Continuity Differentiability for IIT JEE Main and Advanced A compact Booklet (Math Book Book 1) P N Tripathi Beeus

86 Pages·2019·1.76 MB·English

by  P N Tripathi

Continuity &
Diffrentiability
Short Book
JEE MAIN | CBSE
P.N. Tripathi
TOPIC PPPAAAGGGEEE NNNOOO...
♦ THEORY WITH SOLVED EXAMPLES 3 – 28
♦ EXERCISE - I 29 – 38
♦ EXERCISE - II 39 – 42
♦ EXERCISE - III 43 – 47
♦ EXERCISE - IV 48 – 51
♦ EXERCISE - V 52 – 54
♦ ANSWER KEY 55 – 56
JEE Syllabus :
continuity of a function, continuity of the sum, difference, product and quotient of two functions,
continuity of composite functions, intermediate value property of continuous functions
CONTINUITY & DIFFERENTIABILITY Page # 3
A. DEFINITION OF CONTINUITY
Continuity at a Point : – A function f is continuous at c if the following three conditions are met.
lim lim
(i) f(x) is defined. (ii) f(x) exists. (iii) f(x) = f(c).
x→c x→c
Limit
In other words function f(x) is said to be continuous at x = c , if f(x) = f(c).
x→c
Limit Limit
Symbolically f is continuous at x = c if f(c - h) = f(c+h) = f(c).
h→0 h→0
x = a x = a
(continuous) (Discontinuous)
One-sided Continuity :
A function f defined in some neighbourhood of a point c for c ≤ c is said to be continuous at c from the
lim
left if f(x) = f(c).
x→c−
A function f defined in some neigbourhood of a point c for x ≥ c is said to be continuous at c from the
lim
right if f(x) = f(c).
x→c+
One-sided continuity is a collective term for functions continuous from the left or from the right.
If the function f is continuous at c, then it is continuous at c from the left and from the right .
lim
Conversely, if the function f is continuous at c from the left and from the right, then f(x) exists &
x→c
lim
f(x) = f(c).
x→c
The last equality means that f is continuous at c.
lim
If one of the one-sided limits does not exist, then f(x) does not exist either. In this case, the point
x→c
c is a discontinuity in the function, since the continuity condition is not met.
Continuity In An Interval :
(a) A function f is said to be continuous in an open interval (a , b) if f is continuous at each & every
point ∈ (a , b).
(b) A function f is said to be continuous in a closed interval [a,b] if :
(i) f is continuous in the open interval (a , b) &
Limit
(ii) f is right continuous at ‘a’ i.e. f(x) = f(a) = a finite quantity..
x→a+
Limit
(iii) f is left continuous at ‘b’ i.e. f(x) = f(b) = a finite quantity..
x→b−
Page # 4 CONTINUITY & DIFFERENTIABILITY
A function f can be discontinuous due to any of the following three reasons :
Limit Limit Limit
(i) f(x) does not exist i.e. f(x) ≠ f (x)
x→c x→c− x→c+
Limit
(ii) f(x) is not defined at x= c (iii) f(x) ≠ f (c)
x→c
Geometrically, the graph of the function will exhibit a break at x= c.
Ex.1 Test the following functions for continuity
2x5 −8x2 +11 3sin3 x+cos2x+1
(a) (b) f(x) =
x4 +4x3 +8x2 +8x+4 4cosx−2
Sol. (a) A function representing a ratio of two continous functions will be (polynomials in this case)
discontinuous only at points for which the denominator zero. But in this case
(x4 + 4x3 + 8x2 + 8x + 4) = (x2 + 2x + 2)2 = [(x + 1)2 + 1]2 > 0 (always greater than zero)
Hence f(x) is continuous throughout the entire real line.
(b) The function f(x) suffers discontinuities only at points for which the denominator is equal to zero
i.e. 4 cos x – 2 = 0 or cos x = 1/2 ⇒ x = x = ± π/3 + 2nπ (n = 0, ±1, ±2...)
n
Thus the function f(x) is continuous everywhere, except at the point x .
n
2−(256−7x)1/8
Ex.2 The function f(x) = , x ≠ 0 is continuous everywhere then find the value of f(0).
(5x+32)1/5 −2
Sol. Since f(x) is continuous
lim lim
∴ f(0) = R.H.L. = f(x) = f(0 + h)
x→0+ h→0
(256−7h)1/8 −(256)1/8 (256−7h)1/8 −(256)1/8
(256−7h)1/8 −(256)1/8 (256−7h)−256 .(−7h) 7 (256−7h)−256
= – lim = – lim = lim
h→0 (5h+32)1/5 −(32)1/5 h→0 (5h+32)1/5 −(32)1/5 5h→0 (5h+32)1/5 −(32)1/5
.(5h)
(5h+32)−32 (5h+32)−32
1
= 57.81.(.(23526)1)1/5/8−−11 = 87.((22))−−74 = 87.213 = 674 ⇒ f(0) = 674. ∵xli→maxxn −−aan −nan−1
5
 −2sinx if x≤−π/2
 π
Asinx+B if − <x<π/2
Ex.3 Let f(x) =  2 Find A and B so as to make the function continuous.
 cosx if x≥π/2
Sol. At x = – π/2
L.H.L. = Lim (−2sinx) R.H.L. = Lim A sin x + B
π− π+
x→− x→−
2 2
CONTINUITY & DIFFERENTIABILITY Page # 5
π π
Replace x by – – h where h → 0 Replace x by – + h where h → 0.
2 2
 π   π 
Lim − −h Lim − +h
– 2 sin = 2 = A sin + B = B – A
h→0  2  h→0  2 
So B – A = 2 ...(i)
At x = π/2
L.H.L. = Lim A sin x + B R.H.L. = Lim cos x
π− π+
x→ x→
2 2
π π
Replace x by – h Replace x by + h
2 2
where h → 0 where h → 0
π  π 
Lim  −h Limcos +h
= A sin + B = A + B = = 0
h→0 2  h→0 2 
So A + B = 0 ...(ii)
Solving (i) & (ii), B = 1, A = –1
 1 1
 2− + 
(x+1) |x| x , x≠0
Ex.4 Test the continuity of f(x) at x = 0 if f(x) =

