If fx-x2-x-6x-2- x-2k- x-2 is continuous at x-2- then the absolute value of k is

F(x) is continuous at x= 2 ∴lim_x→ 2f(x)=f( 2) ⇒lim_x→ 2x^2 x 6x+2=k ⇒lim_x→ 2(x 3)(x+2)x+2=k ⇒ k= 5 ∴ |k|=5 ...

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Derivative of Some Standard Functions

If fx=x2-x-6x...

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If $f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x^{2} - x - 6}{x + 2}, & x \neq - 2 k, & x = - 2 \end{matrix}$ is continuous at $x = - 2,$ then the absolute value of $k$ is

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f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x^{2} - (a + 2) x + a}{x - 2} & x \neq 2 2 & x = 2 \end{matrix}

x = 2

a

λ

x = - 1

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x^{2} - 2 x - 3}{x + 1}, & x \neq - 1 λ, & x = - 1 \end{matrix}

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \begin{matrix} \frac{x^{5} - 32}{x - 2}, & x \neq 2 \end{matrix} \begin{matrix} k, & x = 2 \end{matrix} \end{matrix}

x = 2

k

k

f

f (x) =

⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \frac{k cos x}{π - 2 x}, x \neq \frac{π}{2} 3, x = \frac{π}{2} \end{matrix}

x =

\frac{π}{2}

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x^{5} - 32}{x - 2} k, x = 2 \end{matrix}, x \neq 2

x = 2

k = . . . . .

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