JEE Main 2026 — Mathematics Calculus

Using the standard expansions for small $x$x, $\cos \theta \approx 1 - \dfrac{\theta^2}{2}$cosθ≈1−2θ2​ and $\sin \theta \approx \theta$sinθ≈θ.

The given limit can be written as:

$$\displaystyle\lim_{x \to 0} \dfrac{1 - \left(1 - \dfrac{\alpha^2 x^2}{2}\right)\left(1 - \dfrac{(\alpha+1)^2 x^2}{2}\right)\left(1 - \dfrac{(\alpha+2)^2 x^2}{2}\right)}{((\alpha+1)x)^2}$$x→0lim​((α+1)x)21−(1−2α2x2​)(1−2(α+1)2x2​)(1−2(α+2)2x2​)​

Neglecting higher powers of $x$x, the numerator simplifies to:

$$1 - \left(1 - \dfrac{x^2}{2}\left(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2\right)\right) = \dfrac{x^2}{2}\left(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2\right)$$1−(1−2x2​(α2+(α+1)2+(α+2)2))=2x2​(α2+(α+1)2+(α+2)2)

Substituting this back into the limit:

$$\displaystyle\lim_{x \to 0} \dfrac{\dfrac{x^2}{2}\left(\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2\right)}{(\alpha+1)^2 x^2} = \dfrac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2}$$x→0lim​(α+1)2x22x2​(α2+(α+1)2+(α+2)2)​=2(α+1)2α2+(α+1)2+(α+2)2​

We are given that this limit is equal to $2$2. Therefore:

$$\dfrac{\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2}{2(\alpha+1)^2} = 2$$2(α+1)2α2+(α+1)2+(α+2)2​=2

$$\alpha^2 + (\alpha+1)^2 + (\alpha+2)^2 = 4(\alpha+1)^2$$α2+(α+1)2+(α+2)2=4(α+1)2

Let $y = \alpha+1$y=α+1. Then $\alpha = y-1$α=y−1 and $\alpha+2 = y+1$α+2=y+1. The equation becomes:

$$(y-1)^2 + y^2 + (y+1)^2 = 4y^2$$(y−1)2+y2+(y+1)2=4y2

$$y^2 - 2y + 1 + y^2 + y^2 + 2y + 1 = 4y^2$$y2−2y+1+y2+y2+2y+1=4y2

$$3y^2 + 2 = 4y^2$$3y2+2=4y2

$$y^2 = 2$$y2=2

Substituting $y = \alpha+1$y=α+1 back into the equation:

$$(\alpha+1)^2 = 2$$(α+1)2=2

$$\alpha^2 + 2\alpha + 1 = 2$$α2+2α+1=2

$$\alpha^2 + 2\alpha - 1 = 0$$α2+2α−1=0

The roots of this quadratic equation represent all possible values of $\alpha$α. The product of the roots is given by $\dfrac{c}{a}$ac​:

$$\text{Product} = \dfrac{-1}{1} = -1$$Product=1−1​=−1

Answer: $-1$−1