JEE Main 2022 (27 June Shift 1)

LEVELJEE Main

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $a$ be an integer such that $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to :

Q: Wait, why does $[1-x]$ become $-6$ when $x \to 7^-$?

This is a classic point of confusion! When $x \to 7^-$, $x$ is slightly less than $7$, so let's use $x=6.9$. Then $1-x = 1-6.9 = -5.9$. The greatest integer function $[t]$ returns the largest integer less than or equal to $t$. Since $-6 \le -5.9 < -5$, the greatest integer of $-5.9$ is indeed $-6$. Always be careful with negative numbers in the greatest integer function!

Select Answer:

Progressive Hints

Hint 1

Hint 1 (The Spark): Visualize the Greatest Integer Function as a staircase. At

x = 7

, the function experiences a jump discontinuity. To determine if the limit exists, you must analyze the behavior of the function in the immediate neighborhood of

x = 7

, specifically checking the values just to the left (

x \to 7^{-}

) and just to the right (

x \to 7^{+}

Hint 2

Hint 2 (The Arsenal): Recall the fundamental condition for the existence of a limit:

lim_{x \to c^{-}} f (x) = lim_{x \to c^{+}} f (x)

. You must evaluate the Left Hand Limit (LHL) and the Right Hand Limit (RHL) independently. Remember that for any

x

in the interval

[n, n + 1)

[x] = n

Hint 3

Hint 3 (The Execution): Start by calculating the LHL:

lim_{x \to 7^{-}} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}

. For

x

slightly less than

7

(e.g.,

x = 6.9

[x] = 6

and

[1 - x] = [1 - 6.9] = [- 5.9] = - 6

. Substitute these values to find the LHL in terms of

a

, then repeat the process for the RHL where

x = 7.1

Visualized Solution

Introduction to the Limit Problem

Given limit: $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$
Condition for existence: Left Hand Limit (LHL) = Right Hand Limit (RHL)
Here, $[t]$ denotes the greatest integer less than or equal to $t$ .

Visualizing GIF Jumps at $x = 7$

The functions $[x]$ and $[1 - x]$ are step functions.
They experience abrupt jumps at integer values.
We must analyze the exact values in the left and right neighborhoods of $x = 7$ .

Calculating LHL: Analyzing $[x]$

For Left Hand Limit (LHL), $x \to 7^{-}$
Let $x = 6.9$ (slightly less than $7$ )
Therefore, $[x] = 6$

Calculating LHL: Analyzing $[1 - x]$

Still for LHL ( $x \to 7^{-}$ )
If $x = 6.9$ , then $1 - x = 1 - 6.9 = - 5.9$
The greatest integer of $- 5.9$ is $- 6$
So, $[1 - x] = - 6$

Calculating LHL: Substitution

Substitute the values into the limit expression:
Numerator: $18 - (- 6) = 24$
Denominator: $6 - 3 a$
$LHL = \frac{24}{6 - 3 a}$

Calculating RHL: Analyzing $[x]$

For Right Hand Limit (RHL), $x \to 7^{+}$
Let $x = 7.1$ (slightly more than $7$ )
Therefore, $[x] = 7$

Calculating RHL: Analyzing $[1 - x]$

Still for RHL ( $x \to 7^{+}$ )
If $x = 7.1$ , then $1 - x = 1 - 7.1 = - 6.1$
The greatest integer of $- 6.1$ is $- 7$
So, $[1 - x] = - 7$

Calculating RHL: Substitution

Substitute the values into the limit expression:
Numerator: $18 - (- 7) = 25$
Denominator: $7 - 3 a$
$RHL = \frac{25}{7 - 3 a}$

Equating LHL and RHL

For the limit to exist, we must have:
$LHL = RHL$
$\frac{24}{6 - 3 a} = \frac{25}{7 - 3 a}$

Solving the Equation: Cross Multiplication

Cross-multiplying the equation:
$24 (7 - 3 a) = 25 (6 - 3 a)$

Expanding the Terms

Expanding both sides:
$24 \times 7 - 24 \times 3 a = 25 \times 6 - 25 \times 3 a$
$168 - 72 a = 150 - 75 a$

Finding the Value of $a$

Rearranging terms to isolate $a$ :
$75 a - 72 a = 150 - 168$
$3 a = - 18$
$a = - 6$

Final Conclusion

Key Takeaway: For limits involving step functions like $[x]$ , always evaluate LHL and RHL separately.
Final Answer: $a = - 6$

00:00 / 00:00

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

Solution Diagram

The Core Challenge

We are tasked with finding an integer

a

such that the limit

x \to 7 lim \frac{18 - [ 1 - x ]}{[ x ] - 3 a}

exists. The existence of a limit at a point

x = 7

requires that the path from the left must meet the path from the right at the exact same destination.

Mathematically, we require

LHL = RHL

. Because the Greatest Integer Function (GIF) jumps at every integer,

x = 7

is a critical point where we must be extremely vigilant.

