Maths Applications of Derivatives

$f\text{'}\left(x\right)=\frac{4x^2-1}{x}$ so f (x) is decreasing in $\left(0,\frac{1}{2}\right)and\left(\frac{1}{2},\infty \right)$ $\Rightarrow a=\frac{1}{2}$  

Tangent at y² = 2x is y = mx + $\frac{1}{2m}$ it is passing through (4, 3) therefore we get m = $\frac{1}{2}or\frac{1}{4}$  

So tangent may be  $y=\frac{1}{2}x+1ory=\frac{1}{4}x+2buty=\frac{1}{2}x+1$  passes through (-2, 0) so rejected.

Equation of normal  $\frac{x}{9}+\frac{y}{36}=1$  

$f\left(x\right)=\{\begin{array}{cc}\hfill x^3-x^2+10x-7,\hfill & \hfill x\le 1\hfill \\ \hfill -2x+lo{g}_2\left(b^2-4\right),\hfill & \hfill x>1\hfill \end{array}$  

If f(x) has maximum value at x = 1 then

$f\left(1\right)\le f\left(1\right)\Rightarrow -2+lo{g}_2\left(b^2-4\right)\le 1-1+10-7$

$\Rightarrow lo{g}_2\left(b^2-4\right)\le 5\Rightarrow 0<b^2-4\le 32$

$\Rightarrow b^2-4>0\Rightarrow b\in \left(-\infty ,-2\right)\cup \left(2,\infty \right)$         …….(i)

$And{b}^2-4\le 32\Rightarrow b\in \left[-6,6\right]$                      …….(ii)

From (i) and (ii) we get  $b\in [-6,-2)\cup (2,6]$  

OP² = x² = y²

y = e^x, y' = e^x,

slope of normal =  $-\frac{1}{e^x}$

$\frac{y}{x}=-\frac{1}{e^x}$    

$-\frac{1}{e^x}=\frac{e^x}{x}\Rightarrow -x=e^{2x}$      

By hit and trial we get  $x=-\frac{2}{5}$

$P\equiv \left(-\frac{2}{5},e^{-2/5}\right)$

$OP=\sqrt[]{\frac{4}{25}+e^{-4/5}}\Rightarrow O{P}^2=\frac{14}{25}=\frac{m}{n}$

$\frac{dy}{dx}=3a{x}^2-2bx$

${\frac{dy}{dx}|}_{x=1}=3a-2b=3$

$a=\frac{2b+3}{3}\in \left[1,2\right]$

$2b+3\in \left[3,6\right]$

$2b\in \left[0,3\right]$

$b\in \left[0,\frac{3}{2}\right]$

Equation of tangent of at $y=\mathrm{l}\mathrm{o}\mathrm{g}?x$  at $\left(e^{\lambda },\lambda \right)$  is $y-\lambda =\frac{1}{e^{\lambda }}\left(x-e^{\lambda }\right)$

It cuts y   axis at $(0,\lambda ,-1)$

Also normal to $y^2=8x$    at $\left(\mathrm{7,4}\right)$    is $y-4=-(x-2)$  

  $y=-x+6\mathrm{}$  Cuts y axis at $\left(\mathrm{0,6}\right)\mathrm{}$  SO ATQ $6=-1+\lambda \Rightarrow \lambda =7$

TSA  of cube $=6a^2$

$\frac{d}{dt}\left(6a^2\right)=4.8;12a\frac{da}{dt}=4.8;a\frac{da}{dt}=0.4;\frac{dv}{dt}=\frac{d}{dt}\left(a^3\right)\Rightarrow 3a\left(a\frac{da}{dt}\right)$

$3*15*0.4=3*15*\frac{04}{10}\Rightarrow 18$

f' (x) = cosx + sinx − k ≤ 0∀x ∈ R
k ≥ √2

c = -5, a = 2, b = 1

f (x) = x? /20 - x? /12 + 5
f' (x) = x? /4 - x³/3 = x³ (x/4 - 1/3)
Local maxima at 0, Local minima at 4/3
f' (x) = x³ - x² = x² (x-1)
x = 1 point of inflection