Differentiation Rules JEE 2024: An Ambiguous Integral Limit

What does the supplied differentiation rules question actually ask?

The supplied differentiation rules jee 2024 question does not establish a unique numerical answer. It leaves a coefficient undefined, contains ambiguous notation and gives conflicting answer claims.

The problem gives a line whose slope is represented using two real parameters:

45 x + 5 y + 3 = 0, slope = 27 r_{1} + \frac{9 r_{2}}{2}, r_{1}, r_{2} \in R .

It asks for the following integral’s limit as the variable approaches 3:

{lim}_{x \to 3} (\int_{x / 3}^{3} \frac{s t^{2}}{\frac{x^{2}}{2} - r_{2} x^{2} - r_{1} 3 - 3 x} d t) .

The question never defines the numerator’s coefficient, s. The supplied worked solution interprets the denominator’s ambiguous term as:

r_{1} 3 ⟼ 3 r_{1} .

All algebra below follows that interpretation. This is a numerical-answer question with no supplied options, so no answer choices can be checked or inferred.

Publication gate: the year field says 2024, the date says 2023-01-24, the answer field says 1, and the worked solution ends with 12. Obtain a corrected primary-source question and solution before presenting this as an authenticated JEE 2024 solved PYQ.

What relation does the slope give between the parameters?

The slope gives one equation for two parameters, not a unique pair of values. Following the supplied solution’s slope-first method lets us eliminate one parameter, but it does not determine the other. No differentiation is needed because the coefficient of the horizontal variable gives the slope once the line is rearranged: $y = - 9 x - \frac{3}{5} .$

Its slope is negative nine, so:

27 r_{1} + \frac{9 r_{2}}{2} = - 9

$54 r_{1} + 9 r_{2} = - 18$ $6 r_{1} + r_{2} = - 2$ $r_{2} = - 2 - 6 r_{1} .$ Substitute this relation into the interpreted denominator:

(\frac{1}{2} - r_{2}) x^{2} - 3 r_{1} - 3 x & = (\frac{1}{2} + 2 + 6 r_{1}) x^{2} - 3 r_{1} - 3 x & = (\frac{5}{2} + 6 r_{1}) x^{2} - 3 r_{1} - 3 x .

The denominator still contains an undetermined parameter. Keep it in the calculation rather than assigning it a convenient value.

Which variable stays constant during integration?

The limit variable stays fixed while we integrate with respect to the integration variable. For each fixed value where the denominator is nonzero, that denominator is constant during integration, even though it varies with the limit variable. Following the supplied solution, we provisionally treat s as constant too, although its value and role remain unspecified.

The differential identifies the integration variable: $d t .$

Thus the integral becomes:

\frac{s}{(\frac{5}{2} + 6 r_{1}) x^{2} - 3 r_{1} - 3 x} \int_{x / 3}^{3} t^{2} d t .

Evaluate the polynomial antiderivative and substitute both endpoints:

\int_{x / 3}^{3} t^{2} d t & = {[\frac{t^{3}}{3}]}_{x / 3}^{3} & = \frac{3^{3} - (x / 3)^{3}}{3} & = \frac{27 - x^{3} / 27}{3} & = \frac{729 - x^{3}}{81} .

The resulting expression is:

\frac{s (729 - x^{3})}{81 [(\frac{5}{2} + 6 r_{1}) x^{2} - 3 r_{1} - 3 x]} .

The variable lower endpoint must be substituted and cubed. It does not change the integration variable or create a differentiation task.

For more practice evaluating polynomial antiderivatives, use Indefinite Integration Practice Questions JEE: 6 Worked Examples. Apply the antiderivative to both endpoints before simplifying.

What does direct substitution actually give?

Direct substitution gives a parameter-dependent expression, provided the denominator tends to a nonzero value. It does not establish either claimed numerical answer. After integration, the numerator and denominator are polynomials, so continuity permits substitution once that denominator condition has been checked.

First evaluate the numerator’s polynomial factor: $729 - 3^{3} = 729 - 27 = 702 .$

The denominator factor becomes:

(\frac{5}{2} + 6 r_{1}) 9 - 3 r_{1} - 9 & = \frac{45}{2} + 54 r_{1} - 3 r_{1} - 9 & = \frac{27}{2} + 51 r_{1} .

