Application of Derivatives

y (x) = ∫? (2t² - 15t + 10)dt
dy/dx = 2x² - 15x + 10.
For tangent at (a, b), slope is m = dx/dy = 1 / (dy/dx) = 1 / (2a² - 15a + 10).
Given slope is -1/3.
2a² - 15a + 10 = -3
2a² - 15a + 13 = 0 (The provided solution has 2a²-15a+7=0, suggesting a different problem or a typo)
Following the image: 2a² - 15a + 7 = 0
(2a - 1) (a - 7) = 0
a = 1/2 or a = 7.
a = 1/2 Rejected as a > 1. So a = 7.
b = ∫? (2t² - 15t + 10)dt = [2t³/3 - 15t²/2 + 10t] from 0 to 7.
6b = [4t³ - 45t² + 60t] from 0 to 7 = 4 (7)³ - 45 (7)² + 60 (7) = 1372 - 2205 + 420 = -413.
|a + 6b| = |7 - 413| = |-406| = 406.

f' (c) = 1 + lnc = e/ (e-1)
lnc = e/ (e-1) - 1 = (e - (e-1)/ (e-1) = 1/ (e-1)
c = e^ (1/ (e-1)

$CD=\sqrt[]{{\left(10+x^2\right)}^2-{\left(10-x^2\right)}^2}=2\sqrt[]{10}\left|x\right|$

Area 

$=\frac{1}{2}*CD*AB=\frac{1}{2}*2\sqrt[]{10}\left|x\right|\left(20-2x^2\right)$

$\Rightarrow 10-x^2=2x$

3x² = 10

 x = k

3k² = 10

By truth table

So F₁ (A, B, C) is not a tautology

Now again by truth table

So      F₂ (A, B) be a tautology.

From option let it be isosceles where AB = AC then

$x=\sqrt[]{r^2-{\left(h-r\right)}^2}$            

=  $\sqrt[]{r^2-h^2-r^2+2rh}$

$x=\sqrt[]{2hr-h^2}........\left(i\right)$

 Now ar $\left(\Delta ABC\right)=\Delta =\frac{1}{2}BC*AL$

$\Delta =\frac{1}{2}*2\sqrt[]{2hr-a^2}*h$        

then  $x=\sqrt[]{2*\frac{3r}{2}*r-\frac{9r^2}{4}}=\frac{\sqrt[]{3}}{2}rfrom\left(i\right)$

$\Rightarrow BC=\sqrt[]{3}r$    

So $AB=\sqrt[]{h^2+x^2}=\sqrt[]{\frac{9r^2}{4}+\frac{3r^2}{4}}=\sqrt[]{3}r$ .

Hence $\Delta$ be equilateral having each side of length $\sqrt[]{3}r.$

Given f(X) = ${\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{\left(1+t\right)}dt............\left(i\right)}$  

So  $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^{1/x}\frac{\lambda lo{g}_{e}t}{1+t}dt}............\left(ii\right)$

put $t=\frac{1}{z}thendt=-\frac{1}{z^2}dz$

$f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\frac{lo{g}_{e}t}{t\left(1+t\right)}}dt............\left(iii\right)$    

(i) + (iii), f(x) + $f\left(\frac{1}{x}\right)={\displaystyle {\int }_1^x\left(\frac{lo{g}_{e}t}{1+t}+\frac{lo{g}_{e}t}{t\left(1+t\right)}\right)dt}$

$={\left[\frac{{\left(lo{g}_{e}t\right)}^2}{2}\right]}_1^x=\frac{{\left(lo{g}_{x}x\right)}^2}{2}$  

Hence f(e) + $f\left(\frac{1}{e}\right)=\frac{1}{2}$

$\underset{x\to a}{lim}\frac{xf\left(a\right)-af\left(x\right)}{x-a}\left(\frac{0}{0}\right)$

By L'hospital Rule

$=\underset{x\to a}{lim}\frac{f\left(a\right)-af\text{'}\left(x\right)}{1}=f\left(a\right)-af\text{'}\left(a\right)$

$=4-2a$

Now equation of line OA be

Given f(x) =   _{${\displaystyle {\int }_e^x{e}^{t}f\left(t\right)dt+e^x..........\left(i\right)}$}

using Leibniz rule then

f'(x) = e^xf(x) + e^x

_{$\Rightarrow \frac{dy}{dx}=e^{x}y+e^{x}wherey=f\left(x\right)then\frac{dy}{dx}=f\text{'}\left(x\right)$}            

P = -e^x, Q = e^x

Solution be y. (I.F.) =  ${\displaystyle \int Q\left(I.F.\right)dx+c}$

I. f. =  $e^{{\displaystyle \int -e^{x}dx}=e^{-e^x}}$

$\Rightarrow y.\left(e^{-e^x}\right)={\displaystyle \int e^x.e^{-e^x}dx+c}$   

$y.e^{-e^x}=-{\displaystyle \int dt+c=-t+c=-e^{-e^x}+c..........\left(ii\right)}$

Put x = 0 , in (i) f (0) = 1

$From\left(ii\right),\frac{1}{e}=-\frac{1}{e}+cgivenc=\frac{2}{e}$   

Hence f(x) = 2. $e^{\left(e^x-1\right)}-1$

Given curves $\frac{x^2}{9}+\frac{y^4}{4}=1.........\left(i\right)$

&     $x^2+y^2=\frac{31}{4}...........\left(ii\right)$      

Equation of any tangent to (i) be y = mx +  $\sqrt[]{9m^2+4}............\left(iii\right)$

For common tangent (iii) also should be tangent to (ii) so by condition of common tangency

$9m^2+4=\frac{31}{4}\left(1+m^2\right)$

OR 36m² + 16 = 31 + 31m²

=>m² = 3