Real-valued Functions JEE 2024: Continuity at Zero

What value makes the supplied function continuous at zero?

The supplied expression gives zero, not minus four. This real-valued functions JEE 2024 record needs source correction before it can be presented as a verified PYQ. Find the parameters and the value at zero that make this function continuous:

f (x) = \frac{\sin (3 x) + α \sin x - β \cos (3 x)}{x^{2}}, x \neq 0 .

The quotient is undefined at zero. The task therefore asks for a continuous extension, not direct substitution into the fraction.

The metadata also conflicts: the record says JEE Main 2024 but lists 2023-01-24, Morning. Neither the year nor the shift should be silently corrected. The supplied solution names option D, minus four, but the complete option list is absent and cannot be reconstructed from this record.

What must you check before applying L’Hôpital’s rule?

Continuity forces the cosine parameter to be zero and the sine parameter to be minus three. These are prerequisites for repeated differentiation, not values to choose from the answer key. Continuity requires a finite two-sided limit:

f (0) = {lim}_{x \to 0} f (x) \in R .

Define the numerator:

N (x) = \sin (3 x) + α \sin x - β \cos (3 x) .

A finite quotient limit requires: $N (x) = x^{2} f (x) ⟶ 0 .$

The numerator is continuous, so:

N (0) = - β = 0 ⟹ β = 0 .

Only now is the original quotient a zero-over-zero form. Differentiating numerator and denominator separately gives:

N^{'} (x) = 3 \cos (3 x) + α \cos x + 3 β \sin (3 x), \frac{d}{d x} (x^{2}) = 2 x .

With the cosine parameter fixed at zero, a nonzero constant in the derivative numerator would give:

\frac{N^{'} (x)}{2 x} ~ \frac{3 + α}{2 x}, 3 + α \neq 0 .

That expression has no finite two-sided limit. To establish necessity for the original quotient, use differentiability:

{lim}_{x \to 0} \frac{N (x)}{x} = N^{'} (0) = 3 + α .

A finite original quotient requires this same limit to be zero:

\frac{N (x)}{x} = x f (x) ⟶ 0 .

Therefore, before applying L’Hôpital a second time:

α = - 3 .

All the functions being differentiated are smooth, and the relevant denominator derivatives are nonzero away from zero. The remaining checks are the indeterminate forms and derivative-quotient limits.

What does the limit equal after both parameters are fixed?

The limit is zero, so that is the value required for continuity. Substitute both parameters before evaluating the derivative quotients:

β = 0, α = - 3,

{lim}_{x \to 0} \frac{\sin (3 x) - 3 \sin x}{x^{2}} .

The original numerator and denominator both approach zero. The first L’Hôpital quotient is:

{lim}_{x \to 0} \frac{3 \cos (3 x) - 3 \cos x}{2 x} .

This is again zero over zero. Differentiating once more gives a quotient with a finite limit, justifying both applications:

{lim}_{x \to 0} \frac{- 9 \sin (3 x) + 3 \sin x}{2} = \frac{0}{2} = 0 .

Thus the continuous extension is:

α = - 3, β = 0, f (0) = 0 .

The supplied solution calls the following expression a third derivative. It is the second derivative of the original numerator:

N^{″} (x) = - 9 \sin (3 x) - α \sin x + 9 β \cos (3 x) .

Its stated intermediate result must use the parameter already forced by continuity:

\frac{9 β}{2} = \frac{9 (0)}{2} = 0 .

The source’s error is treating that parameter as free after assuming the original numerator approaches zero. It cannot then be chosen to fit an option. The supplied answer, minus four, is unsupported by the expression as written; a corrected question or source solution is required before publication as an authenticated PYQ.

Why can’t you choose a parameter to manufacture minus four?

Choosing the cosine parameter from the proposed answer makes the original quotient diverge to positive infinity. A differentiated expression cannot replace the original fraction unless L’Hôpital’s conditions hold.

Error, not a valid solution:

\frac{9 β}{2} = - 4 ⟹ β = - \frac{8}{9} .

