Maxima and Minima Complete JEE Notes

Step 1: Differentiate by the product rule.

$f'(x)=4(x-2)^3(x+1)^3+3(x+1)^2(x-2)^4.$

Step 2: Take out the common factor. Both terms share $(x-2)^3(x+1)^2$:

$f'(x)=(x-2)^3(x+1)^2\big[4(x+1)+3(x-2)\big]=(x-2)^3(x+1)^2(7x-2).$

Step 3: Critical points. $f'(x)=0$ at $x=-1$, $x=\dfrac27$ and $x=2$.

Step 4: Sign chart. Note $(x+1)^2$ has an even power, so it never flips the sign; only $(x-2)^3$ and $(7x-2)$ do.

Step 5: Read it off.

• At $x=\dfrac27$: $+\to-$, so a local maximum. (I) TRUE.
• At $x=2$: $-\to+$, so a local minimum. (II) TRUE.
• At $x=-1$: no sign change, so a point of inflection. (III) TRUE.

Answer: all three are correct.

Step 1: Differentiate.

$\frac{dy}{dx}=\frac ax+2bx+1.$

Step 2: Impose both conditions. An extremum needs $\dfrac{dy}{dx}=0$ there:

$\text{at }x=-1:\quad -a-2b+1=0\ \Longrightarrow\ a+2b-1=0,$ $\text{at }x=2:\quad \frac a2+4b+1=0\ \Longrightarrow\ a+8b+2=0.$

Step 3: Subtract.

$(a+8b+2)-(a+2b-1)=6b+3=0\ \Longrightarrow\ b=-\frac12.$

Step 4: Back-substitute. $a+2\left(-\dfrac12\right)-1=0$ gives $a=2$.

Answer: $\boxed{a=2,\ b=-\dfrac12}$

Step 1: Differentiate by the product rule.

$f'(x)=25x^{24}(1-x)^{75}+x^{25}\cdot 75(1-x)^{74}(-1).$

Step 2: Factor out the common part $25x^{24}(1-x)^{74}$:

$f'(x)=25x^{24}(1-x)^{74}\big[(1-x)-3x\big]=25x^{24}(1-x)^{74}(1-4x).$

Step 3: Critical points on $[0,1]$. $x=0$, $x=\dfrac14$, $x=1$.

Step 4: Sign chart. On $(0,1)$ both $x^{24}$ and $(1-x)^{74}$ are positive (even powers), so the sign of $f'$ is the sign of $1-4x$: positive for $x<\dfrac14$, negative for $x>\dfrac14$.

Step 5: Conclude. $f'$ changes $+\to-$ at $x=\dfrac14$, so that is the maximum. The endpoints give $f(0)=f(1)=0$, so it is the absolute maximum too.

Answer: $\boxed{x=\dfrac14}$

Step 1: Rewrite using the double angle.

$f(x)=\sin x+\sin x\cos x=\sin x+\frac12\sin 2x.$

Step 2: Differentiate twice.

$f'(x)=\cos x+\cos 2x,\qquad f''(x)=-\sin x-2\sin 2x.$

Step 3: Solve $f'=0$. Using $\cos 2x=2\cos^2x-1$,

$2\cos^2x+\cos x-1=0\ \Longrightarrow\ (2\cos x-1)(\cos x+1)=0,$

so $\cos x=\dfrac12$ or $\cos x=-1$, giving $x=\dfrac\pi3,\ \pi,\ \dfrac{5\pi}{3}$ in one period.

Step 4: Classify with $f''$.

$f''\!\left(\frac\pi3\right)=-\sin\frac\pi3-2\sin\frac{2\pi}{3}<0\ \Rightarrow\ \text{maximum},$

Step 1: Differentiate by Newton–Leibnitz.

$f'(x)=\frac{\sin x}{x}.$

Step 2: Critical points. For $x>0$ the denominator never vanishes, so $f'(x)=0$ needs $\sin x=0$, i.e. $x=n\pi$ for $n\in\mathbb N$.

