Q.Prove that the function $f$ given by $f(x) = |x - 1|, x \in \mathbf{R}$ is not differentiable at $x = 1$.

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-AN1 markMCQ
   Q.A function $f: \mathbb{R} \to \mathbb{R}$ defined as $f(x) = \begin{cases} \dfrac{x}{|x|}, & x < -2 \\ \dfrac{x|x|}{4}, & -2 \le x \le 2 \\ \dfrac{|x|}{x}, & x > 2 \end{cases}$ is (A) Differentiable for all real $x$ (B) Differentiable for all real $x$ except for $x = -2, 0, 2$ (C) Continuous for all real $x$ and differentiable for all real $x$ except for $x = -2, 2$ (D) Continuous for all real $x$ except for $x = 0, -2, 2$ and differentiable at $x = -2, 0, 2$

   ›Reveal solutionSolution

   The function is continuous everywhere (the pieces meet exactly at $x=\pm2$), but the slope jumps at $x=\pm2$ (constant slope $0$ outside vs. slope $\pm1$ from the middle piece) while it is smooth through $x=0$.

   Concept and Intuition

   Outside $[-2,2]$ the function reduces to constants ($x/|x|=\mp1$), while inside it is the smooth-looking $x|x|/4$, which is actually $x^2/4$ for $x\ge0$ and $-x^2/4$ for $x<0$ — a function that is itself differentiable everywhere including at $0$ (both one-sided derivatives are $0$ there). The only risk of a kink is at the junctions $x=\pm2$ where the constant pieces meet the quadratic piece.

   Step-by-Step Solution

   1. Continuity at $x=-2$: left piece value $=-1$ (constant); middle piece at $x=-2$: $(-2)|{-2}|/4=(-2)(2)/4=-1$. Equal — continuous.
   2. Continuity at $x=2$: middle piece at $x=2$: $(2)(2)/4=1$; right piece value $=1$. Equal — continuous. So $f$ is continuous for all real $x$.
   3. Differentiability at $x=0$: for $0\le x\le2$, $f=x^2/4$, $f'=x/2\to0$ as $x\to0^+$; for $-2\le x\le0$, $f=-x^2/4$, $f'=-x/2\to0$ as $x\to0^-$. Both one-sided derivatives are $0$ — differentiable at $x=0$.
   4. Differentiability at $x=-2$: left piece (constant $-1$) has derivative $0$; middle piece derivative at $x=-2^+$ is $-x/2=1$. $0\ne1$ — NOT differentiable. …

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2. AP EAPCET 2026Set eng-2026-05-14-FN1 markMCQ
   Q.The set of all the points at which $f(x) = |2 - |x||$ is continuous but not differentiable is (A) $\{0, 1, 2\}$ (B) $\{-1, 0, 2\}$ (C) $\{-2, 0, 2\}$ (D) $\{-2, 1, 2\}$

   ›Reveal solutionSolution

   $|2-|x||$ is continuous everywhere but fails to be differentiable exactly where an absolute value "folds" the graph — at $x=0$ and at $x=\pm2$ where $2-|x|=0$.

   Concept and Intuition

   An absolute value $|g(x)|$ is always continuous if $g$ is continuous (composition with the continuous function $|\cdot|$). But it fails to be differentiable at any point where $g$ itself is not differentiable, and additionally at any point where $g(x)=0$ with $g'\neq 0$ there (because $|g(x)|$ has a sharp corner/fold exactly where $g$ crosses zero). We must check both sources of non-differentiability for $f(x)=|2-|x||$.

   Step-by-Step Solution

   1. Let $g(x) = 2 - |x|$. Then $f(x) = |g(x)|$.
   2. $g(x)$ itself is not differentiable at $x=0$ because of the inner $|x|$ (corner there), and $g(0)=2\neq 0$, so near $x=0$, $g(x)>0$ and $f(x)=g(x)$ — $f$ inherits $g$'s corner at $x=0$.
   3. $g(x)=0$ when $|x|=2$, i.e. at $x=2$ and $x=-2$. Near these points $g$ is differentiable (it's just $2-x$ or $2+x$, linear pieces with nonzero slope $\mp1$), but since $g$ changes sign there, $f=|g|$ has a "V"-shaped corner at each of $x=2$ and $x=-2$. …

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3. AP EAPCET 2026Set eng-2026-05-18-AN1 markMCQ
   Q.If $f(x) = |x - 2|\left(3^{4|x|} - 1\right)$ is a real valued function, then the set of points at which $f$ is not differentiable, is (A) $\{0\}$ (B) $\{2\}$ (C) $\{0, 2\}$ (D) $\varnothing$

   ›Reveal solutionSolution

   Both $|x-2|$ and the $|x|$ hidden inside the exponent create corners; check whether the other factor vanishes at each corner to see if it's smoothed away — here neither is, so both survive.

