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JEE Advanced

Explore popular questions from Limits, Continuity and Differentiability for JEE Advanced. This collection covers Limits, Continuity and Differentiability previous year JEE Advanced questions hand picked by experienced teachers.

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Limits, Continuity and Differentiability

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Q 1. If {tex} f ( x ) = \sqrt { \frac { x - \sin x } { x + \cos ^ { 2 } x } } , {/tex} then {tex} \lim _ { x \rightarrow \infty } f ( x ) {/tex} is

A

{tex} 0 {/tex}

B

{tex} \infty {/tex}

{tex} 1 {/tex}

D

none of these

Explanation

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Q 2. If {tex} f ( a ) = 2 , f ^ { \prime } ( a ) = 1 , g ( a ) = - 1 , g ^ { \prime } ( a ) = 2 , {/tex} then the value of {tex} \lim _ { x \rightarrow a } \frac { g ( x ) f ( a ) - g ( a ) f ( x ) } { x - a } {/tex} is

A

{tex} - 5 {/tex}

B

{tex} \frac { 1 } { 5 } {/tex}

5

D

none of these

Explanation

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Q 3. {tex} \lim _ { n \rightarrow \infty } \left\{ \frac { 1 } { 1 - n ^ { 2 } } + \frac { 2 } { 1 - n ^ { 2 } } + \ldots + \frac { n } { 1 - n ^ { 2 } } \right\} {/tex} is equal to

A

{tex}0{/tex}

{tex} - \frac { 1 } { 2 } {/tex}

C

{tex} \frac { 1 } { 2 } {/tex}

D

none of these

Explanation

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Q 4. {tex} \begin{aligned} \text { If } f ( x ) & = \frac { \sin [ x ] } { [ x ] } , \quad [ x ] \neq 0 \\ & = 0 ,\quad \quad\ \quad [ x ] = 0 \end{aligned} {/tex}
Where {tex} [ x ] {/tex} denotes the greatest integer less than or equal to {tex} x {/tex}. then {tex} \lim _ { x \rightarrow 0 } f ( x ) {/tex} equals {tex} - {/tex}

A

{tex} 1 {/tex}

B

{tex} 0 {/tex}

C

{tex} - 1 {/tex}

none of these

Explanation


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Q 5. The function {tex} f ( x ) = [ x ] ^ { 2 } - \left[ x ^ { 2 } \right] ( \text { where } [ y ] {/tex} is the greatest integer less than or equal to {tex} y ) , {/tex} is discontinuous at

A

all integers

B

all integers except 0 and 1

C

all integers except 0

all integers except 1

Explanation


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Q 6. The function {tex} f ( x ) = \left( x ^ { 2 } - 1 \right) \left| x ^ { 2 } - 3 x + 2 \right| + \cos ( | x | ) {/tex} is NOT differentiable at

A

-1

B

0

C

1

2

Explanation




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Q 7. For {tex} x \in R , \lim _ { x \rightarrow \infty } \left( \frac { x - 3 } { x + 2 } \right) ^ { x } = {/tex}

A

{tex} e {/tex}

B

{tex} e ^ { - 1 } {/tex}

{tex} e ^ { - 5 } {/tex}

D

{tex} e ^ { 5 } {/tex}

Explanation

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Q 8. {tex} \lim _ { h \rightarrow 0 } \frac { f \left( 2 h + 2 + h ^ { 2 } \right) - f ( 2 ) } { f \left( h - h ^ { 2 } + 1 \right) - f ( 1 ) } , {/tex} given that {tex} f ^ { \prime } ( 2 ) = 6 {/tex} and {tex} f ^ { \prime } ( 1 ) = 4 {/tex}

A

does not exist

B

is equal to {tex} - 3 / 2 {/tex}

C

is equal to {tex} 3 / 2 {/tex}

is equal to {tex}3 {/tex}

Explanation

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Q 9. If {tex} \lim _ { x \rightarrow 0 } \left[ 1 + x \ell n \left( 1 + b ^ { 2 } \right) \right] ^ { 1 / x } = 2 b \sin ^ { 2 } \theta , b > 0 {/tex} and {tex} \theta \in ( - \pi , \pi ] {/tex} then the value of {tex} \theta {/tex} is

A

{tex} \pm \frac { \pi } { 4 } {/tex}

B

{tex} \pm \frac { \pi } { 3 } {/tex}

C

{tex} \pm \frac { \pi } { 6 } {/tex}

{tex} \pm \frac { \pi } { 2 } {/tex}

Explanation


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Q 10. Let {tex} \alpha ( a ) {/tex} and {tex} \beta ( a ) {/tex} be the roots of the equation {tex} ( \sqrt [ 3 ] { 1 + a } - 1 ) x ^ { 2 } + ( \sqrt { 1 + a } - 1 ) x + ( \sqrt [ 6 ] { 1 + a } - 1 ) = 0 {/tex} where {tex} a > - 1 . {/tex} Then {tex} \lim _ { a \rightarrow 0 ^ { + } } \alpha ( a ) {/tex} and {tex} \lim _ { a \rightarrow 0 ^ { + } } \beta ( a ) {/tex} are

A

{tex} - \frac { 5 } { 2 } {/tex} and 1

{tex} - \frac { 1 } { 2 } {/tex} and -1

C

{tex} - \frac { 7 } { 2 } {/tex} and 2

D

{tex} - \frac { 9 } { 2 } {/tex} and 3

Explanation


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Q 11. If , where
. Then, is equal to

A

-1

1

C

0

D

Cannot be determined

Explanation

Given,


At

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Q 12. If is continuous at , then is equal to

A

4

B

2

C

1

{tex}\frac{1}{4}{/tex}

Explanation


[by L Hospitals rule]
Since, is continuous at

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Q 13. Let be differentiable for all . If and for , then

A

B

C

Explanation

Since,


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Q 14. If for a continuous function f, and , then is equal to

A

1

2

C

0

D

None of these

Explanation

We have,



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Q 15. If , then is

A

Discontinuous and non-differentiable at and

B

Continuous and differentiable at

Discontinuous at

D

Continuous but not differentiable at

Explanation

We have,


It is evident from the definition that is discontinuous at

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Q 16. Given that is a differentiable function of and that and that . Then, is equal to

10

B

24

C

15

D

None of these

Explanation

We have,







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Q 17. If is continuous at and , then is

A

0

2

C

D

None of these

Explanation

Since is continuous at
(i)
Now, using L Hospitals rule, we have

[Using (i)]

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Q 18. The function , is

A

Continuous but not differentiable at

B

Discontinuous at

Continuous and differentiable at

D

Not defined at

Explanation

Continuity at
LHL=
RHL
LHL=RHL , it is continuous
Differentiability at
LHD

RHD

LHD=RHD
Hence, it is differentiable.