Question

Number of points of non-differentiability of the function

$g(x) = [x^2]\{cos^2(4x)\} + \{x^2\}[cos^2 (4x)] + x^2 sin^2 (4x) + [x^2][cos^2 (4x)] + \{x^2\}\{cos^2 (4x)\}$ in (-50, 50)

where [x] and {x} denotes the greatest integer function and fractional part function of x respectively, is equal to:

• A. 98
• B. 99
• C. 100
• D. 0

Answer 1

Answer: (D)

0

Answer 2

The given function is: $g(x) = [x^2]\{\cos^2(4x)\} + \{x^2\}[\cos^2 (4x)] + x^2 \sin^2 (4x) + [x^2][\cos^2 (4x)] + \{x^2\}\{\cos^2 (4x)\}$

We know that for any real number $y$, $y = [y] + \{y\}$, where $[y]$ is the greatest integer function and $\{y\}$ is the fractional part function.

Let's group the terms in the expression for $g(x)$: The first, second, fourth, and fifth terms form a specific pattern. Let $A = x^2$ and $B = \cos^2(4x)$. The terms are: $[A]\{B\} + \{A\}[B] + [A][B] + \{A\}\{B\}$.

This sum can be factored: $([A] + \{A\})([B] + \{B\})$

Using the identity $y = [y] + \{y\}$, we can substitute back $A$ and $B$: $([x^2] + \{x^2\})([\cos^2(4x)] + \{\cos^2(4x)\})$ This simplifies to: $(x^2)(\cos^2(4x))$ So, the sum of the first, second, fourth, and fifth terms is $x^2 \cos^2(4x)$.

Now, substitute this back into the original expression for $g(x)$: $g(x) = x^2 \cos^2(4x) + x^2 \sin^2(4x)$

Factor out $x^2$ from the expression: $g(x) = x^2 (\cos^2(4x) + \sin^2(4x))$

Using the fundamental trigonometric identity $\sin^2\theta + \cos^2\theta = 1$: $g(x) = x^2 (1)$ $g(x) = x^2$

The function $g(x)$ simplifies to $g(x) = x^2$. This is a polynomial function. Polynomial functions are continuous and differentiable for all real numbers. Therefore, $g(x) = x^2$ is differentiable for all $x \in \mathbb{R}$.

The question asks for the number of points of non-differentiability of $g(x)$ in the interval $(-50, 50)$. Since $g(x)$ is differentiable everywhere, it is differentiable throughout the interval $(-50, 50)$. Thus, the number of points of non-differentiability of $g(x)$ in $(-50, 50)$ is 0.