If -x- denotes a greatest integer not exceeding x and if the function f defined by a-2Cosx - x 2 - if x-0 btan-x-4-if x- 0is continuous

a+2cos xx^2, x lt; 0 for h=0^ b+a (π[x+4]) x gt; 0 LHL=a+2cos hh^2→ does not exist ∴ = f(x) is not continuous ...

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Continuity of a Function

If [x] deno...

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If $[x]$ denotes a greatest integer not exceeding $x$ and if the function $f$ defined by $⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{a + 2 C o s x}{x^{2}}, i f x < 0 b tan (\frac{π}{[x + 4]}), i f x \geq 0 \end{matrix}$ is continuous

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[x]

x

f

f (x) = ⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{a + 2 cos 2 x}{x^{2}}, i f x < 0 b tan (\frac{π}{[x + 4]}), i f x \geq 0 \end{matrix}

x = 0

(a, b)

x

f

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{a + 2 cos x}{x^{2}} & (x < 0) b tan \frac{π}{[x + 4]} & (x \geq 0) \end{matrix}

x = 0

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \begin{matrix} \frac{A + 3 cos x}{x^{2}}, & i f x < 0 B tan (\frac{π}{[x + 3]}), & i f x \geq 0 \end{matrix}

[.]

x = 0

f (x) = {\begin{matrix} x^{2} sin \frac{1}{x}, & i f x \neq 0 0, & i f x = 0 \end{matrix}

f

f (x) =

{\begin{matrix} x^{2} sin \frac{1}{x}, if x \neq 0 0, if x = 0 \end{matrix}

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