Let f and g be two functions defined by f ( x ) = { x + 1 , x 0 | x - 1 | , x 0 and g ( x ) = { x + 1 , x 0 1 , x 0 Then ( g o f ) ( x ) is [2023]

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Q.
$Let f and g be two functions defined by$ $f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$ Then $(g o f) (x)$ is [2023]

1 continuous everywhere but not differentiable exactly at one point

2 not continuous at $x = - 1$

3 continuous everywhere but not differentiable at $x = 1$

4 differentiable everywhere

Ans.
(1)

We have,

$f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$

$f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ x - 1, & x \geq 1 \\ 1 - x, & 0 \leq x < 1 \end{matrix}$ $∴$ $(g o f) (x) =$ ${\begin{matrix} x + 2, & x < - 1 \\ 1, & x \geq - 1 \end{matrix}$

$Hence, (g o f) (x) is not differentiable at x = - 1, but it is continuous everywhere.$

Solution

Q.
$Let f and g be two functions defined by$ $f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$ Then $(g o f) (x)$ is [2023]

1 continuous everywhere but not differentiable exactly at one point

2 not continuous at $x = - 1$

3 continuous everywhere but not differentiable at $x = 1$

4 differentiable everywhere

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Solution

Q. Let f and g be two functions defined by f(x)={x+1,x<0|x-1|, x≥0 and g(x)={x+1,x<01,x≥0 Then (gof)(x) is [2023]

1 continuous everywhere but not differentiable exactly at one point

2 not continuous at x=-1

3 continuous everywhere but not differentiable at x=1

4 differentiable everywhere

Ans. (1) We have, f(x)={x+1,x<0|x-1|,x≥0 and g(x)={x+1,x<01,x≥0 f(x)={x+1,x<0x-1,x≥11-x,0≤x<1 ∴ (gof)(x)={x+2,x<-11,x≥-1 Hence, (gof)(x) is not differentiable at x=-1, but it is continuous everywhere.

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Q.
$Let f and g be two functions defined by$ $f (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ | x - 1 |, & x \geq 0 \end{matrix}$ and $g (x) =$ ${\begin{matrix} x + 1, & x < 0 \\ 1, & x \geq 0 \end{matrix}$ Then $(g o f) (x)$ is [2023]

2 not continuous at $x = - 1$

3 continuous everywhere but not differentiable at $x = 1$