12th Grade > Mathematics
LIMITS CONTINUITY AND DIFFERENTIABILITY MCQs
Total Questions : 45
| Page 5 of 5 pages
Answer: Option A. -> a=−1,b=12
:
A
We have,limx→−∞(√x2−x+1−ax−b)=0[putting x=-y; x→−∞,y→∞]⇒limy→∞(√y2+y+1+ay−b)=0⇒limy→∞[y(1+1y+1y2)1/2+ay−b]=0⇒limy→∞[y{1+12(1y+1y2+.........)}+ay−b]=0⇒limy→∞[y(1+a)+(12−b)+12y+........]=0⇒1+a=0and(1/2)−b=0⇒a=−1andb=1/2
:
A
We have,limx→−∞(√x2−x+1−ax−b)=0[putting x=-y; x→−∞,y→∞]⇒limy→∞(√y2+y+1+ay−b)=0⇒limy→∞[y(1+1y+1y2)1/2+ay−b]=0⇒limy→∞[y{1+12(1y+1y2+.........)}+ay−b]=0⇒limy→∞[y(1+a)+(12−b)+12y+........]=0⇒1+a=0and(1/2)−b=0⇒a=−1andb=1/2
Answer: Option A. -> 0
:
A
limx→0(cosx)1/x4=elimx→0cosx−1x4
elimx→0−2sinx2x2.sinx2x2×14x2=e−∞=0
:
A
limx→0(cosx)1/x4=elimx→0cosx−1x4
elimx→0−2sinx2x2.sinx2x2×14x2=e−∞=0
Answer: Option B. -> a
:
B
limx→αsin(ax2+bx+c)(x−α)2=limx→αsina(x−α)(x−α)(x−α)2
=limx→αsina(x−α)2a(x−α)2×a=a
:
B
limx→αsin(ax2+bx+c)(x−α)2=limx→αsina(x−α)(x−α)(x−α)2
=limx→αsina(x−α)2a(x−α)2×a=a
Answer: Option C. -> (n!)an
:
C
limx→0(1x+2x+3x+...+nxn)a/x(1∞form)
=elimx→0(1x+2x+3x+...+nxn−1).an
=elimx→0(1x+2x+3x+...+nx−nn).an
=elimx→0(1x+2x+3x+...+nx−nx).an
=elimx→0{1x−1x+2x−1x+...+nx−1x}an
We know limx→0{ax−1x}=a
So, =e(log1+log2+log3+...logn)an
ea/n(log(1.2.3...n)) = e(logen!)a/n=(n!)a/n
:
C
limx→0(1x+2x+3x+...+nxn)a/x(1∞form)
=elimx→0(1x+2x+3x+...+nxn−1).an
=elimx→0(1x+2x+3x+...+nx−nn).an
=elimx→0(1x+2x+3x+...+nx−nx).an
=elimx→0{1x−1x+2x−1x+...+nx−1x}an
We know limx→0{ax−1x}=a
So, =e(log1+log2+log3+...logn)an
ea/n(log(1.2.3...n)) = e(logen!)a/n=(n!)a/n
Answer: Option A. -> 2/3
:
A
limx→03√1+sinx−3√1−sinxx=limx→0(1+sinx)−(1−sins)(1+sinx)2/3+(1+sinx)2/3+(1−sinx)2/3(1−sinx)2/3×1x=limx→02(1+sinx)2/3+(1+sinx)1/3+(1−sinx)1/3+(1−sinx)2/3×sin(x)x
= 23.1=23
:
A
limx→03√1+sinx−3√1−sinxx=limx→0(1+sinx)−(1−sins)(1+sinx)2/3+(1+sinx)2/3+(1−sinx)2/3(1−sinx)2/3×1x=limx→02(1+sinx)2/3+(1+sinx)1/3+(1−sinx)1/3+(1−sinx)2/3×sin(x)x
= 23.1=23
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