Q.The function $f(x) = e^{|x|}$ is
(A) continuous everywhere but not differentiable at $x = 0$
(B) continuous and differentiable everywhere
(C) not continuous at $x = 0$
(D) none of these

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2022Set eng-2022-07-07-AN1 markMCQ
   Q.Assertion (A): $f(x) = |x|$ is differentiable at $x = a \ne 0$ and continuous but not differentiable at $x = 0$ Reason (R): If a function is differentiable at a point then it is continuous at that point. But converse is not true. (A) A is correct, R is correct, R is correct explanation of A (B) A is correct, R is correct, but R is not correct explanation of A (C) A is correct, R is false. (D) A is false, R is correct.

   ›Reveal solutionSolution

   Both the assertion and reason are individually true standard facts about $|x|$ and about differentiability vs. continuity, and R does explain why A is true. Answer: (A).

   Concept and Intuition

   $f(x)=|x|$ has slope $-1$ for $x<0$ and slope $+1$ for $x>0$ — these are unequal, so the two one-sided derivatives at $x=0$ never match, and $f$ isn't differentiable there, even though $f$ is perfectly continuous at $x=0$ (no jump or break). This is the textbook example showing that continuity is a necessary but not sufficient condition for differentiability — which is exactly what Reason (R) states in general.

   Step-by-Step Solution

   1. Verify Assertion (A): for $a\neq0$, $|x|$ equals either $x$ or $-x$ in a neighbourhood of $a$, both of which are differentiable, so $f'(a)$ exists. At $x=0$: left derivative $=-1$, right derivative $=+1$; unequal, so not differentiable, but $\lim_{x\to0}|x|=0=f(0)$, so it is continuous. A is TRUE.
   2. Verify Reason (R): this is the standard theorem "differentiable $\Rightarrow$ continuous", and $|x|$ at $0$ is the classic counterexample to the converse. R is TRUE and is a general statement. …

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2. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
   Q.Which of the following is differentiable at x = 0 ? (A) $f(x) = \cos|x| + |x|$ (B) $f(x) = \sin|x| + |x|$ (C) $f(x) = \cos|x| - |x|$ (D) $f(x) = \sin|x| - |x|$

   ›Reveal solutionSolution

   Writing each function separately for $x>0$ and $x<0$ and comparing the one-sided derivatives at $0$, only $f(x)=\sin|x|-|x|$ has equal left- and right-hand derivatives (both $0$), making it the only one differentiable at $x=0$.

   Concept and Intuition

   A function built from $|x|$ typically has a "kink" at $x=0$ because $\dfrac{d}{dx}|x|$ jumps from $-1$ to $+1$. But if the other piece of the function also contributes a matching jump that exactly cancels this discontinuity in slope, the combination can become smooth at $0$ even though $|x|$ alone is not differentiable there. Since $\cos|x|\equiv\cos x$ (cosine is even), only $|x|$'s own kink matters in options (A) and (C). Since $\sin|x|$ is itself non-smooth at $0$ (behaving like $|x|$ for options B and D), we must check each combination directly.

   Step-by-Step Solution

   1. (A) $f=\cos|x|+|x|=\cos x+|x|$. Right derivative at $0$: $-\sin(0)+1=1$. Left derivative: $-\sin(0)-1=-1$. Not equal — not differentiable.
   2. (B) $f=\sin|x|+|x|$. For $x>0$: $f=\sin x+x,\,f'=\cos x+1\to2$. For $x<0$: $|x|=-x,\sin|x|=\sin(-x)=-\sin x$, so $f=-\sin x-x,\,f'=-\cos x-1\to-2$. Not equal.
   3. (C) $f=\cos|x|-|x|=\cos x-|x|$. Right derivative: $-\sin(0)-1=-1$. Left derivative: $-\sin(0)+1=1$. Not equal. …

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3. 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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4. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.The set of all points where the function $f(x) = 2x|x|$ is differentiable is ________ (A) $(-\infty, \infty)$ (B) $(-\infty, 0) \cup (0, \infty)$ (C) $(0, \infty)$ (D) $[0, \infty)$

   ›Reveal solutionSolution

   $f(x)=2x|x|$ is a smoothly-joined piecewise quadratic; it is differentiable at every real number, including $x=0$. Answer: all of $\mathbb R$.

   Concept and Intuition

   Functions built from $|x|$ often fail to be differentiable at $0$ (like $|x|$ itself), but multiplying by an extra factor of $x$ can "soften" the corner into a genuine smooth point — that's exactly what happens here.

