Let g-x-1 - x - -x- and f-x- - -1 amp- x lt- 0 amp- - 0- amp- x - 0 amp- - 1- amp- x gt- 0 amp- -then for all x- f-g-x- is equal to- -IIT 2001-1xf-x-g-x-

The correct option is A 1g(x) = 1 + x [x] is greater than 1 since x [x] gt; 0 fg(x) = 1 since f(x) = 1 for all x gt; 0 Therefore, (b) is the answer. ...

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Standard VIII

Mathematics

Greatest integer function

Let g(x)=1 + ...

Question

Let g(x)=1 + x - [x] and f(x) = $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1 & x < 0 0, & x = 0 1, & x > 0 \end{matrix}$ then for all x, f[g(x)] is equal to:
(IIT 2001)

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f(x)

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g(x)

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Solution

The correct option is B 1
g(x) = 1 + x - [x] is greater than 1 since x - [x] > 0
f{g(x)} = 1 since f(x) = 1 for all x > 0
Therefore, (b) is the answer.

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Similar questions

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g (x) = 1 + x - [x] and f (x) = \{\begin{cases} - 1, & x < 0 \\ 0, & x = 0, \\ 1, & x > 0 \end{cases}

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f (x) = ⎧ ⎨ ⎩ \begin{matrix} \begin{matrix} - 1 & x < 0 \end{matrix} \begin{matrix} 0 & x = 0 \end{matrix} \begin{matrix} 1 & x > 0 \end{matrix} \end{matrix}

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Let g(x)=1 + x - [x] and f(x) =⎧⎪⎨⎪⎩−1x<00,x=01,x>0 then for all x, f[g(x)] is equal to: (IIT 2001)

The correct option is B 1g(x) = 1 + x - [x] is greater than 1 since x - [x] > 0 f{g(x)} = 1 since f(x) = 1 for all x > 0 Therefore, (b) is the answer.

Let g(x)=1 + x - [x] and f(x) = $⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1 & x < 0 0, & x = 0 1, & x > 0 \end{matrix}$ then for all x, f[g(x)] is equal to:
(IIT 2001)

The correct option is B 1
g(x) = 1 + x - [x] is greater than 1 since x - [x] > 0
f{g(x)} = 1 since f(x) = 1 for all x > 0
Therefore, (b) is the answer.