Check that choice in the original numerator:

N (0) = - β = \frac{8}{9} \neq 0 .

The denominator approaches zero through positive values. The numerator approaches a positive constant, regardless of the other parameter:

{lim}_{x \to 0} \frac{\sin (3 x) + α \sin x + \frac{8}{9} \cos (3 x)}{x^{2}} = + \infty .

It cannot approach minus four. The mistake is applying L’Hôpital without an indeterminate form, then using a proposed answer to choose parameters.

No verified distractor list was supplied. This audits the source’s stated D, minus four; it does not identify a known wrong option from an authenticated paper.

Keep this order in your working:

Check the original numerator at zero.
Check the first derivative at zero.
Evaluate the finite limit only after enforcing those conditions.

Can you solve two related continuity questions using the same checks?

These two problems from Limit, Continuity and Differentiability are original practice, not historical JEE questions. The first uses two differentiations; the second uses three. Fix each parameter before evaluating the limit.

How do you make the cosine quotient continuous?

The required parameter is zero, and the extension value is minus two. Find the parameter and the value at zero for:

g (x) = \frac{\cos (2 x) - 1 + a \sin x}{x^{2}}, x \neq 0 .

The numerator already vanishes at zero, giving zero over zero. A finite original quotient requires its first derivative at zero to vanish:

\frac{d}{d x} [\cos (2 x) - 1 + a \sin x] = - 2 \sin (2 x) + a \cos x,

$N_{g}^{'} (0) = a = 0 .$ After substitution, the first derivative quotient is still zero over zero. Differentiate again:

{N_{g}^{'}}^{'} (x) = - 4 \cos (2 x) - a \sin x,

g (0) = \frac{- 4}{2} = - 2 .

How do you make the cubic-denominator quotient continuous?

The required parameter is minus two, and the extension value is minus one. Find both for:

h (x) = \frac{\sin (2 x) + b \sin x}{x^{3}}, x \neq 0 .

The original form is zero over zero. A finite original quotient requires the first numerator derivative at zero to vanish:

N_{h}^{'} (x) = 2 \cos (2 x) + b \cos x, 2 + b = 0, b = - 2 .

The successive derivative quotients are:

\frac{2 \cos (2 x) - 2 \cos x}{3 x^{2}}, \frac{- 4 \sin (2 x) + 2 \sin x}{6 x}, \frac{- 8 \cos (2 x) + 2 \cos x}{6} .

The first two remain zero over zero, allowing further differentiation. The last has a finite limit, completing the justification:

h (0) = \frac{- 8 + 2}{6} = - 1 .

A squared denominator requires cancellation through the linear term; a cubed denominator also requires the quadratic term to vanish. Before every repeated differentiation, check the current numerator and denominator at zero.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

Frequently asked questions

What values of alpha and beta make the supplied function continuous at zero?

For f(x) = [sin(3x) + alpha sin(x) - beta cos(3x)]/x², continuity requires beta = 0 and alpha = -3. These values make the numerator and its first derivative vanish at zero. The resulting limit is zero, so the continuous extension requires f(0) = 0.

Why is the supplied answer of -4 incorrect?

Obtaining -4 from 9 beta/2 requires beta = -8/9, but this makes the original numerator approach 8/9 rather than zero. Since x² approaches zero through positive values, the original quotient then diverges to positive infinity. The supplied expression therefore does not support -4 as a finite limit.

What should I check before applying L'Hopital's rule repeatedly?

For this problem, check that the numerator and denominator both approach zero before each differentiation. First enforce beta = 0, then alpha = -3, as required for the original quotient to have a finite limit. After substitution, both applications of L'Hopital's rule are justified and give a limit of zero.

Is this a verified JEE Main 2024 question?

The supplied record cannot be treated as a verified PYQ without source correction. It is labelled JEE Main 2024 but lists 2023-01-24, Morning, and its stated answer conflicts with the expression. The complete option list is also missing, so neither the shift nor the options should be reconstructed.