12th Grade > Mathematics
LIMITS CONTINUITY AND DIFFERENTIABILITY MCQs
Total Questions : 45
| Page 1 of 5 pages
Answer: Option C. -> 116
:
C
limx→π2cotx−cosx(π−2x)3
Let x=π2+t
If x→π2,t→0
limt→0sint−tant−8t3
=limt→0(t−t33!+t55!+…)−(t+t33+2t515+…)−8t3
=116
We can put x=π2−t and we'll get L.H.L also same.
Since L.H.L = R.H.L the limit exists and is =116
:
C
limx→π2cotx−cosx(π−2x)3
Let x=π2+t
If x→π2,t→0
limt→0sint−tant−8t3
=limt→0(t−t33!+t55!+…)−(t+t33+2t515+…)−8t3
=116
We can put x=π2−t and we'll get L.H.L also same.
Since L.H.L = R.H.L the limit exists and is =116
Answer: Option C. -> −1
:
C
limx→−∞[x4sin(1x)+x21+|x|3]
limx→−∞x31+|x|3[xsin(1x)+1x]
As x→−∞
x31+|x|3=−1
xsin(1x)+1x=1
⇒limx→−∞x31+|x|3(xsin1x+1x)=−1
:
C
limx→−∞[x4sin(1x)+x21+|x|3]
limx→−∞x31+|x|3[xsin(1x)+1x]
As x→−∞
x31+|x|3=−1
xsin(1x)+1x=1
⇒limx→−∞x31+|x|3(xsin1x+1x)=−1
Answer: Option D. -> n+1n
:
D
limx→0((a−n)nx−tanx)sinnxx2=0
limx→0((a−n)n−(tanxx)).sinnxnx.n=0
⇒((a−n)n−1).1.n=0
⇒a=1n+n
:
D
limx→0((a−n)nx−tanx)sinnxx2=0
limx→0((a−n)n−(tanxx)).sinnxnx.n=0
⇒((a−n)n−1).1.n=0
⇒a=1n+n
Answer: Option B. -> e−1
:
B
limx→1(tanxπ4)tanπx2(1∞from)
elimx→1(tanxπ4−1)tanπx2
=elimx→1(sinπx4−cosπx4cosπx4)2sinπx4cosπx4cosπx2
=elimx→12sin2πx4−2sinπx4cosπx4cosπx2
=elimx→1(1−cosπx2−sinπx2cosπx2)
=elimx→1(1−sinπx2cosπx2−1)
=elimx→1(cosπx21+sinπx2−1)
= e−1
:
B
limx→1(tanxπ4)tanπx2(1∞from)
elimx→1(tanxπ4−1)tanπx2
=elimx→1(sinπx4−cosπx4cosπx4)2sinπx4cosπx4cosπx2
=elimx→12sin2πx4−2sinπx4cosπx4cosπx2
=elimx→1(1−cosπx2−sinπx2cosπx2)
=elimx→1(1−sinπx2cosπx2−1)
=elimx→1(cosπx21+sinπx2−1)
= e−1
Answer: Option A. -> 1
:
A
limx→1−f(x)=limx→1−(4x−3)=1
limx→1+f(x)=limx→1+x2=1
Since LHL = RHL
⇒limx→1f(x)=1
:
A
limx→1−f(x)=limx→1−(4x−3)=1
limx→1+f(x)=limx→1+x2=1
Since LHL = RHL
⇒limx→1f(x)=1
Answer: Option B. -> a+b
:
B
Since limit of a function is a + b as x →0, therefore to be continuous at a function, its value must be
a + b at x = 0 ⇒f(0) = a + b.
:
B
Since limit of a function is a + b as x →0, therefore to be continuous at a function, its value must be
a + b at x = 0 ⇒f(0) = a + b.
Answer: Option B. -> 1
:
B
limn→∞cos2nx = {1,x=rπ,r∈I0,x≠rπ,r∈I
Here, for x = 10, limn→∞cos2n(x−10) = 1
and in all other cases it is zero
∴limn→∞20∑x=1cos2n(x−10) = 1
:
B
limn→∞cos2nx = {1,x=rπ,r∈I0,x≠rπ,r∈I
Here, for x = 10, limn→∞cos2n(x−10) = 1
and in all other cases it is zero
∴limn→∞20∑x=1cos2n(x−10) = 1
Answer: Option B. -> 1
:
B
limn→∞an+bnan−bn
a>b>1
=nn→∞1+(ba)n1−(ba)n
=1
:
B
limn→∞an+bnan−bn
a>b>1
=nn→∞1+(ba)n1−(ba)n
=1
Answer: Option D. -> 1
:
D
Given that,
f(x) = {sinxx≠nπ,n=0,±1,±2...2,otherwise
and
g(x) = ⎧⎪⎨⎪⎩x2+1,x≠0,24,x=05,x=2
Then limx→0g[f(x)]=limx→9g(sinx)
=limx→0(sin2x+1)=1
:
D
Given that,
f(x) = {sinxx≠nπ,n=0,±1,±2...2,otherwise
and
g(x) = ⎧⎪⎨⎪⎩x2+1,x≠0,24,x=05,x=2
Then limx→0g[f(x)]=limx→9g(sinx)
=limx→0(sin2x+1)=1
Answer: Option A. -> 1120
:
A
limx→0sinx−x+x36x5 ....... (1)
We know expansion of Sinx.
sinx=x−x33!+x55!+ ........ (2)
Substituting (2) in (1) we get
limx→0x55!.x5
=15!
=1120
:
A
limx→0sinx−x+x36x5 ....... (1)
We know expansion of Sinx.
sinx=x−x33!+x55!+ ........ (2)
Substituting (2) in (1) we get
limx→0x55!.x5
=15!
=1120
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