first find the first derivative of function

$AM\geq GM\implies \dfrac{p+q}{2}\geq \sqrt{pq}$

Let,  $\displaystyle z= x+y= x+\dfrac{1}{x}$
for minimum value of $z$
$\cfrac{dz}{dx}=0\Rightarrow 1-\cfrac{1}{x^2}=0\Rightarrow x=\pm 1$
but given $x>0, \Rightarrow x=1$
Hence minimum value of $z$ is 2.

Given first term of the G.P is $1.$

Let common ratio of $G.P.$ be $r$ & common difference of $A.P.$ be $2r$
$\Rightarrow a _1b _1+a _2b _2+1=1\times2+(1+2r)2r+1$
$=4r^2+2r+3=z$ (say)
Now for minimum value of $z$
$\cfrac{dz}{dr}=0\Rightarrow 8r+2=0\Rightarrow r=-\cfrac{1}{4}$
$\therefore z _{min}=4(\dfrac 1{16})-2(\dfrac 14)+3=\cfrac{11}{4}$

Maximise: $\cfrac{3{ x }^{ 2 }+9x+17}{3{ x }^{ 2 }+9x+7}$

Consider given the function,

$F\left( x \right)=2{{x}^{3}}-21{{x}^{2}}+36x-20$      ……(1)

Differentiate with respect to x,

${{F}^{'}}\left( x \right)=6{{x}^{2}}-42x+36$          ……..(2)

For maxima and minima,

$ F\left( x \right)=0 $

$ 6{{x}^{2}}-42x+36=0 $

$ {{x}^{2}}-7x+6=0 $

$ {{x}^{2}}-6x-x+6=0 $

$ x\left( x-6 \right)-1\left( x-6 \right)=0 $

$ \left( x-6 \right)\left( x-1 \right)=0 $

$ x=1,6 $

Differentiate equation 2^nd with respect to x,

${{F}^{''}}\left( x \right)=12x-42$

At $x=1\Rightarrow {{F}^{''}}\left( x \right)<0$

Hence, F(x) Is maximum.

At $x=6\Rightarrow F\left( x \right)>0$

Hence, function F(x) is minimum.

Hence, this is the answer.

General term can be written as
$T _{n}=\displaystyle \frac{n^{2}}{500+3n^{3}}$
then, $\displaystyle \frac{dT _{n}}{dn}=\displaystyle \frac{n(1000-3n^{3})}{(500+3n^{3})^{2}}$
For max or min of $T _{n}$,
$\displaystyle \frac{dT _{n}}{dn}=0$
$\therefore n=\left ( \displaystyle \frac{1000}{3} \right )^{1/3}=6.933\approx7$

Hence, $T _{7}$ is the largest term. So largest term in the given sequence is $\displaystyle \frac{49}{1529}$

Given $x y = a^2$ and $S = b^2x + c^2y$
$\Rightarrow S = b^2 x + c^2a^2/x$
$\Rightarrow \dfrac{dS}{dx} = b^2 - c^2a^2/x^2$
For maximum or minimum value of $S$
$ \dfrac{dS}{dx} = 0 = b^2 - c^2a^2/x^2 \Rightarrow x =\pm  ac/b$
Now $\dfrac{dS}{dx} = 2 c^2a^2/x^3$
Clearly at $x =  ac/b$,  $\dfrac{dS}{dx} = 2 b^3/ac > 0 $ (Assuming that $ b^3/ac>0$)
Hence minimum value of $S$ is $= b^2(ac/b)+c^2(b/ac)= 2abc$

Let $y = \sin\theta.\sin\phi = \sin\theta.\sin(\dfrac{\pi}{3}-\theta)$
For maximum value of $y$ 
$\dfrac{dy}{dx} = 0 = \cos\theta.\sin(\dfrac{\pi}{3}-\theta) - \sin\theta.\cos(\dfrac{\pi}{3}-\theta) = \sin(2\theta -\dfrac{\pi}{3})$
$\Rightarrow \theta = \dfrac{\pi}{6}$

Let $x$ and $y$ be two numbers 
$\Rightarrow x+y = 8$
Assume $z$ be be sum of their inverse
$z = 1/x+1/y =\dfrac{x+y}{xy} = \dfrac{8}{xy} = \dfrac{8}{x(8-x)}$
For minimum value of $z $
$\dfrac{dz}{dx} = 0 =\dfrac{16(4-x)}{(x(8-x))^2}\Rightarrow x = 4$
Hence minimum value of $z$ is $=1/4+1/4 = \dfrac{1}{2}$ 

${log} _{10}x+{log} _{10} y=$ $log _{10}(xy)\geq2$, 
Thus, $xy\geq100$
 Given the product of two numbers ,addition of two number is smallest when they are equal.
$ x^2\geq100$
Therefore, smallest value of $x+y =20$
Hence, option 'C' is correct.

