Consider the piecewise defined function $f(x) = \begin{cases} \{x\}\sqrt{4x^2 - 12x + 9}, & 1 \leq x \leq 2 \\ \cos\left(\dfrac{\pi}{2}(|x| - \{x\})\right), & -1 \leq x < 1 \end{cases}$ with $\{x\}$ denoting the fractional part of $x$. Which of the following is/are true?
Range of $f(x)$ is equal to $[0, 1]$.
The number of values of $x$ for which function is continuous but not differentiable is 1.
$f(x) = 1$ has two solutions.
Number of values of $x$ for which $f(x)$ is discontinuous is 2.
Step-by-Step Solution
Key Concept: The key idea here is to correctly simplify the piecewise function by meticulously applying the definitions of fractional part $\{x\}$ and absolute value $|y|$ across all sub-intervals, especially at integer boundaries and points where the expression inside the absolute value changes sign. This careful simplification is fundamental for accurately analyzing the function's range, continuity, and differentiability.
Step 1: Simplify $f(x)$ for $1 \leq x \leq 2$. Note that $4x^2 - 12x + 9 = (2x-3)^2$, so $\sqrt{4x^2-12x+9} = |2x-3|$. For $1 \leq x \leq 3/2$, $|2x-3| = 3-2x$, and for $3/2 \leq x \leq 2$, $|2x-3| = 2x-3$. Also $\{x\} = x - 1$ for $1 \leq x < 2$ and $\{2\} = 0$. So $f(x) = (x-1)|2x-3|$ for $1 \leq x \leq 2$.
Step 2: Simplify $f(x)$ for $-1 \leq x < 1$. Note that $|x| - \{x\} = |x| - (x - \lfloor x \rfloor)$. For $0 \leq x < 1$: $|x| - \{x\} = x - x = 0$, so $f(x) = \cos(0) = 1$. For $-1 \leq x < 0$: $|x| = -x$, $\{x\} = x+1$, so $|x| - \{x\} = -x - (x+1) = -2x-1$, giving $f(x) = \cos\left(\dfrac{\pi}{2}(-2x-1)\right)$.
Step 3: Determine the range of $f(x)$. For $0 \leq x < 1$: $f(x) = 1$. For $-1 \leq x < 0$: $f(x) = \cos\left(\dfrac{\pi}{2}(-2x-1)\right)$; as $x$ goes from $-1$ to $0$, $-2x-1$ goes from $1$ to $-1$, so $\dfrac{\pi}{2}(-2x-1)$ goes from $\dfrac{\pi}{2}$ to $-\dfrac{\pi}{2}$, giving $f(x) \in [0, 1]$. For $1 \leq x \leq 2$: $f(x) = (x-1)|2x-3|$ ranges from $0$ (at $x=1$ and $x=3/2$) to maximum. At $x=2$: $f(2) = 0$. At $x=1$: $f(1)=0$. Maximum occurs; checking $x=3/2$: $f(3/2)=0$. Checking endpoints and critical points: for $1 \leq x \leq 3/2$, $f(x)=(x-1)(3-2x)$, max at $x=5/4$: $f(5/4)=(1/4)(1/2)=1/8$. For $3/2 \leq x \leq 2$, $f(x)=(x-1)(2x-3)$, max at $x=2$: $f(2)=0$, or at $x=7/4$: $f(7/4)=(3/4)(1/2)=3/8$. So range for this piece is $[0, 3/8]$. Overall range is $[0,1]$. Option (a) is TRUE.
Step 4: Check continuity and differentiability. At $x=0$: $f(0^-) = \cos(-\pi/2) = 0$, $f(0^+) = 1$. So $f$ is discontinuous at $x=0$. At $x=1$: $f(1^-) = 1$ (from $0 \leq x < 1$ piece), $f(1) = (1-1)|2-3| = 0$. So $f$ is discontinuous at $x=1$. Thus number of discontinuities is 2. Option (d) is TRUE.
Step 5: Check option (b). The function is continuous but not differentiable at $x = -1$ (boundary) and $x = 3/2$ (where $|2x-3|$ has a corner). Checking carefully: at $x=3/2$, $f$ is continuous (both sides give 0) but not differentiable. At $x=-1$: it's an endpoint. So the number of points where $f$ is continuous but not differentiable is 1 (at $x = 3/2$). Option (b) is TRUE.
Step 6: Check option (c). $f(x) = 1$ requires: for $0 \leq x < 1$, $f(x) = 1$ (entire interval, infinitely many solutions, not just two). So option (c) is FALSE.
Correct Answer: 1, 2, 4