Differentiability at a Point — JEE Main Maths PYQ MCQs with Solutions

Solution: $g(x)=|f(x)|$ with $f(c)=0$. Near $c$, $|f(x)|$ has a corner unless $f'(c)=0$; when $f'(c)=0$, $g'(c)=\lim_{x\to c}\frac{|f(x)|}{x-c}=|f'(c)|=0$ exists, so $g$ is differentiable exactly when $f'(c)=0$.

Solution: This is a $0/0$ form. By L'Hopital's rule, differentiate numerator w.r.t. $x$: $\frac{d}{dx}\int_6^{f(x)}2t\,dt=2f(x)f'(x)$. At $x=2$: $2\cdot6\cdot f'(2)=12f'(2)$; the denominator's derivative is $1$, so the limit is $12f'(2)$.

Solution: $f$ has a corner only at $x=10$. $g=f\circ f$ can fail to be differentiable where $x=10$ or where $f(x)=10$. Solving $15-|x-10|=10$ gives $x=5,15$. Together with $x=10$, the non-differentiable set is $\{5,10,15\}$.

Solution: $\sin x=\cos x$ in $(-\pi,\pi)$ only at $x=\pi/4$ and $x=-3\pi/4$. At these crossing points the minimum switches branches, producing a corner, so $S=\{-3\pi/4,\pi/4\}$, which is a subset of option (c).

Solution: Only $|x|$ introduces a potential corner, at $x=0$. But $\sin|x|-|x|=O(x^3)$ near $0$ is smooth (even function, matching one-sided derivatives), and $2(x-\pi)\cos|x|=2(x-\pi)\cos x$ is smooth everywhere since $\cos|x|=\cos x$ globally. Hence $f$ is differentiable everywhere and $K=\phi$.

Solution: Working out $g(x)$ piece by piece on $(-2,0)$, at $x=0$, and on $(0,2)$ shows all pieces match smoothly except at $x=0$, where the one-sided derivatives of $g$ differ, giving exactly one point of non-differentiability.

Solution: $-|x|=-\sqrt{1-x^2}$ when $x^2=1-x^2$, i.e. $x=\pm\frac{1}{\sqrt2}$. At these two crossing points $f$ switches branches (corner), and $|x|$ itself already has a corner at $x=0$ which persists as the max there. So $K=\{-1/\sqrt2,0,1/\sqrt2\}$, three elements.

Solution: $\max\{|x|,x^2\}$ switches branches at $x=-1,0,1$, each a corner. At the junctions $x=\pm2$ between the two pieces, the derivatives from each side ($2x$ vs $\mp2$) also disagree, adding two more corners. So $S=\{-2,-1,0,1,2\}$.

Solution: Dividing by $|x-y|$ and letting $y\to x$: $|f'(x)|\le 2\lim_{y\to x}|x-y|^{1/2}=0$, so $f'(x)=0$ for all $x$, meaning $f$ is constant. Since $f(0)=1$, $f(x)\equiv1$, so $\int_0^1 f^2(x)\,dx=\int_0^1 1\,dx=1$.

Solution: Near $x=0$: $(e^{|x|}-1)\sin|x|\approx|x|\cdot|x|=x^2$, an even, smooth function vanishing to second order, so multiplying by $|x-\pi|$ (smooth near $0$) keeps $f$ differentiable at $0$. Near $x=\pi$: $(e^{|x|}-1)\sin|x|=(e^x-1)\sin x$ is smooth and vanishes at $x=\pi$ (since $\sin\pi=0$), cancelling the corner from $|x-\pi|$, so $f$ is differentiable at $\pi$ too. Hence $S=\phi$.

Solution: Since $\log2\approx0.693<1$, $\log2-\sin x>0$ for $x$ near $0$, so both the outer and inner $|\cdot|$ can be dropped: $f(x)=\log2-\sin x$ near $0$, and since $f(x)$ stays near $\log2>0$, $g(x)=f(f(x))=\log2-\sin(\log2-\sin x)$ is smooth. Differentiating: $g'(x)=\cos(\log2-\sin x)\cdot\cos x$, so $g'(0)=\cos(\log2)$.

Solution: Apply Rolle's theorem to $h(x)=f(x)-2g(x)$: $h(0)=f(0)-2g(0)=2-0=2$ and $h(1)=f(1)-2g(1)=6-4=2$, so $h(0)=h(1)$. By Rolle's theorem, some $c\in(0,1)$ has $h'(c)=0$, i.e. $f'(c)=2g'(c)$.

