JEE MAIN - Mathematics (2023 - 30th January Evening Shift - No. 6)
Let $f, g$ and $h$ be the real valued functions defined on $\mathbb{R}$ as
$f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$
$g(x)=\left\{\begin{array}{cc}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right.$
and $h(x)=2[x]-f(x)$, where $[x]$ is the greatest integer $\leq x$. Then the
value of $\lim\limits_{x \rightarrow 1} g(h(x-1))$ is :
$f(x)=\left\{\begin{array}{cc}\frac{x}{|x|}, & x \neq 0 \\ 1, & x=0\end{array}\right.$
$g(x)=\left\{\begin{array}{cc}\frac{\sin (x+1)}{(x+1)}, & x \neq-1 \\ 1, & x=-1\end{array}\right.$
and $h(x)=2[x]-f(x)$, where $[x]$ is the greatest integer $\leq x$. Then the
value of $\lim\limits_{x \rightarrow 1} g(h(x-1))$ is :
1
$-1$
$\sin (1)$
0
Explanation
$$f(x) = {\mathop{\rm sgn}} (x)$$
$$h(x) = 2[x] - {\mathop{\rm sgn}} (x)$$
If $$x \to {1^ + }$$ then $$h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$$
$$ = 0 - 1 = - 1$$
& if $$x \to {1^ - }$$ then $$h(x - 1) = 2[x] - 2 - {\mathop{\rm sgn}} (x - 1)$$
$$ = - 2 + 1 = - 1$$
$$\therefore$$ $$\mathop {\lim }\limits_{x \to {1^ + }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ + }} {{\sin (h(x + 1) + 1)} \over {h(x - 1) + 1}} = 1$$
$$\mathop {\lim }\limits_{x \to {1^ - }} g(h(x - 1)) = \mathop {\lim }\limits_{x \to {1^ - }} {{\sin (h(x - 1) + 1)} \over {h(x - 1) + 1}} = 1$$
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