Application of Derivatives

The points on the curve are (0, 0), (2, 2) and $\left(3,\frac{21}{2}\right)$

$\therefore \frac{dy}{dx}=2x^3-15x^2+36x-19$

$at\left(0,0\right),\frac{dy}{dx}=-19,at\left(2,2\right),\frac{dy}{dx}=9at\left(3,\frac{21}{2}\right),\frac{dy}{dx}=8$

Hence maximum slope at (2, 2) is 9.

y = x³

${\frac{dy}{dx}=3x^2\Rightarrow \frac{dy}{dx}|}_{\left(t,t^3\right)}=3t^2$        

Equation of tangent y – t³ = 3t² (x – t)  

Let again meet the curve at $Q\left(t_1,t_1^3\right)$  

$\Rightarrow t_1^3-t^3=3t^2\left(t_1-t\right)$           

  $t_1^2+t{t}_1+t^2=3t^2\left[?t_1\ne t\right]$          

$t_1^2+t{t}_1-2t^2=0$            

->t₁ = -2t

Required ordinate =    $\frac{2t^3+t_1^3}{3}=\frac{2t^3-8t^3}{3}=-2t^3$

  $\left|\begin{array}{cc}\hfill f\left(X\right)\hfill & \hfill f\text{'}\left(x\right)\hfill \\ \hfill f\text{'}\left(x\right)\hfill & \hfill f\text{'}\text{'}\left(x\right)\hfill \end{array}\right|=0$         

$f\left(x\right)f\text{'}\text{'}\left(x\right)=f\text{'}\left(x\right)f\text{'}\left(x\right)$

$\frac{f\text{'}\text{'}\left(x\right)}{f\text{'}\left(x\right)}=\frac{f\text{'}\left(x\right)}{f\left(x\right)},$ Integrating on both sides

$\frac{f\text{'}\left(x\right)}{f\left(x\right)}=2,$ again integrating on both side

ln f(x) = 2^X + k

f(x) = e^{2x + k}

f(0) = e^k = e^k = 1-> k = 0

$\therefore f\left(x\right)=e^{2x}\left[\therefore e=2.718\right]$           

$\therefore e^2\in \left(6,9\right)$           

Let f(x) = x⁶ + ax⁵ + bx⁴ + ax³ + dx² + ex + f

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$                   Non zero finite

So, d = e = f = 0

f(x)  = x⁶ + ax⁵ + bx⁴ + cx³

$\underset{x\to 0}{lim}\frac{f\left(x\right)}{x^3}=1$          Non zero finite

f'(x) = $6x^5+5a{x}^4+4b{x}^3+3x^2$  

f'(1) = 0

6 + 5a + 4b + 3 = 0

5a + 4b = - 9 .(i)

f'(-1) = 0

-6 + 5a – 4b + 3 = 0 .(ii)

Solving (i) and (ii)

a  -3/5, b = -3/2

$\therefore f\left(x\right)=x^6+\left(\frac{-3}{5}\right)x^5+\left(\frac{-3}{2}\right)x^4+x^3$

$\therefore 5.f\left(2\right)=144$  

  $\frac{x^2}{a}+\frac{y^2}{b}=1,\frac{x^2}{c}-\frac{y^2}{\left(-d\right)}=1$

Ellipse and Hyperbola are orthogonal so these will be confocal.

$\sqrt[]{a-b}=\sqrt[]{c+\left(-d\right)}$            

a – b = c – d

$f\left(x\right)=x^3-a{x}^2+bx-4$

f (1) = f (2)

->1 – a + b – 4 = 8 – 4a + 2b – 4

->3a – b = 7 . (i)

8a = 16 + 3b . (ii)

(i) and (ii) -> (a, b) = (5, 8)

Let equation of normal PQ is

$y=-tx+3t+\frac{3}{2}{t}^3$ and it is passing through

$P\left(3,\frac{3}{2}\right)\therefore t=1$

$\therefore Q\left(\frac{3}{2},3\right)\Rightarrow a=\frac{3}{2}andb=3$     

2 (a + b) = 9

The projection of $\overrightarrow{BA}on\overrightarrow{BC}=cos\angle ABC=7\left|\frac{7^2+3^2-5^2}{2*7*3}\right|=\frac{11}{2}$

$f\left(x\right)=x^3-3x^2-\frac{3f\text{'}\text{'}\left(2\right)}{2}x+f\text{'}\text{'}\left(1\right)$ . (i)

f' (x) = 3x² – 6x -   $\frac{3f\text{'}\text{'}\left(2\right)}{2}$ . (ii)

$f\text{'}\text{'}\left(x\right)=6x-6$ . (iii)

  $\therefore f\text{'}\text{'}\left(2\right)=12-6=6$           

 Hence f (x) is local maxima at x = -1

and f (x) is local minima at x = 3

$\therefore$ from (i) local minimum value at x = 3 is f (3) = -27

$\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -1\hfill & \hfill 4\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -3\hfill \\ \hfill 6\hfill & \hfill 5\hfill & \hfill k\hfill \end{array}\right|=0\Rightarrow k=-5$