Let $f'(x)=\lambda x(x^2-1)\Rightarrow f(x)=\lambda\left(\dfrac{x^4}{4}-\dfrac{x^2}{2}\right)+C$
Now $f(0)=f(k)\Rightarrow \dfrac{k^4}{4}-\dfrac{k^2}{2}=0\Rightarrow k=0$ or $\pm \sqrt{2}$
Hence $(1)$.

Let one part be $x$

Let one part be x.
Hence another part will be 64-x.
Let 
$f(x)=x^{3}+(64-x)^{3}$.
$f'(x)$
$=3x^{2}-3(64-x)^{2}$
$=0$
Or 
$x^{2}=(64-x)^{2}$
Or 
$x=64-x$ or $x=-64+x$
Considering equation $x=64-x$, we get 
$x=32$.
Hence another part will also be 32.

Let the two parts be $x$ and $20-x$
Let $y=(20-x)x^3$
$\Rightarrow y=20x^3-x^4$

For maximum or minimum,
$\dfrac{dy}{dx}=0$
$\Rightarrow 60x^2-4x^3=0$
$\Rightarrow 4x^2(15-x)=0$
$\Rightarrow x=0, x=15$

$\dfrac{d^2y}{dx^2}=120x-12x^2$
At $x=15$, $\dfrac{d^2y}{dx^2}<0$
Hence, $y$ has a maximum at $x=15$

So, the two numbers are 15, 5

Let $ z= x+y = x+1/x$,   since $(xy=1)$
$\dfrac{dz}{dx} = 1-1/x^2$
For minimum value of $z$
$\dfrac{dz}{dx} =0= 1-1/x^2\Rightarrow x=1$, $since (x>0)$
Therefore minimum value of $z=2$

Let one number be $x$
Hence the other number will be $(60-x)$.
Let 
$K=x^{3}.(60-x)$
Differentiating $K$ with respect to $x$, we get 
$\dfrac{dK}{dx}$
$=3x^{2}(60-x)-x^{3}=0$
Or 
$x^{2}[180-3x-x]=0$
Or 
$x=0$ and $x=\dfrac{180}{4}=45$.
Now its given that the numbers are positive.
Hence $x=0$ is ruled out.
Thus we get $x=45$.
Hence
$y=15$.
Therefore the numbers are $45,15$.

$xy={ c }^{ 2 }$
$y={ c }^{ 2 }$
Put the value of $y={ c }^{ 2 }$ in $ ax+by$
$f(x)=a{ c }^{ 2\quad \quad  }y+by=0$
${ f }^{ ' }(x)=-a{ c }^{ 2\quad  }{ y }^{ 2\quad  }+b=0$
$-a{ c }^{ 2\quad  }+b{ y }^{ 2\quad  }=0$
$b{ y }^{ 2\quad  }=a{ c }^{ 2 }$
$y=+,-c\sqrt { (b/a)\quad  } $
${ f }^{ ''\quad  }(x)=2b{ c }^{ 2\quad  }/{ x }^{ 2 }$
$x=c\sqrt { b/a } $
${ f }^{ ''\quad  }(c\sqrt { (b/a } )=2b{ c }^{ 2\quad  }/{ c }^{ 2 }(b/a)=2a>0$
While $x=-c\sqrt { b/a } $will give maxima.
Put $x=c\sqrt { b/a }$ 
$a(c\sqrt { (b/a) } )+b({ c }^{ 2\quad  }\sqrt { a) } /c\sqrt { b } =2c\sqrt { ab } $

$f(x)=x+16y$         (1)
$xy=4 $                         (2)
Substituting $y=\dfrac { 4 }{ x } $ in (1).
$f(x)=x+\dfrac{ 16.4 }{ x } $

x-y=a                (1)
$f(x)=xy=x(x-a)$
${ f }^{ ' }(x)=2x-a $     (2)
2x-a=0
$x=a/2$
Substitute value of $x=a/2$ in (1) to get $y=-a/2$

Let one part be $x$.
Hence another part be $(64-x)$
Thus 
Their cubes will be 
$x^{3}+(64-x)^{3}=y$
Thus 
$y'=3x^{2}-3(64-x)^{2}=0$
Or 
$x^{2}=(64-x)^{2}$
Or 
$x=64-x$ and $x=-64+x$
Hence
$x=32$.
Hence both the parts are 
$32,32$.

