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

If sgn (x³ – x² – 6x) = 1, then number of integral value(s) of x satisfying the equation is 2

Step-by-step explanation:

We know that

$\text{sgn} f(x)=\frac{|f(x)|}{f(x)}$

And

$|f(x)|=f(x)$ for all x for which $f(x)\ge0$

$|f(x)|=-f(x)$ for all x for which $f(x)<0$

Here

$f(x)=x^3-x^2-6x$

$\implies f(x)=x(x^2-x-6)$

$\implies f(x)=x(x^2-3x+2x-6)$

$\implies f(x)=x[x(x-3)+2(x-3)]$

$\implies f(x)=x(x+2)(x-3)$

Now,

$f(x)<0$

$\implies x(x+2)(x-3)<0$

$\implies x\in (-\infty,-2)\cup(0,3)$

And

$f(x)\ge0$

$\implies x(x+2)(x-3)\ge0$

$\implies x\in [-2,0]\cup[3,\infty)$

Now given in the question that

$x\in (-3,5)$

i.e. $x^3-x^2-6x$ will be negative for $(-3,-2)\cup(0,3)$ and positive or equal to zero for $[-2,0]\cup[3,5)$

$f(x)$ will be strictly positive for $(-2,0)\cup(3,5)$

Now,

$\text{sgn} (x^3-x^2-6x)=\frac{|x^3-x^2-6x|}{x^3-x^2-6x}$

$\text{sgn} (x^3-x^2-6x)$ will be 1 only when $|x^3-x^2-6x|=x^3-x^2-6x$

Which will be possible for all x where $x^3-x^2-6x>0$ which is for $x\in(-2,0)\cup(3,5)$

Therefore, the integral values of x will be -1 and 4

The number of integral values  = 2

Hope this answer is helpful.

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