12th Grade > Mathematics
LIMITS CONTINUITY AND DIFFERENTIABILITY MCQs
Total Questions : 45
| Page 2 of 5 pages
Answer: Option B. -> 4
:
B
Given that f(9) = 9, f’(9) = 4
limx→9√f(x)−3√x−3
On rationalisation we get -
=limx→9(√f(x)−3√x−3)(√x+3√x+3)(√f(x)+3√f(x)+3)
=limx→9f(x)−f(9)x−9 .66
=limx→9f(x)−f(9)x−9
=f′(9)
=4
:
B
Given that f(9) = 9, f’(9) = 4
limx→9√f(x)−3√x−3
On rationalisation we get -
=limx→9(√f(x)−3√x−3)(√x+3√x+3)(√f(x)+3√f(x)+3)
=limx→9f(x)−f(9)x−9 .66
=limx→9f(x)−f(9)x−9
=f′(9)
=4
Answer: Option D. -> 1
:
D
limx→0g{f(x)}=limx→0⎧⎪⎨⎪⎩(f(x))2+1,f(x)≠0,24,f(x)=05,f(x)=2=limx→0⎧⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎩sin2x+1,sinx≠0,2andx≠nπ4,sinx=0andx≠nπ5,sinx=2andx≠nπ5,2≠0,2andx≠nπ4,2=0andx=nπ5,2=2andx=nπ=limx→0{1+sin2x,x≠nπ5,x=nπ=1+sin20=1+0=1
:
D
limx→0g{f(x)}=limx→0⎧⎪⎨⎪⎩(f(x))2+1,f(x)≠0,24,f(x)=05,f(x)=2=limx→0⎧⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎨⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪
⎪⎩sin2x+1,sinx≠0,2andx≠nπ4,sinx=0andx≠nπ5,sinx=2andx≠nπ5,2≠0,2andx≠nπ4,2=0andx=nπ5,2=2andx=nπ=limx→0{1+sin2x,x≠nπ5,x=nπ=1+sin20=1+0=1
Answer: Option B. -> exists for every real k
:
B
limx→2+f(x)=limx→2+k(x2−4)2−x=limx→2+k(−2−x)=−4klimx→2+f(x)=limx→2+k(x2−4)2−x=limx→2−k(−2−x)=−4k
Hence limits exists for every real value of k.
:
B
limx→2+f(x)=limx→2+k(x2−4)2−x=limx→2+k(−2−x)=−4klimx→2+f(x)=limx→2+k(x2−4)2−x=limx→2−k(−2−x)=−4k
Hence limits exists for every real value of k.
Answer: Option B. -> 9(2log3)
:
B
limx→01−cos3xx(3x−1)
cos3x=1−9x22!+…
1−cos3x=9x22!+… ......... (1)
limx→03x−1x=log3 .......... (2)
limx→01−cos3xx2(3x−1x)
From (1) and (2) we get
=92log3
:
B
limx→01−cos3xx(3x−1)
cos3x=1−9x22!+…
1−cos3x=9x22!+… ......... (1)
limx→03x−1x=log3 .......... (2)
limx→01−cos3xx2(3x−1x)
From (1) and (2) we get
=92log3
Answer: Option A. -> 1
:
A
Since x→1 means x→1-, ∵ sin-1 x is not defined for x>1
limx→1[sinsin−1x]=limx→1−[sinsin−1x]=1
:
A
Since x→1 means x→1-, ∵ sin-1 x is not defined for x>1
limx→1[sinsin−1x]=limx→1−[sinsin−1x]=1
Answer: Option A. -> 1/2
:
A
=limx→2{1√x+2+√x−√2√x2−4}Limit=12+limx→2x−2√x+√21√(x+2)(x−2)=12+limx→21√x+√2√x−2x+2=12
:
A
=limx→2{1√x+2+√x−√2√x2−4}Limit=12+limx→2x−2√x+√21√(x+2)(x−2)=12+limx→21√x+√2√x−2x+2=12
Answer: Option B. -> In 3
:
B
3=limx→0(1+asinx)cosecx
[1∞form]⇒limx→0ecosecx.asinx=ea
∴ea=3⇒a=loge3=ln3.
:
B
3=limx→0(1+asinx)cosecx
[1∞form]⇒limx→0ecosecx.asinx=ea
∴ea=3⇒a=loge3=ln3.
Answer: Option D. -> limx→00x
:
D
∵limx→ax3−a3x−a=(00inderminant form)
limx→asinxcosecx=(0×∞inderminant form)
limx→axx=(00inderminant form)
limx→a0x=0(not inderminant form)
:
D
∵limx→ax3−a3x−a=(00inderminant form)
limx→asinxcosecx=(0×∞inderminant form)
limx→axx=(00inderminant form)
limx→a0x=0(not inderminant form)
Answer: Option C. -> 3/2
:
C
limx→01−cos3xxsinxcosx=limx→0(1−cosx)(1+cosx+cos2x)xsinxcosx=limx→02sin2(x2)2sin(x2)cos(x2).x×(1+cosx+cos2x)cosx=limx→0sin(x2)2(x2)×1+cosx+cos2xcos(x2)cosx=12×3=32
:
C
limx→01−cos3xxsinxcosx=limx→0(1−cosx)(1+cosx+cos2x)xsinxcosx=limx→02sin2(x2)2sin(x2)cos(x2).x×(1+cosx+cos2x)cosx=limx→0sin(x2)2(x2)×1+cosx+cos2xcos(x2)cosx=12×3=32
Answer: Option A. -> π4
:
A
limx→∞n.cos(π4n)sin(π4n)
=limn→∞n2sinπ2n
=π4
:
A
limx→∞n.cos(π4n)sin(π4n)
=limn→∞n2sinπ2n
=π4
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