Maxima and Minima — JEE Main Maths PYQ MCQs with Solutions

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: Since extrema are at $-1,0,1$, $f'(x)=k(x^3-x)$, so $f(x)=k\left(\dfrac{x^4}{4}-\dfrac{x^2}{2}\right)+c$. Solving $f(x)=f(0)=c$ gives $x^2(x^2-2)=0$, so $x=0,\ \pm\sqrt2$ — one rational ($0$) and two irrational ($\pm\sqrt2$) numbers.

Solution: For a cylinder of height $h$, radius $r$ in a sphere of radius $R$: $r^2=R^2-h^2/4$. $V=\pi h\left(R^2-\dfrac{h^2}{4}\right)$; $\dfrac{dV}{dh}=\pi\left(R^2-\dfrac{3h^2}{4}\right)=0\Rightarrow h=\dfrac{2R}{\sqrt3}=2\sqrt3$.

Solution: $f'(x)=36x(x+2)(x-1)$. Sign analysis shows $f'$ changes $-\to+$ at $x=-2$ and $x=1$ (local minima), and $+\to-$ at $x=0$ (local maximum). So $S_1=\{-2,1\}$, $S_2=\{0\}$.

Solution: A point on the curve is $(y^2+2,y)$; its distance to the line $x-y=0$ is $d=\dfrac{|y^2-y+2|}{\sqrt2}$. Minimizing $y^2-y+2$ gives $y=\dfrac12$, value $\dfrac74$, so $d=\dfrac{7}{4\sqrt2}$.

Solution: With vertices $(\pm x,12-x^2)$, area $A=2x(12-x^2)=24x-2x^3$. $A'(x)=24-6x^2=0\Rightarrow x=2$, giving $A(2)=48-16=32$.

Solution: $S$: $x^2-11x+30\le0\Rightarrow x\in[5,6]$. $f'(x)=9(x-1)(x-3)>0$ on $[5,6]$, so $f$ is increasing there; the maximum is at $x=6$: $f(6)=648-648+162-40=122$.

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: With slant height $3$: $r^2+h^2=9$. $V=\dfrac13\pi r^2h=\dfrac13\pi(9-h^2)h$; $\dfrac{dV}{dh}=\dfrac13\pi(9-3h^2)=0\Rightarrow h=\sqrt3$, $r^2=6$. $V=\dfrac13\pi(6)(\sqrt3)=2\sqrt3\pi$.

Solution: $f(x)=g(x)^2+2$, so $h(x)=g(x)+\dfrac{2}{g(x)}$. For $g(x)>0$, by AM-GM $h(x)\ge2\sqrt2$, with equality when $g(x)=\sqrt2$, so the local minimum value is $2\sqrt2$.

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: Constraint: $4x+2\pi r=2$. Minimizing $A=x^2+\pi r^2$ using $x=\dfrac{1-\pi r}{2}$, $\dfrac{dA}{dr}=2x\cdot\left(-\dfrac{\pi}{2}\right)+2\pi r=0\Rightarrow x=2r$.

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: $f'(x)=\dfrac{\alpha}{x}+2\beta x+1=0$ at $x=-1,2$: $-\alpha-2\beta+1=0$ and $\dfrac{\alpha}{2}+4\beta+1=0$. Solving gives $\alpha=2,\beta=-\dfrac12$.

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: On $(-3,-1]$, $f(x)=(2+x)^3$ increases with no interior turning point; at $x=-1$ it meets $f(x)=x^{2/3}$, which decreases to $0$ at $x=0$ then increases. This gives a local maximum at $x=-1$ and a local minimum at $x=0$: total $2$.

Solution: $\min f=2c^2-b^2$ (at the vertex) and $\max g=c^2+b^2$. The condition $2c^2-b^2>c^2+b^2$ gives $c^2>2b^2$, i.e. $|c|>|b|\sqrt2$.

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: $f(x)=1-\dfrac{2}{x^2+1}$. Since $x^2+1\ge1$, $\dfrac{2}{x^2+1}\le2$, so $f(x)\ge-1$, with equality at $x=0$. The minimum value is $-1$.

Solution: The maximum value $2$ requires $\cos x=1$ and $\cos(\sqrt2 x)=1$ simultaneously, i.e. $x=2m\pi$ and $\sqrt2x=2n\pi$ for integers $m,n$, forcing $\sqrt2=n/m$, which is impossible unless $x=0$. So the maximum is attained at exactly one point.

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: 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: $P$ is an even function; $P'(x)=2a_1x+4a_2x^3+\cdots$ is odd with $P'(0)=0$. Since all coefficients are positive, $P'(x)>0$ for $x>0$ and $P'(x)<0$ for $x<0$, so $P$ has exactly one minimum, at $x=0$.

Solution: $y'=\dfrac{a}{x}+2bx+1=0$ at $x=1,2$: $a+2b+1=0$ and $\dfrac{a}{2}+4b+1=0$. Solving gives $a=2,b=-\dfrac12$.

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: 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 $(-\pi,\pi)$.