Application of Derivatives — JEE Main Maths PYQ MCQs with Solutions

Solution: Line $-x+y+4=0$ has slope $1$, so tangent slope $=-1$. $y'=3x^2+a$, at $x=1$: $3+a=-1 \Rightarrow a=-4$. Since $(1,-5)$ is on the curve: $-5=1-4+b \Rightarrow b=-2$. Curve: $y=x^3-4x-2$. Checking $(2,-2)$: $8-8-2=-2$ ✓.

Solution: Intersection: $10-x^2=2+x^2 \Rightarrow x^2=4 \Rightarrow x=2$ (taking one point). Slopes: $m_1=-2x=-4$, $m_2=2x=4$. $|\tan\theta|=\left|\dfrac{m_1-m_2}{1+m_1m_2}\right|=\left|\dfrac{-8}{-15}\right|=\dfrac{8}{15}$.

Solution: The normal's slope is $\tan\dfrac{3\pi}{4}=-1$. The tangent slope $f'(3)$ is the negative reciprocal of the normal slope, so $f'(3)=1$.

Solution: Perimeter $2r+r\theta=20$, area $A=\dfrac12r^2\theta=\dfrac12r(20-2r)=10r-r^2$. $\dfrac{dA}{dr}=10-2r=0\Rightarrow r=5$, giving $A=50-25=25$.

Solution: For $x$ near $0$ but not equal to $0$, $f(x)=|x|$ is close to $0$, which is less than $f(0)=1$. So $f$ has a local maximum at $x=0$ (not a global one, since $f$ reaches values up to $2$ elsewhere).

Solution: Taking logs, $\ln y=25\ln x+75\ln(1-x)$, so $\dfrac{y'}{y}=\dfrac{25}{x}-\dfrac{75}{1-x}=0\Rightarrow25(1-x)=75x\Rightarrow x=\dfrac14$.

Solution: This is a standard identity: for two reals, subtracting half the gap $|p-q|$ from the average $\dfrac{p+q}{2}$ gives the smaller one, i.e. $\min(p,q)=\dfrac12(p+q-|p-q|)$.

Solution: $D^2=(x-\tfrac12)^2+x^3$ (since $y-7=x^{3/2}$). $\dfrac{d(D^2)}{dx}=2(x-\tfrac12)+3x^2=0 \Rightarrow 3x^2+2x-1=0 \Rightarrow (3x-1)(x+1)=0 \Rightarrow x=\tfrac13$. $D^2=(-\tfrac16)^2+\tfrac1{27}=\tfrac{1}{36}+\tfrac1{27}=\tfrac{7}{108}$, so $D=\sqrt{\dfrac{7}{108}}=\dfrac16\sqrt{\dfrac73}$.

Solution: $\sqrt{\dfrac{1+\sin x}{1-\sin x}}=\tan\left(\dfrac{\pi}{4}+\dfrac{x}{2}\right)$, so $f(x)=\dfrac{\pi}{4}+\dfrac{x}{2}$ and $f'(x)=\dfrac12$. At $x=\dfrac{\pi}{6}$, $f\left(\dfrac{\pi}{6}\right)=\dfrac{\pi}{3}$. Normal slope $=-2$: $y-\dfrac{\pi}{3}=-2\left(x-\dfrac{\pi}{6}\right) \Rightarrow y=-2x+\dfrac{2\pi}{3}$, which passes through $\left(0,\dfrac{2\pi}{3}\right)$.

Solution: The homogeneous curve factors as $(x-y)(x+3y)=0$, i.e. the lines $y=x$ and $y=-\dfrac{x}{3}$; $(1,1)$ lies on $y=x$. Differentiating gives slope $1$ at $(1,1)$, so the normal is $y=-x+2$. It meets $y=-\dfrac{x}{3}$ at $x=3,y=-1$, i.e. $(3,-1)$, which lies in the fourth quadrant.

Solution: Differentiating: $y'=\dfrac{-6x}{3y^2-12}$. A vertical tangent needs $3y^2-12=0 \Rightarrow y=\pm2$ (with $x \neq 0$). For $y=2$: $8+3x^2=24 \Rightarrow x=\pm\dfrac{4}{\sqrt3}$; for $y=-2$ there is no real $x$. So the point is $\left(-\dfrac{4}{\sqrt3},2\right)$ (or its mirror).

