Q.If $f(x) = |\cos x|$, then $f'\left(\dfrac{\pi}{4}\right) =$ __________.

Showing the 12 most recent of 22 on this concept.

1. 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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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 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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4. 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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5. 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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6. AP EAPCET 2023Set eng-2023-05-16-FN1 markMCQ
   Q.Let $f(x)=\text{Max}\{\cos x,\sin x,0\}$. If the number of points at which $f(x)$ is not differentiable in $(0,2024\pi)$ is $1012k$, then $k=$ (A) $3/2$ (B) $6$ (C) $3$ (D) $2$

   ›Reveal solutionSolution

   Careful case analysis of $\max(\sin x,\cos x,0)$ over one period shows exactly 3 genuine corners per $2\pi$; over $1012$ periods that's $3036=1012k$, giving $k=3$.

   Concept and Intuition

   $f(x)=\max(\sin x,\cos x,0)$ is non-differentiable only where the identity of the maximum actually switches from one function to another (a mere crossing of two functions underneath the current maximum causes no corner). We must track, across one period, which of the three functions is on top, and find only the switch-points.

   Step-by-Step Solution

   1. Critical crossing points in $[0,2\pi)$: $\sin x=\cos x$ at $\pi/4,5\pi/4$; $\sin x=0$ at $0,\pi$; $\cos x=0$ at $\pi/2,3\pi/2$.
   2. $(0,\pi/4)$: $\cos x$ is the max (positive, larger than $\sin x$ and $0$).
   3. $(\pi/4,\pi)$: $\sin x$ becomes the max (it overtakes $\cos x$ at $\pi/4$, and stays $>0>\cos x$ or $>0$ through to $\pi$; $\cos x=0$ at $\pi/2$ doesn't matter since $\sin x$ is already on top). Corner at $\pi/4$ (slopes $-\sin(\pi/4)$ vs $\cos(\pi/4)$ differ).
   4. At $x=\pi$: $\sin x=0$ crosses down to negative while $\cos x=-1<0$, so the max switches to the constant $0$. Corner at $\pi$ (slope $\cos\pi=-1$ vs slope $0$ differ).
   5. $(\pi,3\pi/2)$: both $\sin x,\cos x<0$ throughout this open interval, so $\max=0$ flat — the crossing $\sin x=\cos x$ at $5\pi/4$ happens underneath the max and causes no corner.
   6. At $x=3\pi/2$: $\cos x$ crosses up through $0$ becoming positive while $\sin x$ stays negative, so the max switches from $0$ to $\cos x$. Corner at $3\pi/2$ (slope $0$ vs slope $-\sin(3\pi/2)=1$ differ). …

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7. 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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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-2022-07-08-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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10. 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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11. 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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12. AP EAPCET 2022Set eng-2022-07-08-FN1 markMCQ
    Q.If $f(x) = \begin{cases} \tan^{-1}x, & \text{when } |x| \le 1 \\ \dfrac{1}{2}(|x|-1), & \text{when } |x| > 1 \end{cases}$, then the domain of $\dfrac{d}{dx}f(x)$ is (A) $\mathbf{R} - \{-1, 1\}$ (B) $\mathbf{R} - (-1, 1)$ (C) $\mathbf{R} - [-1, 1]$ (D) $\mathbf{R} - \{-1\}$

    ›Reveal solutionSolution

    $f$ is discontinuous at $x=\pm1$, so its derivative exists everywhere except there: domain $\mathbf{R}-\{-1,1\}$ — option (A).

    Concept and Intuition

    A function must at least be continuous to be differentiable. At the points where the piecewise definition switches ($x=\pm1$), check whether the two pieces meet. If the left and right values differ, there is a jump discontinuity and the derivative cannot exist there.

    Step-by-Step Solution

    1. At $x=1$: inner branch $\tan^{-1}(1)=\pi/4$; outer branch $\tfrac12(1-1)=0$. Since $\pi/4 \ne 0$, $f$ is discontinuous at $x=1$.
    2. At $x=-1$: inner branch $\tan^{-1}(-1)=-\pi/4$; outer branch $\tfrac12(|-1|-1)=0$. Since $-\pi/4 \ne 0$, $f$ is discontinuous at $x=-1$.
    3. Discontinuity ⇒ non-differentiability at both $x=1$ and $x=-1$.
    4. Everywhere else ($|x|<1$ smooth $\tan^{-1}x$; $|x|>1$ smooth linear pieces) the derivative exists.
    5. Hence the domain of $\frac{d}{dx}f(x)$ is $\mathbf{R}-\{-1,1\}$ → option (A).

    Common Mistakes …

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