Q.It is given that at $x = 1$, the function $x^4 - 62x^2 + ax + 9$ attains its maximum value, on the interval $[0, 2]$. Find the value of $a$.

Concept and Intuition

On $[0,1]$, $u=x^{20}$ also ranges over $[0,1]$ monotonically, so the problem reduces to finding the extrema of $g(u)=u^2-u$ on $[0,1]$. Since $g$ is an upward parabola, its vertex is a MINIMUM, and the maximum must be sought at the endpoints.

Step-by-Step Solution

1. Let $u=x^{20}$; as $x$ ranges over $[0,1]$, so does $u$ (monotonically increasing). $f=x^{40}-x^{20}=u^2-u$.
2. $g(u)=u^2-u$, $g'(u)=2u-1=0\Rightarrow u=\tfrac12$; $g''(u)=2>0$, so $u=1/2$ gives a local MINIMUM: $g(1/2)=1/4-1/2=-1/4$.
3. Check endpoints: $g(0)=0$, $g(1)=1-1=0$.
4. So on $[0,1]$, $g$ ranges between $-1/4$ (min, at $u=1/2$) and $0$ (max, at both endpoints $u=0,1$, i.e. $x=0,1$). …

Concept and Intuition

For $y=x(\log x)^2$ (domain $x>0$), differentiate using the product rule, then classify critical points via the sign change of $y'$ around them.

Step-by-Step Solution

1. $y'=(\log x)^2+x\cdot 2(\log x)\cdot\dfrac{1}{x}=(\log x)^2+2\log x=\log x(\log x+2)$.
2. Setting $y'=0$: $\log x=0\Rightarrow x=1$, or $\log x=-2\Rightarrow x=e^{-2}$.
3. $y'$ is a quadratic in $t=\log x$ opening upward: $t(t+2)$. It is negative for $t\in(-2,0)$ and positive outside.
4. So for $x<e^{-2}$ ($t<-2$), $y'>0$ (increasing); for $e^{-2}<x<1$ ($-2<t<0$), $y'<0$ (decreasing); for $x>1$, $y'>0$ (increasing again). …

Concept and Intuition

At an interior maximum of a differentiable function, the derivative must vanish. This turns the geometric condition ("attains maximum at $x=\pi/3$") into a simple algebraic equation.

Step-by-Step Solution

1. $f(x) = a\sin x + \dfrac13\sin3x$.
2. $f'(x) = a\cos x + \cos3x$.
3. Since $f$ has a maximum at $x=\pi/3$: $f'(\pi/3)=0$.
4. $\cos(\pi/3)=\dfrac12$, $\cos(3\cdot\pi/3)=\cos\pi=-1$.
5. $a\left(\dfrac12\right) + (-1) = 0 \Rightarrow \dfrac{a}{2}=1\Rightarrow a=2$. …

Concept and Intuition

The function $y=x(\log x)^2$ has domain $x>0$, dips down near $x=0^+$ (approaching $0$), rises to a local max, dips to a local min at $x=1$ (where $y=0$), then increases without bound as $x\to\infty$. So the question is really asking for the local maximum value, which is the meaningful turning point before the function heads back down toward the min at $x=1$.

Step-by-Step Solution

1. Differentiate using the product rule: $y'=(\log x)^2 + x\cdot 2\log x\cdot\dfrac1x = (\log x)^2+2\log x = \log x(\log x+2)$.
2. Set $y'=0$: $\log x=0 \Rightarrow x=1$, or $\log x=-2 \Rightarrow x=e^{-2}$.
3. Differentiate again: $y'' = \dfrac{2\log x}{x}+\dfrac{2}{x} = \dfrac{2}{x}(\log x+1)$.
4. At $x=1$: $y''=2(0+1)=2>0$, so $x=1$ is a local minimum ($y=1\cdot 0=0$). …

Concept and Intuition

At an extremum (local maximum or minimum), the derivative of the function vanishes (assuming the function is differentiable there, which it is away from $x=0$). Given that extrema occur at two specified points, we simply impose $y'=0$ at each point — this produces a system of two linear equations in the two unknown constants $a$ and $b$.

Step-by-Step Solution

1. $y = a\log|x| + bx^2 + x$.
2. $y' = \dfrac{a}{x} + 2bx + 1$.
3. Extremum at $x=-1$: $y'(-1) = \dfrac{a}{-1} + 2b(-1) + 1 = -a-2b+1 = 0 \Rightarrow a+2b=1$. — (i)
4. Extremum at $x=2$: $y'(2) = \dfrac{a}{2} + 4b + 1 = 0$. Multiply by 2: $a + 8b + 2 = 0 \Rightarrow a+8b=-2$. — (ii)
5. Subtract (i) from (ii): $(a+8b)-(a+2b) = -2-1 \Rightarrow 6b = -3 \Rightarrow b=-\dfrac12$. …

Concept and Intuition

First we must find the actual domain $A$ from the given inequality — this restricts where $f$ is even defined. Then, instead of blindly checking critical points of $f$, we check whether those critical points actually fall inside the restricted domain; if $f$ has no critical points inside $[4,5]$, the maximum must occur at an endpoint.

Step-by-Step Solution

1. Solve for the domain $A$:

So $A=[4,5]$.

2. Differentiate $f(x)=2x^3-15x^2+36x-48$:

3. The critical points are $x=2,3$ — both outside $[4,5]$. For $x\in[4,5]$, both $(x-2)>0$ and $(x-3)>0$, so $f'(x)=6(x-2)(x-3)>0$ throughout the interval: $f$ is strictly increasing on $[4,5]$.

