JEE Main 2026 Apr 2 Shift 1, Mathematics Q18: Maxima & Minima

JEE Main2026Apr 2, Shift 1Mathematics

The number of critical points of the function $f (x) = ⎩ ⎨ ⎧ \frac{sin x}{x}, x \neq = 0 1, x = 0$ in the interval $(- 2 π, 2 π)$ is equal to :

Solution

$\displaystyle \lim_{x\to 0}\left|\frac{\sin x}{x}\right| = 1 = f(0)$

$\Rightarrow f (x)$ is continuous.

Now,

$\displaystyle \frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{x\cos x-\sin x}{x^2} $

$\Rightarrow f^{'} (x) = 0$

$\Rightarrow tan x = x$

There are $3$ solutions of $tan x = x$ in the interval $(- 2 π, 2 π)$ .

Also, $f^{'} (x)$ does not exist at $x = - π, π$ .

Therefore, total number of points $= 3 + 2 = 5$ .

$5$

Concept behind this question

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