Question

Consider $f:R\to R,f\left(x\right)=|x^2-5|x|+6|$. Then which of the following is NOT true?

• A. f(x) is non differentiable at 5 points
• B. f(x) has local minima at x = 0
• C. f(x) = R has 4 solutions for $R\in (\frac{1}{4},6)\cup$ {0}
• D. f(x) = R has 8 solutions for $R\in (0,\frac{1}{3})$

Answer 1

The correct option is C f(x) = R has 4 solutions for $R\in (\frac{1}{4},6)\cup$ {0}

f(x) = |(|x| – 2)(|x| – 3)|
= g(|x|) where g(x) = |(x – 2)(x – 3)|
g(x) is the upward parabola y = (x – 2)(x – 3) whose negative interval (2, 3) is reflected about x-axis to obtain local maxima at $\frac{5}{2}$
g(x) has y-intercept = 6. To obtain f(x) = g(|x|), replace g(x) on negative x-axis by the reflection of g(x) on positive x-axis about x = 0.

From graph we find that f(x) is non differentiable at x = –3, –2, 0, 2, 3. K is the line $y=\frac{1}{4}$ which touches the two local maxima at $x=\pm \frac{5}{2}$. Any line M between y = 6 (y-intercept) and K intersects the graph at 4 distinct points. Also any line L between y = 0(x-axis) and K intersects the graph at 8 distinct points .
So (A), (C), (D) are correct.
But x = 0 is not a local minima, it is a local maxima.