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CBSE Questions for Class 11 Commerce Applied Mathematics Limits And Continuity Quiz 1 - MCQExams.com
CBSE
Class 11 Commerce Applied Mathematics
Limits And Continuity
Quiz 1
Evaluate
lim
x
→
3
4
√
x
3
using the properties of limits.
Report Question
0%
28
1
/
4
0%
25
1
/
4
0%
27
1
/
4
0%
26
1
/
4
Explanation
lim
x
→
3
4
√
x
3
=
(
3
3
)
1
/
4
=
(
27
)
1
/
4
I
f
A
i
=
x
−
a
i
|
x
−
a
i
|
,
i
=
1
,
2
,
3
,
.
.
.
.
.
n
and
a
1
<
a
2
<
a
3
.
.
.
.
<
a
n
,
t
h
e
n
lim
x
→
a
m
(
A
1
A
2
.
.
.
.
.
.
A
n
)
,
1
≤
m
≤
n
Report Question
0%
is equal to
(
−
1
)
m
0%
is equal to
(
−
1
)
m
+
1
0%
is equal to
(
−
1
)
n
−
m
0%
is equal to
(
−
1
)
n
−
m
−
1
lim
x
→
π
2
cot
x
−
cos
x
(
π
2
−
x
)
3
Report Question
0%
−
1
2
0%
1
2
0%
2
0%
1
f
(
x
)
=
x
sin
1
x
,
f
o
r
x
≠
0
=
0
,
f
o
r
x
=
0
Then.
Report Question
0%
f
′
(
0
+
)
e
x
i
t
b
u
t
f
′
(
0
−
)
does not exit
0%
f
′
(
0
+
)
a
n
d
f
′
(
0
−
)
do not exit
0%
f
′
(
0
+
)
=
f
′
(
0
−
)
0%
none of these
lim
x
→
0
sin
−
1
x
−
tan
−
1
x
x
3
is equal to
Report Question
0%
0
0%
1
0%
−
1
0%
1
2
If
lim
x
→
∞
(
x
2
+
x
+
1
x
+
1
−
a
x
−
b
)
=
4
,then
Report Question
0%
a
=
1
,
b
=
4
0%
a
=
1
,
b
=
−
4
0%
a
=
2
,
b
=
−
3
0%
a
=
2
,
b
=
3
lim
n
→
∞
(
tan
θ
+
1
2
tan
θ
2
+
1
2
2
tan
θ
2
2
+
.
.
.
+
1
2
n
tan
θ
2
n
)
equals?
Report Question
0%
1
θ
0%
1
θ
−
2
cot
2
θ
0%
2
cot
2
θ
0%
None of these
Let p=
lim
x
→
0
+
(
1
+
t
a
n
2
√
x
)
1
2
x
then log p is equal to :
Report Question
0%
1
0%
1
2
0%
1
4
0%
2
lim
x
→
π
4
(
sin
2
x
)
sec
2
2
x
is equal to
Report Question
0%
−
1
2
0%
1
2
0%
e
−
1
2
0%
e
1
2
lim
θ
→
π
/
2
1
−
sin
θ
(
π
/
2
−
θ
)
cos
θ
is equal to
Report Question
0%
1
0%
−
1
0%
1
/
2
0%
−
1
/
2
If
l
i
m
x
→
∞
(
1
+
a
x
−
4
x
2
)
2
x
=
e
3
, then 'a' is equal to :
Report Question
0%
2
0%
3
2
0%
2
3
0%
1
2
The value of
l
i
m
x
→
0
s
i
n
(
π
c
o
s
2
x
)
x
2
equals
Report Question
0%
−
π
0%
π
0%
π
2
0%
2
π
l
i
m
x
→
1
x
t
a
n
(
x
−
[
x
]
)
x
−
1
is:
Report Question
0%
1
0%
0
0%
-1
0%
does not exist
The value of
l
i
m
θ
→
π
2
(
s
e
c
θ
−
t
a
n
θ
)
equals
Report Question
0%
0
0%
1
0%
2
0%
∞
The value of
lim
x
→
0
1
−
cos
3
x
x
sin
x
cos
x
is
Report Question
0%
2
5
0%
3
5
0%
3
2
0%
3
4
The value of
lim
x
→
∞
a
x
sin
(
b
a
x
)
where
a
>
1
is
Report Question
0%
b
log
a
0%
a
log
b
0%
b
0%
a
The value of
lim
x
→
3
(
x
3
+
27
)
log
e
(
x
−
2
)
x
2
−
9
is
Report Question
0%
9
0%
18
0%
27
0%
1
3
The value of
lim
x
→
x
4
4
√
2
−
(
cos
x
+
sin
x
)
5
1
−
sin
2
x
is
Report Question
0%
0
0%
√
2
0%
5
√
2
0%
3
lim
x
→
0
(
[
−
5
sin
x
x
]
+
[
6
sin
x
x
]
)
(where
[
.
