Question

If a differentiable function 'f' satisfies the relation $f(x + y) - f(x - y) = 4xy - 10y \forall$ $x, y \in R$ and $f(1) = 2$, then find the number of points of non-derivability of $g(x) = \left| \left| f(|x|) \right| - \frac{1}{4} \right|$.

Answer 1

9

Answer 2

First, we need to find the function $f(x)$. Given the relation $f(x + y) - f(x - y) = 4xy - 10y$ for all $x, y \in R$. Since $f$ is a differentiable function, we can differentiate this relation.

Differentiate with respect to $y$: $\frac{\partial}{\partial y} [f(x + y) - f(x - y)] = \frac{\partial}{\partial y} [4xy - 10y]$ $f'(x + y) \cdot (1) - f'(x - y) \cdot (-1) = 4x - 10$ $f'(x + y) + f'(x - y) = 4x - 10$ (Equation 1)

Differentiate with respect to $x$: $\frac{\partial}{\partial x} [f(x + y) - f(x - y)] = \frac{\partial}{\partial x} [4xy - 10y]$ $f'(x + y) \cdot (1) - f'(x - y) \cdot (1) = 4y$ $f'(x + y) - f'(x - y) = 4y$ (Equation 2)

Now we have a system of two linear equations in terms of $f'(x+y)$ and $f'(x-y)$:

1. $f'(x + y) + f'(x - y) = 4x - 10$
2. $f'(x + y) - f'(x - y) = 4y$

Add Equation 1 and Equation 2: $2f'(x + y) = (4x - 10) + 4y$ $2f'(x + y) = 4x + 4y - 10$ $f'(x + y) = 2x + 2y - 5$

Let $t = x + y$. Then, we have $f'(t) = 2t - 5$. So, $f'(x) = 2x - 5$.

Now, integrate $f'(x)$ to find $f(x)$: $f(x) = \int (2x - 5) dx = x^2 - 5x + C$

We are given the condition $f(1) = 2$. Use this to find the value of $C$: $f(1) = (1)^2 - 5(1) + C = 2$ $1 - 5 + C = 2$ $-4 + C = 2$ $C = 6$

Thus, the function is $f(x) = x^2 - 5x + 6$. We can factorize $f(x)$ as $f(x) = (x-2)(x-3)$. The roots of $f(x)$ are $x=2$ and $x=3$. The vertex of the parabola $f(x)$ is at $x = -\frac{-5}{2(1)} = \frac{5}{2} = 2.5$. The minimum value of $f(x)$ is $f(2.5) = (2.5)^2 - 5(2.5) + 6 = 6.25 - 12.5 + 6 = -0.25 = -\frac{1}{4}$.

Now, we need to find the number of points of non-derivability of $g(x) = \left| \left| f(|x|) \right| - \frac{1}{4} \right|$.

Let's break down the function $g(x)$ step by step:

Step 1: Analyze $h(x) = f(|x|)$ $h(x) = \begin{cases} f(x) = x^2 - 5x + 6 & \text{if } x \ge 0 \\ f(-x) = (-x)^2 - 5(-x) + 6 = x^2 + 5x + 6 & \text{if } x < 0 \end{cases}$

Check differentiability of $h(x)$ at $x=0$: $h'(x) = \begin{cases} 2x - 5 & \text{if } x > 0 \\ 2x + 5 & \text{if } x < 0 \end{cases}$ Left-hand derivative at $x=0$: $h'(0^-) = 2(0) + 5 = 5$ Right-hand derivative at $x=0$: $h'(0^+) = 2(0) - 5 = -5$ Since $h'(0^-) \ne h'(0^+)$, $h(x)$ is not differentiable at $x=0$. This is one point of non-derivability.

The roots of $h(x)$ are: For $x \ge 0$: $x^2 - 5x + 6 = 0 \implies (x-2)(x-3)=0 \implies x=2, x=3$. For $x < 0$: $x^2 + 5x + 6 = 0 \implies (x+2)(x+3)=0 \implies x=-2, x=-3$. So, $h(x)=0$ at $x \in \{-3, -2, 2, 3\}$.

Step 2: Analyze $k(x) = |h(x)| = |f(|x|)|$ A function $|u(x)|$ is non-differentiable at points where $u(x)$ is non-differentiable, or where $u(x)=0$ and $u'(x) \ne 0$. We already know $h(x)$ is non-differentiable at $x=0$. So $k(x)$ is non-differentiable at $x=0$. Now check the roots of $h(x)$: $x \in \{-3, -2, 2, 3\}$. At $x=2$: $h'(2) = 2(2)-5 = -1 \ne 0$. So $x=2$ is a non-derivability point. At $x=3$: $h'(3) = 2(3)-5 = 1 \ne 0$. So $x=3$ is a non-derivability point. At $x=-2$: $h'(-2) = 2(-2)+5 = 1 \ne 0$. So $x=-2$ is a non-derivability point. At $x=-3$: $h'(-3) = 2(-3)+5 = -1 \ne 0$. So $x=-3$ is a non-derivability point. So far, $k(x)$ is non-differentiable at $x \in \{-3, -2, 0, 2, 3\}$. (5 points)

Step 3: Analyze $m(x) = k(x) - \frac{1}{4} = |f(|x|)| - \frac{1}{4}$ Subtracting a constant does not change the differentiability points. So, $m(x)$ is non-differentiable at $x \in \{-3, -2, 0, 2, 3\}$.

Step 4: Analyze $g(x) = |m(x)| = \left| \left| f(|x|) \right| - \frac{1}{4} \right|$ $g(x)$ is non-differentiable where $m(x)$ is non-differentiable, or where $m(x)=0$ and $m'(x) \ne 0$. The points where $m(x)$ is non-differentiable are $x \in \{-3, -2, 0, 2, 3\}$. These 5 points are non-derivability points for $g(x)$.

