JEE MainMathematics

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Limit of Greatest Integer Function Involving a Trigonometric Form

Q: Can we interchange the limit and the greatest integer function: is lim of [f(x)] = [lim of f(x)]?

Not in general, because [·] is discontinuous at integers! However, lim of f(x) = 1/2, which is strictly between 0 and 1 (not an integer). In an open neighborhood around 2p, f(x) takes values in (0, 1), so [f(x)] = 0 identically. Thus the limit is 0.

Let x=2 be a root of the equation x^2 + px + q = 0 and f(x) = (1-cos(x^2 - 4px + q^2 + 8q + 16))/((x-2p)^4), x ≠ 2p; 0, x = 2p lim as x→ 2p^(+) of [f(x)] where [·] denotes greatest integer function, is

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A0B-1C2D1

Step-by-step solutionView

Correct answer

Using $x = 2$ as a root gives $q = - (2 p + 4)$ , simplifying the argument inside the cosine to $(x - 2 p)^{2}$ , which yields $\lim_{x \to 2 p^{+}} f (x) = \frac{1}{2}$ and hence $[f (x)] = 0$ .

Option analysis

Why each option works or fails

A · 0

The student correctly evaluates $\lim_{x \to 2 p^{+}} f (x) = \frac{1}{2}$ and takes the floor to get $0$ . Correctly substitute $q = - 2 p - 4$ to rewrite the argument inside the cosine as $(x - 2 p)^{2}$ , apply the standard limit $\lim_{θ \to 0} \frac{1 - \cos θ}{θ^{2}} = \frac{1}{2}$ , and evaluate $[1 / 2] = 0$ as $f (x)$ approaches $1 / 2$ strictly from below or above within $(0, 1)$ .

B · -1

The student mistakenly believes that $f (x)$ approaches a negative value near $0$ , causing the floor function to yield $- 1$ . Note that $1 - \cos (u) \geq 0$ for all real $u$ , and $(x - 2 p)^{4} > 0$ , so $f (x) > 0$ everywhere near $x = 2 p$ . Thus $[f (x)]$ cannot be negative.

C · 2

The student inverted the standard limit, using $\frac{1 - \cos θ}{θ^{2}} \to 2$ instead of $\frac{1}{2}$ . Recall the standard Taylor expansion or half-angle identity: $1 - \cos θ = 2 \sin^{2} (θ / 2) \approx \frac{θ^{2}}{2}$ , so the limit is $\frac{1}{2}$ , not $2$ .

D · 1

The student rounds the limit $\frac{1}{2}$ up to $1$ , or evaluates the limit after mistakenly concluding the value equals $1$ . The greatest integer function $[\cdot]$ rounds down to the nearest integer less than or equal to the input; for values strictly between $0$ and $1$ , $[f (x)] = 0$ .

Reviewed route

Solution

StepWorking

01Given

$x = 2$ is a root of $x^{2} + p x + q = 0$ , and $f (x) = \frac{1 - \cos (x^{2} - 4 p x + q^{2} + 8 q + 16)}{(x - 2 p)^{4}}$ for $x \neq 2 p$ .

02Goal

Evaluate the limit $\lim_{x \to 2 p^{+}} [f (x)]$ , where $[\cdot]$ is the greatest integer function.

03Approach

Use the root condition to express $q^{2} + 8 q + 16$ in terms of $p$ . Then, recognize the argument of the cosine as a perfect square $(x - 2 p)^{2}$ , apply the standard limit $\lim_{θ \to 0} \frac{1 - \cos θ}{θ^{2}} = \frac{1}{2}$ , and determine the floor of the approaching values.

04Execute

Since $x = 2$ is a root: $2^{2} + 2 p + q = 0 ⟹ 4 + 2 p + q = 0 ⟹ q + 4 = - 2 p$ . Squaring both sides yields $(q + 4)^{2} = q^{2} + 8 q + 16 = 4 p^{2}$ .