   Concept and Intuition

   $|g(x)|$-type expressions are non-differentiable exactly where $g(x)=0$ (a corner), unless multiplied by a factor that is itself zero there with enough smoothness to cancel the kink. Here there are two potential kink locations — $x=2$ from $|x-2|$, and $x=0$ hidden inside $3^{4|x|}$ — so each must be checked independently against the other factor.

   Step-by-Step Solution

   1. At $x=2$: near $x=2$, $f(x)=\pm(x-2)(3^{4|x|}-1)$, a corner from $|x-2|$ times a smooth nonzero factor ($3^{4\cdot2}-1=3^8-1\ne0$ at $x=2$). Computing one-sided derivatives: for $x\to2^-$, $f'(2^-)=-(3^8-1)$; for $x\to2^+$, $f'(2^+)=+(3^8-1)$. These differ, so $f$ is not differentiable at $2$.
   2. At $x=0$: here $|x-2|=2-x$ is smooth (equals $2$ at $x=0$, nonzero). Expand $3^{4|x|}-1$ near $0$: for $x>0$, $3^{4x}-1\approx4x\ln3$; for $x<0$, $3^{-4x}-1\approx-4x\ln3=4|x|\ln3$. So $3^{4|x|}-1\approx(4\ln3)|x|$ — itself a corner (like $c|x|$), not smoothed to a higher power. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.Match the functions in Column I with their properties in Column II. In the following [x] denotes the greatest integer less than or equal to x. Column I: A) $x|x|$ B) $\sqrt{|x|}$ C) $x + [x]$ D) $|x-1|+|x+1|+|x|$ Column II: I. Strictly increasing and continuous in (-1, 1) II. Continuous but not differentiable in (-1,1) III. Differentiable in (-1,1) IV. Differentiable in $(-1, 0)\cup(0,1)$ V. Strictly increasing and not differentiable in (-1,1) The correct match is (A) A-III, B-V, C-II, D-I (B) A-II, B-III, C-I, D-V (C) A-I, B-II, C-V, D-IV (D) A-IV, B-I, C-V, D-III

   ›Reveal solutionSolution

   A continuity/differentiability matching problem — reduce each function to its explicit form on $(-1,1)$ and classify by monotonicity, continuity, and differentiability. Answer: A-I, B-II, C-V, D-IV.

   Concept and Intuition

   Differentiability $\Rightarrow$ continuity, but not conversely. Kinks (unequal finite one-sided derivatives) and cusps (infinite one-sided derivatives) both break differentiability but keep continuity; jump discontinuities break both. Monotonicity is checked directly from the explicit piecewise form.

   Step-by-Step Solution

   1. A) $f(x)=x|x|$. On $x\ge0$: $f=x^2$; on $x<0$: $f=-x^2$. Both pieces meet smoothly at $0$: $f'(x)=2|x|$, continuous, and $f'(0)=0$ (both one-sided derivatives are $0$). So $f$ is differentiable on all of $(-1,1)$, continuous, and strictly increasing (since $x_1<x_2\Rightarrow x_1|x_1|<x_2|x_2|$ always). The description that uniquely fits it among the five (once the others are assigned) is I: strictly increasing and continuous.
   2. B) $f(x)=\sqrt{|x|}$. Continuous at $0$ ($f(0)=0$). Differentiability at $0$: $\frac{f(x)-f(0)}{x}=\frac{\sqrt{|x|}}{x}$; as $x\to0^+$ this is $\frac{1}{\sqrt x}\to+\infty$ — no finite derivative, a cusp. Also not monotonic on $(-1,1)$ (falls on $(-1,0)$, rises on $(0,1)$). So it is continuous but not differentiable → II.
   3. C) $f(x)=x+[x]$. On $[-1,0)$: $[x]=-1\Rightarrow f=x-1$ (rising from near $-2$ to just below $-1$). At $x=0$: $[0]=0\Rightarrow f=0$ — a jump upward from the left-limit $-1$. On $[0,1)$: $[x]=0\Rightarrow f=x$. Every piece is increasing and the jump itself is upward, so $f$ is strictly increasing on the whole interval despite the discontinuity — and being discontinuous, it certainly is not differentiable. This matches V: strictly increasing and not differentiable. …

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5. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.Assertion (A): If $y = f(x) = (|x| - |x-1|)^2$, then $\left(\frac{dy}{dx}\right)_{x=1} = 1$. Reason (R): If $\displaystyle\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ exist, then it is called derivative of f(x) at x = a. Then (A) (A) is true, (R) is true, (R) is correct explanation to (A) (B) (A) is true, (R) is true, (R) is not the correct explanation to (A) (C) (A) is true, (R) is false (D) (A) is false, (R) is true

   ›Reveal solutionSolution

   Splitting the modulus function on both sides of $x=1$ shows the left- and right-hand derivatives are $4$ and $0$ respectively — they disagree, so the derivative doesn't exist at $x=1$ (certainly not $1$), making (A) false; (R) is the correct textbook definition, so it is true.

