Q.Examine the differentiability of $f$, where $f$ is defined by $f(x) = \begin{cases} 1 + x, & x \le 2 \\ 5 - x, & x > 2 \end{cases}$ at $x = 2$.

We need to analyze the behavior of the function $h(x) = |f(x)|\, g(x)$ at the zeros $a_1, a_2, \dots, a_n$ of $f$. The question tests two things: continuity and differentiability of a product involving an absolute value, and the logical connection between the assertion and the reason.

Concept and intuition:

The absolute value function $|x-a|$ is continuous everywhere but not differentiable at $x=a$ (its graph has a sharp corner). Multiplying by another function can sometimes "fix" differentiability if that function also vanishes at $a$ — because then the corner gets flattened. Here $f(x)$ vanishes at all $a_i$, so $|f(x)|$ has corners at each $a_i$. The factor $g(x)$ may or may not vanish at those points. If $g(a_i) \neq 0$, the corner remains; if $g(a_i)=0$, the product might become differentiable. Since $g$ shares only three zeros with $f$, at the other $n-3$ zeros $g$ is nonzero, so $h$ is not differentiable there. Hence Assertion (A) is false.

Now let’s go step by step.

1. Continuity of $h(x) = |f(x)|\,g(x)$ at any $a_i$

   Since $f(a_i)=0$, we have $|f(a_i)|=0$, so $h(a_i)=0$.

   As $x \to a_i$, $|f(x)| \to 0$ because $f$ is continuous. Multiplying by $g(x)$ (which is finite) gives $h(x) \to 0$.

   Thus $\lim_{x\to a_i} h(x) = h(a_i)$, so $h$ is continuous at every $a_i$.

   So far, the continuity part of Assertion (A) is true.

2. Differentiability of $h$ at a zero $a_i$

   Differentiability requires the limit of the difference quotient to exist:

Near $a_i$, $f(x) \approx f'(a_i)(x-a_i)$ (since $f(a_i)=0$ and $f$ is a polynomial). So $|f(x)| \approx |f'(a_i)|\,|x-a_i|$.

Hence the difference quotient behaves like

The factor $\frac{|x-a_i|}{x-a_i}$ equals $1$ for $x>a_i$ and $-1$ for $x<a_i$.

Therefore, as $x \to a_i$, the limit exists if and only if $g(a_i)=0$. Why? Because if $g(a_i) \neq 0$, the left-hand and right-hand limits are $|f'(a_i)|\,g(a_i)$ and $-|f'(a_i)|\,g(a_i)$, which are different (nonzero). If $g(a_i)=0$, then the product $g(x)\frac{|x-a_i|}{x-a_i}$ might still have a limit if $g$ vanishes to at least first order — but here $g$ is a polynomial, so if $g(a_i)=0$, then $g(x) \approx g'(a_i)(x-a_i)$, and the difference quotient becomes $|f'(a_i)|\,g'(a_i)\,|x-a_i|$, which tends to $0$ from both sides. So differentiability holds exactly at those $a_i$ where $g$ also vanishes.

3. Apply to the given situation …

The key idea is that differentiability on the whole real line forces two conditions at each point where the definition changes: the function must be continuous there (no jumps), and the left-hand and right-hand derivatives must be equal. Here the boundaries are $x = -1$ and $x = 1$. The function is already smooth on each separate piece — the only possible trouble is at these two join points.

Let’s work through it systematically.

1. Write the pieces clearly.

   For $x \leq -1$ or $x \geq 1$, $f(x) = \dfrac{a}{|x|}$.

   Since $|x| = -x$ when $x \leq -1$ (negative), and $|x| = x$ when $x \geq 1$ (positive), we can rewrite:

   • For $x \leq -1$: $f(x) = \dfrac{a}{-x} = -\dfrac{a}{x}$.
   • For $x \geq 1$: $f(x) = \dfrac{a}{x}$. For $-1 < x < 1$, $f(x) = x^2 + b$.

2. Continuity at $x = 1$.

   The left-hand limit (from inside $(-1,1)$) is $\lim_{x \to 1^-} (x^2 + b) = 1 + b$.

   The right-hand limit (from $x \geq 1$) is $\lim_{x \to 1^+} \dfrac{a}{x} = a$.