]
denotes greatest integer function) is equal to
Report Question
0%
0
0%
−
12
0%
1
0%
2
If
l
i
m
x
→
0
x
3
√
a
+
x
(
b
x
−
s
i
n
x
)
=
1
,
a > 0, then a + b is equal to
Report Question
0%
36
0%
37
0%
38
0%
40
lim
x
→
0
sin
x
5
sin
4
x
=
Report Question
0%
0
0%
1
0%
−
1
0%
∞
lim
x
→
0
(
1
−
cos
2
x
)
sin
5
x
x
2
sin
3
x
=
?
Report Question
0%
10
/
3
0%
3
/
10
0%
6
/
5
0%
56
l
i
m
x
→
0
1
−
cos
2
x
cos
2
x
−
cos
8
x
is equal to
Report Question
0%
−
1
/
15
0%
1
/
10
0%
1
/
15
0%
15
lim
x
→
−
1
cos
2
−
cos
2
x
x
2
−
|
x
|
is equal to :
Report Question
0%
0
0%
cos
2
0%
2
sin
2
0%
None
lim
x
→
0
x
.
10
x
−
x
1
−
c
o
s
x
=
Report Question
0%
log
10
0%
2
log
10
0%
3
log
10
0%
4
log
10
lim
x
→
0
3
√
1
+
sin
x
−
3
√
1
−
sin
x
x
=
Report Question
0%
0
0%
1
0%
2
3
0%
3
2
l
i
m
x
→
0
s
i
n
(
6
x
2
)
I
n
c
o
s
(
2
x
2
−
x
)
=
Report Question
0%
12
0%
-12
0%
6
0%
-6
lim
x
→
1
(
log
3
3
x
)
log
x
3
=
?
Report Question
0%
e
−
1
0%
e
0%
−
1
0%
1
Let
f
(
β
)
=
lim
α
→
β
sin
2
α
−
sin
2
β
α
2
−
β
2
,
then
f
(
π
4
)
is greater than-
Report Question
0%
lim
x
→
0
1
−
cos
3
x
x
sin
2
x
0%
lim
x
→
π
/
2
cot
x
−
cos
x
(
π
−
2
x
)
3
0%
lim
x
→
∞
(
cos
√
x
+
1
−
cos
√
x
)
0%
lim
x
→
a
√
a
+
2
x
−
√
3
x
√
3
a
+
x
−
2
√
x
where
a
>
0
l
i
m
x
→
−
∞
(
3
x
4
+
2
x
2
)
s
i
n
(
1
x
)
+
|
x
|
3
+
5
|
x
|
3
+
|
x
|
2
+
|
x
|
+
1
=
Report Question
0%
2
0%
1
0%
-2
0%
-3
The value f
lim
x
→
π
/
4
√
1
−
√
sin
2
x
π
−
4
x
=
Report Question
0%
−
1
4
0%
1
4
0%
1
2
0%
N
o
n
e
o
f
t
h
e
s
e
If
L
=
l
i
m
x
→
0
a
s
i
n
x
−
s
i
n
2
x
t
a
n
3
x
is finite, then the value of L is :
Report Question
0%
1
0%
2
0%
3
0%
-1
If
lim
x
→
0
a
e
−
x
−
b
cos
x
−
1
2
c
x
x
cos
x
=
2
then the value of
a
+
b
+
c
is-
Report Question
0%
4
0%
−
4
0%
2
0%
−
2
lim
x
→
0
sin
2
x
+
3
x
2
x
+
sin
3
x
is equal to
Report Question
0%
1
0%
1
5
0%
2
0%
Does not exist
If
f
(
x
)
is continuous and
f
(
9
2
)
=
2
9
, then
lim
x
→
0
f
(
1
−
cos
3
x
x
2
)
is equal to :
Report Question
0%
9
2
0%
2
9
0%
0
0%
8
9
Explanation
lim
x
→
0
1
−
c
o
s
3
x
x
2
=
lim
x
→
0
2
sin
2
(
3
x
2
)
(
3
x
2
)
2
×
9
4
=
9
2
⇒
lim
x
→
0
f
(
1
−
c
o
s
3
x
x
2
)
=
f
(
9
2
)
=
2
9
.
