x dydx

=⇒ v + x dydx

= 4x−3vx3x−2vx

=⇒ v + x dvdx

= 4−3v3−2v

=⇒ x dvdx

= 4−3v3−2v − v

=⇒ x dvdx

= 4−6v+2v2

3−2v

Using variable seperable form

3−2v4−6v+2v2

dv = 1xdx

ddv

(4− 6v + 2v2) = −6 + 4v = −2(3− 2v)

=⇒ −2(3−2v)−2(4−6v+2v2)

dv = 1xdx

=⇒ −12

( −6+4v4−6v+2v2

dv = 1xdx)

=⇒ −12

∫ −6+4v2v2−6v+4

dv =∫

1xdx

=⇒ −12

log(2v2 − 6v + 4) = log x− log c

=⇒ log(2v2 − 6v + 4) = −2 log x+ 2 log c =⇒ log(2v2 − 6v + 4) = log x−2c2

=⇒ 2v2 − 6v + 4 = x−2c2

Replacing v by yx

2( yx)2 − 6 y

x+ 4 = c2

x2

Multiplying by x2 on both sides, we get

2y2 − 6xy + 4x2 = c2

2.(y sin( y

x) + x

)− x sin( y

x) dydx

= 0



Ans: y sin( yx) + x = x sin( y

x) dydx

y sin( yx)+x

x sin( yx)= dy

dx

yx

+ 1sin( y

x)

= dydx

Put y = vx.

dydx

= v + x dydx

v + 1sin v

= v + x dvdx

1sin v

= x dvdx

Using variable seperable form

1xdx = sin vdv

=⇒∫

1xdx =

∫sin vdv

=⇒ log x = − cos v + c

Replacing v by yx,

log x = − cos( yx) + c

3. x dydx− y =

√x2 − y2

Ans: x dydx− y =

√x2 − y2

=⇒ x dydx

= y +√x2 − y2

=⇒ dydx

=y+√x2−y2x

Put y = vx

dydx

= v + x dvdx

=⇒ v + x dvdx

= vx+√x2−v2x2x

=⇒ v + x dvdx

= v +√

1− v2

=⇒ x dvdx

=√

1− v2

Using variable seperable form,

1√1−v2dv = 1

xdx



By taking integration of both sides∫1√

1−v2 dv =∫

1xdx

sin−1 v = log x+ c

By replacing v = yx

sin−1 yx

= log x+ c

4. Show that the differential equations is (2x3 − xy2 − 2y + 3)dx − (x2y + 2x)dy = 0exact and solve it.

Ans: M = 2x3 − xy2 − 2y + 3 and N = −x2y − 2x

=⇒ ∂M∂y

= −2xy − 2 and ∂N∂x

= −2xy − 2

=⇒ ∂M∂y

= ∂N∂x

Therefore the given differential equation is exact. The solution is given by∫y const

Mdx+∫(terms in N free from x) dy = c

=⇒∫

y const

(2x3 − xy2 − 2y + 3)dx+∫

0dy = c

=⇒ 2x4

4− x2y2

2− 2xy + 3x = c

=⇒ x4

2− x2y2

2− 2xy + 3x = c

=⇒ x4 − x2y2 − 4xy + 6x = c

5. Find order and degree of the following differential equations.

a) dydx

+xy = x3 b) d2ydx2

+ ( dydx

)2 + 5y = 0 c)(1 + ( dy

dx)2) 3

2 = d2ydx2

Ans: a) dydx

+ xy = x3

The order is 1 and degree is 1.

b) d2ydx2

+ ( dydx

)2 + 5y = 0


c)(1 + ( dy

dx)2) 3

2 = d2ydx2




6. Which of the following functions are homogeneous?

a) f(x, y) = x2+y2 b) g(x, y) =√x2 + y2 c) h(x, y) = x2+xy2

Ans: a) f(tx, ty) = t2x2 + t2y2 = t2(x2 +y2) = t2f(x, y). Therefore f is homogeneousfunction of degree 2.

b) g(tx, ty) =√t2x2 + t2y2 =

√t2(x2 + y2) = t

√x2 + y2 = tg(x, y). Therefore g

is homogeneous function of degree 1.

c) h(tx, ty) = t2x2 + txt2y2 = t2(x2 + txy2) 6= t2h(x, y). Therfore this is not ho-mogeneous differential equation.