   For continuity, these must be equal:

3. Differentiability at $x = 1$. The left-hand derivative: for $x < 1$, $f'(x) = 2x$, so at $x = 1^-$ it is $2$. The right-hand derivative: for $x > 1$, $f'(x) = -\dfrac{a}{x^2}$, so at $x = 1^+$ it is $-a$. For differentiability, these must match:

4. Continuity at $x = -1$. The left-hand limit (from $x \leq -1$) is $\lim_{x \to -1^-} \left(-\dfrac{a}{x}\right) = -\dfrac{a}{-1} = a$. The right-hand limit (from inside $(-1,1)$) is $\lim_{x \to -1^+} (x^2 + b) = 1 + b$. Continuity gives: $a = 1 + b. \qquad(2)$ …

We have a function defined by two different rules on either side of $x = 3$. For it to be differentiable at that point, it must first be continuous there (no jump), and the slopes from left and right must agree. That gives us two equations in the unknowns $K$ and $m$.

1. Continuity at $x = 3$ The left-hand limit as $x \to 3^-$ uses $K\sqrt{x+1}$:

The right-hand limit as $x \to 3^+$ uses $mx+2$:

For continuity, these must be equal:

2. Differentiability at $x = 3$ The derivative from the left: for $0 \le x \le 3$, $g(x) = K\sqrt{x+1}$.

The derivative from the right: for $3 < x \le 5$, $g(x) = mx + 2$.

For differentiability, these must match:

3. Solve the system Substitute $m = \frac{K}{4}$ from (2) into (1):

Multiply through by 4:

The key idea is that a function is continuous at a point if the value of the function at that point equals the limit from both sides. Here, the function is defined piecewise, so we must compute the left-hand limit (as $x \to 0^-$), the right-hand limit (as $x \to 0^+$), and set them equal to $p = f(0)$. This yields equations for $p$ and $q$, and then we sum them.

1. Left-hand limit ($x \to 0^-$) For $x < 0$, $f(x) = \frac{\cos 3x - \cos x}{x \sin x}$. Use the identity $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)$:

So the expression becomes:

Now use $\sin(2x) \sim 2x$ as $x \to 0$:

Thus the left-hand limit is $-4$.

2. Right-hand limit ($x \to 0^+$) For $x > 0$, $f(x) = \frac{\log(1 + q \sin x)}{x}$. As $x \to 0$, $\sin x \sim x$, so $q \sin x \to 0$. Use the standard limit $\log(1+u) \sim u$ for $u \to 0$:

Hence:

So the right-hand limit is $q$.

3. Continuity condition …

Concept and Intuition

The function involves two nested absolute values. Each absolute value creates a potential "corner" (a cusp or a sharp turn) where the derivative does not exist. The outer absolute value acts on $|x|-1$, so we need to find where the inner expression changes sign and where the outer expression changes sign. The key is to break the domain into intervals where the expressions inside the absolute values have constant sign, then check differentiability at the boundaries.

Step-by-step reasoning

1. Rewrite the function piecewise Start with the inner absolute value:

Then $f(x) = \big||x|-1\big|$ becomes:

• For $x \ge 0$: $f(x) = |x-1|$.
• For $x < 0$: $f(x) = |-x-1| = |-(x+1)| = |x+1|$.

2. Further break each piece
   • On $x \ge 0$, $|x-1|$ has a corner at $x=1$:

• On $x < 0$, $|x+1|$ has a corner at $x=-1$:

3. Combine into a full piecewise definition

4. Check differentiability at the transition points
   • At $x = -1$: Left derivative: for $x < -1$, $f'(x) = -1$. Right derivative: for $-1 < x < 0$, $f'(x) = 1$. Since $-1 \neq 1$, the derivative does not exist at $x = -1$.
   • At $x = 0$: Left derivative: for $-1 < x < 0$, $f'(x) = 1$. Right derivative: for $0 < x < 1$, $f'(x) = -1$. Since $1 \neq -1$, the derivative does not exist at $x = 0$. …

The key idea: absolute value functions create sharp corners (cusps) where their argument is zero. For $f(x) = ||x| - 1|$, there are two layers of absolute values. The inner layer $|x|$ has a corner at $x = 0$. The outer layer $|\, \cdot \, - 1|$ has a corner wherever its argument $|x| - 1$ equals zero — that is, when $|x| = 1$, so at $x = 1$ and $x = -1$. At each of these three points, the function changes slope abruptly, so it is not differentiable there. Everywhere else, the function is a straight line (or a combination of straight-line pieces), hence differentiable.

Let’s walk through it step by step.

1. Break the function into pieces.

   Start from the inside: $|x|$ is $x$ for $x \ge 0$ and $-x$ for $x < 0$.

   So $f(x) = ||x| - 1|$ becomes:

   • For $x \ge 0$: $f(x) = |x - 1|$.
   • For $x < 0$: $f(x) = |-x - 1| = |-(x+1)| = |x+1|$ (since absolute value removes the minus sign).

