JEE Main 2026 — Mathematics Calculus

The given function is $\displaystyle f(x) = \lim_{y \to 0} \dfrac{(1 - \cos(xy)) \tan(xy)}{y^3}$f(x)=y→0lim​y3(1−cos(xy))tan(xy)​.

Multiplying and dividing by $x^3$x3, we get:

$$f(x) = \lim_{y \to 0} \dfrac{1 - \cos(xy)}{(xy)^2} \cdot \dfrac{\tan(xy)}{xy} \cdot x^3$$f(x)=y→0lim​(xy)21−cos(xy)​⋅xytan(xy)​⋅x3

Using the standard limits $\displaystyle \lim_{t \to 0} \dfrac{1 - \cos t}{t^2} = \dfrac{1}{2}$t→0lim​t21−cost​=21​ and $\displaystyle \lim_{t \to 0} \dfrac{\tan t}{t} = 1$t→0lim​ttant​=1, we obtain:

$$f(x) = \dfrac{1}{2} \cdot 1 \cdot x^3 = \dfrac{x^3}{2}$$f(x)=21​⋅1⋅x3=2x3​

We need to find the number of solutions for the equation $f(x) = \sin x$f(x)=sinx, which is $\dfrac{x^3}{2} = \sin x$2x3​=sinx.

Let $g(x) = \dfrac{x^3}{2} - \sin x$g(x)=2x3​−sinx.

Differentiating with respect to $x$x:

$$g'(x) = \dfrac{3x^2}{2} - \cos x$$g′(x)=23x2​−cosx

$$g''(x) = 3x + \sin x$$g′′(x)=3x+sinx

For $x > 0$x>0, $g''(x) > 0$g′′(x)>0, which implies that $g'(x)$g′(x) is strictly increasing on $(0, \infty)$(0,∞).

We have $g'(0) = -1 < 0$g′(0)=−1<0 and $g'(\pi/2) = \dfrac{3\pi^2}{8} > 0$g′(π/2)=83π2​>0. Thus, $g'(x) = 0$g′(x)=0 has exactly one root in $(0, \infty)$(0,∞), say at $x = x_0$x=x0​.

This means $g(x)$g(x) decreases on $(0, x_0)$(0,x0​) and increases on $(x_0, \infty)$(x0​,∞).

Since $g(0) = 0$g(0)=0 and $g(x)$g(x) initially decreases, $g(x_0) < 0$g(x0​)<0. As $x \to \infty$x→∞, $g(x) \to \infty$g(x)→∞. Therefore, $g(x) = 0$g(x)=0 has exactly one positive root.

Since $g(x)$g(x) is an odd function ($g(-x) = -g(x)$g(−x)=−g(x)), it must also have exactly one negative root.

Including the trivial root $x = 0$x=0, the equation $g(x) = 0$g(x)=0 has exactly $3$3 solutions.

Answer: $3$3