Test: Limits And Derivatives - 1 - JEE MCQ

# Test: Limits And Derivatives - 1 - JEE MCQ

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## 25 Questions MCQ Test - Test: Limits And Derivatives - 1

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Test: Limits And Derivatives - 1 - Question 1

### is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 1

We have:
limx→0 ln(1+ax)]/x=a
lim x→0 ln(1+ax)/ax=a
and then:
lim x→0 [ln(1+ax)-ln(1+bx)]/x
= a-b

Test: Limits And Derivatives - 1 - Question 2

### is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 2

On rationalizing,
Lt x-->0 (1-Cosx)/[√(1+x) +1]
= lt x-->0 2sin2(x/2)[√(1+x)+1] /x
= lt x-->0 [2sin(x/2)/x].sinx/2 [√(1+x) +1 ]
= 2.1/2 .0[1+1]=0

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Test: Limits And Derivatives - 1 - Question 3

### Let f (x) = x sin 1/x, x ≠ 0, then the value of the function at x = 0, so that f is continuous at x = 0, is

Detailed Solution for Test: Limits And Derivatives - 1 - Question 3

f(0) = lim(x→0) x sin (1/x)
​We know ∀ x ∈ R,sin (1/x) ∈ [−1,1]
Hence, f(0) = lim(x→0) xsin(1/x)
= 0

Test: Limits And Derivatives - 1 - Question 4

The positive integer n so that limx→3 (xn – 3n)/(x – 3) = 108 is

Detailed Solution for Test: Limits And Derivatives - 1 - Question 4

We know that,

limx→3 (xn – 3n)/(x – 3) = n(3)n-1

Thus, n(3)n-1 = 108 {from the given}

n(3)n-1 = 4(27) = 4(33) = 4(3)4-1

Therefore, n = 4

Test: Limits And Derivatives - 1 - Question 5

is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 5

lim x→0 (1−cosx)/(xlog(1+x))
= lim x→0 {2sin2 x/2}/[xlog(1+x)]
= lim x→0 {{2sin2x/2/(x/2)2}*/1/4}/[1/xlog(1+x)]
= lim x→0 [log(1+x)/x]1/2
= 2​

Test: Limits And Derivatives - 1 - Question 6

is equal to

Test: Limits And Derivatives - 1 - Question 7

Detailed Solution for Test: Limits And Derivatives - 1 - Question 7

Consider the given function.
lim x→0 ((1-x)n −1)/x)
This is 0/0 form.
So, apply L-Hospital rule,
lim x→0 [nx-1x(1-x)n-1 −0)]/1
lim x→ 0(-n(1-x)n-1)
=-n

Test: Limits And Derivatives - 1 - Question 8

is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 8

lim(x → 0) (tanx-x)/x2 tanx
As we know that tan x = sinx/cosx
lim(x → 0) (sinx/cosx - x)/x2(sinx/cosx)
lim(x → 0) (sinx - xcosx)/(x2 sinx)
lim(x → 0) cosx - (-xsinx + cosx)/(x2cosx + sinx (2x))
lim(x → 0) (cosx + xsinx - cosx)/x2cosx + 2xsinx)
lim(x → 0) sinx/(xcosx + 2sinx)
Hence it is 0/0 form, apply L hospital rule
lim(x → 0) cosx/(-xsinx + cosx + 2cosx)
⇒ 1/(0+1+2)
= 1/3

Test: Limits And Derivatives - 1 - Question 9

Detailed Solution for Test: Limits And Derivatives - 1 - Question 9

lim x-->0 ((x2cosx-)/1-cosx)
Differentiating numerator and denominator
lim x-->0 (2xcosx - x2sinx)/sinx
Differentiating it for a second time
lim x-->0 (2cos - 2xsinx - 2xsinx - x2cosx)/cosx
Substituting x = 0
= lim x-->0  2/1
= 2

Test: Limits And Derivatives - 1 - Question 10

is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 10

Suppose that the Reqd. Limit L = lim x→0 (sinx − x)/x3.
Substiture x = 3y , so that, as x→0 , y→0.
∴ L = lim y→0 sin3y − 3y/(3y)3,
= lim y→0 (3siny − 4sin3y)− 3y)/27y3,
= lim y→0 {3(siny − y)/27y3) − (4sin3y/27y3)},
⇒ L = lim y→0 1/9 * (siny − y)/y3) − 4/27* (siny/y)3...(∗)
.Note that, here,
= lim y→0(siny − y)/y3)
= lim x→0 (sinx−x)/x3) = L.
Therefore, (∗) ⇒ L = 1/9*L − 4/27
or,  8/9L = −4/27
Hence, L = −4/27*9/8
=−1/6

Test: Limits And Derivatives - 1 - Question 11

is equal to

Test: Limits And Derivatives - 1 - Question 12

then for f to be continuous at x = 0, f (0) must be equal to

Test: Limits And Derivatives - 1 - Question 13

Test: Limits And Derivatives - 1 - Question 14

where a > 0, is equal to

Test: Limits And Derivatives - 1 - Question 15

(x=n) , is equal to

Test: Limits And Derivatives - 1 - Question 16

is equal to

Test: Limits And Derivatives - 1 - Question 17

is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 17

lim(h → 0) [sin2(x+h) - sin2x]/h
l = lim(h→0) = [2sin(x+h)cos(x+h)]/1
{As we know that sin2x = 2sinxcosx
= sin2x

Test: Limits And Derivatives - 1 - Question 18

If a is a real number, then

Test: Limits And Derivatives - 1 - Question 19

If k be a integer, then  is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 19

- lim x-->k+ = x-{x}
Lim x→k+   k-{k}
= 0

Test: Limits And Derivatives - 1 - Question 20

is equal to

Detailed Solution for Test: Limits And Derivatives - 1 - Question 20

Correct Answer :- C

Explanation : lt(x->a-) -|x-a|/x-a

= - |-a-a|/(-a-a)

= -|-2a|/(-2a)

= -2a/(-2a)

= 1

Test: Limits And Derivatives - 1 - Question 21

is equal to

Test: Limits And Derivatives - 1 - Question 22

Maximum value of x3−3x+2in [0,2] is

Test: Limits And Derivatives - 1 - Question 23

Detailed Solution for Test: Limits And Derivatives - 1 - Question 23

√x / √(16+√x) - 4

= (√x * √(16+√x) + 4) / ( √(16+√x) - 4) * (√(16+√x) + 4)

= (√x * √(16+√x) + 4) / 16+√x - 16

= √(16+√x) + 4

now at lim x->0

= √16 + 4 = 8

Test: Limits And Derivatives - 1 - Question 24

is equal to

Test: Limits And Derivatives - 1 - Question 25

Detailed Solution for Test: Limits And Derivatives - 1 - Question 25

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