12th Grade > Mathematics
LIMITS CONTINUITY AND DIFFERENTIABILITY MCQs
Total Questions : 45
| Page 3 of 5 pages
Answer: Option B. -> sin2β(2β)
:
B
limα→βsin2α−sin2βα2−β2=limα→βsin(α−β)sin(α+β)(α−β)(α+β)=sin2β(2β)
:
B
limα→βsin2α−sin2βα2−β2=limα→βsin(α−β)sin(α+β)(α−β)(α+β)=sin2β(2β)
Answer: Option B. -> 1
:
B
limx→∞√x+sinxx−cosx
=limx→∞x12√1+sinxxx12√1−cosxx
We know, that for any value of x, sinx and cosx will be [-1,1]
So,=limx→∞Sinxx=0
And =limx→∞Cosxx=0
=x12√1+sinxx√1+sinxx√1+cosxx
=limx→∞11
=1
:
B
limx→∞√x+sinxx−cosx
=limx→∞x12√1+sinxxx12√1−cosxx
We know, that for any value of x, sinx and cosx will be [-1,1]
So,=limx→∞Sinxx=0
And =limx→∞Cosxx=0
=x12√1+sinxx√1+sinxx√1+cosxx
=limx→∞11
=1
Answer: Option C. -> x2−17x+66=0
:
C
f(x)={x2−3,2<x<32x+5,3<x<4
∴limx→3−f(x)=limx→3−(x2−3)=6
and limx→3+f(x)=limx→3+(2x+5)=11
Hence, the required equation will be
x2 - (sum of roots)x + (Products of roots) = 0
:
C
f(x)={x2−3,2<x<32x+5,3<x<4
∴limx→3−f(x)=limx→3−(x2−3)=6
and limx→3+f(x)=limx→3+(2x+5)=11
Hence, the required equation will be
x2 - (sum of roots)x + (Products of roots) = 0
Answer: Option A. -> 1
:
A
limx→π2[sin−1sinx]
= limx→π2[x]
= 1
:
A
limx→π2[sin−1sinx]
= limx→π2[x]
= 1
Answer: Option A. -> −1
:
A
limx→∞(2+x)40(4+x)5(2−x)45
=limx→∞x45(2x+1)40(4x+1)5x45(2x−1)45
=(1)40(1)5(−1)45
=(1)45(−1)45
=−1
:
A
limx→∞(2+x)40(4+x)5(2−x)45
=limx→∞x45(2x+1)40(4x+1)5x45(2x−1)45
=(1)40(1)5(−1)45
=(1)45(−1)45
=−1
Answer: Option C. -> tan−12
:
C
limn→∞n∑r=1tan−1(44r2+3)
=limn→∞n∑r=1tan−1(1r2−14+1)
=limn→∞n∑r=1tan−1((r+12)−(r−12)1+(r+12)(r−12))
=limn→∞n∑r=1tan−1{tan−1(r+12)−tan−1(r−12)}
=limn→∞{tan−1(n+12)−tan−1(12)}
= π2−tan−112
= tan−12
:
C
limn→∞n∑r=1tan−1(44r2+3)
=limn→∞n∑r=1tan−1(1r2−14+1)
=limn→∞n∑r=1tan−1((r+12)−(r−12)1+(r+12)(r−12))
=limn→∞n∑r=1tan−1{tan−1(r+12)−tan−1(r−12)}
=limn→∞{tan−1(n+12)−tan−1(12)}
= π2−tan−112
= tan−12
Answer: Option D. -> 12√6
:
D
limx→1−√25−x2−(−√24)x−1
=limx→1√24−√25−x2x−1×√24+√25−x2√24+√25−x2
limx→1x2−1(x−1)[√24+√25−x2]=22√24=12√6
:
D
limx→1−√25−x2−(−√24)x−1
=limx→1√24−√25−x2x−1×√24+√25−x2√24+√25−x2
limx→1x2−1(x−1)[√24+√25−x2]=22√24=12√6
Answer: Option A. -> 12
:
A
limx→2√x−2+√x−√2√x2−4
=limx→2(√x−2√x+2√x−2+√x−√2√x2−4)
On rationalisation -
=limx→2(1√x+2+x−2√x2−4(√x+√2))
=limx→21√x+2+limx→2√x−2x+2×1√x+√2
=12
:
A
limx→2√x−2+√x−√2√x2−4
=limx→2(√x−2√x+2√x−2+√x−√2√x2−4)
On rationalisation -
=limx→2(1√x+2+x−2√x2−4(√x+√2))
=limx→21√x+2+limx→2√x−2x+2×1√x+√2
=12
Answer: Option B. -> −34
:
B
limn→∞2.3n−3.5n3.3n+4.5n
=limn→∞5n(2(35)n−3)5n(3(35)n+4)
Asn→∞,(35)n→0
=−34
:
B
limn→∞2.3n−3.5n3.3n+4.5n
=limn→∞5n(2(35)n−3)5n(3(35)n+4)
Asn→∞,(35)n→0
=−34
Answer: Option A. -> e52
:
A
limx→0e1x2(log(1+2x)−log(1+3x))+1xelimx→0(log(1+2x)−log(1+3x))+xx2
Using the expansion we get -
=e52
:
A
limx→0e1x2(log(1+2x)−log(1+3x))+1xelimx→0(log(1+2x)−log(1+3x))+xx2
Using the expansion we get -
=e52
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