   Concept and Intuition

   A function built from absolute values changes its algebraic form at the points where the inner expressions vanish (here $x=0$ and $x=1$). To differentiate at a 'corner' point like $x=1$, you must examine the one-sided formulas separately — a derivative exists only if both one-sided derivatives agree.

   Step-by-Step Solution

   1. In a neighbourhood of $x=1$, $x>0$, so $|x|=x$ throughout.
   2. For $0<x<1$: $|x-1|=1-x$, so $y=(x-(1-x))^2=(2x-1)^2$.
   3. For $x>1$: $|x-1|=x-1$, so $y=(x-(x-1))^2=1^2=1$ (a constant).
   4. Left-hand derivative at $x=1$: $\frac{d}{dx}(2x-1)^2=4(2x-1)$, which at $x=1$ gives $4(1)=4$.
   5. Right-hand derivative at $x=1$: the derivative of the constant $1$ is $0$.
   6. Since $4\neq 0$, $y$ is not differentiable at $x=1$ — the assertion's claim that $(dy/dx)_{x=1}=1$ is false. …

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6. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
   Q.If $y = |\cos x - \sin x| + |\tan x - \cot x|$, then $\left(\frac{dy}{dx}\right)_{x=\frac{\pi}{3}} + \left(\frac{dy}{dx}\right)_{x=\frac{\pi}{6}} =$ (A) $1$ (B) $-1$ (C) $2$ (D) $0$

   ›Reveal solutionSolution

   Removing the moduli using the correct sign of each expression near $x=\pi/3$ and $x=\pi/6$ gives two derivative expressions that are exact negatives of one another, so their sum is $0$.

   Concept and Intuition

   With modulus terms, always check the sign of the inner expression at the point in question (and its neighbourhood) before differentiating — the derivative formula only applies to the branch that is actually active there.

   Step-by-Step Solution

   1. At $x=\pi/3$: $\cos x=\tfrac12,\ \sin x=\tfrac{\sqrt3}{2}$, so $\cos x-\sin x=\tfrac{1-\sqrt3}{2}<0 \Rightarrow |\cos x-\sin x|=\sin x-\cos x$. Also $\tan x=\sqrt3,\ \cot x=\tfrac1{\sqrt3}$, so $\tan x-\cot x=\tfrac{2}{\sqrt3}>0\Rightarrow|\tan x-\cot x|=\tan x-\cot x$.
   2. So near $x=\pi/3$: $y=(\sin x-\cos x)+(\tan x-\cot x)$, and $\dfrac{dy}{dx}=(\cos x+\sin x)+(\sec^2x+\csc^2x)$ (using $\frac{d}{dx}(-\cot x)=\csc^2x$).
   3. At $x=\pi/6$: $\cos x=\tfrac{\sqrt3}{2},\ \sin x=\tfrac12$, so $\cos x-\sin x>0\Rightarrow|\cos x-\sin x|=\cos x-\sin x$. Also $\tan x=\tfrac1{\sqrt3},\ \cot x=\sqrt3$, so $\tan x-\cot x<0\Rightarrow|\tan x-\cot x|=\cot x-\tan x$.
   4. So near $x=\pi/6$: $y=(\cos x-\sin x)+(\cot x-\tan x)$, and $\dfrac{dy}{dx}=-(\sin x+\cos x)-(\sec^2x+\csc^2x)$. …

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7. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.If $f(x) = 2 + |\sin^{-1} x|$ and $A = \{x \in \mathbb{R} \mid f'(x) \text{ exists}\}$, then $A =$ (A) $\{0\}$ (B) $[-1, 1]$ (C) $(-\infty, -1) \cup (1, \infty)$ (D) $(-1, 0) \cup (0, 1)$

   ›Reveal solutionSolution

   $f(x)=2+|\sin^{-1}x|$ fails to be differentiable at $x=0$ (a corner from the absolute value) and at $x=\pm1$ (vertical tangent of $\sin^{-1}x$); it's differentiable everywhere else in its domain.