The function
f
:
(
R
−
0
)
→
R given by
f
(
x
)
=
1
x
−
2
e
2
x
−
1
can be made continuous at
x
=
0
by defining
f
(
0
)
as
Report Question
0%
2
0%
−
1
0%
0
0%
1
Explanation
Given
f
(
x
)
=
1
x
−
2
e
2
x
−
1
⇒
f
(
0
)
=
lim
x
→
0
{
1
x
−
2
e
2
x
−
1
}
=
lim
x
→
0
e
2
x
−
1
−
2
x
x
(
e
2
x
−
1
)
.
.
.
.
.
.
.
[
0
0
f
o
r
m
]
∴
usingL'Hospital rule
f\displaystyle \left( 0 \right) =\lim _{ x\rightarrow 0 }{ \frac { 2{ e }^{ 2x }\quad -2\quad }{ (e^{ 2x }\quad -\quad 1\quad +2x{ e }^{ 2x }) } }
\displaystyle f(0) =\lim _{ x\rightarrow 0 }{ \frac { 4{ e }^{ 2x } }{ 4{ xe }^{ 2x }+2{ e }^{ 2x }+2{ e }^{ 2x } } } \quad \quad =\frac { 4.{ e }^{ 0 } }{ 4\left( 0+{ e }^{ 0 } \right) } =1
If the function
f(x)= \left\{\begin{matrix}\dfrac {\sqrt {2+\cos x}-1}{(\pi-x)^2} & x\neq \pi \\ k & x=\pi \end{matrix}\right.
is continuous at
x=\pi
, then
k
equals :
Report Question
0%
0
0%
\dfrac {1}{2}
0%
2
0%
\dfrac {1}{4}
Explanation
\displaystyle\lim _{ x\rightarrow \pi }{ \frac { \sqrt { 2+\cos { x } } -1 }{ (\pi -x)^{ 2 } } } =\lim _{ x\rightarrow \pi }{ \frac { 1+\cos { x } }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } }
[Rationalize numerator and denominator ]
\displaystyle =\lim _{ x\rightarrow \pi }{ \frac { 1-\cos { (\pi -x } ) }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } }
=\displaystyle \lim _{ x\rightarrow \pi }{ \frac { 2\sin ^{ 2 }{ \frac { (\pi -x) }{ 2 } } }{ (\pi -x)^{ 2 }(\sqrt { 2+\cos { x } } +1) } } =\frac { 1 }{ 4 }
So,
k =\dfrac{1}{4}
.
If
f:R\rightarrow R
is a function defined by
f(x)=[x]\displaystyle \cos\left(\frac{2x-1}{2}\pi\right)
, where
[{x}]
denotes the greatest integer function, then
{f}
is:
Report Question
0%
continuous for every real
x
0%
discontinuous only at
x = 0
0%
discontinuous only at non-zero integral values of
x
0%
continuous only at
x = 0
Explanation
For
n\in I
\displaystyle\lim _{ x\rightarrow n+ }{ f\left( x \right) } =\lim _{ x\rightarrow n+ }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 } \right) \pi } } =n\cos { \left( \frac { 2n-1 }{ 2 } \right) \pi } =0
And
\displaystyle\lim _{ x\rightarrow n- }{ f\left( x \right) } =\lim _{ x\rightarrow n- }{ \left[ x \right] \cos { \left( \frac { 2x-1 }{ 2 } \right) \pi } } =\left( n-1 \right) \cos { \left( \frac { 2n-1 }{ 2 } \right) \pi } =0
Thus
f
is continuous for
x=n\in I
........ (1)
Since the function
g\left( x \right)=\left[ x \right]
and
\displaystyle h\left( x \right) =\cos { \left( \frac { 2x-1 }{ 2 } \right) \pi }
are continuous on
x\in R-I
, ......... (2)
From, (1) and (2), we get
f
is continuous everywhere
\displaystyle \lim_{x\rightarrow 0}x^{2}\displaystyle \sin\frac{\pi}{x}=
Report Question
0%
1
0%
0
0%
does not exist
0%
\infty
Explanation
sin\ \frac{\pi }{x} \text{ behaves\ as\ oscillatory\ function\ at}
x\rightarrow 0
but\ have\ finite\ value = 0
0\times\ finite\ value\ =0
\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } }
is equal to
Report Question
0%
3
0%
1
0%
0
0%
2
Explanation
\displaystyle \lim _{ x\rightarrow 0 }{ \cfrac { x{ e }^{ x }-\sin { x } }{ x } }
=\displaystyle \lim_{x\to 0}\dfrac{xe^x}{x}-\lim_{x\to 0}\dfrac{\sin x}{x}
, split up
=\displaystyle \lim_{x\to 0}e^x-\lim_{x\to 0}\dfrac{\sin x}{x}
=e^0-1=1-1=0
Use limit properties to evaluate
\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x
Report Question
0%
12
0%
14
0%
16
0%
18
Explanation
\displaystyle\lim_{x\to4}\dfrac{3x^2\tan \dfrac {\pi}{x}}x =\lim_{x\to 4}\left(3x\tan\dfrac{\pi}{x}\right)
, simplify
=3(4)\tan\dfrac{\pi}{4}=12(1)=12
, [substitute the limit ]
The from of the limit is not indeterminate, so we can substitute the limit value
For every integer
n
, let
a_{n}
and
b_{n}
be real numbers. Let function
f: IR \rightarrow IR
be given by
f(x)=\left\{\begin{array}{l} a_{n}+\sin\pi x,\ for\ x\in[2n,\ 2n+1]\\ b_{n}+\cos\pi x,\ for\ x\ \in(2n-1,2n) \end{array}\right.