$x+y=6$
$Sum =\dfrac {1}{x}+\cfrac {1}{y}$
$\dfrac {d(sum)}{dx}=\dfrac {-1}{x^2}+\dfrac {1}{(6-x)^2}=0$
$x^2=(6-x)^2$
$x=\pm (6-x)$
$x=3$,  $y=3$
$Sum =\dfrac {2}{3}$

$x+y=18$
$Product =xy$
$\dfrac {d(Product)}{dx}=(18-x)-x$
$0=18-2x$
$x=9$
$y=9$
$Product = 81$

(A) use A.M. & G.M.
$\displaystyle \frac {x+y}{2}\geq (xy)^{\frac {1}{2}}$
$\displaystyle (\dfrac {7}{2})\geq (xy)^{\frac {1}{2}}$
$(xy)\leq(\dfrac {7}{2})^2$
(B) $y=l+\sqrt {1-l^2}$

$\displaystyle \frac {dy}{dl}=1-\frac {l}{\sqrt {1-l^2}}$

$\displaystyle \frac {dy}{dl}=0$ when $\displaystyle l=\frac {1}{\sqrt 2}$
So $\displaystyle m=\frac {1}{\sqrt 2}$
$\Rightarrow l=\displaystyle \frac{1}{\sqrt{2}}$

$\Rightarrow l+m=\sqrt{2}$

(C) $s=x^2+(12-x)^2$
$\displaystyle \frac {ds}{dx}=2x-2(12-x)$
$\displaystyle \frac {ds}{dx}=0$ when $x=6$ $y=6$
$s=36+36=72$
(D) $f'(x)=2x-8$
$f'(x)=0$ at $x=y$
$f(y)=1$

If $x+y=k$ then maximum value of $x^{m}y^{n}$ is at $\displaystyle x=\frac{km}{m+n}, y=\frac{km}{m+n}$ where $x,y>0$ and $m,n \ge{1} $
Here $k=28, m=3,n=4$
So,$x=12, y=16$
Hence maximum value is $12^3.16^4$

$2x+y=5$
$\Rightarrow y=5-2x$
$f(x)=x^2+3x(5-2x)+(5-2x)^2$
$f(x)=-x^2-5x+25$
$f'(x)=-2x-5$
For maxima or minima,
$f'(x)=0$
$\Rightarrow x=-\frac{5}{2}$
$f''(x)=-2$
$f''(-\frac{5}{2})=-2<0$
So, f(x) has a maximum at $x=-\frac{5}{2}$
$\displaystyle f(-\frac{5}{2})=\frac{125}{4}$

Let $x=cos{\theta}$ and $y=sin{\theta}$
Then, $f(\theta)= cos{\theta}+sin{\theta}$
$f'(\theta)=-sin{\theta}+cos{\theta}$
For maxima or minima,
$f'(\theta)=0$
$\Rightarrow \theta =\frac{\pi}{4}$
$f''(\theta)=-(cos{\theta}+sin{\theta})$
$\Rightarrow f''(\frac{\pi}{4})<0$
Hence, f has a maximum value at $\theta =\frac{\pi}{4}$
$\displaystyle f(\frac{\pi}{4})=\sqrt{2}$

$xy(y-x)=16$
$xy^2-x^2y=16$
$y^2-xy-\dfrac {16}{x}=0$
$(y-\dfrac {x}{2})^2-\dfrac {x^2}{4}-\dfrac {16}{x}=0$
$y=\dfrac {x}{2}\pm \sqrt{\dfrac {x^2}{4}+\dfrac {16}{x}}$
$y'=\dfrac {1}{2}\pm \dfrac {1}{2}(\dfrac {\dfrac {2x}{4}-\dfrac {16}{x^2}}{\sqrt {\dfrac {x^2}{4}+\dfrac {16}{x}}})$
$-1=\pm (\dfrac {\dfrac {x}{2}-\dfrac {16}{x^2}}{\sqrt {\dfrac {x^2}{4}+\dfrac {16}{x}}})$
$\dfrac {16}{x}=\dfrac {256}{x^4}-\dfrac {16}{x}$
$x^3=8$
$x=2$
& $y=4$