Therefore, under the condition

\frac{27}{2} + 51 r_{1} \neq 0 ⟺ r_{1} \neq - \frac{9}{34},

the limit is:

\frac{702 s}{81 (\frac{27}{2} + 51 r_{1})} = \frac{26 s}{3 (\frac{27}{2} + 51 r_{1})} .

This is where the justified calculation stops. The slope relation does not exclude the exceptional parameter value, and the expression still depends on unspecified parameters.

The answer field’s 1 and the worked solution’s 12 are source claims, not established mathematical conclusions. No supplied condition makes the expression a unique number. Choosing convenient parameters, inventing a missing condition or appending a step to force either claim would change the question, not solve it.

Why is “finite, therefore nonzero” an invalid step?

A finite number can equal zero. The supplied solution calls the denominator’s limiting value finite and then asserts that it is nonzero, without proving this.

The exceptional value is compatible with the slope relation: $r_{1} = - \frac{9}{34}$

r_{2} = - 2 - 6 (- \frac{9}{34}) = - 2 + \frac{27}{17} = - \frac{7}{17} .

For this pair:

\frac{27}{2} + 51 (- \frac{9}{34}) = 0 .

Direct substitution into a quotient requires a nonzero limiting denominator, not merely a finite one. When that check fails, the limit needs separate analysis; the quotient cannot be evaluated by dividing the substituted values.

Parameters cannot be assigned just to match an answer key. Since no options were supplied, this is an invalid solution step, not a reason to select or reject an invented option.

How do I choose between substitution, factorisation and differentiation?

Choose the operation from what the question asks, then check the expression’s form. An integral limit does not become a differentiation problem just because its endpoint varies. The following two questions from Limit, Continuity and Differentiability are original practice, not verified PYQs.

How do I evaluate this limit when substitution gives zero over zero?

In original practice 1, factorisation removes the indeterminate form. Evaluate the limit, noting what direct substitution gives first:

{lim}_{x \to 3} \frac{x^{2} - 9}{x - 3} ⟶ \frac{0}{0} .

Use the difference of squares:

x^{2} - 9 = (x - 3) (x + 3) .

Cancellation is valid only away from the limiting point:

\frac{(x - 3) (x + 3)}{x - 3} = x + 3, x \neq 3 .

The limit concerns nearby values, so:

{lim}_{x \to 3} (x + 3) = 6 .

How do I find the derivative of this quotient at three?

In original practice 2, use the quotient rule because the task asks for a derivative. For the following function, find its derivative at 3:

f (x) = \frac{x^{2} + 1}{x + 1}, x \neq - 1 .

Apply the rule, then simplify:

f^{'} (x) & = \frac{2 x (x + 1) - (x^{2} + 1)}{(x + 1)^{2}} & = \frac{x^{2} + 2 x - 1}{(x + 1)^{2}} .

Now substitute:

f^{'} (3) = \frac{9 + 6 - 1}{16} = \frac{14}{16} = \frac{7}{8} .

Keep the tasks separate:

Supplied integral limit: integrate first, then substitute only under the nonzero-denominator parameter condition.
Original practice 1: remove the indeterminate form by factorisation.
Original practice 2: apply the quotient rule because a derivative is requested.

Before using a rule, write down the operation requested and check the denominator. Do not differentiate unless the task requires it.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Frequently asked questions

Is the answer to this differentiation rules JEE 2024 question 1 or 12?

Neither number is established by the supplied conditions. The coefficient s is undefined, the denominator contains ambiguous notation, and the slope relation leaves a free parameter. A corrected primary source is needed before treating this as an authenticated JEE 2024 solved question.

Why is x treated as constant when integrating with respect to t?

The differential dt identifies t as the integration variable, so x stays fixed during integration. The denominator can therefore be taken outside the integral wherever it is nonzero. Substitute both endpoints into the antiderivative before taking the limit in x.

Does a finite denominator mean I can use direct substitution?

No: a finite limiting denominator can still equal zero. Under the article's interpretation of the denominator, its limit is 27/2 + 51r1, which vanishes at r1 = -9/34. Direct substitution into the quotient is valid only when this limiting denominator is nonzero; the exceptional case requires separate analysis.

Should I differentiate an integral just because its lower limit contains x?

No: a variable endpoint does not by itself make the task a differentiation problem. For the supplied integral limit, integrate with respect to t first, then check the denominator before substituting x = 3. In the article's separate derivative example, the quotient rule is appropriate because a derivative is explicitly requested.