4. Since $f$ is increasing on a closed interval with no interior critical points, its maximum occurs at the right endpoint $x=5$: …

Concept and Intuition

$f(x)=3x+\dfrac{12}{x}$ is unbounded overall on $\mathbb R-\{0\}$ (it goes to $\pm\infty$), so "its maximum value" refers to the local maximum produced by the turning points of the curve. We locate these via $f'(x)=0$ and classify with the second derivative.

Step-by-Step Solution

1. Differentiate: $f'(x)=3-\dfrac{12}{x^2}$.
2. Set $f'(x)=0$: $3=\dfrac{12}{x^2}\Rightarrow x^2=4\Rightarrow x=\pm2$.
3. Second derivative: $f''(x)=\dfrac{24}{x^3}$.
   • At $x=2$: $f''(2)=3>0$ — local minimum, value $f(2)=6+6=12$.
   • At $x=-2$: $f''(-2)=-3<0$ — local maximum, value $f(-2)=-6-6=-12$.
4. So $M=-12$ (the local maximum value). …

Concept and Intuition

$f(x)=xe^{-x}$ is the classic "rises then falls" shape: for small $x$ the linear

factor $x$ dominates and $f$ grows, but as $x$ grows the exponential decay

$e^{-x}$ eventually wins and drags $f$ back to $0$. The turning point between

these two regimes is the unique critical point, found by the product rule.

Step-by-Step Solution

1. $f(x)=xe^{-x}$. By the product rule: $f'(x)=e^{-x}+x(-e^{-x})=e^{-x}(1-x)$.
2. Set $f'(x)=0$: since $e^{-x}\neq0$ for all real $x$, we need $1-x=0\Rightarrow x=1$.
3. Second-derivative check: $f''(x)=\dfrac{d}{dx}\left[e^{-x}(1-x)\right]=-e^{-x}(1-x)-e^{-x}=e^{-x}(x-2)$.
4. At $x=1$: $f''(1)=e^{-1}(1-2)=-e^{-1}<0$, confirming a local maximum.
5. So $k=1$. …

Concept and Intuition

When $f$ is expressed purely in terms of $\sin x$, it is often easier to substitute $s=\sin x$ and minimise with respect to $s$ on its natural domain, then convert back to $x$ (or $\cos x$) at the end using $\cos^2x=1-\sin^2x$.

Step-by-Step Solution

1. Let $s=\sin x\in(0,1)$ for $x\in\left(0,\tfrac{\pi}{2}\right)$. Then $f(s)=\dfrac{4}{s}+\dfrac{1}{1-s}$.
2. $\dfrac{df}{ds}=-\dfrac{4}{s^2}+\dfrac{1}{(1-s)^2}$. Setting this to $0$: $\dfrac{1}{(1-s)^2}=\dfrac{4}{s^2}\Rightarrow s^2=4(1-s)^2$.
3. Taking positive square roots (both sides positive on $(0,1)$): $s=2(1-s)\Rightarrow 3s=2\Rightarrow s=\tfrac23$. (The other sign choice gives $s=2$, outside the domain.)
4. So $\sin k=\tfrac23$, and $\cos k=\sqrt{1-\tfrac49}=\sqrt{\tfrac59}=\dfrac{\sqrt5}{3}$. …

Concept and Intuition

For a displacement function, maxima/minima of position occur where velocity $dS/dt=0$; the sign of the acceleration $d^2S/dt^2$ at that instant distinguishes a maximum (acceleration negative) from a minimum (acceleration positive).

Step-by-Step Solution

1. $\dfrac{dS}{dt}=3t^2-32t+64$.
2. Set $3t^2-32t+64=0$. Discriminant $=32^2-4\cdot3\cdot64=1024-768=256$, $\sqrt{256}=16$.
3. $t=\dfrac{32\pm16}{6}$, giving $t=8$ or $t=\dfrac{8}{3}$.
4. $\dfrac{d^2S}{dt^2}=6t-32$.
5. At $t=8$: $6(8)-32=16>0$ — this is a minimum.
6. At $t=8/3$: $6(8/3)-32=16-32=-16<0$ — this is a maximum. …

Concept and Intuition

Local extrema occur where $f'=0$ and the sign of $f'$ changes; factoring $f'$ into

simple pieces makes both the location AND the type (max/min) of each critical point

easy to read off from a sign chart, without needing the second derivative.

Step-by-Step Solution

1. $f(x)=x-\dfrac{x^2}{2}-xe^{2-x}$. Differentiate the last term via product rule: $\dfrac{d}{dx}\left[xe^{2-x}\right]=e^{2-x}-xe^{2-x}=e^{2-x}(1-x)$.
2. So $f'(x)=1-x-e^{2-x}(1-x)=(1-x)\big(1-e^{2-x}\big)$.
3. Zeros: $x=1$, or $e^{2-x}=1\Rightarrow x=2$.
4. Sign chart: for $x<1$, $(1-x)>0$ and $(1-e^{2-x})<0$ (since $2-x>1\Rightarrow e^{2-x}>e>1$) so $f'<0$. For $1<x<2$, $(1-x)<0$, $(1-e^{2-x})<0$ (since $0<2-x<1 \Rightarrow e^{2-x}>1$) so $f'>0$. For $x>2$, $(1-x)<0$, $(1-e^{2-x})>0$ (since $2-x<0\Rightarrow e^{2-x}<1$) so $f'<0$. …