JEE MAIN - Mathematics (2023 - 11th April Evening Shift - No. 11)
Let $$f$$ and $$g$$ be two functions defined by
$$f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ |x-1|, & x \geq 0\end{array}\right.$$ and $$\mathrm{g}(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ 1, & x \geq 0\end{array}\right.$$
Then $$(g \circ f)(x)$$ is :
continuous everywhere but not differentiable at $$x=1$$
differentiable everywhere
not continuous at $$x=-1$$
continuous everywhere but not differentiable exactly at one point
Explanation
$$
\begin{aligned}
& \text { Sol. } f(x)=\left\{\begin{array}{c}
x+1, x<0 \\\
1-x, 0 \leq x<1 \\
x-1,1 \leq x
\end{array}\right. \\\\
& g(x)=\left\{\begin{array}{c}
x+1, x<0 \\
1, x \geq 0
\end{array}\right. \\\\
& g(f(x))=\left\{\begin{array}{c}
x+2, x<-1 \\
1, x \geq-1
\end{array}\right.
\end{aligned}
$$
_11th_April_Evening_Shift_en_11_1.png)
$\therefore \mathrm{g}(\mathrm{f}(\mathrm{x}))$ is continuous everywhere
$\mathrm{g}(\mathrm{f}(\mathrm{x}))$ is not differentiable at $\mathrm{x}=-1$ Differentiable everywhere else.
$\therefore \mathrm{g}(\mathrm{f}(\mathrm{x}))$ is continuous everywhere
$\mathrm{g}(\mathrm{f}(\mathrm{x}))$ is not differentiable at $\mathrm{x}=-1$ Differentiable everywhere else.
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