Q.Prove that the greatest integer function defined by $f(x) = [x]$, $0 < x < 3$, is not differentiable at $x = 1$ and $x = 2$.

Step 1 — Solve the quadratic.

$[x]^2 - 5[x] + 6 = 0 \;\Rightarrow\; ([x]-2)([x]-3) = 0 \;\Rightarrow\; [x] = 2 \text{ or } [x] = 3$

Step 2 — Convert each to an $x$-interval. …

Step 1 — Continuity.

$f(x)=|\cos x|=g(h(x))$ where $h(x)=\cos x$ (continuous on $\mathbb{R}$) and $g(t)=|t|$ (continuous on $\mathbb{R}$). The composition of continuous functions is continuous, so

This alone kills option (C), which claims discontinuity.

Step 2 — Where can differentiability fail?

Where $\cos x\neq0$, $\cos x$ keeps a constant sign in a neighbourhood, so locally $f(x)=\pm\cos x$ — a smooth function, and $f'(x)=\mp\sin x$ exists. So the only suspect points are the zeros of $\cos x$:

This already rules out option (D) ("not differentiable everywhere") — $f$ is differentiable at, say, $x=0$, where $f'(0)=0$.

Step 3 — Test $x=\dfrac{\pi}{2}$ explicitly.

Just left of $\pi/2$, $\cos x>0$ so $f(x)=\cos x$ and $f'(x)=-\sin x$:

Just right of $\pi/2$, $\cos x<0$ so $f(x)=-\cos x$ and $f'(x)=+\sin x$:

$Rf'\!\left(\frac{\pi}{2}\right)=\sin\frac{\pi}{2}=+1$ …

Step 1 — Understand the integrand

$[x]$, the greatest integer function (floor), returns the largest integer $\le x$. It is a step function: it holds a constant value on each interval $[k,\,k+1)$, namely $k$.

We cannot integrate it by an antiderivative formula; instead we exploit its piecewise-constant nature.

Step 2 — Split the interval $[0,8]$ into unit pieces

Using the additivity of the definite integral:

Step 3 — Evaluate each piece

On $[k,\,k+1)$ the integrand equals the constant $k$, and the interval has length $1$:

(The single point $x=k+1$, where the value jumps, has zero measure and does not affect the integral.) …

The key here is that the greatest integer function $[x]$ only takes integer values. So the condition for the denominator to be defined and non-zero is not about $x$ directly, but about which integer $[x]$ equals for a given $x$.

We need $[x]^2 - [x] - 6 > 0$ because:

• The square root requires the inside to be positive (not zero, since it's in the denominator).
• The denominator itself cannot be zero.

So treat $[x]$ as an integer variable, say $n = [x]$, and solve $n^2 - n - 6 > 0$.

1. Factor the quadratic in $n$:

   $n^2 - n - 6 = (n - 3)(n + 2)$

   So the inequality is $(n - 3)(n + 2) > 0$.

2. Solve the integer inequality:

   For a quadratic with roots at $n = -2$ and $n = 3$, the product is positive when $n < -2$ or $n > 3$.

   Since $n$ is an integer, this means:

   $n \leq -3$ or $n \geq 4$.

3. Translate back to $x$:

   • If $[x] \leq -3$, then $x < -2$ (because the greatest integer less than or equal to $x$ is at most $-3$ only when $x$ is strictly less than $-2$).

     For example, $x = -2.1$ gives $[x] = -3$, which works. But $x = -2$ gives $[x] = -2$, which does NOT work.

     So this gives $x \in (-\infty, -2)$.

   • If $[x] \geq 4$, then $x \geq 4$ (because the greatest integer is at least $4$ only when $x$ is at least $4$).

     For example, $x = 4$ gives $[x] = 4$, which works.

     So this gives $x \in [4, \infty)$.

4. Combine the intervals: …

1. The condition for continuity. $f$ is continuous at $x=0$ precisely when

So everything reduces to evaluating the limit of the $x \neq 0$ branch.

2. Evaluate the limit. Both numerator and denominator $\to 0$ as $x \to 0$, so this is a $\tfrac00$ indeterminate form. Force the two standard limits to appear by inserting the matching $x$ factors:

Now use the two limits every student should have memorised:

As $x\to0$ we have $3x\to0$ and $2x\to0$, so both bracketed ratios tend to 1, leaving

3. Cross-check with L'Hôpital's rule. Differentiating numerator and denominator separately:

The two methods agree, so the limit is definitely $\tfrac32$.

4. Impose continuity. The value assigned at $x=0$ must equal this limit:

$f(0) = \frac{3}{2} \quad\Longrightarrow\quad k = \frac{3}{2}$ …