7. Solve the differential equation dydx

= x+yx+2y−3

Ans: Compairing given differential equation with dydx

= a1x+b1y+c1a2x+b2y+c2

We get a1 = 1, b1 =1, a2 = 1, b2 = 2, c1 = 0, c2 = 1

buta1a2

= 116= b1

b2= 1

2Therefore put x = X + h, andy = Y + k =⇒ dx

dX= 1

and dydY

= 1 =⇒ dydx

= dYdX

Therefore given differential becomes dYdX

= X+h+Y+kX+h+2(Y+k)−3

=⇒ dYdX

= X+Y+(h+k)X+2Y+(h+2k−3)

Put h + k = 0 and h + 2k − 3 = 0. We obtain by solving h = −3 and k = 3.Therefore dY

dX= X+Y

X+2Y

Put Y = V X. Then dYdX

= V +X dVdX

V +X dVdX

= X+V XX+2V X

= !+V1+2V

X dVdX

= 1+V1+2V

= −V = 1+V−V−2V 2

1+2V

=⇒ X dVdX

= 1−2V 2

1+2V

=⇒ 1+2V1−2V 2dV = 1

XdX

=⇒ 11−2V 2dV + 2V

1−2V 2dV = 1XdX

=⇒ 12( 1

2−V 2)

dV − (−2V )1−2V 2dV = 1

XdX



=⇒ 12( 1

2−V 2)

dV − 12(−4V )1−2V 2dV = 1

XdX

By taking integration of both sides, we get

12

∫1

2( 12−V 2)

dV − 12

∫ (−4V )1−2V 2 dV =

∫1X

dX

12× 1

2

[log(

V− 1√2

V+√2)]− 1

2log(1− 2V 2) = logX

Multiplying by 4 on both sides, we get[log(

V− 1√2

V+√2)]− 2 log(1− 2V 2) = 4 logX

=⇒[log(

V− 1√2

V+√2× 1

(1−2V 2)2)]

= logX4 + logC

=⇒V− 1√

2

V+√2× 1

(1−2V 2)2= X4C

=⇒√2v−1√2v+1× 1

(1−√2V )2(1+

√2V )2

= X4C

=⇒√2v−1√2v+1× 1

(√2V−1)2(1+

√2V )2

= X4C

=⇒ 1(√2V+1)(

√2V−1)(1+

√2V )2

= X4C

=⇒ 1(2V 2−1)(1+

√2V )2

= X4C

Put V = YX

1

(2 Y2

X2−1)(1+√2 YX)2

= X4C

On Simplifying, we get

(2Y 2 −X2)(X +√

2Y )2 = 1C

Replace X by x+ 3 and Y by y − 3

(2(y − 3)2 − (x+ 3)2)((x+ 3) +√

2(y − 3))2 = 1C

= d

On Simplifying we get, (2y2 − 12y − x2 − 6x+ 9)(x+√

2y + 3(1−√

2))2 = d

8. Solve (1 + x2) dydx

+ 2xy − 1 = 0

Ans: (1 + x2) dydx

+ 2xy − 1 = 0

=⇒ dydx

+ 2x1+x2

y = 11+x2

Compairing given differential equation with dydx

+ P (x)y = Q(x)



We get P (x) = 2x1+x2

and Q(x) = 11+x2

Therefore integrating factor is

I.F. = e∫P (x)dx = e

∫2x

1+x2dx

= elog(1+x2) = (1 + x2)

Therefore solution is y.I.F. =∫(I.F.)Q(x)dx

=⇒ y.(1 + x2) =∫(1 + x2) 1

1+x2dx

=⇒ y.(1 + x2) =∫

dx

=⇒ y.(1 + x2) = x+ c

=⇒ y = x+c1+x2

9. Solve (1 + y2)dx = (tan−1 y − x)dy

Ans: (1 + y2) = (tan−1 y − x) dydx

(1 + y2)dxdy

= (tan−1 y − x)

dxdy

+ 11+y2

x = tan−1 y1+y2

−

Comparing given differential equation with dxdy

+ P (y)x = Q(y)