05Execute

Substitute $q^{2} + 8 q + 16 = 4 p^{2}$ into $f (x)$ : $x^{2} - 4 p x + q^{2} + 8 q + 16 = x^{2} - 4 p x + 4 p^{2} = (x - 2 p)^{2}$ . Therefore, $f (x) = \frac{1 - \cos ((x - 2 p)^{2})}{((x - 2 p)^{2})^{2}}$ .

06Execute

Let $θ = (x - 2 p)^{2}$ . As $x \to 2 p^{+}$ , $θ \to 0^{+}$ . Using the standard limit, $\lim_{θ \to 0^{+}} \frac{1 - \cos θ}{θ^{2}} = \frac{1}{2}$ . For small $θ > 0$ , since $\cos θ = 1 - \frac{θ^{2}}{2} + \frac{θ^{4}}{24} - \dots$ , we have $f (x) = \frac{1}{2} - \frac{θ^{2}}{24} + \dots < \frac{1}{2}$ , and clearly $f (x) > 0$ . Thus, $0 < f (x) < 1$ , which gives $[f (x)] = 0$ . Hence $\lim_{x \to 2 p^{+}} [f (x)] = [0] = 0$ .

✓Verify

Check with a simple test value: choose $p = - 2$ , then $q = - 4 - 2 (- 2) = 0$ . Root condition: $2^{2} - 2 (2) + 0 = 0$ , verified. Then $2 p = - 4$ . $f (x) = \frac{1 - \cos (x^{2} + 8 x + 16)}{(x + 4)^{4}} = \frac{1 - \cos (x + 4)^{2}}{(x + 4)^{4}} \to 1 / 2$ . For all $x$ near $- 4$ , $f (x) \in (0, 1 / 2)$ , so $[f (x)] = 0$ .

Hints that build this answer step by step

Since $x = 2$ is a root of $x^{2} + p x + q = 0$ , how can $q$ be expressed in terms of $p$ ?
$q = - 2 p - 4$
Substitute $q = - (2 p + 4)$ into the expression inside the cosine: $E = x^{2} - 4 p x + q^{2} + 8 q + 16$ . What does it simplify to?
$(x - 2 p)^{2}$
With $E = (x - 2 p)^{2}$ , what is $\lim_{x \to 2 p^{+}} f (x) = \lim_{x \to 2 p^{+}} \frac{1 - \cos ((x - 2 p)^{2})}{(x - 2 p)^{4}}$ ?
$\frac{1}{2}$
Since $f (x) \to \frac{1}{2}$ as $x \to 2 p^{+}$ , what is the value of $\lim_{x \to 2 p^{+}} [f (x)]$ ?
$0$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 9 September 2026

Quick checks

Students also ask

Can we interchange the limit and the greatest integer function: is

\lim [f (x)] = [\lim f (x)]

Not in general, because $[\cdot]$ is discontinuous at integers! However, $\lim f (x) = 1 / 2$ , which is strictly between 0 and 1 (not an integer). In an open neighborhood around $2 p$ , $f (x)$ takes values in $(0, 1)$ , so $[f (x)] = 0$ identically. Thus the limit is 0.

Answer

Using $x = 2$ as a root gives $q = - (2 p + 4)$ , simplifying the argument inside the cosine to $(x - 2 p)^{2}$ , which yields $\lim_{x \to 2 p^{+}} f (x) = \frac{1}{2}$ and hence $[f (x)] = 0$ .

