Question

Let $f(x) = (x^2 - 3x + 2)|x^3 - 6x^2 + 11x - 6| + |\sin(x+\frac{\pi}{4})|$. The number of non-differentiable points of the function $y = f(x)$ in $[0, 2\pi]$ equals

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (C)

3

Answer 2

The function is $f(x) = (x^2 - 3x + 2)|x^3 - 6x^2 + 11x - 6| + |\sin(x+\frac{\pi}{4})|$. Let $g(x) = (x^2 - 3x + 2)|x^3 - 6x^2 + 11x - 6|$. Factorizing, we get $x^2 - 3x + 2 = (x-1)(x-2)$ and $x^3 - 6x^2 + 11x - 6 = (x-1)(x-2)(x-3)$. So, $g(x) = (x-1)(x-2)|(x-1)(x-2)(x-3)|$. The term $|(x-1)(x-2)(x-3)|$ is non-differentiable at $x=1, 2, 3$. At $x=1$ and $x=2$, $(x-1)(x-2) = 0$. Let $u(x) = (x-1)(x-2)$ and $v(x) = (x-1)(x-2)(x-3)$. Then $g(x) = u(x)|v(x)| = u(x)|u(x)(x-3)| = |u(x)|^2|x-3|$. Since $|u(x)|^2$ is differentiable and $(x-3)$ is differentiable, $g(x)$ is differentiable at $x=1$ and $x=2$. At $x=3$, $(x-1)(x-2) = (3-1)(3-2) = 2 \neq 0$. Since $|(x-1)(x-2)(x-3)|$ has a simple root at $x=3$, it is non-differentiable there. As $(x-1)(x-2)$ is non-zero at $x=3$, $g(x)$ is non-differentiable at $x=3$.

Let $h(x) = |\sin(x+\frac{\pi}{4})|$. $|\sin(\theta)|$ is non-differentiable when $\sin(\theta)=0$. This occurs when $\theta = n\pi$. So, $x+\frac{\pi}{4} = n\pi$, which means $x = n\pi - \frac{\pi}{4}$. In the interval $[0, 2\pi]$: For $n=1$, $x = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$. For $n=2$, $x = 2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$. These are points where $h(x)$ is non-differentiable.

The non-differentiable points are $x=3$ (from $g(x)$) and $x=\frac{3\pi}{4}, \frac{7\pi}{4}$ (from $h(x)$). These points are distinct. The total number of non-differentiable points is $1 + 2 = 3$.