If lim x->2 sin(x^3 - 5x^2 + ax + b) / ((sqrt(x-1) - 1) log_e(x-1)) = m, then a+b+m is equal to :

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q16 – JEE-MAIN 2026

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Home Maths Limits and DerivativesQuestion #250

lim_{x \to 2} \frac{\sin (x^{3} - 5 x^{2} + a x + b)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)} = m

, then

a + b + m

is equal to :

Clue 1 of 3: Indeterminate Form Analysis

For the limit to be finite and non-zero, the expression must be of the indeterminate form $\frac{0}{0}$ as $x \to 2$ . This implies that the argument of the sine function in the numerator must approach zero.

🧠 Full Solution Path

Step 1: Analyze Indeterminate Form and Initial Conditions✦ Active

As $x \to 2$ , the denominator $(\sqrt{x - 1} - 1) \log_{e} (x - 1)$ approaches $(\sqrt{2 - 1} - 1) \log_{e} (2 - 1) = (1 - 1) \log_{e} (1) = 0 \cdot 0 = 0$ . For the limit to be finite and non-zero, the numerator $\sin (x^{3} - 5 x^{2} + a x + b)$ must also approach $0$ . This implies that the argument of the sine function must approach $0$ . Let $P (x) = x^{3} - 5 x^{2} + a x + b$ . Then $P (2) = 2^{3} - 5 (2^{2}) + a (2) + b = 8 - 20 + 2 a + b = 2 a + b - 12$ . Setting $P (2) = 0$ gives the first condition: $2 a + b = 12$ .

Step 2: Apply L'Hopital's Rule and Determine

a

and

b

○ Expand

The limit is of the form $\frac{0}{0}$ . We can use the standard limit $lim_{z \to 0} \frac{\sin z}{z} = 1$ . So, the given limit becomes $m = lim_{x \to 2} \frac{P (x)}{(\sqrt{x - 1} - 1) \log_{e} (x - 1)}$ . Let $h = x - 2$ , so $x = h + 2$ . As $x \to 2$ , $h \to 0$ . The denominator can be approximated using standard limits: $(\sqrt{h + 1} - 1) \log_{e} (h + 1) \approx (\frac{1}{2} h) (h) = \frac{1}{2} h^{2} = \frac{1}{2} (x - 2)^{2}$ . For the limit to be finite and non-zero, the numerator $P (x)$ must also have $(x - 2)^{2}$ as its lowest power term. This means $P (2) = 0$ (already established) and $P^{'} (2) = 0$ . We find $P^{'} (x) = 3 x^{2} - 10 x + a$ . Setting $P^{'} (2) = 0$ : $3 (2^{2}) - 10 (2) + a = 12 - 20 + a = a - 8 = 0 ⟹ a = 8$ . Substitute $a = 8$ into $2 a + b = 12$ : $2 (8) + b = 12 ⟹ 16 + b = 12 ⟹ b = - 4$ .

💡 Teacher's Secret Hint

Remember that for a limit of the form $\frac{0}{0}$ to be finite and non-zero, the order of the zero in the numerator and denominator must be the same.

Step 3: Calculate

m

and the Final Sum○ Expand

Since $P (2) = 0$ and $P^{'} (2) = 0$ , $P (x)$ can be written using Taylor expansion around $x = 2$ as $P (x) = \frac{P^{″} (2)}{2!} (x - 2)^{2} + higher order terms$ . We find $P^{″} (x) = 6 x - 10$ . So, $P^{″} (2) = 6 (2) - 10 = 12 - 10 = 2$ . Thus, $P (x) = \frac{2}{2} (x - 2)^{2} + \dots = (x - 2)^{2} + higher order terms$ . Now, substitute this back into the limit for $m$ :

m = lim_{x \to 2} \frac{(x - 2)^{2} + higher order terms}{\frac{1}{2} (x - 2)^{2} + higher order terms} = \frac{1}{1 / 2} = 2

Finally, calculate $a + b + m = 8 + (- 4) + 2 = 4 + 2 = 6$ .

💡 Teacher's Secret Hint

Ensure you correctly identify the leading terms in the Taylor expansion for both the numerator and denominator.

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Maths Question 16 – JEE-MAIN 2026

Clue 1 of 3: Indeterminate Form Analysis

🧠 Full Solution Path