Phase 1

The Left Hand Limit (LHL)

Imagine standing on the number line just to the left of

7

, represented as

x \to 7^{-}

, or roughly

x = 6.9

First, we evaluate

[x]

. Since

6 \leq 6.9 < 7

, the greatest integer

[6.9]

6

Next, we evaluate

[1 - x]

. If

x = 6.9

, then

1 - x = 1 - 6.9 = - 5.9

. The greatest integer of

- 5.9

- 6

Substituting these into our expression, the numerator becomes

18 - (- 6) = 24

, and the denominator becomes

6 - 3 a

. Thus:

LHL = \frac{24}{6 - 3 a}

Phase 2

The Right Hand Limit (RHL)

Now, we step to the right of

7

, represented as

x \to 7^{+}

, or roughly

x = 7.1

Here,

[x]

[7.1] = 7

. For the second part,

1 - x = 1 - 7.1 = - 6.1

. The greatest integer of

- 6.1

- 7

Substituting these into our expression, the numerator becomes

18 - (- 7) = 25

, and the denominator becomes

7 - 3 a

. Thus:

RHL = \frac{25}{7 - 3 a}

Phase 3

The Bridge

For the limit to exist, our LHL and RHL must be equal. We set up the equation:

\frac{24}{6 - 3 a} = \frac{25}{7 - 3 a}

Now, we cross-multiply to solve for

a

24 (7 - 3 a) = 25 (6 - 3 a)

Expanding this, we get:

168 - 72 a = 150 - 75 a

Rearranging the terms, we find:

75 a - 72 a = 150 - 168

This simplifies to

3 a = - 18

. Finally, dividing by

3

, we arrive at the final result:

$a = - 6$

Common Traps & FAQs

Wait, why does

[1 - x]

become

- 6

when

x \to 7^{-}

This is a classic point of confusion! When

x \to 7^{-}

x

is slightly less than

7

, so let's use

x = 6.9

. Then

1 - x = 1 - 6.9 = - 5.9

. The greatest integer function

[t]

returns the largest integer less than or equal to

t

. Since

- 6 \leq - 5.9 < - 5

, the greatest integer of

- 5.9

is indeed

- 6

. Always be careful with negative numbers in the greatest integer function!

Why can't we just plug in

x = 7

directly into the expression?

Great question. The greatest integer function is discontinuous at every integer. If you plug in

x = 7

directly, you are ignoring the jump that occurs at that point. The limit exists only if the function approaches the same value from both sides. By plugging in

7

directly, you miss the distinct values the function takes on the left and right sides of

7

How do I know if the limit exists?

The limit exists if and only if the left-hand limit equals the right-hand limit. In this problem, we calculated the LHL as

\frac{24}{6 - 3 a}

and the RHL as

\frac{25}{7 - 3 a}

. By setting them equal, we ensure the function bridges the gap at

x = 7

, allowing the limit to exist.

Animated Solution for Mathematics - Limits, Continuity and Differentiability: Let $a$ be an integer such that $lim_{x \to 7} \frac{18 - [ 1 - x ]}{[ x ] - 3 a}$ exists, where $[t]$ is greatest integer $\leq t$ . Then $a$ is equal to :

Select Answer:

Visualized Solution

Introduction to the Limit Problem

Visualizing GIF Jumps at $x = 7$

Calculating LHL: Analyzing $[x]$

Calculating LHL: Analyzing $[1 - x]$

Calculating LHL: Substitution

Calculating RHL: Analyzing $[x]$

Calculating RHL: Analyzing $[1 - x]$

Calculating RHL: Substitution

Equating LHL and RHL

Solving the Equation: Cross Multiplication

Expanding the Terms

Finding the Value of $a$

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

The Core Challenge

Phase 1

Phase 2

Phase 3

Similar Questions

Let $[t]$ denote the greatest integer $\leq t$ . If for some $λ \in R - {0, 1}$ , $lim_{x \to 0} \frac{1 - ∣ x ∣ + x}{λ - ∣ x ∣ + [ x ]} = L$ , then $L$ is equal to :

Let $[t]$ denote the greatest integer $\leq t$ and ${t}$ denote the fractional part of $t$ . Then integral value of $α$ for which the left hand limit of the function $f (x) = [1 + x] + \frac{α ^{2 [x] + {x}} + [ x ] - 1}{2 [ x ] + { x }}$ at $x = 0$ is equal to $α - \frac{4}{3}$ is

For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then $lim_{x \to 0^{+}} x ([\frac{1}{x}] + [\frac{2}{x}] + \dots + [\frac{15}{x}])$

For each $t \in R$ , let [t] be the greatest integer less than or equal to t. Then, $lim_{x \to 1^{+}} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ])}{∣1 - x ∣ [ 1 - x ]}$

$lim_{x \to 0} \frac{l o g x ^{n} - [ x ]}{[ x ]}, n \in N$ , ( $[x]$ denotes greatest integer less than or equal to $x$ )

The largest value of non-negative integer $a$ for which $lim_{x \to 1} {\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}}^{\frac{1 - x}{1 - x}} = \frac{1}{4}$ is .........

