Limit, Continuity and Differentiability JEE 2026: Key Audit

Does option B, 7, follow from the supplied question?

The listed answer 7 does not follow from the printed expression in this limit, continuity and differentiability JEE 2026 record. The expression and supplied solution are internally inconsistent, so this is a source audit, not a completed official PYQ solution set.

The supplied record identifies JEE Main, January 2026, slot 1, with the question bank’s hard tag 4 and listed answer B. Which listed value of the parameter sum satisfies this condition?

{lim}_{x \to 0} \frac{x^{a} + b^{x} + c^{x}}{x \cos x - \log (1 + x)} = 2, find a + b + c .

The audit checks the denominator, numerator cancellation and claimed arithmetic. A corrected source is needed before presenting this as a conventional worked PYQ.

Why does the denominator start with a quadratic term?

The linear terms cancel, leaving a leading coefficient of one-half at quadratic order. This part of the supplied expansion method is valid. Interpret the logarithm as the natural logarithm, consistently with the supplied solution.

Expand cosine, then multiply by the variable:

\cos x = 1 - \frac{x^{2}}{2} + O (x^{4}) .

x \cos x = x - \frac{x^{3}}{2} + O (x^{5}) .

The logarithm expansion is:

\log (1 + x) = x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + O (x^{4}) .

Subtract the entire logarithm expression, keeping track of every sign:

x \cos x - \log (1 + x) & = (x - \frac{x^{3}}{2} + O (x^{5})) - (x - \frac{x^{2}}{2} + \frac{x^{3}}{3} + O (x^{4})) & = (x - x) + \frac{x^{2}}{2} - (\frac{1}{2} + \frac{1}{3}) x^{3} + O (x^{4}) & = \frac{x^{2}}{2} - \frac{5 x^{3}}{6} + O (x^{4}) & = x^{2} [\frac{1}{2} - \frac{5 x}{6} + O (x^{2})] .

For the quotient to approach 2, the numerator must satisfy:

x^{a} + b^{x} + c^{x} = x^{2} + o (x^{2}) .

Merely containing a quadratic term is not enough. Every lower-order contribution must disappear, and the quadratic coefficient must be 1. The remainder, divided by the square of the variable, must tend to zero.

Why can the printed numerator not produce the stated limit?

Under the real-domain assumptions needed for the supplied exponential expansions, the numerator cannot approach zero. Its constant contributions do not cancel, so coefficient matching cannot produce the stated limit.

The exponential bases must be positive:

b > 0, c > 0 .

Their expansions give:

b^{x} + c^{x} = 2 + x (\ln b + \ln c) + \frac{x^{2}}{2} [(\ln b)^{2} + (\ln c)^{2}] + O (x^{3}) .

Approach zero from the positive side. This avoids assuming that an arbitrary real power is defined for negative inputs. Failure of this one-sided limit already disproves the stated finite two-sided limit.

For every real exponent:

{lim}_{x \to 0^{+}} (x^{a} + b^{x} + c^{x}) = {\begin{matrix} 2, & a > 0, \\ 3, & a = 0, \\ + \infty, & a < 0 . \end{matrix}

The denominator is positive sufficiently close to zero and satisfies:

x \cos x - \log (1 + x) ~ \frac{x^{2}}{2} .

Consequently, in all three cases:

{lim}_{x \to 0^{+}} \frac{x^{a} + b^{x} + c^{x}}{x \cos x - \log (1 + x)} = + \infty .

The supplied solution introduces unsupported equations:

c + 1 = 0, a - c + 1 = 0 .

Neither comes from the printed numerator. The resulting negative base also invalidates the real exponential expansion near zero: $c = - 1 .$

There are two further arithmetic failures. The claimed quadratic equation gives square roots, not 2: $2 b^{2} = 4 ⟹ b = \pm \sqrt{2} .$

Only the positive root meets the positive-base requirement. Separately, the supplied values sum to negative one:

(- 2) + 2 + (- 1) = - 1 \neq 7 .

We cannot infer an intended replacement expression or assign a corrected option. 7 remains only the listed key.

How could an invalid shortcut lead to option C, 3?