Now, find the points where $m(x)=0$: $|f(|x|)| - \frac{1}{4} = 0 \implies |f(|x|)| = \frac{1}{4}$. This implies $f(|x|) = \frac{1}{4}$ or $f(|x|) = -\frac{1}{4}$.

Case A: $f(|x|) = -\frac{1}{4}$ Let $t = |x|$. $t^2 - 5t + 6 = -\frac{1}{4}$ $t^2 - 5t + \frac{25}{4} = 0$ $\left(t - \frac{5}{2}\right)^2 = 0$ $t = \frac{5}{2} = 2.5$. So, $|x| = 2.5 \implies x = \pm 2.5$.

Let's check $m'(x)$ at $x=\pm 2.5$. For $x=2.5$, $f(|x|) = f(2.5) = -1/4$. Since $f(x)$ is negative for $x \in (2,3)$, $|f(x)| = -(x^2-5x+6)$. So, for $x \in (2,3)$, $m(x) = -(x^2-5x+6) - 1/4 = -x^2+5x-6-1/4$. $m'(x) = -2x+5$. At $x=2.5$, $m'(2.5) = -2(2.5)+5 = -5+5 = 0$. Since $m(2.5)=0$ and $m'(2.5)=0$, $x=2.5$ is a point where the graph of $m(x)$ touches the x-axis smoothly. Therefore, $|m(x)|$ is differentiable at $x=2.5$.

Similarly, for $x=-2.5$, $f(|x|) = f(2.5) = -1/4$. Since $f(-x)$ is negative for $x \in (-3,-2)$, $|f(-x)| = -(x^2+5x+6)$. So, for $x \in (-3,-2)$, $m(x) = -(x^2+5x+6) - 1/4 = -x^2-5x-6-1/4$. $m'(x) = -2x-5$. At $x=-2.5$, $m'(-2.5) = -2(-2.5)-5 = 5-5 = 0$. Since $m(-2.5)=0$ and $m'(-2.5)=0$, $x=-2.5$ is also a point where $g(x)$ is differentiable. So, $x=\pm 2.5$ are NOT non-derivability points.

Case B: $f(|x|) = \frac{1}{4}$ Let $t = |x|$. $t^2 - 5t + 6 = \frac{1}{4}$ $t^2 - 5t + \frac{23}{4} = 0$ The discriminant is $\Delta = (-5)^2 - 4(1)\left(\frac{23}{4}\right) = 25 - 23 = 2$. The roots are $t = \frac{5 \pm \sqrt{2}}{2}$. So, $|x| = \frac{5 \pm \sqrt{2}}{2}$. This gives four values for $x$: $x_1 = \frac{5 + \sqrt{2}}{2}$ $x_2 = -\frac{5 + \sqrt{2}}{2}$ $x_3 = \frac{5 - \sqrt{2}}{2}$ $x_4 = -\frac{5 - \sqrt{2}}{2}$

Let's check $m'(x)$ at these points. At these points, $f(|x|) = 1/4 > 0$. So, $|f(|x|)| = f(|x|)$. Thus, $m(x) = f(|x|) - 1/4$. If $x > 0$, $m(x) = f(x) - 1/4 = x^2 - 5x + 6 - 1/4$. $m'(x) = 2x - 5$. For $x_1 = \frac{5 + \sqrt{2}}{2}$: $m'(x_1) = 2\left(\frac{5 + \sqrt{2}}{2}\right) - 5 = 5 + \sqrt{2} - 5 = \sqrt{2} \ne 0$. So $x_1$ is a non-derivability point. For $x_3 = \frac{5 - \sqrt{2}}{2}$: $m'(x_3) = 2\left(\frac{5 - \sqrt{2}}{2}\right) - 5 = 5 - \sqrt{2} - 5 = -\sqrt{2} \ne 0$. So $x_3$ is a non-derivability point.

If $x < 0$, $m(x) = f(-x) - 1/4 = x^2 + 5x + 6 - 1/4$. $m'(x) = 2x + 5$. For $x_2 = -\frac{5 + \sqrt{2}}{2}$: $m'(x_2) = 2\left(-\frac{5 + \sqrt{2}}{2}\right) + 5 = -5 - \sqrt{2} + 5 = -\sqrt{2} \ne 0$. So $x_2$ is a non-derivability point. For $x_4 = -\frac{5 - \sqrt{2}}{2}$: $m'(x_4) = 2\left(-\frac{5 - \sqrt{2}}{2}\right) + 5 = -5 + \sqrt{2} + 5 = \sqrt{2} \ne 0$. So $x_4$ is a non-derivability point.

These 4 points ($x = \pm \frac{5 \pm \sqrt{2}}{2}$) are additional non-derivability points.

Total number of non-derivability points for $g(x)$:

1. From $m(x)$ being non-differentiable: $x \in \{-3, -2, 0, 2, 3\}$ (5 points)
2. From $m(x)=0$ and $m'(x) \ne 0$: $x \in \left\{\pm \frac{5 + \sqrt{2}}{2}, \pm \frac{5 - \sqrt{2}}{2}\right\}$ (4 points)

All these points are distinct. $\sqrt{2} \approx 1.414$. $\frac{5+\sqrt{2}}{2} \approx \frac{6.414}{2} = 3.207$. $\frac{5-\sqrt{2}}{2} \approx \frac{3.586}{2} = 1.793$. These values are not $0, \pm 2, \pm 3$. So, the total number of points of non-derivability is $5 + 4 = 9$.