Define g(x)= _ -3 ^3 f(x-y) f(y) d y, for all real x, where f(t)=… - Vidyadip

MCQ

Define $g(x)=\int \limits_{-3}^3 f(x-y) f(y) d y$, for all real $x$, where $f(t)=\left\{\begin{array}{ll}1, & 0 \leq t \leq 1 \\ 0, & \text { else where }\end{array}\right.$ Then,

A
$g(x)$ is not continuous everywhere
B
$g(x)$ is continuous everywhere but differentiable nowhere
C
$g(x)$ is continuous everywhere and differentiable everywhere except at $x=0,1$
D
$g(x)$ is continuous everywhere and differentiable everywhere except at $x=0,1,2$

✓

Answer

We have,
$ g(x)=\int \limits_{-3}^3 f(x-y) f(y) d y$
$f(t)=\left\{\begin{array}{cc} 1, \quad 0 \leq t \leq 1 \\ 0, \quad \text { else where } \end{array}\right. $
$ g(x)=\int \limits_{-3}^0 g(x-y) f(y) d y$
$+\int \limits_0^1 f(x-y) f(y) d y$
$+\int \limits_1^3 f(x-y) f(y) d y$
Put $x-y=t $
$\Rightarrow-d y=d t$
$\Rightarrow g(x)=-\int \limits_x^{x-1} f(t) d t$
$ \Rightarrow g(x)=\int \limits_{x-1}^x f(t) d t$
Put $x-y=t$
$ \Rightarrow-d y=d t$
$ \Rightarrow g(x)=-\int_x^{x-1} f(t) d t $
$\Rightarrow g(x)=\int_{x-1}^x f(t) d t$
$\Rightarrow g(x)=\left\{\begin{array}{cc} 0, & x \leq 0 \\ x, & 0 < x < 1 \\ 2-x, & 1 \leq x \leq 2 \\ 0, & x > 2 \end{array}\right.$
Clearly, $g(x)$ is continues at for all $x$
$g^{\prime}(x)= 0, x \leq 0$
$1, 0 < x < 1$
$-1, 1 \leq x \leq 2$
$0, x > 2 .$
$g(x)$ is not differentiable at $0,1,2$

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