 0 , x=0
Sol. For x < 0,
 1 1 
2− + 
L.H.L. = lim f(x) = lim f(0 – h) = lim (0−h+1) |0−h| (0−h) = lim (1 – h)2 = (1 – 0)2 = 1
x→0− h→0 h→0 h→0
2−1+1 2−2
f(0) = 0. & R.H.L = lim f(x) = lim f(0 + h) = lim (h+1) |h| h = lim(h+1) h = 1 – ∞ = 1
x→0+ h→0 h→0 h→0
∴ L.H.L. = R.H.L. ≠ f(0) Hence f(x) is discontinuous at x = 0.
Ex.5 If f(x) be continuous function for all real values of x and satisfies;
x2 + {f(x) – 2} x + 2 3 – 3 – 3 . f(x) = 0, ∀ x ∈ R. Then find the value of f( 3).
Sol. As f(x) is continuous for all x ∈ R.
x2 −2x+2 3 −3
lim
Thus, f(x) = f( 3) where f(x) = , x ≠ 3
x→ 3 3 −x
x2 −2x+2 3 −3 (2− 3 −x)( 3 −x)
lim f(x) = lim = lim =2(Ι− 3)
x→ 3 x→ 3 3 −x x→ 3 ( 3 −x)
f( 3) = 2(1− 3).
Page # 6 CONTINUITY & DIFFERENTIABILITY
 π
(1+|sinx|a/(|sinx|) , − <x<0
 6
Ex.6 Let f(x) =  b , x=0 Determine a and b such that f is continuous at x = 0.

etan2x/tan3x , x>0


Sol. For x < 0,
lim lim lim
L.H.L. = f(x) = f(0 – h) = (1 + |sin (0 – h)|)a/(|sin(0 – h)|)
x→0− h→0 h→0
lim lim(1+sinh−1).a/(sinh)
= (1 + sin h)a/(sin h) = eh→0 =ea
h→0
For x = 0, f(0) = b
For x > 0,R.H.L. = lim f(x) = lim f(0 + h) = limtan2h = limtan2h/2h.2 = e(1/1) . (2/3) = e2/3
x→0+ h→0 eh→0tan3h eh→0tan3h/3h3
∵ f(x) is continuous at x = 0 ∴∴∴∴ L.H.L. = R.H.L. = f(0)
⇒ ea = e2/3 = b Hence a = 2/3 and b = e2/3.
Acosx+Bxsinx−5
Ex.7 If f (x) = (x ≠ 0) is continuous at x = 0, then find the value of A and B . Also find f
x4
(0).
Limit
Sol. For continuity f (x) = f (0)
x→0
Limit Acosx+Bxsinx−5
Now as x → 0 ; Numerator → A − 5 and Denominator → 0.
x→0 x4
Hence A − 5 = 0 ⇒ A = 5
Limit Bxsinx−5(1−cosx) Limit B.sixnx−1+c5osx sixn22x
Hence =
x→0 x4 x→0 x2
5 5
as x → 0 ; Numerator → B − and Denominator → 0 ⇒ B =
2 2
5 xsinx−2(1−cosx) 5 2xsinxcosx−4sin2x
Hence l = Limit = Limit 2 2 2
2 x→0 x4 2 x→0 x4
5 2sinx xcosx−2cosx
Limit 2 Limit 2 2
= Let x = 2 θ
2 x→0 x x→0 x3
5 2θcosθ−2sinθ 5 (θ−tanθ) 5 θ−tanθ 5  1 5
= Limit = Limit 2 cos θ = Limit = −  = −
16 θ→0 θ3 16 θ→0 θ3 8 θ→0 θ3 8  3 24
CONTINUITY & DIFFERENTIABILITY Page # 7
a2[x]+{x} −1
 , x≠0
 2[x]+{x}
Ex.8 Discuss the continuity of the function f(x) = (a ≠ 1)