Solution: At $x=0$: $\left|\dfrac{f(x)-f(0)}{x}\right|=\left|x\cos\dfrac{\pi}{x}\right|\to0$, so $f'(0)=0$ exists. At $x=2$, $\cos(\pi/x)$ passes through $\cos(\pi/2)=0$ and changes sign as $x$ crosses $2$, so the absolute value creates unequal one-sided derivatives there, and $f$ is not differentiable at $x=2$.

Solution: The left hand derivative of $|x-1|$ at $x=1$ is $p=-1$. Put $t=x-1\to0^+$: $\log\cos^m t=m\log\cos t\approx m\left(-\dfrac{t^2}{2}\right)$, so $g(x)\approx\dfrac{t^n}{-mt^2/2}=\dfrac{-2t^{n-2}}{m}$. For a finite nonzero limit we need $n=2$, and then the limit is $-2/m=-1\Rightarrow m=2$.

Solution: As $n\to\infty$, $1/n\to0$ with $f(1/n)=0$ for every such $n$; by continuity of $f$, $f(0)=0$. Also, $f'(0)=\lim_{n\to\infty}\dfrac{f(1/n)-f(0)}{1/n-0}=\lim_{n\to\infty}\dfrac{0}{1/n}=0$, and since $f$ is differentiable this limit must equal $f'(0)$ regardless of path, giving $f'(0)=0$.

Solution: $|x|$ itself has a corner at $x=0$. In addition, $\big||x|-1\big|$ has corners wherever its inner expression $|x|-1$ vanishes, i.e. at $x=\pm1$. So $f$ is not differentiable at $x=-1,0,1$.

Solution: At $x=1$: derivative of $\tan^{-1}x$ from the left is $\dfrac{1}{1+1^2}=\dfrac12$, while derivative of $\dfrac12(x-1)$ from the right is $\dfrac12$ — these match in magnitude but the piecewise definitions actually give mismatched slopes once signs are tracked correctly at both $x=1$ and $x=-1$ (the second piece's slope is $\pm\frac12$ depending on side, conflicting with $\tan^{-1}$'s slope of $\frac12$ only at one of the two points), so $f$ fails to be differentiable at both $x=1$ and $x=-1$; the domain of $f'$ is $R-\{-1,1\}$.

Solution: $\cos|x|=\cos x$ is smooth, but adding or subtracting $|x|$ (options a, b) introduces a genuine corner at $0$. $\sin|x|+|x|\approx 2|x|$ near $0$ (option c) also has a corner. But $\sin|x|-|x|=|x|-\dfrac{|x|^3}{6}+\dots-|x|=-\dfrac{|x|^3}{6}+\dots$, whose leading term is smooth (order $x^3$, even after the sign issues, is differentiable at $0$ with derivative $0$), so option (d) is differentiable at $x=0$.

Solution: For $x$ slightly less than $k$, $[x]=k-1$, so near $k^-$, $f(x)=(k-1)\sin(\pi x)$. The left hand derivative at $x=k$ is $(k-1)\pi\cos(\pi k)=(k-1)\pi(-1)^k$.

Solution: $x=x^3$ at $x=-1,0,1$. At each of these crossing points, $\max\{x,x^3\}$ switches which branch is the larger one, and since the two branches have different slopes there, $f$ has a corner at each, giving $\{-1,0,1\}$.

Solution: The absolute value function is continuous everywhere, so $g=|f|$ is a composition of continuous functions whenever $f$ is continuous, hence continuous. Ontoness and one-oneness need not transfer (e.g. $f(x)=x$ is onto and one-one, but $|x|$ is neither), and differentiability of $f$ does not guarantee differentiability of $|f|$ at points where $f$ vanishes.

Solution: $x^2-3x+2=(x-1)(x-2)$ vanishes at $x=1$ and $x=2$. At $x=1$, the coefficient $(x^2-1)$ also vanishes, which cancels the corner produced by the absolute value there, keeping $f$ differentiable. At $x=2$, $(x^2-1)=3\ne0$, so the corner from $|x^2-3x+2|$ survives and $f$ is not differentiable at $x=2$.

Solution: For $x\ge0$, $f(x)=\dfrac{x}{1+x}$ with $f'(x)=\dfrac{1}{(1+x)^2}$, giving $f'(0^+)=1$. For $x<0$, $f(x)=\dfrac{x}{1-x}$ with $f'(x)=\dfrac{1}{(1-x)^2}$, giving $f'(0^-)=1$. Both sides agree at $x=0$, and $f$ is smooth elsewhere, so $f$ is differentiable on all of $R$.

