Q.If $f(x) = |x|^3$, show that $f''(x)$ exists for all real $x$ and find it.

Showing the 12 most recent of 22 on this concept.

1. AP EAPCET 2022Set eng-2022-07-08-AN1 markMCQ
   Q.If $f(x) = |x^2 - 3x + 2|$, then $\dfrac{df}{dx} =$ (A) $2x - 3$, when $1 < x < 2$ (B) $3 - 2x$, when $x > 2$ (C) $2x - 3$, when $x > 2$ (D) $3 + 2x$, when $1 < x < 2$

   ›Reveal solutionSolution

   Removing the modulus needs the sign of $(x-1)(x-2)$ on each interval, which flips between $(1,2)$ and $x>2$.

   Concept and Intuition

   $|g(x)|=g(x)$ where $g\ge0$ and $=-g(x)$ where $g<0$; differentiating a modulus requires splitting the domain by the sign of the inside expression.

   Step-by-Step Solution

   1. $x^2-3x+2=(x-1)(x-2)$; this is negative exactly on $(1,2)$ and positive for $x<1$ or $x>2$.
   2. On $(1,2)$: $f(x)=-(x^2-3x+2)=-x^2+3x-2\Rightarrow f'(x)=-2x+3=3-2x$.
   3. On $x>2$: $f(x)=x^2-3x+2\Rightarrow f'(x)=2x-3$. …

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2. 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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3. 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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4. 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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5. 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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6. 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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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 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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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 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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11. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
    Q.Let $f(x) = |x-3|+|x+5|$ and $A = \left\{a \in \mathbb{R} : \lim_{x\to a} \dfrac{f(x)-f(a)}{x-a} \text{ exists}\right\}$. Then the number of real numbers which are in $(-\infty,-3)\cup(5,\infty)$ but not in A is (A) 2 (B) 0 (C) 1 (D) 3

    ›Reveal solutionSolution

    The only non-differentiable points of $f(x)=|x-3|+|x+5|$ are the two kinks $x=3,-5$; checking which of these fall inside $(-\infty,-3)\cup(5,\infty)$ shows only $x=-5$ does, giving the count $1$.

    Concept and Intuition

    $f$ is piecewise linear with corners exactly where each absolute-value term changes sign, i.e. at $x=3$ and $x=-5$. Everywhere else it's a sum of linear pieces, hence differentiable. So $A=\mathbb{R}\setminus\{-5,3\}$, and we just need to check how many of the excluded points $\{-5,3\}$ actually lie in the specified region $(-\infty,-3)\cup(5,\infty)$.

    Step-by-Step Solution

    1. $f(x)=|x-3|+|x+5|$ is non-differentiable exactly at $x=3$ and $x=-5$ (the two "kink" points of the absolute values).
    2. So the points not in A (points where the limit does not exist) are $\{-5, 3\}$.
    3. Check $x=-5$: is $-5\in(-\infty,-3)\cup(5,\infty)$? Since $-5<-3$, yes, $-5\in(-\infty,-3)$. …

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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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