Application of Derivatives

$f\left(x\right)=a{x}^3+b{x}^2+cx+d$   

  $f\text{'}\left(x\right)=3a{x}^2+2bx+c$              

$f\text{'}\text{'}\left(x\right)=6ax+2b$

$f\text{'}\text{'}\left(-1\right)=0$       

-6a + 2b = 0

=> b = 3a

$f\text{'}\left(1\right)=0$      

 -5 + d = -10 . (i)

 f (-1) = 6

11a + d = 6 . (ii)

$a=1,d=-5,b=3,c=-9$

$f\left(3\right)=27+27-27-5=22$

f (x) = x + a sin x

$wherea={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosyf\left(y\right)dy}$                

$a={\displaystyle \underset{0}{\overset{\frac{\pi }{2}}{\int }}cosy\left(y+asiny\right)dy}$                

-> a = p - 2

f (0) = 0, f (1) = 1, f (2) = 2

h (x) = f (x) – x has three roots

$\Rightarrow h\text{'}\left(x\right)=f\text{'}\left(x\right)-1$ , has at least two roots

$\Rightarrow h\text{'}\text{'}\left(x\right)=f\text{'}\text{'}\left(x\right)$ has at least one root.

Required area

$A=2{\displaystyle {\int }_{}^{}\left(\left(sinx+cosx\right)-\left(cosx-sinx\right)\right)dx}$

$=2{\displaystyle {\int }_0^{\pi /4}sinxdx}$

$=2\sqrt[]{2}\left(\sqrt[]{2}-1\right)$    

$b^2{x}^2+a^2{y}^2=a^2{b}^2$

$\left(b^2{x}^2+a^2\left(ab-x^2\right)\right)=a^2{b}^2$

$x^2=\frac{b{a}^2\left(b-a\right)}{b^2-a^2},y^2=\frac{a{b}^2}{a+b}$

points of intersection

$\left(a\sqrt[]{\frac{b}{a+b}},b\sqrt[]{\frac{a}{a+b}}\right)$

$tan\theta =\left|\frac{m_1-m_2}{1+m_1.m_2}\right|$

$=\left|\frac{a-b}{\sqrt[]{ab}}\right|$

$f\left(x\right)=x^2+ax+1\Rightarrow f\text{'}\left(x\right)=2x+a$

for increasing   $f\text{'}\left(x\right)\ge 0$

$\therefore 2x+a\ge 0\Rightarrow a\ge -2x\Rightarrow a\ge -2*2\Rightarrow a\ge -4\therefore R=-4$

And for decreasing   $f\text{'}\left(x\right)\le 0\Rightarrow a\le -2x\Rightarrow a\le -2*1\therefore S=-2$

|R – S| = 2

Applying Leibniz theorem,  

$\sqrt[]{1-{\left(f\text{'}\left(x\right)\right)}^2}=f\left(x\right)\Rightarrow \frac{f\text{'}\left(x\right)}{\sqrt[]{1-{\left(f\left(x\right)\right)}^2}}=1$ on integrating both sides, we get

$f\left(x\right)=sinx+C$ put x = 0 and f (0) = 0 we get C = 0

$Now\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x}{\int }}f\left(t\right)dt}}{x^2},\left(\frac{0}{0}\right)$ by L' Hospital rule $\underset{x\to 0}{lim}\frac{f\left(x\right)}{2x}=\underset{x\to 0}{lim}\frac{sinx}{2x}=\frac{1}{2}$

Breadth = b – 2x

& height = x

Let volume V = $\left(a-2x\right)\left(b-2x\right)x$  

For minimum volume $\frac{dv}{dx}=0$  

$\left(a-2x\right)\left(b-2x\right)-2\left(a-2x\right)x-2\left(b-2x\right)x=0$              

Since x =  $\left\{\left(a+b\right)+\sqrt[]{a^2+b^2-ab}\right\}/6$ not possible because maxima occurs

Let square is made with piece of length x metre & hexagon with piece of length y metre

x + y = 20 .(i)

$a=\frac{x}{4}\&b=\frac{y}{6}$       

Now let A = area of square + area of hexagon

$A=\frac{x^2}{16}+6*\frac{\sqrt[]{3}}{4}*\frac{y^2}{36}=\frac{x^2}{16}+\frac{\sqrt[]{3}{y}^2}{24}=\frac{x^2}{16}+\frac{\sqrt[]{3}}{24}{\left(20-x\right)}^{2}from\left(i\right)$ ,

for minimum area

$x=\frac{80\sqrt[]{3}}{6+4\sqrt[]{3}}=\frac{80\sqrt[]{3}}{2\sqrt[]{3}\left(\sqrt[]{3}+2\right)}=\frac{40}{2+\sqrt[]{3}}$

$x=40\left(2-\sqrt[]{3}\right)$

=> side of hexagon = $\frac{y}{6}=\frac{20\sqrt[]{3}\left(2-\sqrt[]{3}\right)}{6}=\frac{20\sqrt[]{3}}{6\left(2+\sqrt[]{3}\right)}=\frac{10\sqrt[]{3}}{3\left(2+\sqrt[]{3}\right)}=\frac{10}{2\sqrt[]{3}+3}$

Side of square = a

Radius of circle = r

Given : 4a + 2pr = 36

S = a² + pr²

^{$=\left(\frac{{\pi }^2}{4}+\pi \right)r^2-9\pi r+81$}

^{$\frac{ds}{dr}=0\Rightarrow r=\frac{18}{\pi +4}$}