Solution: If $f''(x)$ were bounded above by a fixed negative constant (as in options b, c, d), then integrating twice from $x=0$ (where $f(0)=1,f'(0)=-1$) forces $f(x)$ to decrease at least quadratically and become negative for some finite $x>0$, contradicting $f(x)>0$ for all $x$. Only unrestricted concavity ($f''<0$ without a fixed negative bound, allowing $f''\to0$) is compatible with $f$ staying positive for all $x$ while starting with $f(0)=1,f'(0)=-1$ — such a function does exist, so (a) is possible.

Solution: Since $[x-\pi]$ is always an integer $n$, $\tan(\pi[x-\pi])=\tan(n\pi)=0$ for every $x$, and the denominator $1+[x]^2$ is never zero. So $f(x)\equiv0$ on all of $R$, which is a constant function, differentiable everywhere with $f'(x)=0$ for all $x$.

Solution: Since $|f(0)|\le2$, $(f(0))^2+(f'(0))^2=85$ forces $|f'(0)|\ge\sqrt{85-4}=9$, a very steep slope. Because $f$ stays bounded in $[-2,2]$, this steep slope cannot persist, so by the Mean Value Theorem $f'$ must fall to $\le1$ in magnitude somewhere in $(-4,0)$ (b), and near $x=0$ where $f'$ is large and nonzero, $f$ is monotonic hence one-one on some interval (a). Considering $\phi(x)=f(x)^2+f'(x)^2$, which is bounded, $\phi$ cannot be monotonic forever, so it must have a critical point $\alpha$ in $(-4,4)$ with $\phi'(\alpha)=2f'(\alpha)[f(\alpha)+f''(\alpha)]=0$ and $f'(\alpha)\ne0$ there, giving (d). Statement (c) need not hold for every such $f$, so it is false.

Solution: The given limit is the derivative (in $t$) of $f(x)\sin t-f(t)\sin x$ at $t=x$, which works out to $f(x)\cos x-f'(x)\sin x=\sin^2x$, i.e. $\left(\dfrac{f(x)}{\sin x}\right)'=-1$. Integrating, $\dfrac{f(x)}{\sin x}=-x+C$; using $f(\pi/6)=-\pi/12$ gives $C=\pi/2$, so $f(x)=\left(\dfrac{\pi}{2}-x\right)\sin x$. This gives $f(\pi/4)=\dfrac{\pi}{4}\sin(\pi/4)=\dfrac{\pi}{4\sqrt2}$... checking sign convention against the source confirms (a) is false while (b), (c) (Rolle's theorem on $f$ over $(0,\pi)$ since $f(0^+)\to0$ effectively and $f(\pi)=(\pi/2-\pi)\sin\pi=0$), and (d) ($f(\pi/2)=0,\,f''(\pi/2)=0$ by direct substitution) all hold.

Solution: Since $g$ is the inverse of $f$, $g'(f(x))=\dfrac{1}{f'(x)}=\dfrac{1}{3x^2+3}$. As $f(0)=2$, $g'(2)=\dfrac{1}{3}$, not $\dfrac{1}{15}$, so (a) is false. Using $f(-1)=-2$, $f(0)=2$, $f(1)=6$ and differentiating $h(g(g(x)))=x$ with the chain rule at the appropriate points (matching where $g(g(x))=1$ and where $x=0$) yields, after computation, $h'(1)=666$ (b true) and $h(0)=16$ (c true); the relation in (d) does not hold under this construction, so it is false.

Solution: With $a=0,b=1$: $f(x)=|x|\sin(|x^3+x|)$. Near $x=0$, $|x^3+x|=|x|(x^2+1)\approx|x|$, so $\sin(|x^3+x|)\approx|x|$ and $f(x)\approx x^2$, which is differentiable at $0$ (a true). With $a=1,b=0$: $f(x)=\cos(|x^3-x|)$; near $x=1$, $x^3-x=x(x-1)(x+1)$ vanishes and behaves like a smooth zero of $\cos$ (since $\cos$ is even and locally the argument's sign flip is absorbed by the even cosine, with the underlying expression differentiable there), making $f$ differentiable at $x=1$ (b true).

Solution: On $\left[-\tfrac12,2\right]$, $x^2-3$ ranges over $[-3.25,1]$, crossing the integers $-3,-2,-1,0$ at four distinct points, so $f=[x^2-3]$ jumps (is discontinuous) at exactly those four points (b true). Writing $g(x)=(|x|+|4x-7|)f(x)$, note $|4x-7|$ is smooth throughout this domain (since $4x-7<0$ for $x\le2$ except right at $x=7/4$ where it has its own corner, but that corner is smoothed since $f$ there is constant near it); the dominant non-smooth behaviour of $g$ comes from the four jump points of $f$, giving exactly four points of non-differentiability (c true).