Question

Please Explain the question as well as the Solution. Please answer it quickly Please ​

Answer 1

The function $f:[0,\ 10]\to\mathbb{R}$ is defined as,

$\longrightarrow f(x)=\left\{\begin{array}{ll}x^2,&0\leq x\leq 3\\3x,&3\leq x\leq 10\end{array}\right.$

Taking LHL as $x$ tends to 3,

$\displaystyle\longrightarrow\lim_{x\to3^-}f(x)=\lim_{x\to3^-}x^2$

$\displaystyle\longrightarrow\lim_{x\to3^-}f(x)=9\quad\quad\dots(1)$

Taking RHL as $x$ tends to 3,

$\displaystyle\longrightarrow\lim_{x\to3^+}f(x)=\lim_{x\to3^+}3x$

$\displaystyle\longrightarrow\lim_{x\to3^+}f(x)=9\quad\quad\dots(2)$

From (1) and (2),

$\displaystyle\longrightarrow\lim_{x\to3^-}f(x)=\lim_{x\to3^+}f(x)$

So $f(x)$ is continuous at $x=3$ and $f(3)$ is unique.

$f(x)$ has, therefore, a unique value for each and every $x\in[0,\ 10].$

Hence $f(x)$ is a function.

The function $g:[0,\ 10]\to\mathbb{R}$ is defined as,

$\longrightarrow g(x)=\left\{\begin{array}{ll}x^2,&0\leq x\leq 2\\3x,&2\leq x\leq 10\end{array}\right.$

Taking LHL as $x$ tends to 2,

$\displaystyle\longrightarrow\lim_{x\to2^-}g(x)=\lim_{x\to2^-}x^2$

$\displaystyle\longrightarrow\lim_{x\to2^-}g(x)=4\quad\quad\dots(3)$

Taking RHL as $x$ tends to 3,

$\displaystyle\longrightarrow\lim_{x\to2^+}g(x)=\lim_{x\to2^+}3x$

$\displaystyle\longrightarrow\lim_{x\to2^+}g(x)=6\quad\quad\dots(4)$

From (3) and (4),

$\displaystyle\longrightarrow\lim_{x\to2^-}g(x)\neq\lim_{x\to2^+}g(x)$

So $g(x)$ is discontinuous at $x=2$ and so $g(2)$ is not unique, i.e., $g(2)$ has more than one value.

$\longrightarrow g(2)\in\{4,\ 6\}$

Definition of a function is that all elements in domain has exactly one image in range.

But $g(x)$ for $x=2$ has two images, 4 and 6, not exactly one.

Hence $g(x)$ is not a function.