Question

Number of integral values of b for which the equation $\frac{x^3}{3}-x=b$ has 3 distinct solutions is

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

The correct option is B 1

Let $\text{f(x)}=\frac{x^3}{3}-x-b$
$\Rightarrow \text{f'(x)}=x^2-1=0\Rightarrow x=\pm 1$
Now for f to have 3 distinct solutions, we must have,
$\text{f'(-1)}\cdot \text{f'(1)}<0$
$\Rightarrow (-\frac{1}{3}+1-b)(\frac{1}{3}-1-b)<0$
$\Rightarrow (\frac{2}{3}-b)(-\frac{2}{3}-b)<0$
$\Rightarrow (b-\frac{2}{3})(b+\frac{2}{3})<0$
$\Rightarrow b\in (-\frac{2}{3},\frac{2}{3})$
Hence there is only one integral value, $b=0$