An invalid shortcut replaces all three numerator terms by 1, obtains 3 and selects option C. This is an illustrative reasoning error, not a documented student response or evidence about how the distractor was designed.

The positive-base exponential terms approach 1. The power term does not generally approach 1; it does so here only when its exponent is zero:

b^{x} \to 1, c^{x} \to 1, x^{0} = 1 .

Even then, a numerator approaching 3 over this denominator gives divergence, not the stipulated finite quotient. A numerator limit is also not the requested parameter sum.

Use this sequence instead:

Check the real domains.
Inspect constant terms.
Establish that all lower-order terms cancel.
Match the first surviving coefficients.

How do you apply the method to two solvable questions?

The expansion method works when the numerator has the required cancellation. Both questions below are original related practice, not additional verified JEE questions.

How do you evaluate the related exponential limit?

The limit is 4, because the numerator’s leading quadratic coefficient is 2 and the denominator’s is one-half. Evaluate:

{lim}_{x \to 0} \frac{e^{2 x} - 1 - 2 x}{x \cos x - \log (1 + x)} .

Expand the exponential:

e^{2 x} = 1 + 2 x + 2 x^{2} + \frac{4}{3} x^{3} + O (x^{4}) .

The subtractions produce genuine cancellation:

e^{2 x} - 1 - 2 x = 2 x^{2} + O (x^{3}) .

Using the denominator already established:

{lim}_{x \to 0} \frac{2 x^{2} + O (x^{3})}{x^{2} / 2 + O (x^{3})} & = {lim}_{x \to 0} \frac{2 + O (x)}{1 / 2 + O (x)} & = 4 .

Which value makes the logarithmic function continuous, and what is its derivative at zero?

Continuity requires negative one-half, and the derivative at zero is one-third. Find both for the function defined near zero by:

f (x) = {\begin{matrix} \frac{\log (1 + x) - x}{x^{2}}, & x \neq 0, \\ k, & x = 0 . \end{matrix}

Expand before dividing:

\log (1 + x) - x = - \frac{x^{2}}{2} + \frac{x^{3}}{3} + O (x^{4}) .

f (x) = - \frac{1}{2} + \frac{x}{3} + O (x^{2}) .

Hence continuity requires:

k = {lim}_{x \to 0} f (x) = - \frac{1}{2} .

Use the derivative definition with that value:

\frac{f (h) - f (0)}{h} = \frac{(- \frac{1}{2} + \frac{h}{3} + O (h^{2})) - (- \frac{1}{2})}{h} = \frac{1}{3} + O (h) .

f^{'} (0) = {lim}_{h \to 0} \frac{f (h) - f (0)}{h} = \frac{1}{3} .

Choose the truncation order from what the question asks:

First limit: quadratic terms determine the answer.
Continuity value: the quadratic logarithm term is sufficient.
Derivative: retain the cubic logarithm term before dividing.

Neither practice problem repairs or verifies the supplied key. Before spending more time trying to obtain 7, obtain a corrected question and a matching solution.

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 option B, 7, correct for the printed limit question?

The listed answer 7 does not follow from the printed expression. With positive exponential bases and any real exponent a, the quotient diverges to positive infinity as x approaches zero from the right. A corrected question and matching solution are needed before assigning an option.

What is the expansion of x cos x - log(1+x) near zero?

Taking log as the natural logarithm, the expansion is x²/2 - 5x³/6 + O(x⁴). The linear terms cancel, so the leading term is x²/2. For a quotient with this denominator to approach 2, its numerator must equal x² + o(x²).

Why can't x^a + b^x + c^x approach zero?

For positive bases b and c, both b^x and c^x approach 1 as x approaches zero. From the positive side, x^a approaches 0 when a > 0, equals 1 when a = 0, and diverges to positive infinity when a < 0. The numerator therefore approaches 2, 3 or positive infinity, never zero.

How do I find k and f'(0) for f(x) = [log(1+x)-x]/x² with f(0) = k?

Using the natural logarithm, expand the expression for nonzero x to obtain f(x) = -1/2 + x/3 + O(x²). Continuity at zero requires k = -1/2, and the derivative definition then gives f'(0) = 1/3. Retain the cubic term in the logarithm expansion before dividing by x² to calculate the derivative.