Why each option works or fails

A · correct: 0 - The student correctly evaluates $\lim_{x \to 2 p^{+}} f (x) = \frac{1}{2}$ and takes the floor to get $0$ . Correctly substitute $q = - 2 p - 4$ to rewrite the argument inside the cosine as $x = 2$ , apply the standard limit $\lim_{θ \to 0} \frac{1 - \cos θ}{θ^{2}} = \frac{1}{2}$ , and evaluate $[1 / 2] = 0$ as $f (x)$ approaches $1 / 2$ strictly from below or above within $(0, 1)$ .
B: -1 - The student mistakenly believes that $f (x)$ approaches a negative value near $0$ , causing the floor function to yield $- 1$ . Note that $1 - \cos (u) \geq 0$ for all real $u$ , and $(x - 2 p)^{4} > 0$ , so $f (x) > 0$ everywhere near $x = 2 p$ . Thus $[f (x)]$ cannot be negative.
C: 2 - The student inverted the standard limit, using $\frac{1 - \cos θ}{θ^{2}} \to 2$ instead of $\frac{1}{2}$ . Recall the standard Taylor expansion or half-angle identity: $1 - \cos θ = 2 \sin^{2} (θ / 2) \approx \frac{θ^{2}}{2}$ , so the limit is $\frac{1}{2}$ , not $2$ .
D: 1 - The student rounds the limit $(x - 2 p)^{4} > 0$ up to $1$ , or evaluates the limit after mistakenly concluding the value equals $1$ . The greatest integer function $[\cdot]$ rounds down to the nearest integer less than or equal to the input; for values strictly between $0$ and $1$ , $[f (x)] = 0$ .

Step-by-step solution

Given: $x = 2$ is a root of $x^{2} + p x + q = 0$ , and $f (x) = \frac{1 - \cos (x^{2} - 4 p x + q^{2} + 8 q + 16)}{(x - 2 p)^{4}}$ for $x \neq 2 p$ .
Goal: Evaluate the limit $\lim_{x \to 2 p^{+}} [f (x)]$ , where $0$ is the greatest integer function.
Approach: Use the root condition to express $q^{2} + 8 q + 16$ in terms of $p$ . Then, recognize the argument of the cosine as a perfect square $(x - 2 p)^{2}$ , apply the standard limit $\lim_{θ \to 0} \frac{1 - \cos θ}{θ^{2}} = \frac{1}{2}$ , and determine the floor of the approaching values.
Execute: Since $x = 2$ is a root: $2^{2} + 2 p + q = 0 ⟹ 4 + 2 p + q = 0 ⟹ q + 4 = - 2 p$ . Squaring both sides yields $(q + 4)^{2} = q^{2} + 8 q + 16 = 4 p^{2}$ .
Execute: Substitute $q^{2} + 8 q + 16 = 4 p^{2}$ into $f (x)$ : $x^{2} - 4 p x + q^{2} + 8 q + 16 = x^{2} - 4 p x + 4 p^{2} = (x - 2 p)^{2}$ . Therefore, $f (x) = \frac{1 - \cos ((x - 2 p)^{2})}{((x - 2 p)^{2})^{2}}$ .
Execute: Let $θ = (x - 2 p)^{2}$ . As $x \to 2 p^{+}$ , $θ \to 0^{+}$ . Using the standard limit, $\lim_{θ \to 0^{+}} \frac{1 - \cos θ}{θ^{2}} = \frac{1}{2}$ . For small $θ > 0$ , since $\cos θ = 1 - \frac{θ^{2}}{2} + \frac{θ^{4}}{24} - \dots$ , we have $f (x) = \frac{1}{2} - \frac{θ^{2}}{24} + \dots < \frac{1}{2}$ , and clearly $f (x) > 0$ . Thus, $0 < f (x) < 1$ , which gives $[f (x)] = 0$ . Hence $\lim_{x \to 2 p^{+}} [f (x)] = [0] = 0$ .
Verify: Check with a simple test value: choose $p = - 2$ , then $q = - 4 - 2 (- 2) = 0$ . Root condition: $2^{2} - 2 (2) + 0 = 0$ , verified. Then $2 p = - 4$ . $f (x) = \frac{1 - \cos (x^{2} + 8 x + 16)}{(x + 4)^{4}} = \frac{1 - \cos (x + 4)^{2}}{(x + 4)^{4}} \to 1 / 2$ . For all $x$ near $- 4$ , $f (x) \in (0, 1 / 2)$ , so $[f (x)] = 0$ .

Limit of Greatest Integer Function Involving a Trigonometric Form

What feels right?

Using $x = 2$ as a root gives $q = - (2 p + 4)$ , simplifying the argument inside the cosine to $(x - 2 p)^{2}$ , which yields $\lim_{x \to 2 p^{+}} f (x) = \frac{1}{2}$ and hence $[f (x)] = 0$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

✓Verify

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