Let f and g be two functions defined by fx= x+1, x<0 x-1, x≥0 and gx= x+1, x<0 1, x≥0 .Then gofx is

Correct answer: (A) Continuous everywhere but not differentiable exactly at one point. Given, fx= x+1, x 1 & gx= 1+x, x<0 1, x≥0 Now, for gofx gofx= 1+fx, fx<0, x<0 1, fx≥0, x≥0 ⇒ gofx= x+2, x<-1 1, x≥-1 Now plotting the diagram we get, Now from above diagram we can say that function is continuous but not differentiable at x=-1 due to sharp corner.

Q44 / 90 · JEE Main 2023 (11 Apr Shift 2) · Mathematics

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

Correct answer
A
Continuous everywhere but not differentiable exactly at one point
B
Continuous everywhere but not differentiable at $x = 1$
C
Differentiable everywhere
D
Not continuous at $x = 1$

Solution

Given,

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

$\Rightarrow f (x) = \{\begin{array}{lc} x + 1, & x < 0 \\ 1 - x, & 0 \leq x < 1 \\ x - 1, & x > 1 \end{array} & g (x) = \{\begin{matrix} 1 + x, & x < 0 \\ 1, & x \geq 0 \end{matrix}$

Now, for $g o f (x)$

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

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

Now plotting the diagram we get,

Now from above diagram we can say that function is continuous but not differentiable at $x = - 1$ due to sharp corner.

Hence, the correct answer is option: 'A'.

Let f and g be two functions defined by fx=x+1, x<0x-1, x≥0and gx=x+1, x<01, x≥0.Then gofx is

Solution

Relevant Questions

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