The set of all values of $a$ for which $lim_{x \to a} ([x - 5] - [2 x + 2]) = 0$ , where $[x]$ denotes the greater integer less than or equal to $x$ is equal to

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

For each $x \in R$ , let $[x]$ be the greatest integer less than or equal to $x$ . Then $lim_{x \to 0^{+}} \frac{x ([ x ] + ∣ x ∣ ) s i n [ x ]}{∣ x ∣}$ is equal to

Let $f (x) = {x - 1, 2 x, x is even, x is odd, x \in N$ . If for some $a \in N, f (f (f (a))) = 21$ , then $lim_{x \to a^{-}} {\frac{∣ x ∣ ^{3}}{a} - [\frac{x}{a}]}$ , where $[t]$ denotes the greatest integer less than or equal to $t$ , is equal to :

Select Answer:

Visualized Solution

Introduction to the Limit Problem

Visualizing GIF Jumps at .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=7

Calculating LHL: Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x]

Calculating LHL: Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [1−x]

Calculating LHL: Substitution

Calculating RHL: Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [x]

Calculating RHL: Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } [1−x]

Calculating RHL: Substitution

Equating LHL and RHL

Solving the Equation: Cross Multiplication

Expanding the Terms

Finding the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Final Conclusion

The Sigma Insight: Evaluation of Limits & L'Hopital's Rule

The Core Challenge

Phase 1

Phase 2

Phase 3

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let [t] denote the greatest integer ≤t. If for some λ∈R−{0,1}, limx→0​λ−∣x∣+[x]1−∣x∣+x​=L, then L is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For each t∈R, let [t] be the greatest integer less than or equal to t. Then limx→0+​x([x1​]+[x2​]+⋯+[x15​])

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For each t∈R, let [t] be the greatest integer less than or equal to t. Then, limx→1+​∣1−x∣[1−x](1−∣x∣+sin∣1−x∣)sin(2π​[1−x])​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } limx→0​[x]logxn−[x]​,n∈N, ([x] denotes greatest integer less than or equal to x)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The largest value of non-negative integer a for which limx→1​{x+sin(x−1)−1−ax+sin(x−1)+a​}1−x​1−x​=41​ is .........

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For each x∈R, let [x] be the greatest integer less than or equal to x. Then limx→0+​∣x∣x([x]+∣x∣)sin[x]​ is equal to

Visualizing GIF Jumps at $x = 7$

Calculating LHL: Analyzing $[x]$

Calculating LHL: Analyzing $[1 - x]$

Calculating RHL: Analyzing $[x]$

Calculating RHL: Analyzing $[1 - x]$

Finding the Value of $a$

Let $[t]$ denote the greatest integer $\leq t$ . If for some $λ \in R - {0, 1}$ , $lim_{x \to 0} \frac{1 - ∣ x ∣ + x}{λ - ∣ x ∣ + [ x ]} = L$ , then $L$ is equal to :

For each $t \in R$ , let $[t]$ be the greatest integer less than or equal to $t$ . Then $lim_{x \to 0^{+}} x ([\frac{1}{x}] + [\frac{2}{x}] + \dots + [\frac{15}{x}])$

For each $t \in R$ , let [t] be the greatest integer less than or equal to t. Then, $lim_{x \to 1^{+}} \frac{( 1 - ∣ x ∣ + s i n ∣1 - x ∣ ) s i n ( \frac{π}{2} [ 1 - x ])}{∣1 - x ∣ [ 1 - x ]}$

$lim_{x \to 0} \frac{l o g x ^{n} - [ x ]}{[ x ]}, n \in N$ , ( $[x]$ denotes greatest integer less than or equal to $x$ )

The largest value of non-negative integer $a$ for which $lim_{x \to 1} {\frac{- a x + s i n ( x - 1 ) + a}{x + s i n ( x - 1 ) - 1}}^{\frac{1 - x}{1 - x}} = \frac{1}{4}$ is .........

The set of all values of $a$ for which $lim_{x \to a} ([x - 5] - [2 x + 2]) = 0$ , where $[x]$ denotes the greater integer less than or equal to $x$ is equal to

The value of $lim_{x \to 0^{+}} \frac{c o s ^{- 1} ( x - [ x ] ^{2} ) \cdot s i n ^{- 1} ( x - [ x ] ^{2} )}{x - x ^{3}}$ , where $[x]$ denotes the greatest integer $\leq x$ is :

For each $x \in R$ , let $[x]$ be the greatest integer less than or equal to $x$ . Then $lim_{x \to 0^{+}} \frac{x ([ x ] + ∣ x ∣ ) s i n [ x ]}{∣ x ∣}$ is equal to