2. Now break each piece further at the points where the inner expression changes sign.

   • For $x \ge 0$: $|x-1|$ is $1-x$ when $0 \le x \le 1$, and $x-1$ when $x \ge 1$.
   • For $x < 0$: $|x+1|$ is $-(x+1)$ when $x \le -1$ (since $x+1 \le 0$ there), and $x+1$ when $-1 \le x < 0$.

   So the full piecewise definition is:

3. Check differentiability at the boundary points. A function is differentiable at a point if the left-hand derivative equals the right-hand derivative there. Let’s test each candidate:
   • At $x = -1$: Left side ($x \to -1^-$): derivative of $-(x+1)$ is $-1$. Right side ($x \to -1^+$): derivative of $x+1$ is $+1$. They differ, so not differentiable at $x = -1$.
   • At $x = 0$: Left side ($x \to 0^-$): derivative of $x+1$ is $+1$. Right side ($x \to 0^+$): derivative of $1-x$ is $-1$. They differ, so not differentiable at $x = 0$.
   • At $x = 1$: Left side ($x \to 1^-$): derivative of $1-x$ is $-1$. …

Concept. At a point where a piecewise function switches branches, check continuity first: $f$ is continuous at $x = 1$ iff $\lim_{x\to 1^-} f(x) = f(1) = \lim_{x\to 1^+} f(x)$. Differentiability at a point requires continuity there, so a discontinuous function is automatically non-differentiable.

Step 1 — value and left limit. For $0 < x \le 1$, $f(x) = \dfrac{8}{x^3} - 6x$. So

Step 2 — right limit. For $x > 1$,

$f(x) = \frac{x-1}{\sqrt{x-1}} = \sqrt{x-1} \;\xrightarrow[x\to 1^+]{}\; 0.$ …

The core idea here is the relationship between continuity and differentiability. If a function is differentiable at a point, it must be continuous there — but a continuous function need not be differentiable. This asymmetry is the key to checking each statement.

Let’s examine them one by one.

1. Statement (a): “If a function is differentiable at a point ‘p’ then it is not continuous at ‘p’.”

   This is the exact opposite of the truth. Differentiability at a point implies continuity at that point. So (a) is false.

2. Statement (b): “If a function is not continuous at $x = a$, then it is not differentiable at $x = a$.”

   This is the contrapositive of “differentiable ⇒ continuous”. Since the original implication is true, its contrapositive is also true. A break in the graph means no tangent line can exist. So (b) is correct.

3. Statement (c): “If $f(x) = |x|$ then $f(x)$ is not differentiable but continuous on $\mathbb{R}$.”

   $|x|$ is continuous everywhere — no jumps, no holes. At $x=0$, the left-hand derivative is $-1$ and the right-hand derivative is $+1$, so the derivative does not exist. Everywhere else it is differentiable. So the statement is correct.

4. Statement (d): “If $f(x) = x - \lfloor x \rfloor$, then $f'(1) = 1$.” …

The core idea here is that a rational function can be discontinuous at a point where the denominator vanishes, but if the numerator also vanishes at that same point, the discontinuity might be removable — meaning we can redefine the function at that point to make it continuous. The value we assign is simply the limit of the function as $x$ approaches that point.

For $f$ to be continuous at $x = -2$, we need $f(-2) = \lim_{x \to -2} f(x)$. So the problem reduces to finding that limit.

1. Check what happens at $x = -2$.

   The denominator is $x^2 + x - 2$. At $x = -2$:

   $(-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0$.

   The numerator is $x^3 + 2x^2 + x + 2$. At $x = -2$:

   $(-2)^3 + 2(4) + (-2) + 2 = -8 + 8 - 2 + 2 = 0$.

   Both numerator and denominator are zero — this is a $\frac{0}{0}$ indeterminate form, so the discontinuity is removable if the factors cancel.

2. Factor both polynomials.

   Denominator: $x^2 + x - 2 = (x + 2)(x - 1)$.

   Numerator: try synthetic division by $(x + 2)$.

   Coefficients: $1, 2, 1, 2$.

   Bring down $1$, multiply by $-2$ → $-2$, add to $2$ → $0$, multiply by $-2$ → $0$, add to $1$ → $1$, multiply by $-2$ → $-2$, add to $2$ → $0$.

   The quotient is $x^2 + 0x + 1 = x^2 + 1$, with remainder $0$. So

   $x^3 + 2x^2 + x + 2 = (x + 2)(x^2 + 1)$.

3. Cancel the common factor.

   For $x \neq -2$ and $x \neq 1$ (the other point where denominator vanishes),

4. Take the limit as $x \to -2$. …