Question

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Answer 1

$\large\text{\underline{Question}}$

Find $k$ that satisfies $(f\circ f)(k)=5$, where $f(x)=2x-1$.

$\large\text{\underline{Basic Concepts}}$

Function composition.

$(f\circ f)(x)$ is a function that the range of $f(x)$ corresponds to the domain of

We can also write it as $f^{2}(x)$.

$\large\text{\underline{Solution}}$

$y=(f\circ f)(x)$

$\hookrightarrow y=f(2x-1)$

$\hookrightarrow y=2(2x-1)-1$

$\hookrightarrow y=4x-3$

$\therefore (f\circ f)(x)=4x-3$

Then,

$\hookrightarrow (f\circ f)(k)=5$

$\hookrightarrow4k-3=5$

$\hookrightarrow 4k=8\implies\therefore k=2$

$\large\text{\underline{Answer}}$

Hence, the required value is 2.

Answer 2

Question :-

• Find k if f.f (k)=5 where f(k)= 2k-1

$\sf\tt\large{\green {\underline {\underline{⚘\;Answer:}}}}$

$\sf\tt\large{\purple {\underline {\underline{⚘\; The \;value \;of \;k \;is \;2:}}}}$

$\sf\tt\large{\red {\underline {\underline{⚘\;Given:}}}}$

• f.f (k)=5
• where f(k)=2k-1

$\sf\tt\large{\red {\underline {\underline{⚘\;To find:}}}}$

• The value of k .

$\sf\tt\large{\red {\underline {\underline{⚘\;Solution:}}}}$

• f(k)=2k-1
• fof=5

• f (f (k)m=f (2k-1)=5

• 4k-2-1=5

• 4k-3=5

• 4k=5+3

• 4k=8

• $k = \frac{8}{4} = 2$

$\sf\tt\large{\red {\underline {\underline{⚘\;Therefore:}}}}$

• Value of k =2

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