 log a , x=0
e
at x = 0, where [x] and {x} are the greatest integer function and fractional part of x respectively.
Sol. Value of function = f(0) = log a
e
L.H.L. = lim f(x) = lim f(0 – h) = lim a2[0−h]+{0−h} −1 = lim a2[0−h]+{−1+(1−h)} −1 = a−1−1=1− 1
x→0− h→0 h→02[0−h]+{0−h} h→02[0−h]+{−1+(1−h)} −1 a
a2[0+h]+{0+h}−1 a0+h −1 ah −1
and R.H.L. = lim f(x) = lim f(0 + h)= lim = lim = lim = log a
x→0+ h→0 h→02[0+h]+{0+h} h→0 0+h h→0 h e
∴∴∴∴ L.H.L. ≠ R.H.L. = f(0) Hence f(x) is discontinuous at x = 0.
1+acos2x+bcos4x
 if x≠0
 x2sin2x
Ex.9 Let f (x) =  If f (x) is continuous at x = 0, then find the value of (b + c)3


c if x=0
– 3a.
1+acos2x+bcos4x
Sol. Lim as x → 0, Nr → 1 + a + b Dr → 0
x→0 x4
acos2x+bcos4x−(a+b)
Lim
for existence of limit a + b + 1 = 0 ∴ c = .....(2)
x→0 x4
a(1−cos2x) b(1−cos4x)
+
= – Lim x2 x2
x→0 x2
limit of Nr ⇒ 2a + 8b = 0 ⇒ a = – 4b
1 4
hence – 4b + b = – 1 ⇒ b = and a = –
3 3
4(1−cos2x)−(1−cos4x) 8sin2x−2sin22x 8sin2x−8sin2xcos2x
hence c = Lim = =
x→0 3x2 3x4 3x4
8 2 2 8
· ·
2 2
Page # 8 CONTINUITY & DIFFERENTIABILITY

a(1−xsinx)+bcosx+5
 x<0
 x2
Ex.10 Let f(x) =  3 x=0 . If f is continuous at x = 0, then find the values of a,
 1
 1+cx+dx3 x x>0
   x2 
b, c & d .
a(1− x sinx)+ b cosx + 5
Sol. f (0 −) = Limit for existence of limit a + b + 5 = 0
x→0 x2
a(1− x sinx)− (a + 5)cosx + 5
a(1− cosx)+ 5(1− cosx) − ax sinx
Limit x2 Limit
= =
x→0 x→0 x2
a 5
= + − a = 3 ⇒ a = − 1 ⇒ b = − 4
2 2
 x (c + dx2)1/x
f (0+) = Limit 1+  for existence of limit c = 0
x→0  x2 
Limit (1 + dx)1/x = eLxi→mi0t x1 dx = ed = 3 ⇒ d = ln 3
x→0
e2x−1−x(e2x+1)
Ex.11 The function , f (x) = is not defined at x = 0 . What should be the value of f (x) so
x3
that f (x) is continuous at x = 0 .
e2x−1−x(e2x+1)
Sol. l = Limit Put x = 3 t
x→0 x3
Limit e6t−1−3t(e6t+1) Limit (e2t−1)3+3e2t(e2t−1)−3t(e6t+1)
= =
t→0 27t3 t→0 27t3
[ ] [ ]
(e2t−1)3+3e2t e2t−1−t(e2t+1) +3t e2t(e2t+1)−e6t−1
Limit
=
t→0 27t3
(e2t−1)3 1 1
= Limit + Limit e2t × l − Limit (e2t − 1) (e4t − 1)
t→0 27t3 9 t→0 t→0 9t2
8(cid:2) 8 8 e2t−1 e4t−1 1 2
⇒ = − Limit   ×   ⇒ l = − 1 = –
9 27 9 t→0  2t   4t  3 3
CONTINUITY & DIFFERENTIABILITY Page # 9
max{f(t):x+1≤t≤x+2,−3≤x<0}
Ex.12 Let f(x) = x3 – 3x2 + 6∀ x ∈ R and g(x) = 
1−x, for x≥0
Test continuity of g (x) for x ∈ [–3, 1].
Sol. Since f(x) = x3 – 3x2 + 6 ⇒ f’(x) = 3x2 – 6x = 3x (x – 2)
for maxima and minima f′(x) = 0
∴ x = 0, 2
f″ (x) = 6x – 6
f″ (0) = –6 < 0 (local maxima at x = 0)
f″ (2) = 6 > 0 (local minima at x = 2)
x3 – 3x2 + 6 = 0 has maximum 2 positive and 1 negative real roots. Y
f(0) = 6. 6
Now graph of f(x) is :
Clearly f(x) is increasing in (– ∞, 0) ∪ (2, ∞) and decreasing in (0, 2)
y = f(x)
⇒ x + 2 < 0 ⇒ x < – 2 ⇒ –3 ≤ x < – 2
⇒ –2 ≤ x + 1 < –1 and –1 ≤ x + 2 < 0
f(x) = 2
in both cases f(x) increases (maximum) of g(x) = f(x + 2)
2
∴ g(x) = f(x + 2); –3 ≤ x < – 2 ...(1)
and if x + 1 < 0 and 0 ≤ x + 2 < 2 X
–1 O 1 2 3
⇒ – 2 ≤ x < –1 then g(x) = f(0)
Now for x + 1 ≥ 0 and x + 2 < 2 ⇒ –1 ≤ x < 0, g(x) = f(x + 1)
f(x+2) ; −3≤x<−2