We get P (y) = 11+y2

and Q(y) = tan−1 y1+y2

Therefore integrating factor is

I.F. = e∫P (y)dy = e

∫1

1+y2dy

= etan−1 y

Therefore solution is x.I.F. =∫(I.F.)Q(y)dy

x.I.F. =∫etan

−1 y(tan−1 y1+y2

)dy

Put tan−1 = y =⇒ sec2 udu = dy

Therefore xetan−1 y =

∫etan

−1 tanu(tan−1 tanu1+tan2 u

)sec2 udu

=⇒ xetan−1 y =

∫euudu

Using product rule for integration we get

=⇒ xetan−1 y = u

∫eudu−

∫∂u∂u

(∫eudu)du



=⇒ xetan−1 y = u

∫eudu−

∫(∫eudu)du

=⇒ xetan−1 y = ueu − eu + C

By repacing u by tan−1 y we get

=⇒ xetan−1 y = tan−1 yetan

−1 y − etan−1 y + C

=⇒ x = tan−1 y + C

etan−1 y− 1

=⇒ x = tan−1 y + Ce− tan−1 y − 1

10. Solve (x+ 2y3) dydx

= y

Ans: (x+ 2y3) dydx

= y

=⇒ y dxdy

= x+ 2y3

=⇒ dxdy− x

y= 2y2 . . . (I)

The above equation is of the form dydx

+ P.x = Q where P and Q are functionsof y.

Therefore

I.F. = e∫Pdy = e

∫ −1ydy=e− log y= 1

y

Multiplying equation (I) by I.F.,

1y(dxdy− x

y) = 2y2 1

y

=⇒ ddy

(x 1y) = 2y

Integrating w.r.t. y

x 1y

=∫

2ydy + c

=⇒ xy

= y2 + c

=⇒ x = y(c+ y2)

This is the general solution of given differential equation.

11. Find orthogonal trajectories of the family x2 + y2 = a2

Ans: Differentiate the family x2 + y2 = a2 with respect to x

=⇒ 2x+ 2y dydx

= 0



=⇒ x+ y dydx

= 0

=⇒ dydx

= −xy

By replacing dydx

by −dxdy

we get

=⇒ −dxdy

= −xy

=⇒ dxdy

= xy

By using variable separable form

=⇒ 1ydy = 1

xdx

Taking integration on both sides∫1ydy =

∫1xdx

log y = log x+ log c

=⇒ log y = log cx

=⇒ y = cx

Therefore orthogonal trajectories of the family x2 + y2 = a2 is the family of straightline pasiing through origine i.e. y = cx where c is a slope of the line



Practical 7: Miscellaneous

1. By knowing the graph of f(x) = ex, draw the graphs of a) f(x) = −ex b) f(x) = e−x

Ans: a) Graph of −f(x) ≡ mirror image of Graph of f(x) about X axis

=⇒ Graph of −ex ≡ mirror image of Graph of ex about X axis

b) Graph of f(−x) ≡ mirror image of Graph of f(x) about Y axis

=⇒ Graph of e−x ≡ mirror image of Graph of ex about Y axis



2. By knowing the graph of f(x) = sin x, draw the graphs of a) f(x) = sin(x) + 2 b)f(x) = sin(x)− 2 c) f(x) = sin(x− 2) d) f(x) = sin(x+ 2)

Ans: a) Graph of f(x) + b ≡ Graph of f(x) displaced along Y axis by amount b

=⇒ Graph of sin(x) + 2 ≡ Graph of sinx displaced along Y axis by amount 2

b) Graph of f(x) + b ≡ Graph of f(x) displaced along Y axis by amount b

=⇒ Graph of sin(x)− 2 ≡ Graph of sinx displaced along Y axis by amount −2

c) Graph of f(x− a) ≡ Graph of f(x) displaced along X axis by amount a



=⇒ Graph of sin(x− 2) ≡ Graph of sinx displaced along X axis by amount 2

d) Graph of f(x− a) ≡ Graph of f(x) displaced along X axis by amount a

=⇒ Graph of sin(x+ 2) ≡ Graph of sinx displaced along X axis by amount −2

3. Evaluate the integral∫

x2+3(x−1)(x2+4)

dx

Ans: Partial fractions are

x2+3(x−1)(x2+4)

= Ax−1 + Bx+c

x2+4. . . (I)

=⇒ x2 + 3 = A(x2 + 4) + (Bx+ C)(x− 1)

=⇒ x2 + 3 = (A+B)x2 + (C −B)x+ (4A− C)



