In the following [x] denotes the...

MathematicsRelationship Between Continuity and DifferentiabilityJEE Advanced 2007Moderate

In the following $[x]$ denotes the greatest integer less than or equal to $x$ . Match the functions in Column I with the properties in Column II and indicate your answer by darkening the appropriate bubbles in the $4 \times 4$ matrix given in the ORS.

List-I

(P)

x ∣ x ∣

(Q)

∣ x ∣

(R)

x + [x]

(S)

∣ x - 1∣ + ∣ x + 1∣

List-II

(1)

continuous in

(- 1, 1)

(2)

differentiable in

(- 1, 1)

(3)

strictly increasing in

(- 1, 1)

(4)

not differentiable at least at one point in

(- 1, 1)

Answer:P → NaN, P → 4, R → 4, P → 2

Visualized Solution (Hindi)

Introduction to the Problem

Objective: Match functions in Column I with properties in Column II for $x \in (- 1, 1)$ .
Properties to check:
(p) Continuous in $(- 1, 1)$
(q) Differentiable in $(- 1, 1)$
(r) Strictly increasing in $(- 1, 1)$
(s) Not differentiable at least at one point in $(- 1, 1)$

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x) = x ∣ x ∣

$f (x) = x ∣ x ∣ = {- x^{2} x^{2} x < 0 x \geq 0$
Continuity: $lim_{x \to 0^{-}} f (x) = 0 = lim_{x \to 0^{+}} f (x) = f (0)$ . Continuous.
Differentiability: $f^{'} (x) = {- 2 x 2 x x < 0 x \geq 0$ . At $x = 0$ , $L H D = 0 = R H D$ . Differentiable.
Monotonicity: $f^{'} (x) = 2∣ x ∣ \geq 0$ , and $f^{'} (x) = 0$ only at $x = 0$ . Strictly increasing.

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x) = ∣ x ∣

$f (x) = ∣ x ∣ = {- x x x < 0 x \geq 0$
Continuity: Continuous at $x = 0$ as $lim_{x \to 0} ∣ x ∣ = 0$ .
Differentiability: $f^{'} (x) = \frac{sgn ( x )}{2 ∣ x ∣}$ for $x \neq = 0$ . As $x \to 0$ , $∣ f^{'} (x) ∣ \to \infty$ . Not differentiable at $x = 0$ .
Monotonicity: Decreasing for $x < 0$ and increasing for $x > 0$ . Not strictly increasing in $(- 1, 1)$ .

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x) = x + [x]

$f (x) = x + [x] = {x - 1 x - 1 < x < 0 0 \leq x < 1$
Continuity: $lim_{x \to 0^{-}} f (x) = - 1$ , but $f (0) = 0$ . Discontinuous at $x = 0$ .
Differentiability: Not differentiable at $x = 0$ due to discontinuity.
Monotonicity: Both $x - 1$ and $x$ are strictly increasing, and $f (0) > lim_{x \to 0^{-}} f (x)$ . Strictly increasing in $(- 1, 1)$ .

Analyzing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f (x) = ∣ x - 1∣ + ∣ x + 1∣

For $x \in (- 1, 1)$ :
$(x - 1) < 0 ⟹ ∣ x - 1∣ = - (x - 1) = 1 - x$
$(x + 1) > 0 ⟹ ∣ x + 1∣ = x + 1$
$f (x) = (1 - x) + (x + 1) = 2$
Properties: Constant function is continuous (p) and differentiable (q). Not strictly increasing (r).

Final Matching Matrix

Final Match:
(A) $x ∣ x ∣ \to$ (p, q, r)
(B) $∣ x ∣ \to$ (p, s)
(C) $x + [x] \to$ (r, s)
(D) $∣ x - 1∣ + ∣ x + 1∣ \to$ (p, q)
Key Takeaway: Continuity is necessary for differentiability, but not sufficient (as seen in $∣ x ∣$ ).

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