   Concept and Intuition

   An absolute value $|g(x)|$ is non-differentiable wherever $g(x)=0$ and $g$ changes sign there (a corner point), unless $g'$ also vanishes there smoothly. Also, $\sin^{-1}x$ itself is only defined on $[-1,1]$ and has an infinite (vertical) derivative at the endpoints, so no extension of it can be differentiable there.

   Step-by-Step Solution

   1. The domain of $f$ is the domain of $\sin^{-1}x$, namely $[-1,1]$.
   2. On $(0,1]$, $\sin^{-1}x>0$, so $f(x)=2+\sin^{-1}x$, and $f'(x)=\dfrac{1}{\sqrt{1-x^2}}$ — defined for $x\in(0,1)$ but $\to\infty$ as $x\to1^-$.
   3. On $[-1,0)$, $\sin^{-1}x<0$, so $f(x)=2-\sin^{-1}x$, and $f'(x)=-\dfrac{1}{\sqrt{1-x^2}}$ — defined for $x\in(-1,0)$ but $\to-\infty$ as $x\to-1^+$.
   4. At $x=0$: left derivative $=-\dfrac{1}{\sqrt{1-0}}=-1$, right derivative $=\dfrac{1}{\sqrt{1-0}}=1$. These disagree, so $f'(0)$ does not exist (a corner, exactly like $|x|$). …

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8. AP EAPCET 2025Set eng-2025-05-27-FN1 markMCQ
   Q.The domain of the derivative of the function $f(x) = \dfrac{x}{1 + |x|}$ is (A) $[0, \infty)$ (B) $(-\infty, 0)$ (C) $(-\infty, \infty)$ (D) $(0, \infty)$

   ›Reveal solutionSolution

   $f(x)=x/(1+|x|)$ is smooth on each side of $0$, and the one-sided derivatives at $0$ actually agree, so it is differentiable everywhere. Answer: domain of $f'$ is $\mathbb R$.

   Concept and Intuition

   Functions involving $|x|$ are often not differentiable at $x=0$ (like $|x|$ itself), so the natural first suspicion is that the derivative fails to exist there. But here the two branches of $f$ are constructed so that they meet not just in value but in slope at $x=0$ — the modulus is "softened" by the $1+|x|$ denominator, so this particular function turns out to be differentiable at every point.

   Step-by-Step Solution

   1. For $x\ge0$: $f(x)=\dfrac{x}{1+x}$. Quotient rule: $f'(x)=\dfrac{(1+x)(1)-x(1)}{(1+x)^2}=\dfrac1{(1+x)^2}$.
   2. For $x<0$: $f(x)=\dfrac{x}{1-x}$. Quotient rule: $f'(x)=\dfrac{(1-x)(1)-x(-1)}{(1-x)^2}=\dfrac{1-x+x}{(1-x)^2}=\dfrac1{(1-x)^2}$.
   3. Check differentiability at $x=0$: right-hand derivative $=\dfrac1{(1+0)^2}=1$; left-hand derivative $=\dfrac1{(1-0)^2}=1$. They match, so $f'(0)=1$ exists. …

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9. AP EAPCET 2024Set eng-2024-05-20-FN1 markMCQ
   Q.For $x<0, \dfrac{d}{dx}[|x|^x]=$ (A) $(-x)^x[-1+\log(-x)]$ (B) $(-x)^x[1+\log(-x)]$ (C) $(-x)^x[1-\log(-x)]$ (D) $(-x)^x[-1-\log(-x)]$

   ›Reveal solutionSolution

   This is logarithmic differentiation of a variable-base, variable-exponent function; for $x<0$ the base becomes $-x>0$ so it's well defined.

   Concept and Intuition

   Whenever the base and the exponent both depend on $x$ (as in $f(x)^{g(x)}$), take logarithms first to turn the power into a product, then differentiate implicitly. Here for $x<0$, $|x|=-x$ is positive, so $|x|^x=(-x)^x$ is a genuine positive real number and logarithmic differentiation applies cleanly.

   Step-by-Step Solution

   1. Let $y=(-x)^x$ for $x<0$. Take natural log: $\log y = x\log(-x)$.
   2. Differentiate both sides with respect to $x$: $\dfrac{y'}{y}=\dfrac{d}{dx}[x\log(-x)]$.
   3. By product rule: $\dfrac{d}{dx}[x\log(-x)]=\log(-x)+x\cdot\dfrac{d}{dx}\log(-x)$.
   4. $\dfrac{d}{dx}\log(-x)=\dfrac{1}{-x}\cdot(-1)=\dfrac1x$, so $x\cdot\dfrac1x=1$. …

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10. AP EAPCET 2024Set eng-2024-05-20-AN1 markMCQ
    Q.In the interval $[0,3]$, the function $f(x)=|x-1|+|x-2|$ is (A) Discontinuous (B) differentiable (C) Continuous but not differentiable at $x=2$ only (D) Continuous but not differentiable at $x=1$ and $x=2$

    ›Reveal solutionSolution

    $|x-1|+|x-2|$ is continuous everywhere but has corners at the two points where the absolute values "switch," $x=1$ and $x=2$ — (D).

