Determine the value of - k - for which the following function is continuous at x -3 - f x - Veft matrix -fracx-3- 2-36x-3- - x neq 3 1k - x -3 -endmatrix -right-

Here,f(3)=k, and lim_x→ 3f(x)=lim_x→ 3(x+3)^2 36/x 3=lim_x+3→ 6(x+3)^2 6/(x+3) 6=2× 6^2 1=12. As f is continuous at x=3 so, ∴ k=12 ...

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Standard XII

Mathematics

Definition of Function

Determine the...

Question

Determine the value of 'k' for which the following function is continuous at x=3 :

$f (x) = {\begin{matrix} \frac{(x + 3)^{2} - 36}{x - 3}, & x \neq 3 k, x = 3 \end{matrix}$

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Solution

Here,f(3)=k, and ${lim}_{x \to 3} f (x) = {lim}_{x \to 3} \frac{(x + 3)^{2} - 36}{x - 3} = {lim}_{x + 3 \to 6} \frac{(x + 3)^{2} - 6}{(x + 3) - 6} = 2 \times 6^{2 - 1} = 12.$
$As f is continuous at x=3 so,$ $∴$ $k = 12$

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Determine the value of 'k' for which the following function is continuous at x=3 : f(x)={(x+3)2−36x−3,x≠3k,x=3

Here,f(3)=k, and limx→3f(x)=limx→3(x+3)2−36x−3=limx+3→6(x+3)2−6(x+3)−6=2×62−1=12. As f is continuous at x=3 so, ∴ k=12

Determine the value of 'k' for which the following function is continuous at x=3 :

$f (x) = {\begin{matrix} \frac{(x + 3)^{2} - 36}{x - 3}, & x \neq 3 k, x = 3 \end{matrix}$

Here,f(3)=k, and ${lim}_{x \to 3} f (x) = {lim}_{x \to 3} \frac{(x + 3)^{2} - 36}{x - 3} = {lim}_{x + 3 \to 6} \frac{(x + 3)^{2} - 6}{(x + 3) - 6} = 2 \times 6^{2 - 1} = 12.$
$As f is continuous at x=3 so,$ $∴$ $k = 12$