Let g x -1- x - x - and f x - left beginmatrix -1 - x0 -endmatrix right- then for all x- f-gx- is equal to-IIT 2001A- xB- 1C- f x D- gx

The correct option is B 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. ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard IX

Mathematics

Multiplication of Matrices

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)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

f(x)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

g(x)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

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.

Suggest Corrections

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

x

f [g (x)]

is equal to:

Q. Let

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

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

x, f (g (x))

is equal to
(a) x
(b) 1
(c) f(x)
(d) g(x)

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

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}

. Then for all

x, f [g (x)]

is equal to

Join BYJU'S Learning Program

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.