Step 3: Second derivative by the quotient rule.

$f''(x)=\frac{x\cos x-\sin x}{x^2}.$

Step 4: Evaluate at $x=n\pi$. There $\sin n\pi=0$ and $\cos n\pi=(-1)^n$, so

$f''(n\pi)=\frac{n\pi(-1)^n-0}{(n\pi)^2}=\frac{(-1)^n}{n\pi}.$

Step 5: Read the sign. For $n$ odd, $f''(n\pi)<0$ — a local maximum. For $n$ even, $f''(n\pi)>0$ — a local minimum.

Answer: maximum or minimum according as $n$ is odd or even respectively.

Step 1: Differentiate. By Newton–Leibnitz, replace $t$ by $x$:

$f'(x)=(\sin x-\cos x)\left(e^{x}-2\right)(x-1)^3(x-2)^5.$

Step 2: Find every zero in $(0,4]$.

• $\sin x=\cos x\ \Rightarrow\ \tan x=1\ \Rightarrow\ x=\dfrac\pi4\approx0.785$ and $x=\dfrac{5\pi}{4}\approx3.927$;
• $e^{x}=2\ \Rightarrow\ x=\log_e 2\approx0.693$;
• $(x-1)^3=0\ \Rightarrow\ x=1$;
• $(x-2)^5=0\ \Rightarrow\ x=2$.

Step 3: Check the multiplicities. Every factor here has odd multiplicity — the two trigonometric and exponential zeros are simple, and the powers $3$ and $5$ are odd. So $f'$ changes sign at all five points.

Step 4: Fix the sign on one interval, then alternate. Take $x=0.5$, which is left of every zero:

$(\sin 0.5-\cos 0.5)<0,\quad \left(e^{0.5}-2\right)<0,\quad (0.5-1)^3<0,\quad (0.5-2)^5<0,$

four negatives, so $f'(0.5)>0$. With five sign changes the pattern across the five points is

$+\;\big|\;-\;\big|\;+\;\big|\;-\;\big|\;+\;\big|\;-$

Step 5: Count the $+\to-$ transitions. They occur at $x=\log_e 2$, $x=1$ and $x=\dfrac{5\pi}{4}$ — three of them. (The $-\to+$ transitions at $\dfrac\pi4$ and $2$ are minima.)

Answer: $\boxed{3}$ points of local maximum.

Step 1: Pull the $x$ out of the integral. The factor $e^{x-t}=e^x e^{-t}$, so

Step 1: Check continuity. Both pieces give $0$ at $x=0$: the left limit is $0$, the right limit is $-3\sin 0=0$, and $f(0)=0$. So $f$ is continuous there.

Step 2: Left-hand derivative.

$f'(0^-)=\lim_{h\to0}\frac{f(-h)-f(0)}{-h}=\lim_{h\to0}\frac{-h^3+h^2-10h}{-h}=10.$

Step 3: Right-hand derivative.

$f'(0^+)=\lim_{h\to0}\frac{f(h)-f(0)}{h}=\lim_{h\to0}\frac{-3\sin h}{h}=-3.$

Step 4: Classify. The two are unequal, so $f$ is not differentiable at $0$ — but that does not matter. What matters is the sign: $f'(0^-)>0$ means $f$ is rising into $0$, and $f'(0^+)<0$ means it falls away after. The value at $0$ beats both sides.

Answer: $x=0$ is a point of local maximum, even though $f'(0)$ does not exist.

Step 1: Work with the inside. $\log_{10}$ is strictly increasing, so $f$ is largest exactly where $g(x)=4x^3-12x^2+11x-3$ is largest.

Step 2: Factorise $g$. Since $g(1)=4-12+11-3=0$, $(x-1)$ divides it:

$g(x)=(x-1)\left(4x^2-8x+3\right)=(x-1)(2x-1)(2x-3).$

Step 3: Check for critical points inside $[2,3]$. $g'(x)=12x^2-24x+11$, whose discriminant is $576-528=48>0$, with roots $x=1\pm\dfrac{1}{\sqrt3}\approx0.42,\ 1.58$ — both outside $[2,3]$. So $g$ has no critical point in the interval and is monotonic there.