, for\ all\ integers\ n.
lf
f
is continuous, then which of the following hold(s) for all
n
?
Report Question
0%
a_{n-1}-b_{n-1}=0
0%
a_{n}-b_{n}=1
0%
a_{n}-b_{n+1}=1
0%
a_{n-1}-b_{n}=-1
Explanation
Check the continuity at
x=2n
L.H.L =\underset {x\rightarrow 2n^-}{\lim} [f(x)]
=\underset {h\rightarrow 0}{\lim}[f(2n-h)]
=\underset {h\rightarrow 0}{\lim}[b_n+\cos \pi (2n-h)]
=\underset {h\rightarrow 0}{\lim}[b_n+\cos \pi h]
=b_n+1
R.H.L=\underset {x\rightarrow 2n^+}{\lim}[f(x)]=\underset {h\rightarrow 0}{\lim}[f(2n+h)]
=\underset {h\rightarrow 0}{\lim}[a_n+\sin \pi(2n+h)]=a_n
f(2n)=a_n+\sin 2\pi n=a_n
For continuity,
\underset {x\rightarrow 2n^-}{\lim}f(x)=\underset {x\rightarrow 2n^+}{\lim} f(x)=f(2n)
So,
a_n=b_n+1\Rightarrow a_n-b_n=1
Now, check for continuity at
x=2n+1
L.H.L.=\underset {h\rightarrow 0}{\lim}(a_n+\sin \pi (2n+1-h))=a_n
R.H.L.=\underset {h\rightarrow 0}{\lim}(b_{n+1}+\cos (\pi (2n+1-h)))=b_{n+1}-1
f(2n+1)=a_n
.
For continuity,
a_n=b_{n+1}-1\Rightarrow a_{n-1}=b_n-1
Both, option B and option D are correct.
If
\lim _{ x\rightarrow \infty }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =A
and
\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =B
, then which one of the following is correct?
Report Question
0%
A=1
and
B=0
0%
A=0
and
B=1
0%
A=0
and
B=0
0%
A=1
and
B=1
Explanation
As given
A=\lim _{ x\rightarrow \infty }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } =\lim _{ x\rightarrow \infty }{ \cfrac { \sin { \left( \cfrac { 1 }{ x } \right) } }{ \left( \cfrac { 1 }{ x } \right) } }
Let
t=\cfrac { 1 }{ x } ;x\rightarrow \alpha ,t\rightarrow 0
\Rightarrow A=\lim _{ t\rightarrow \infty }{ \cfrac { \sin { t } }{ t } } =1\left[ \because \lim _{ t\rightarrow 0 }{ \cfrac { \sin { x } }{ x } =1 } \right]
and
B=\lim _{ x\rightarrow 0 }{ x\sin { \left( \cfrac { 1 }{ x } \right) } } \Rightarrow B=0\quad
Therefore,
A=1
and
B=0
is correct
The function
f\left( x \right)=\left[ x \right] ,
at
{ x }=5
is:
Report Question
0%
left continuous
0%
right continuous
0%
continuous
0%
cannot be determined
Explanation
We know that the greatest integer function is discontinuous at integer values. For a value of
x
just smaller than 5, the greatest integer smaller than it is 4. So,
\lim _{ x\rightarrow { 5 }^{ - } }{ [x] } = 4
f\left( x=5 \right)=\left[ 5 \right] =5
For a value of
x
just bigger than 5, the greatest integer smaller than it is 5. So,
\lim _{ x\rightarrow { 5 }^{ + } }{ [x] } = 5
Since
\lim _{ x\rightarrow { 5 }^{ - } }{ f\left( x \right) } \neq f\left( 5 \right)
, the function is not left continuous.
