Fx-x3-x2-16x-20x-22 if x- 2 k if x-2 fx is continuous at x-2 then f2-

The correct option is D k=7 Since, f is continuous at x=2 Therefore, lim _ x→ 2 x^ 3 +x^ 2 16x+20 (x 2)^ 2 #160; =f( 2 ) it is ...

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Theorems for Continuity

fx=x3+x2-16x+...

Question

$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} \frac{x^{3} + x^{2} - 16 x + 20}{(x - 2)^{2}} & i f x \neq 2 k & i f x = 2 \end{matrix}$
$f (x)$ is continuous at $x = 2$ then $f (2) =$

k = 3

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k = 5

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k = 7

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k = 9

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Similar questions

f (x) = \frac{x^{3} + x^{2} - 16 x + 20}{(x - 2)^{2}}, x \neq 2

= k, x = 2

f (x) = ⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩ \begin{matrix} \begin{matrix} \frac{(x^{3} + x^{2} - 16 x + 20)}{{(x - 2)}^{2}}, & x \neq 2 \end{matrix} \begin{matrix} k & x = 2 \end{matrix} \end{matrix}

f (x)

x

k

{\begin{matrix} \frac{x^{3} + x^{2} - 16 x + 20}{(x - 2)^{2}}, & i f x \neq 2 k, & i f x = 2 \end{matrix} .

f (x) = \frac{x^{3} + x^{2} - 16 x + 20}{{(x - 2)}^{2}} i f x \neq 2, = k, i f x = 2.

f (x)

k = b \times 7

f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2 x - 1 & i f & x > 2 k & i f & x = 2 x^{2} - 1 & i f & x < 2 \end{matrix}

x = 2

k =

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