Limits Continuity And Differentiability(11th And 12th > Mathematics ) Questions and Answers
Question 1.
f(x)={4x−3,x<1x2x≥1, then limx→1f(x)=
1
−1
0
0
Explanation:-
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
Question 2.
limn→∞2.3n−3.5n3.3n+4.5n=
23
−34
1
0
Explanation:-
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
Question 3.
limα→βsin2α−sin2βα2−β2 is equal to
sin3β(4β)
sin2β(2β)
sin8β(7β)
sin6β(4β)
Explanation:-
Answer: Option B. -> sin2β(2β) : B limα→βsin2α−sin2βα2−β2=limα→βsin(α−β)sin(α+β)(α−β)(α+β)=sin2β(2β)
Question 4.
limx→∞(2+x)40(4+x)5(2−x)45
−1
1
16
32
Explanation:-
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
Question 5.
limx→∞√x+sinxx−cosx=
0
1
−1
does not exist
Explanation:-
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
Question 6.
limx→2√x−2+√x−√2√x2−4 is equal to
12
1
2
0
Explanation:-
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
Question 7.
The value of limx→0xa[bx] is, ([.]→G.I.F)
0
∞
ba
does not exist
Explanation:-
Answer: Option C. -> ba : C limx→0xa[bx] =limx→0xa(bx−{bx}) Since {bx}ϵ[0,1) =limx→0xa.{bx}=0 limx→0xa[bx]=limx→0(xa)(bx) =limx→0.ba =ba
Question 8.
limx→02sinx−sin2xx3 is equal to
1
−1
0
does not exist
Explanation:-
Answer: Option A. -> 1 : A limx→02sinx−sin2xx3 =limx→02sinx(1−cosx)(1+cosx)x3(1+cosx) =limx→02sin3xx3×11+cosx =2×(1)3×11+1=1
Question 9.
If G(x) = - √25−x2 then limx→1G(x)−G(1)x−1 has the value
124
15
−√24
12√6
Explanation:-
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
Question 10.
If limx→0((a−n)nx−tanx)sinnxx2=0 where n is nonzero real number, then a is equal to
0
n+1n
n
n+1n
Explanation:-
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