=⇒ A+B = 1, C −B = 0, 4A− C = 3

=⇒ B = C,A+ C = 1, 4A− C = 3

=⇒ 5A = 4, C = 15

= B

=⇒ A = 45, B = 1

5, C = 1

5

So from (I),∫x2+3

(x−1)(x2+4)dx

=∫( 4

5

x−2 +15(x+1)

x2+4

)dx

= 45

∫1

x−1dx+ 15

∫x+1x2−4dx

= 45

log |x− 1|+ 110

∫( 2xx2+4

+ 2x2+4

)dx

= 45

log |x− 1|+ 110

∫2xx2+4

dx+ 15

∫1

x2+4dx

= 45

log |x− 1|+ 110

log |x2 + 4|+ 110

tan−1 x2

+ c

4. If y = (sin−1 x)2, show that (1− x2)yn+2 − (2n+ 1)xyn+1 − n2yn = 0

Ans: y = (sin−1 x)2

=⇒ y1 = 2 sin−1 x 1√1−x2

=⇒ (1− x2)y21 − 4 sin−1 x = 0

=⇒ (1− x2)y21 − 4y = 0

=⇒ (1− x2)2y1y2 − 2xy21 − 4y1 = 0

=⇒ 2y1[(1− x2)y2 − xy1 − 2] = 0

=⇒ (1− x2)y2 − xy1 − 2 = 0

Differentiating n times by using Leibnitz’s Theorem, we get

[nC0(1− x2)yn+2 +n C1(−2x)yn+1 +n C2(−2)yn]− [nC0xyn+1 +n C1yn] = 0

=⇒ (1− x2)yn+2 − 2nxyn+1 − n(n− 1)yn − xyn+1 − nyn = 0

=⇒ (1− x2)yn+2 − (2n+ 1)xyn+1 − n2yn = 0

5. If y = tan−1 x, show that (1 + x2)yn+2 + [(2n+ 1)x− 1]yn+1 + n(n+ 1)yn = 0



Ans: y = tan−1 x

=⇒ y1 = etan−1 x 1

1+x2

=⇒ (1 + x2)y1 = y

Differentiating (n+ 1) times by using Leibnitz’s Theorem, we get

[(1 + x2)y1]n+1 = yn+1

=⇒ (n+1)C0(1 + x2)yn+2 +(n+1) C1yn+1(2x) +(n+1) C2yn(2) = yn+1

=⇒ (1 + x2)yn+2 + 2(n+ 1)xyn+1 + (n+1)n1.2

yn(2) = yn+1

=⇒ (1 + x2)yn+2 + [(2n+ 1)x− 1]yn+1 + n(n+ 1)yn = 0

6. Solve (2x− 2y + 3)dx− (x− y + 1)dy = 0. Given that y = 1 when x = 0.

Ans: (2x− 2y + 3)dx− (x− y + 1)dy = 0

=⇒ (x− y + 1)dy = (2x− 2y + 3)dx

=⇒ dydx

= 2x−2y+3x−y+1

= 2(x−y)+3(x−y)+1

= 2v+3v+1

. . . (I)

where v = x− y

Now,

v = x− y

=⇒ dvdx

= 1− dydx

=⇒ 1− dvdx

= dydx

=⇒ 1− dvdx

= 2v+3v+1

from (I)

=⇒ 1− 2v+3v+1

= dvdx

=⇒ dvdx

= v+1−2v−3v+1

= −v+2v+1

=⇒ v+1v+2

dv = −dx

=⇒∫

v+1v+2

dv = −∫dx+ c

=⇒∫ (

1− 1v+2

)dv = −x+ c

=⇒ v − log |v + 2| = −x+ c



=⇒ x− y − log |x− y + 2| = −x+ c

=⇒ 2x− y − log |x− y + 2| = c . . . (II)

This is the general solution. To find particular solution, y = 1 when x = 0. Therefore

2(0)− 1− log |0− 1 + 2| = c

=⇒ −1 = c

=⇒ c = −1

=⇒ 2x− y − log |x− y + 2| = −1 from (II)

=⇒ 2x− y − log |x− y + 2|+ 1 = 0


Practical 1: Graphs of Elementary Functionsmathskthm.6te.net/F.Y.B.Sc. Calculus Practical.pdfDepartment of Mathematics, K.T.H.M. College, Nashik F.Y.B.Sc. Calculus Practical (Academic

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