Let f x -1- x -1- 1 - x - 3 and g x -2- x -1-2 - x - 2- then fog x is equal tox - 0- x leq 2lendmatrix -right- -1- x -2 x 0 x -1 - 0- x leq 2x -1-2 x 0 1x -1 - 0- x leq 2D- None of these

The correct option is D None of these(fog)(x)=f(x+3), 2≤ x+3≤ 1 f( x+1), 1≤ x+2≤ 2 ={ f(x+3), amp; 2≤ x+3≤ 1 f( x+1), amp; 1≤ x+1≤ 1 f( x+1), ...

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Byju's Answer

Standard IX

Mathematics

Composite Function

Let f x =-1+|...

Question

Let $f (x) = - 1 + | x - 1 |, 1 \leq x \leq 3$ and $g (x) = 2 - | x + 1 |, - 2 \leq x \leq 2$ , then (fog)(x) is equal to

{\begin{matrix} x + 1 - 2 \leq x \leq 0 x - 10 < x \leq 2 \end{matrix}

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{\begin{matrix} x - 1 - 2 \leq x \leq 0 x + 0 < x \leq 2 \end{matrix}

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{\begin{matrix} - 1 - x - 2 \leq x \leq 0 x - 10 < x \leq 2 \end{matrix}

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N o n e o f t h e s e

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Solution

The correct option is D $N o n e o f t h e s e$
$(f o g) (x) = f (x + 3), - 2 \leq x + 3 \leq - 1 f (- x + 1), - 1 \leq - x + 2 \leq 2$
$= ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} f (x + 3), & - 2 \leq x + 3 \leq - 1 f (- x + 1), & - 1 \leq - x + 1 \leq 1 f (- x + 1), & 1 \leq - x + 1 \leq 2 \end{matrix}$
We can see that f(x) will not be defined when $1 \leq - x + 1 \leq 2.$
$= {\begin{matrix} f (x + 3), & - 2 \leq x + 3 \leq - 1 f (- x + 1), & - 1 \leq - x + 1 \leq 1 \end{matrix}$
$= ⎧ ⎪ ⎨ ⎪ ⎩ \begin{matrix} x + 1, & - 2 \leq x \leq - 1 - x - 1, & - 1 \leq x \leq 0 x - 1, & 0 \leq x \leq 1 \end{matrix}$

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Let f(x)=−1+|x−1|,1≤x≤3 and g(x)=2−|x+1|,−2≤x≤2, then (fog)(x) is equal to

Let $f (x) = - 1 + | x - 1 |, 1 \leq x \leq 3$ and $g (x) = 2 - | x + 1 |, - 2 \leq x \leq 2$ , then (fog)(x) is equal to