   Step-by-Step Solution

   1. Write $f(x)=2x|x|=\begin{cases}2x^2,&x\ge0\\-2x^2,&x<0\end{cases}$.
   2. For $x>0$: $f'(x)=4x$. For $x<0$: $f'(x)=-4x$.
   3. Check differentiability at $x=0$ directly from the definition: $f'(0)=\lim_{h\to0}\dfrac{f(h)-f(0)}{h}=\lim_{h\to0}\dfrac{2h|h|}{h}=\lim_{h\to0}2|h|=0$. …

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5. 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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6. 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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7. 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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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 2022Set eng-2022-07-06-AN1 markMCQ
   Q.Let $f(x) = \begin{cases} |x|, & -\infty < x < 2 \\ |2x-4|, & 2 \le x \le 20 \end{cases}$. $x = a$ is a point where $f(x)$ is continuous but not differentiable and $x = b$ is a point where $f(x)$ is not differentiable $(a \ne b)$. Then $a + b =$ (A) 1 (B) 2 (C) -2 (D) 0

   ›Reveal solutionSolution

   The piecewise function has exactly one genuine "corner" (continuous but non-differentiable) at $x=0$ from the $|x|$ piece, and a jump discontinuity (hence non-differentiable) at the junction $x=2$; their sum is $2$.

   Concept and Intuition

   A function can fail to be differentiable at a point either because it has a sharp corner while still being continuous there (like $|x|$ at $0$), or because it's outright discontinuous there (any jump automatically kills differentiability too, but that's a stronger failure). This problem asks us to identify one of each type.

   Step-by-Step Solution

   1. For $-\infty<x<2$: $f(x)=|x|$. This has its classic corner at $x=0$ — continuous there ($f(0)=0$ from both sides) but the left and right derivatives are $-1$ and $+1$, unequal, so not differentiable. This is our point $a=0$.
   2. Check continuity at the junction $x=2$: from the left, using the first piece, $\lim_{x\to2^-}|x|=2$. From the definition at $x=2$ (second piece applies since $2\le x\le20$): $f(2)=|2(2)-4|=|0|=0$.
   3. Since $2\ne0$, $f$ has a jump discontinuity at $x=2$ — so $f$ is not differentiable there (discontinuity is a stronger failure than a mere corner). This is our point $b=2$. …

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10. 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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11. 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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12. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
    Q.If $\alpha\in\mathbb{R}-\{-1\}$ and $f(x)=\big|(|x|+\alpha)(|x|-1)\big|$, then the number of points at which f(x) is not differentiable, is (A) 3, when $\alpha<0$ (B) 5, when $\alpha>0$ (C) 4, when $\alpha>0$ (D) 5, when $\alpha<0$

    ›Reveal solutionSolution

    Writing $f$ as $|x^2+(\alpha-1)|x|-\alpha|$ and counting its zero-crossings plus the origin kink shows $f$ fails to be differentiable at exactly 5 points whenever $\alpha<0$ (and only 3 points when $\alpha>0$).

    Concept and Intuition

    $f(x)=|(|x|+\alpha)(|x|-1)|$ is built by composing an absolute value of $x$ (kink at $x=0$ unless its coefficient vanishes) with another outer absolute value (kinks wherever the inner expression crosses zero with nonzero slope). We must count both kinds of kink.

    Step-by-Step Solution

    1. Expand: $(|x|+\alpha)(|x|-1)=|x|^2+(\alpha-1)|x|-\alpha=x^2+(\alpha-1)|x|-\alpha=g(x)$.
    2. Substituting $u=|x|\ge0$: $g=u^2+(\alpha-1)u-\alpha$. Its discriminant is $(\alpha-1)^2+4\alpha=(\alpha+1)^2$, so roots are $u=\dfrac{(1-\alpha)\pm|\alpha+1|}{2}$, giving $u=1$ and $u=-\alpha$.
    3. $u=1$ is always a valid root ($x=\pm1$). $u=-\alpha$ is valid (i.e. $\ge0$) only when $\alpha\le0$.
    4. For $\alpha<0$: both roots valid, distinct from each other and from $0$ (since $\alpha\ne0,-1$), giving zero-crossings at $x=1,-1,\alpha,-\alpha$ — four points. Checking $g'$ at each (using $g'(x)=2x\pm(\alpha-1)$ on the two branches) shows the slope is nonzero there in all cases, so each is a genuine corner of $|g|$. …

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