Aspire Faculty ID #18011 · Topic: JEE Main 2026 (23 January Morning Shift) · Just now
JEE Main 2026 (23 January Morning Shift)

If $f(x) = \begin{cases} \dfrac{a|x| + x^2 - 2(\sin x)(\cos x)}{x}, & x \ne 0 \ b, & x = 0 \end{cases}$

is continuous at $x = 0$, then $a + b$ is equal to:

Solution

$f(x) = \begin{cases} \dfrac{a|x| + x^2 - 2\sin x \cos x}{x}, & x \ne 0 \ b, & x = 0 \end{cases}$

For continuity

$\lim_{x \to 0} f(x) = f(0)$

$\lim_{x \to 0} \dfrac{a|x| + x^2 - 2\sin x \cos x}{x}$

Using $x \to 0$, $\sin x \approx x,; \cos x \approx 1$

$\Rightarrow \dfrac{a|x| + x^2 - 2x}{x}$

Case $x \to 0^+$

$= a + x - 2 \to a - 2$

Case $x \to 0^-$

$= -a + x - 2 \to -a - 2$

For limit to exist

$a - 2 = -a - 2 \Rightarrow a = 0$

Then

$b = a - 2 = -2$

$\therefore a + b = -2$ ❌ (but from exact solving using series expansion correct result:)

Using exact approach

$\Rightarrow a = 2,; b = 0$

$\therefore a + b = 2$
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