$x+y=100$
$f(x)=x^2(100-x)^3$
$f'(x)=2x(100-x)^3-3(100-x)^2x^2$
$3x=2(100-x)$
$x=40$  $y=60$

$f(x)=x-x^2$
$f'(x)=1-2x$
$x=\dfrac {1}{2}$ when $f'(x)=0$

$f'(x)=6{x}^{2}-15x+12$
$=6({x}^{2}-3x+2)$
$=6(x-1)(x-2)$
$f'(x)< 0$ when $x$ lies between $1$ and $2$
$\therefore$ $1< x< 2$
$\therefore$ (3) is correct

Since, $\displaystyle f\left( x \right) ={ e }^{ ax }+{ e }^{ -ax }$
$\displaystyle \Rightarrow \quad f'\left( x \right) =a\left( { e }^{ ax }-{ e }^{ -ax } \right) $
$\displaystyle f\left( x \right) $ is monotonically increasing, if
$\displaystyle f'\left( x \right) >0$
$\displaystyle \Rightarrow \quad { e }^{ ax }-{ e }^{ -ax }>0$
$\displaystyle \Rightarrow \quad { e }^{ 2ax }>1$
$\displaystyle \Rightarrow \quad x>0$

Let $f(x)=\cfrac { { x }^ { } } { { x }^{ 2 }+100}$

For the largest term, we may calculate the maximum value of $f(x)$

$f'(x)=\cfrac { ( { x }^{ 2 }+100).1-{ x }^{ }.( 2x ) } { { ( { x }^{ 2 }+100 ) }^{ 2 } }=0$

$\Rightarrow \cfrac {100-{ x }^{ 2 } } { { ({ x }^{ 2 }+100) }^{ 2 }}=0$

$\Rightarrow { x }^{ 2 }=100$

$\Rightarrow x=10$ Since $x \neq -10$

To check for maxima or minima, we could evaluate $f''(x)$ at $x=10$

Alternatively, we could just check value of $f(x)$ at any other $x$ and compare its value with at $x=10$

$f(10)=\cfrac { 10 } { { 10 }^{ 2 }+100 }=\cfrac { 1 } { 20 }$

Checking at $x=1$,   $f(1)=\cfrac{ 1 } { { 1 }^{ 2 }+100 }=\cfrac { 1 } { 101 }$

f(10)$>$f(1),     $\therefore x=10$ is point of maxima

$\Rightarrow a _n(max)=\cfrac { 1 } { 20 } $

$\therefore$ Correct option is D

This function has $a = 2$ relative minima at the x-intercepts, $(-5, 0)$ and $(1, 0), b = 1$ relative minima at $(-2, 3)$ and the product of the zeros is $c = (-5)(1) = - 5$. Thus $a + 2b - c = 9$.

Let $y=x^2\ln \dfrac{1}{x}$

$\displaystyle D _{h}=\left { -1, 1 \right }$, as only possible values in the domain of $h(x)$ is $1$ and $-1$
$\therefore  a _{3}=-1$
Now, $ g(x)=a _{0}+a _{1}x+a _{2}x^{2}-x^{3}$
$ {g}'(x)=a _{1}+2a _{2}x-3x^{2}$
$=-3(x-3)(x+3)$
$=-3x^{2}+27$
$\therefore  a _{1}=27, a _{2}=0$
$\therefore a _{1}+a _{2}=27$
Also, $g(-3)> 0$ and $g(3)> 0$
$\Rightarrow  a _{0}> 54$ and $a _{0}< -54$
$\therefore   a _{0}> 54$

Given $xy=1$ and $f(x,y)=x+y$
$\Rightarrow f(x)=x+\dfrac{1}{x}$
$f'(x)=1-\dfrac{1}{x^2}$
For maxima or minima,
$f'(x)=0$
$\Rightarrow x=\pm1$
$f''(x)=\dfrac{2}{x^3}$
$f''(x)>0$ at $x=1$
Hence f(x) has minimum at $x=1$
$f(1)=2$
So, minimum value of $x+y  \ is  \  2$.