Solution: Differentiating implicitly: $2(y-x^5)(y'-5x^4) = (1+x^2)^2 + 4x^2(1+x^2)$. At $(1,3)$: $y-x^5=2$, $1+x^2=2$. LHS $=4(y'-5)$; RHS $=4+8=12$. So $4(y'-5)=12 \Rightarrow y'=8$.

Solution: $h(x)=e^{x^2-x}-(x^2-x)$, so $h'(x)=(2x-1)\left(e^{x^2-x}-1\right)$, which has the same sign as $(2x-1)(x^2-x)$ since $e^t-1$ has the sign of $t$. Solving $(2x-1)(x^2-x)\ge0$ gives $x\in\left[0,\tfrac12\right]\cup[1,\infty)$.

Solution: For any nonzero constant $c$, the function $t\mapsto \dfrac{t}{\sqrt{c^2+t^2}}$ has derivative $\dfrac{c^2}{(c^2+t^2)^{3/2}}>0$, so it is strictly increasing in $t$. Each term of $f$ is such a function of a linear (increasing) expression in $x$, so $f'(x)=\dfrac{a^2}{(a^2+x^2)^{3/2}}+\dfrac{b^2}{(b^2+(x-d)^2)^{3/2}}>0$ for all $x$; hence $f$ is increasing.

Solution: $f'(x)=\dfrac{\sin x-x\cos x}{\sin^2x}$; with $h(x)=\sin x-x\cos x$, $h(0)=0$ and $h'(x)=x\sin x>0$, so $h(x)>0$ and $f$ is increasing. For $g(x)=x\cot x$, $g'(x)=\dfrac{\tfrac12\sin2x-x}{\sin^2x}$; with $k(x)=\tfrac12\sin2x-x$, $k(0)=0$ and $k'(x)=\cos2x-1\le0$, so $k(x)<0$ for $x>0$, making $g'(x)<0$, i.e. $g$ is decreasing. So only $f$ is increasing.

Solution: With $u=e+x>e$ and $k=\pi-e>0$, $f=\dfrac{\ln(u+k)}{\ln u}$. Its derivative has the sign of $\psi(u)-\psi(u+k)$, where $\psi(u)=u\ln u$. Since $\psi'(u)=\ln u+1>0$ for $u>e$, $\psi$ is increasing there, so $\psi(u)<\psi(u+k)$, making the derivative negative. Hence $f$ is decreasing on $(0,\infty)$.

Solution: Since $f$ is increasing and $g$ is decreasing, $h=f\circ g$ is decreasing on $[0,\infty)$, so $h(0)$ is its maximum value, i.e. $h(x)\le h(0)=0$ for all $x\ge0$. But $h(x)=f(g(x))\ge0$ since $f$'s range is $[0,\infty)$. So $h(x)=0$ for every $x\ge0$, meaning $h(x)-h(1)=0$ always -- neither always negative, always positive, nor strictly increasing. The answer is 'None of these'.

Solution: Let $g(x)=f(x)e^{-2x}$. Then $g'(x)=e^{-2x}(f'(x)-2f(x))>0$, so $g$ is increasing; with $g(0)=1$, $g(x)>1$ for $x>0$, giving $f(x)>e^{2x}$. Then $f'(x)>2f(x)>2e^{2x}>e^{2x}$. Also $f'(x)>2f(x)>0$ (since $f(x)>e^{2x}>0$), so $f$ is increasing on $(0,\infty)$.

Solution: Swapping limits directly gives $f(1/x)=-f(x)$, so (c) holds. By Leibniz's rule, $f'(x)=\dfrac{2}{x}\cosh\!\left(x-\dfrac1x\right)>0$ for all $x>0$, so $f$ is strictly increasing on all of $(0,\infty)$ -- so (a) holds but (b) fails. For $\phi(x)=f(2^x)$, $\phi(-x)=f(2^{-x})=f(1/2^x)=-f(2^x)=-\phi(x)$, so $\phi$ is odd, confirming (d).

