Practical 1: Graphs of Elementary Functionsmathskthm.6te.net/F.Y.B.Sc. Calculus...
Transcript of Practical 1: Graphs of Elementary Functionsmathskthm.6te.net/F.Y.B.Sc. Calculus...
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
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
F.Y.B.Sc. Calculus Practical, Page: 2
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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.
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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 .
F.Y.B.Sc. Calculus Practical, Page: 8
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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.
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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)
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Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 12
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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].
F.Y.B.Sc. Calculus Practical, Page: 13
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 14
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 15
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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
F.Y.B.Sc. Calculus Practical, Page: 16
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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.Y.B.Sc. Calculus Practical, Page: 17
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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
F.Y.B.Sc. Calculus Practical, Page: 18
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
2. Evaluate the following integrals.a)∫
1x3−1dx b)
∫x2
x4+5x2+6dx
F.Y.B.Sc. Calculus Practical, Page: 19
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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)
F.Y.B.Sc. Calculus Practical, Page: 20
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
3. Evaluate the following integrals.a)∫
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
F.Y.B.Sc. Calculus Practical, Page: 21
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
4. Evaluate the following integrals.a)∫
x2
x4+1dx b)
∫√3x2 − 4x+ 1dx
F.Y.B.Sc. Calculus Practical, Page: 22
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 23
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
= (3x−2)6
√3x2 − 4x+ 1−
√3
18log((3x− 2) +
√(3x− 2)2 − 1) + c
5. Evaluate the following integrals.a)∫
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
F.Y.B.Sc. Calculus Practical, Page: 24
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=√
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
6. Evaluate the following integrals.a)∫
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
F.Y.B.Sc. Calculus Practical, Page: 25
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 26
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
= 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
F.Y.B.Sc. Calculus Practical, Page: 27
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 28
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 29
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
The order is 2 and degree is 1.
c)(1 + ( dy
dx)2) 3
2 = d2ydx2
The order is 2 and degree is 2.
F.Y.B.Sc. Calculus Practical, Page: 30
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 31
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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)
F.Y.B.Sc. Calculus Practical, Page: 32
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 33
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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
F.Y.B.Sc. Calculus Practical, Page: 34
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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
F.Y.B.Sc. Calculus Practical, Page: 35
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 36
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 37
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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)
F.Y.B.Sc. Calculus Practical, Page: 38
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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
F.Y.B.Sc. Calculus Practical, Page: 39
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
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
F.Y.B.Sc. Calculus Practical, Page: 40
Department of Mathematics, K.T.H.M. College, NashikF.Y.B.Sc. Calculus Practical (Academic Year 2016-17)
=⇒ 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
F.Y.B.Sc. Calculus Practical, Page: 41