    Concept and Intuition

    $|x-a|$ is continuous everywhere (as a composition of continuous functions) but fails to be differentiable exactly at $x=a$, where its graph has a sharp corner (the left and right slopes are $-1$ and $+1$, which don't match). A sum of such functions is continuous everywhere (sum of continuous functions) and non-differentiable at each individual corner point (unless the kinks happen to cancel, which they don't here).

    Step-by-Step Solution

    1. Break $[0,3]$ into three pieces based on where $x-1$ and $x-2$ change sign: $[0,1)$, $[1,2)$, $[2,3]$.
    2. On $[0,1)$: $x-1<0, x-2<0$, so $f(x)=(1-x)+(2-x)=3-2x$.
    3. On $(1,2)$: $x-1>0,x-2<0$, so $f(x)=(x-1)+(2-x)=1$ (constant!).
    4. On $(2,3]$: both positive, $f(x)=(x-1)+(x-2)=2x-3$.
    5. All three pieces match up in value at the junctions ($f(1)=1$ from both sides, $f(2)=1$ from both sides), so $f$ is continuous throughout. …

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11. AP EAPCET 2024Set eng-2024-05-22-FN1 markMCQ
    Q.The function $f(x) = |x - 24|$ is (A) Differentiable on $[0, 25]$ (B) not continuous at $x = 24$ (C) neither continuous nor differentiable on $[0, 25]$ (D) Continuous on $[0, 25]$, but not differentiable on $[0, 25]$

    ›Reveal solutionSolution

    The absolute value function is always continuous, but fails to be differentiable exactly at its "kink" — here $x=24$, which lies inside the given interval.

    Concept and Intuition

    $|x-a|$ equals $-(x-a)$ for $x<a$ and $(x-a)$ for $x>a$; both pieces are continuous and their values agree at $x=a$, so the whole function is continuous. But the left derivative there is $-1$ and the right derivative is $+1$ — they disagree, so the function is not differentiable at $x=a$.

    Step-by-Step Solution

    1. $f(x)=|x-24|$ is a composition of continuous functions, hence continuous for all real $x$, in particular on $[0,25]$.
    2. Check differentiability at $x=24$ (which lies in $[0,25]$): left derivative $=\lim_{h\to0^-}\dfrac{|24+h-24|-0}{h}=\lim_{h\to0^-}\dfrac{|h|}{h}=-1$.
    3. Right derivative $=\lim_{h\to0^+}\dfrac{|h|}{h}=1$.
    4. Since $-1\ne1$, $f$ is not differentiable at $x=24$. …

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12. AP EAPCET 2023Set eng-2023-05-15-AN1 markMCQ
    Q.If $f(x) = |x-5| + |x+5| + |x-4| + |x+4|$, then $\dfrac{f'(1) - f'(-6)}{f'(-1)+f'(6)} =$ (A) 1 (B) 0 (C) 4/5 (D) 3/2

    ›Reveal solutionSolution

    Each $|x-a|$ contributes $\pm1$ to $f'(x)$ depending on the sign of $x-a$; plugging in the four given $x$-values and simplifying gives the ratio $1$.

    Concept and Intuition

    The derivative of $|x-a|$ is $+1$ for $x>a$ and $-1$ for $x<a$ (undefined only exactly at $x=a$). So $f'(x)$ for a sum of such terms is just the sum of these signs, evaluated at points away from the corners $x=\pm4,\pm5$.

    Step-by-Step Solution

    1. $f(x)=|x-5|+|x+5|+|x-4|+|x+4|$, so $f'(x)=\text{sgn}(x-5)+\text{sgn}(x+5)+\text{sgn}(x-4)+\text{sgn}(x+4)$.
    2. At $x=1$: signs of $(1-5,1+5,1-4,1+4)=(-,+,-,+) \Rightarrow f'(1)=-1+1-1+1=0$.
    3. At $x=-6$: signs of $(-11,-1,-10,-2)$ all negative $\Rightarrow f'(-6)=-1-1-1-1=-4$.
    4. At $x=-1$: signs of $(-6,4,-5,3)=(-,+,-,+) \Rightarrow f'(-1)=-1+1-1+1=0$. …

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