Let g x -1- x - x - and f x - left beginmatrix -1 - x0 -lendmatrix right- then for all x - f - g x - is equal to-IIT 2001A- 1B- xC- f x D- gx

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. ...

Byju's Answer

Standard XI

Mathematics

Greatest Integer Function

Let g x =1+ x...

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 A 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

Q. 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)

Q. 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

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g (x) = 1 + x - [x] a n d f (x) = ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} - 1, & x < 0 0, & x = 0 1, & x > 0 \end{matrix}

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, where [x] denotes the greatest integer less than or equal to x. Then for all

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Q. Let

g (x) = 1 + x - [x]

and

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 A 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 A 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.