Tag: maths

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 $\displaystyle \frac{49}{1529}$

For a local minimum at x=1, the function value at x=1 must be less than or equal to the values in its immediate neighborhood. Evaluating the left limit gives |1-1|+a = a, and the right limit gives 2(1)+3 = 5. Thus, we require a <= 5.

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.