Solution: Since $f''\ge0$, $f'$ is non-decreasing (convex $f$). By the Mean Value Theorem on $\left[\tfrac12,1\right]$, there is $c\in\left(\tfrac12,1\right)$ with $f'(c)=\dfrac{f(1)-f(1/2)}{1-1/2}=1$. As $f'$ is non-decreasing, $f'(1)\ge f'(c)=1$; since $c<1$ and equality throughout would force $f$ linear (contradicting the strict convexity needed to pin $c$ inside the open interval), the inequality is strict: $f'(1)>1$.

Solution: $f$ is maximum when $|x-2|=0$, so $\alpha=2$; $g$ is minimum when $|x+1|=0$, so $\beta=-1$. Thus $-\alpha\beta=2$. Factoring, $\dfrac{(x-1)(x-2)(x-3)}{(x-2)(x-4)}=\dfrac{(x-1)(x-3)}{x-4}$. At $x=2$: $\dfrac{(1)(-1)}{-2}=\dfrac12$.

Solution: $V=\begin{vmatrix}1&\lambda&1\\0&1&\lambda\\\lambda&0&1\end{vmatrix}=1+\lambda^3-\lambda$. Let $h(\lambda)=\lambda^3-\lambda+1$; $h'(\lambda)=3\lambda^2-1=0\Rightarrow\lambda=\pm\dfrac{1}{\sqrt3}$. Since $h''(\lambda)=6\lambda$, $\lambda=\dfrac1{\sqrt3}$ gives $h''>0$, a local minimum of the (positive) volume.

Solution: $f(x)^2=kx^3-x^4=h(x)$; $h'(x)=x^2(3k-4x)\ge0$ on $[0,3]$ needs $k\ge\dfrac{4x}{3}$ for all $x\in[0,3]$, so $k\ge4$, giving $m=4$. At $k=4$, $f$ is increasing throughout $[0,3]$, so its max is at $x=3$: $f(3)=3\sqrt{12-9}=3\sqrt3=M$.

Solution: Take $C=(t^2,2t)$ on the parabola. Writing the area of $\triangle ACB$ as a function of $t$ and maximizing gives $t=-\dfrac12$, i.e. $C=\left(\dfrac14,-1\right)$, and substituting back gives the maximum area $31\dfrac14$ sq units.

Solution: Need $a\ge\phi(x)=\dfrac{x-1}{4x^3}$ for all $x>0$. $\phi'(x)=0$ gives $x=\dfrac32$, and $\phi\left(\dfrac32\right)=\dfrac{1}{27}$, so the least $a$ is $\dfrac1{27}$.

Solution: The finite limit forces $f(0)=0,f'(0)=0$ and $f(x)=2x^2+ax^3+bx^4$. Using $f'(1)=0$ and $f'(2)=0$: $4+3a+4b=0$, $8+12a+32b=0$, giving $a=-2,b=\tfrac12$. So $f(2)=8-16+8=0$.

Solution: Let $g(x)=x^2-x\sin x-\cos x$, an even function with $g(0)=-1<0$ and $g(x)\to\infty$ as $x\to\pm\infty$. Since $g'(x)=x(2-\cos x)$ and $2-\cos x>0$ always, $g'$ changes sign only at $x=0$, so $g$ decreases then increases, giving exactly one root on each side of $0$ — $2$ points total.

Solution: On $[0,1]$, $e^{x^2}$ increases and $e^{-x^2}$ decreases, so each of $f,g,h$ attains its maximum at $x=1$; there $x=x^2=1$, so all three functions equal $e+\dfrac1e$ at $x=1$, giving $a=b=c$.

Solution: With $P=(a\cos\theta,b\sin\theta)$, $OP=\sqrt{a^2\cos^2\theta+b^2\sin^2\theta}$ and $ON=\dfrac{ab}{\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}}$. Maximizing the area of $\triangle PON$ leads to $\tan\theta=\pm\dfrac{b}{a}$, giving $P=\left(\pm\dfrac{a^2}{\sqrt{a^2+b^2}},\mp\dfrac{b^2}{\sqrt{a^2+b^2}}\right)$.

Solution: Expanding the determinant and simplifying using trigonometric identities shows $f(0)=1$ is the largest value $f$ attains, so $f$ has its maximum at $x=0$. Since the simplified $f(x)$ is a combination of several cosine harmonics, $f'(x)=0$ at more than three points in $(-\pi,\pi)$.