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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
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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