Step 4: Evaluate at the end points.

$g(2)=(1)(3)(1)=3,\qquad g(3)=(2)(5)(3)=30.$

Step 5: Conclude. The maximum of $g$ on $[2,3]$ is $30$, at $x=3$, so

$f_{\max}=\log_{10}30=\log_{10}(3\times10)=\boxed{1+\log_{10}3}$

Step 1: Simplify with double angles.

$16\sin\frac x2\cos^3\frac x2=8\left(2\sin\frac x2\cos\frac x2\right)\cos^2\frac x2=8\sin x\cdot\frac{1+\cos x}{2}=4\sin x\,(1+\cos x).$

Step 2: Expand and differentiate. Writing $E=4\sin x+2\sin 2x$,

$\frac{dE}{dx}=4\cos x+4\cos 2x=8\cos\frac{3x}{2}\cos\frac x2.$

Step 3: Critical points on $[0,\pi]$. $\cos\dfrac{3x}{2}=0$ gives $x=\dfrac\pi3$; $\cos\dfrac x2=0$ gives $x=\pi$.

Step 4: Evaluate at critical points and end points.

$E(0)=0,\qquad E(\pi)=0,\qquad E\!\left(\frac\pi3\right)=4\cdot\frac{\sqrt3}{2}\cdot\frac32=3\sqrt3.$

Step 5: Pick the largest.

$\boxed{3\sqrt3}$

Step 1: Find the turning points. With $f(x)=x^3-3x+a$,

$f'(x)=3x^2-3=3(x-1)(x+1),$

so $\alpha=-1$ (local maximum) and $\beta=1$ (local minimum).

Step 2: Evaluate the turning values.

$f(-1)=-1+3+a=a+2,\qquad f(1)=1-3+a=a-2.$

Step 3: Apply the condition. Three distinct real roots need $f(\alpha)f(\beta)<0$:

$(a+2)(a-2)<0.$

Step 4: Solve. A product of two brackets is negative when they have opposite signs, giving

$\boxed{a\in(-2,2)}$

Check. At $a=0$ the equation $x^3-3x=0$ has roots $0,\pm\sqrt3$ — three distinct. At $a=2$ it becomes $(x-1)^2(x+2)=0$, only two distinct. At $a=3$ there is one real root. The boundary is exactly right.

Step 1: Set up the geometry. The cone has base radius $R=30\tan 30^\circ=\dfrac{30}{\sqrt3}=10\sqrt3$.

Step 2: Relate the cylinder's height to its radius. By similar triangles, a cylinder of radius $r$ has height

$y=30\left(1-\frac{r}{10\sqrt3}\right).$

Step 3: Write the volume as one variable.

$V=\pi r^2y=30\pi r^2-\sqrt3\,\pi r^3.$

Step 4: Differentiate and solve.

$\frac{dV}{dr}=60\pi r-3\sqrt3\pi r^2=3\pi r\left(20-\sqrt3\,r\right)=0\ \Longrightarrow\ r=\frac{20}{\sqrt3}.$

(This is less than $R=10\sqrt3\approx17.3$, so it is admissible.)

Step 5: Confirm it is a maximum and evaluate. $\dfrac{d^2V}{dr^2}=60\pi-6\sqrt3\pi r$, which at $r=\dfrac{20}{\sqrt3}$ is $60\pi-120\pi<0$. The height is $y=30\left(1-\dfrac23\right)=10$, so

$V=\pi\left(\frac{20}{\sqrt3}\right)^2(10)=\pi\cdot\frac{400}{3}\cdot10=\boxed{\frac{4000\pi}{3}\ \text{cm}^3}$

Check. A numerical sweep over $r$ peaks at $4188.79$, and $\dfrac{4000\pi}{3}=4188.79$. Agrees.