Question:medium

Let $[y]$ represent the greatest integer less than or equal to $y$. Then the set of all $x$ at which $f(x)=\cos^{-1}[4x+3]$ is differentiable is

Show Hint

Functions inside greatest integer brackets are constant over step-intervals; differentiability fails at the transition boundary points where values jump.
Updated On: Jun 3, 2026
  • $\mathbb{R}$
  • $[-1,1]$
  • $[-1,-\frac{1}{4})-\{-\frac{3}{4},-\frac{1}{2}\}$
  • $(-\frac{3}{4},\infty)$
Show Solution

The Correct Option is C

Solution and Explanation

Step 1: See what $[4x+3]$ can be.
The greatest integer $[4x+3]$ is always a whole number. Since $\cos^{-1}(u)$ only allows $-1\le u\le 1$, the only possible values are $-1,0,1$.
Step 2: Find the matching $x$ ranges.
$[4x+3]=-1$ gives $-1\le x<-\tfrac34$; $[4x+3]=0$ gives $-\tfrac34\le x<-\tfrac12$; $[4x+3]=1$ gives $-\tfrac12\le x<-\tfrac14$.
Step 3: Find $f$ on each piece.
$\cos^{-1}(-1)=\pi$, $\cos^{-1}(0)=\tfrac{\pi}{2}$, $\cos^{-1}(1)=0$. So $f$ is a flat (constant) value on each strip.
Step 4: Differentiate the flat pieces.
A constant has slope $0$, so $f$ is differentiable inside each open strip.
Step 5: Look at the jumps.
At $x=-\tfrac34$ and $x=-\tfrac12$ the value of $f$ jumps from one constant to another. A jump is not even continuous, so it cannot be differentiable there.
Step 6: Write the differentiable set.
So $f$ is differentiable on $[-1,-\tfrac14)$ except the two jump points: \[ \left[-1,-\tfrac14\right)-\left\{-\tfrac34,-\tfrac12\right\}. \] \[ \boxed{\left[-1,-\tfrac14\right)-\left\{-\tfrac34,-\tfrac12\right\}} \]
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