f x - left matrix2 - x -2-16 -if - x neq 2 Weginmatrix 4 - X -16 1klend matrix - if -X-2lend matrix right- at -If f x is continuous at x -2- then find the value of k-

We have, f(x)={2^x+2 16/4^x 16 amp;if x≠ 2 k amp; if x=2 . Since, f(x) is continuous at x=2. ∴ LHL = RHL =f(2) At x=2, x→ 2lim2^x.2^2 2^4/4^x 4^2=x→ ...

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Standard XII

Mathematics

Onto Function

f x = left ma...

Question

$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2^{x + 2} - 16 & i f x \neq 2 \begin{matrix} 4^{x} - 16 k \end{matrix} & i f x = 2 \end{matrix} a t x = 2$

If f(x) is continuous at x=2, then find the value of k.

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Solution

We have, $f (x) = {\begin{matrix} \frac{2^{x + 2} - 16}{4^{x} - 16} & i f x \neq 2 k & i f x = 2 \end{matrix}$

Since, f(x) is continuous at x=2.
$∴$ LHL = RHL =f(2)

At x=2,
$l i m x \to 2 \frac{2^{x} {.2}^{2} - 2^{4}}{4^{x} - 4^{2}} = l i m x \to 2 \frac{4. (2^{x} - 4)}{(2^{x})^{-} (4)^{2}} = l i m x \to 2 \frac{4. (2^{x} - 4)}{(2 x^{-} 4) (2^{x} + 4)} [∵ a^{2} - b^{2} = (a + b) (a - b)] = l i m x \to 2 \frac{4}{(2^{x} + 4} = \frac{4}{8} = \frac{1}{2}$

But f(2)=k
$∴ k = \frac{1}{2}$

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f(x)=⎧⎪⎨⎪⎩2x+2−16if x≠24x−16kif x=2at x=2 If f(x) is continuous at x=2, then find the value of k.

We have, f(x)={2x+2−164x−16if x≠2kif x=2 Since, f(x) is continuous at x=2. ∴ LHL = RHL =f(2) At x=2, limx→22x.22−244x−42=limx→24.(2x−4)(2x)−(4)2=limx→24.(2x−4)(2x−4)(2x+4) [∵a2−b2=(a+b)(a−b)]=limx→24(2x+4=48=12 But f(2)=k ∴k=12

$f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} 2^{x + 2} - 16 & i f x \neq 2 \begin{matrix} 4^{x} - 16 k \end{matrix} & i f x = 2 \end{matrix} a t x = 2$

If f(x) is continuous at x=2, then find the value of k.