Since
\lim _{ x\rightarrow { 5 }^{ + } }{ f\left( x \right) } =f\left( 5 \right)
, the function is right continuous.
Correct answer: Option B
f(x)=\left\{\begin{matrix} 2x-1& if &x>2 \\ k & if &x=2 \\ x^{2}-1& if & x<2\end{matrix}\right.
is continuous at
x= 2
then
k =
Report Question
0%
1
0%
2
0%
3
0%
4
Explanation
Since, the function is continuous at x = 2,
\displaystyle\lim_{ x \rightarrow {2}^{-}} f(x) = f(2) = \displaystyle\lim _{x \rightarrow {2}^{+} }f(x)
\displaystyle\lim _{x \rightarrow {2}^{-}} 2x - 1 = f(2) = \displaystyle\lim_{ x \rightarrow {2}^{+}} {x}^{2} - 1
k = 3
Hence, k = 3
Evaluate
\displaystyle \lim_{x \rightarrow -2} \displaystyle \frac{x^2 - 1}{2x + 4}
.
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0%
0
0%
1
0%
2
0%
\infty
Explanation
Consider,
\displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}
Substitute
x=-2+h
\implies h\rightarrow 0 \text{ as } x\rightarrow -2
\implies \displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}
\displaystyle =\lim _{ h\rightarrow 0 } \frac { (h-2)^{ 2 }-1 }{ 2(h-2)+4 }
\displaystyle =\lim _{ h\rightarrow 0 } \frac { h^{ 2 }-4h+3 }{ 2h }
=\infty
Hence,
\displaystyle \lim_{x \rightarrow -2^+} \displaystyle \frac{x^2 - 1}{2x + 4}=\infty
If
f(x)=\displaystyle \frac{\sin x}{x},x\neq 0
is to be continuous at
{x}=0
then
\mathrm{f}({0})=
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0%
0
0%
1
0%
-1
0%
2
Explanation
From the property of limits,
\underset{ x \rightarrow 0}{Lim} \dfrac{sin x}{x} = 1
Thus, for the function to be continuous at
x = 0
f(0) = 1
Say true or false.
Every continuous function is always differentiable.
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0%
True
0%
False
Explanation
If a function
f(x)
is differentiable at a point
a
, then it is continuous at the point
a
. But the converse is not true.
f(x)=|x|
is contineous but not differentiable at
x=0.
Therefore, the given statement is false.
\displaystyle \lim_{x\rightarrow \infty} \sin x
equals
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0%
1
0%
0
0%
\infty
0%
does not exist
Explanation
Let
L = \displaystyle \lim_{x\rightarrow \infty} \sin x
Assume
\displaystyle y = \frac{1}{x}
so as
x\to \infty, y \to 0
\Rightarrow L = \displaystyle \lim_{y\rightarrow 0} \sin \frac{1}{y}
We know
\sin x
lie between -1 to 1
so let
p =\sin x
as
x\to \infty
Thus left hand limit
=L^+=\displaystyle \lim_{y\rightarrow 0^+} \sin \frac{1}{y}=p
and right hand limit
=L^-=\displaystyle \lim_{y\rightarrow 0^-} \sin \frac{1}{y}=-p
Clearly L.H.L
\neq
R.H.L
\Rightarrow
Required limit does not exist.
The function
\displaystyle \mathrm{f}({x})=\frac{|x-3|}{x-3}
at
{x}=3
, is
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0%
Left continuous
0%
Right continuous
0%
continuous
0%
discontinuous
Explanation
we just concentrate on the function
\dfrac{|x|}{x}
LHL = lim
x \rightarrow {0}^{-} \dfrac{|x|}{x} = \dfrac{-x}{x} = -1
RHL = lim
x \rightarrow {0}^{+} \dfrac{|x|}{x} = \dfrac{x}{x} = 1
Let ,
y = x - 3
Now, if
x \rightarrow 3, then\ , y \rightarrow 0
Applying the above derived result for y,
Since, the LHL is not equal to the RHL,
Hence the function is discontinuous at y = 0 or x = 3
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Practice Class 11 Commerce Applied Mathematics Quiz Questions and Answers
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