Question

If sgn (x3 – x2 – 6x) = 1, then number of integral value(s) of x satisfying the equation is (if x (–3, 5))

Answer 1

Correct Question:-

If $\sf{sgn(x^3-x^2-6x)=1,}$ then find no. of integral value(s) of $\sf{x}$ satisfying the equation for $\sf{x\in(-3,\ 5),}$ where $\sf{sgn(x)}$ represents signum function in $\sf{x.}$

Solution:-

Signum function $\sf{sgn(x):\mathbb{R}\to\left\{-1,\ 0,\ 1\right\}}$ is defined as,

$\longrightarrow\sf{sgn(x)=\left\{\begin{array}{rl}\sf{-1,}&\sf{x\ \textless\ 0}\\\sf{0,}&\sf{x=0}\\\sf{1,}&\sf{x\ \textgreater\ 0}\end{array}\right.}$

Therefore, according to the question,

$\longrightarrow\sf{sgn(x^3-x^2-6x)=1}$

$\Longrightarrow\sf{x^3-x^2-6x\ \textgreater\ 0}$

$\longrightarrow\sf{x(x^2-x-6)\ \textgreater\ 0}$

$\longrightarrow\sf{x(x^2-3x+2x-6)\ \textgreater\ 0}$

$\longrightarrow\sf{x(x(x-3)+2(x-3))\ \textgreater\ 0}$

$\longrightarrow\sf{x(x-3)(x+2)\ \textgreater\ 0\quad\quad\dots(1)}$

So the equality $\sf{x(x-3)(x+2)=0}$ holds true for $\sf{x\in\{-2,\ 0,\ 3\}.}$

We see the inequality (1) holds true for $\sf{x\ \textgreater\ 3.}$

By wavy curve method, we get,

• (1) holds true for $\sf{x\in(3,\ \infty).}$

• (1) does not hold true for $\sf{x\in(0,\ 3).}$

• (1) holds true for $\sf{x\in(-2,\ 0).}$

• (1) does not hold true for $\sf{x\in(-\infty,\ -2).}$

Therefore, since $\sf{x\in(-3,\ 5),}$

$\longrightarrow\sf{x\in[(-2,\ 0)\cup(3,\ \infty)]\cap(-3,\ 5)}$

$\longrightarrow\sf{x\in(-2,\ 0)\cup(3,\ 5)}$

Hence integral values of $\sf{x}$ are given by,

$\longrightarrow\sf{x\in\{-1,\ 4\}}$

Thus the no. of integral values of $\sf{x}$ is $\bf{2.}$