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Question

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A. ${x^4}$ for $x \leqslant - 1$

B. ${x^2}$ for $ - 1 < x \leqslant 0$

C. $f(\dfrac{1}{2}) = \dfrac{1}{2}$

D. $f(\dfrac{1}{2}) = \dfrac{1}{4}$

Answer

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In such questions first we need to understand the function. Let us understand it in a simpler function which says $f(x) = \max \{ x,{x^2}\} $.

Here, for any value in the domain the higher value among $x$ and ${x^2}$ is the value of the function. For example if we have to find $f(2)$ we get 2 and 4 when put into the function. The higher one among them is 4 so the value of function turns out to be 4.

Similarly, if have to find out \[f(\dfrac{1}{3})\] we get \[\dfrac{1}{3}\] and \[\dfrac{1}{9}\]. The higher one among them is \[\dfrac{1}{3}\] so the value of function turns out to be \[\dfrac{1}{3}\].

Here, we are given a function $f(x) = \max \{ x,{x^2},{x^3},{x^4}\} $. This is a little more complex than the above example function. These types of complex functions can be easily solved by graphical methods.

For graphical methods first we need to make a graph of every sub-function and then find the maximum of it.

Purple line = ${x^4}$

Green line = ${x^3}$

Blue line = ${x^2}$

Red line = ${x^{}}$

Now, from the graph it can be seen that in $x \leqslant - 1$ ,${x^4}$ is maximum. Therefore option (a) is correct.

Similarly, from the graph it can be seen that in $ - 1 < x \leqslant 0$ , ${x^2}$is maximum. Therefore option (b) is correct.

Now, the graph shows that in the domain $0 < x \leqslant 1$ the function is maximum in the red line which follows the function $f(x) = x$. Therefore, $f(\dfrac{1}{2}) = \dfrac{1}{2}$ and option (c) is correct and (d) incorrect.

Hence, option (a), (b) and (c) are correct.