JEE Main 2026 — Mathematics Calculus

Let $x - 2 = t$x−2=t. As $x \to 2$x→2, $t \to 0$t→0.

The given limit can be written as:

$$\lim_{t \to 0} \dfrac{\sin((t+2)^3 - 5(t+2)^2 + a(t+2) + b)}{(\sqrt{t+1} - 1)\log_e(t+1)} = m$$t→0lim​(t+1​−1)loge​(t+1)sin((t+2)3−5(t+2)2+a(t+2)+b)​=m

The denominator can be approximated for small $t$t as:

$$\lim_{t \to 0} \dfrac{\sqrt{t+1} - 1}{t} \cdot \dfrac{\log_e(t+1)}{t} \cdot t^2 = \dfrac{1}{2} \cdot 1 \cdot t^2 = \dfrac{t^2}{2}$$t→0lim​tt+1​−1​⋅tloge​(t+1)​⋅t2=21​⋅1⋅t2=2t2​

For the limit to be finite, the argument of the sine function in the numerator must have $t^2$t2 as its lowest degree term in its expansion.

Let $P(t) = (t+2)^3 - 5(t+2)^2 + a(t+2) + b$P(t)=(t+2)3−5(t+2)2+a(t+2)+b

$$P(t) = t^3 + 6t^2 + 12t + 8 - 5(t^2 + 4t + 4) + at + 2a + b$$P(t)=t3+6t2+12t+8−5(t2+4t+4)+at+2a+b

$$P(t) = t^3 + t^2 + (a - 8)t + (2a + b - 12)$$P(t)=t3+t2+(a−8)t+(2a+b−12)

For the limit to exist, the constant and linear terms must be zero:

$$a - 8 = 0 \Rightarrow a = 8$$a−8=0⇒a=8

$$2a + b - 12 = 0 \Rightarrow 16 + b - 12 = 0 \Rightarrow b = -4$$2a+b−12=0⇒16+b−12=0⇒b=−4

Substituting $a$a and $b$b back into $P(t)$P(t):

$$P(t) = t^3 + t^2$$P(t)=t3+t2

The limit becomes:

$$m = \lim_{t \to 0} \dfrac{\sin(t^3 + t^2)}{t^2 / 2}$$m=t→0lim​t2/2sin(t3+t2)​

$$m = \lim_{t \to 0} \dfrac{\sin(t^2(t+1))}{t^2(t+1)} \cdot \dfrac{t^2(t+1)}{t^2 / 2} = 1 \cdot 2 = 2$$m=t→0lim​t2(t+1)sin(t2(t+1))​⋅t2/2t2(t+1)​=1⋅2=2

Therefore, $a + b + m = 8 - 4 + 2 = 6$a+b+m=8−4+2=6.

Answer: $6$6