Free Objective Test 01 Practice Test - 11th and 12th
Question 1
If limx→0(1+asinx)cosecx=3, then a is
SOLUTION
Solution : B
3=limx→0(1+asinx)cosecx
[1∞form]⇒limx→0ecosecx.asinx=ea
∴ea=3⇒a=loge3=ln3.
Question 2
limx→0sinx−x+x36x5 is equal to
SOLUTION
Solution : A
limx→0sinx−x+x36x5 ....... (1)
We know expansion of Sinx.
sin x=x−x33!+x55!+ ........ (2)
Substituting (2) in (1) we get
limx→0 x55!.x5
=15!
=1120
Question 3
limn→∞an+bnan−bn, where a>b>1, is equal to
SOLUTION
Solution : B
limn→∞an+bnan−bn
a>b>1
=nn→∞1+(ba)n1−(ba)n
=1
Question 4
limx→π2cotx−cosx(π−2x)3 is equal to
SOLUTION
Solution : 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
Question 5
limx→−∞{x4sin(1x)+x21+|x|3} is equal to
SOLUTION
Solution : C
limx→−∞[x4sin(1x)+x21+|x|3]
limx→−∞x31+|x|3[xsin(1x)+1x]
As x→−∞
x31+|x|3=−1
x sin(1x)+1x=1
⇒limx→−∞x31+|x|3(x sin1x+1x)=−1
Question 6
If f(9) = 9, f'(9) = 4, then limx→9√f(x)−3√x−3 equals
SOLUTION
Solution : 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→9 f(x)−f(9)x−9 . 66
=limx→9 f(x)−f(9)x−9
=f′(9)
=4
Question 7
If 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)} is
SOLUTION
Solution : 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
Question 8
limn→∞n.cos(π4n).sin(π4n) is equal to
SOLUTION
Solution : A
limx→∞n.cos(π4n)sin(π4n)
=limn→∞n2 sinπ2n
=π4
Question 9
limx→01−cos 3xx(3x−1)=
SOLUTION
Solution : B
limx→01−cos3xx(3x−1)
cos 3x=1−9x22!+…
1−cos 3x=9x22!+… ......... (1)
limx→03x−1x=log 3 .......... (2)
limx→01−cos 3xx2(3x−1x)
From (1) and (2) we get
=92 log 3
Question 10
Which of the following limit is not in the indeterminant form ?
limx→ax3−a3x−a
limx→asin x cosec x
limx→0xx
limx→00x
SOLUTION
Solution : D
∵limx→ax3−a3x−a=(00 inderminant form)
limx→asin x cosec x=(0×∞ inderminant form)
limx→axx=(00 inderminant form)
limx→a0x=0 (not inderminant form)
Question 11
Let f(x)={x2k(x2−4)2−xwhen x is an int eger otherwise then limx→2f(x)
exists only when k=1
exists for every real k
exists for every real k = 1
does not exist
SOLUTION
Solution : 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.
Question 12
limx→2√x−2+√x−√2√x2−4is equal to
1/2
1
2
-1
SOLUTION
Solution : 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
Question 13
The value of limx→1[sin sin−1x] is
1
does not exist
π2
0
SOLUTION
Solution : A
Since x→1 means x→1-, ∵ sin-1 x is not defined for x>1
limx→1[sin sin−1x]=limx→1−[sin sin−1x]=1
Question 14
The value oflimx→01−cos3xxsin xcosx
2/5
3/5
3/2
3/4
SOLUTION
Solution : C
limx→01−cos3xxsin xcosx=limx→0(1−cos x)(1+cos x+cos2x)xsin xcos x=limx→02sin2(x2)2sin(x2)cos(x2).x×(1+cosx+cos2x)cos x=limx→0sin(x2)2(x2)×1+cosx+cos2xcos(x2)cos x=12×3=32
Question 15
If f(x)={sinx,x≠nπ,nϵI2,otherwise and g(x)=⎧⎪⎨⎪⎩x2+1,x≠0,24,x=05,x=2, then limx→0g{f(x)}
SOLUTION
Solution : 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≠0,2and x≠nπ4,sinx=0andx≠nπ5,sinx=2 and x≠nπ5,2≠0,2and x≠nπ4,2=0and x=nπ5,2=2and x=nπ=limx→0{1+sin2x,x≠nπ5,x=nπ=1+sin20=1+0=1