Question

$f\left(x\right)=|9-x^2|-|x-a|$ then which of the following statements is/are true?

• A. For a = 8, number of distinct real roots of f(x)=0 is 4
• B. For a = 8, number of distinct real roots of f(x)=0 is 3
• C. For a = 3, number of distinct real roots of f(x)=0 is 4
• D. For a = 3, number of distinct real roots of f(x)=0 is 3

Answer 1

Answer: (A), (D)

For a = 3, number of distinct real roots of f(x)=0 is 3

$|9-x^2|=|x-a|$

Find extreme cases where |x-a| touches $|9-x^2|.$
For Right side $9-x^2=-x+a$
$x^2-x+a-9=0\Rightarrow D=0$
1 = 4(a-9)
$a=\frac{37}{4}$
but for $a=3\to$ three solutions exists.
From the figure we can conclude that for
$aϵ(\frac{-37}{4},-3)\cup (-3.3)\cup (3,\frac{37}{4})\to 4$ real roots.
For $aϵ\{\frac{-37}{4},-3,3,\frac{37}{4}\}\to f\left(x\right)=0$ has 3 real roots.