 f(0) ; −2≤x<−1
Hence g(x) =  Hence g(x) is continuous in the interval [–3, 1].
f(x+1) ; −1≤x<+0

 1−x ; x≥0
Ex.13 Let y = f(x) be defined parametrically as y = t2 + t | t |, x = 2t – | t |, t ∈ R Then at x = 0, find f(x)
and discuss continuity.
Sol. As, y = t2 + t | t | and x = 2t – | t | Y
y = 2x2
Thus when t ≥ 0
⇒ x = 2t – t = t, y = t2 + t2 = 2t2 ∴ x = t and y = 2t2
⇒ y = 2x2 ∀ x ≥ 0
again when, t < 0
X
⇒ x = 2t + t = 3t and y = t2 – t2 = 0 ⇒ y = 0 for all x < 0. y = 0 0
2x2, x≥0
Hence, f(x) =  which is clearly continuous for all x as show graphically..
 0, x<0
Page # 10 CONTINUITY & DIFFERENTIABILITY
 1 1 1 
Ex.14 Given the function, f (x) = x  + + +......upto∞.
x(1+x) (1+x)(1+2x) (1+2x)(1+3x) 
Find f (0) if f (x) is continuous at x = 0 .
( )
1 (1+2x)−(1+x) (1+3x)−(1+2x) (1+nx)− 1+n−1x
Sol. f (x) = + + + ...... +
1+x (1+x)(1+2x) (1+2x)(1+3x) (1+n−1x)(1+nx)
 2
if x ≠0 andn→∞
2 1 1+x
f (x) = − upto n terms when x ≠ 0. Hence f (x) =
1+ x 1+ nx  2 if x =0 for continuity.
Ex.15 Let f : R → R be a function which satisfies f(x + y3) = f(x) + (f(y))3 ∀ x, y ∈ R. If f is continuous at x
= 0, prove that f is continuous every where.
lim
Sol. To prove f(x + h) = f(x).
h→0
Put x = y = 0 in the given relation f(0) = f(0) + (f(0))3 ⇒ f(0) = 0
lim
Since f is continuous at x = 0, f(h) = f(0) = 0.
h→0
lim lim lim
Now, f(x + h) = f(x) + (f(h))3 = f(x) + (f(h))3 = f(x) + 0 = f(x).
h→0 h→0 h→0
Hence f is continuous for all x ∈ R.
B. CLASSIFICATION OF DISCONTINUITY
Definition :– Let a function f be defined in the neighbourhood of a point c, except perhaps at c itself.
lim lim lim lim
Also let both one–sided limits f(x) and f(x) exist, where f(x) ≠ f(x).
x→c− x→c+ x→c+ x→c−
Then the point c is called a discontinuity of the first kind in the function f(x).
lim
In more complicated case f(x) may not exist because
x→c
one or both one-sided limits do not exist. Such condition
is called a discontinuity of the second kind.
x2 +1 for x<0,

 5 for x=0,
The function y =

−x for x>0,

has a discontinuity of the first kind at x = 0
The function y = |x| /x is defined for all x ∈ R, x ≠ 0;
but at x = 0 it has a discontinuity of the first kind.
lim
The left-hand limit is y = –1, while the
x→0−
lim
right-hand limit is y = 1
x→0+