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${\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2xdx}}-{\displaystyle \underset{0}{\overset{2}{\int }}\sqrt[]{2x-x^{2}dx}}={\displaystyle \underset{0}{\overset{1}{\int }}dy}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-y^2}dy-{\displaystyle \underset{0}{\overset{2}{\int }}\frac{y^2}{2}dy}+{\displaystyle \underset{1}{\overset{2}{\int }}2dy+I}}$

$\Rightarrow \frac{8}{3}-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}=1-\frac{8}{6}+2+I$

$I=1-{\displaystyle \underset{0}{\overset{1}{\int }}\sqrt[]{1-t^2}dt}$

$\frac{sin{\theta }_C}{V_B}=\frac{sin90°}{V_A}\Rightarrow sin{\theta }_C=\frac{V_B}{V_A}=\frac{1.5\times 10^{10}}{2.0\times 10^{10}}=\frac{3}{4}$

According to question, we can write

  $\theta >\theta =si{n}^{-1}\left(\frac{3}{4}\right)$

At maximum height velocity (v) = 0

So, momentum = mv = m * 0 = 0

$f\left(x\right)=x+x{\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt}-{\displaystyle \underset{0}{\overset{1}{\int }}{t}_{0}f\left(t\right)dt}$

 Let $1+{\displaystyle \underset{0}{\overset{1}{\int }}f}\left(t\right)dt=\alpha$  

${\displaystyle \underset{0}{\overset{1}{\int }}tf\left(t\right)dt=\beta }$  

So, f(x) = ax - b

Now,  $\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t\right)dt+1}$  

$\alpha ={\displaystyle \underset{0}{\overset{1}{\int }}\left(at-\beta \right)dt+1}$  

$\beta ={\displaystyle \underset{0}{\overset{1}{\int }}t\cdot f\left(t\right)dt}$  

$\beta =\frac{4}{13},\alpha =\frac{18}{13}$  

f(x) = ax – b

$=\frac{18x-4}{13}$

option (D) satisfies

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$f\text{'}\left(x\right)=\frac{n_1\cdot f\left(x\right)}{x-3}+n_2\cdot \frac{f\left(x\right)}{x-5}$

$=\frac{f\left(x\right)\cdot \left(n_1+n_2\right)}{\left(x-3\right)\left(x-5\right)}\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_2}\right)$

$f\text{'}\left(x\right)={\left(x-3\right)}^{n_1-1}\cdot {\left(x-5\right)}^{n_2-1}\cdot \left(n_1+n_2\right)\left(x-\frac{\left(5n_1+3n_2\right)}{n_1+n_{2l}}\right)$  

option (C) is incorrect, there will be minima.

.B.D of person w.r.t. elevator

mg – N = ma

N = mg – ma

N decreases so, person experiences weight loss

According to question, we can write

    $F_{CA}=F_{CB}=\frac{KqQ}{x^2+{\left(\frac{d}{2}\right)}^2}\Rightarrow F=2F_{CA}cos\theta =\frac{2KqQx}{{\left[x^2+{\left(\frac{d}{2}\right)}^2\right]}^{\frac{3}{2}}}$                    

For maxima of force   $\frac{dF}{dx}=0,so$

$x=\frac{d}{2\sqrt[]{2}}$

$x^4-2x^3+2x-1={\left(x-1\right)}^2\left(x^2-1\right)$

$sin\pi x=sin(\pi \left(1-x\right)$

$=-sin\left(sin\pi \left(x-1\right)\right)$

$\underset{x\to 1}{lim}\frac{\left(x^2-1\right)\cdot si{n}^2\pi x}{\left(x^2-1\right){\left(x-1\right)}^2}=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(x-1\right)}^2}$

$=\underset{x\to 1}{lim}\frac{si{n}^2\left(\pi \left(x-1\right)\right)}{{\left(\pi \left(x-1\right)\right)}^2}\cdot {\pi }^2$

=2

$P=\frac{\alpha }{\beta }lo{g}_e\left(\frac{kt}{\beta x}\right)$

$\frac{kt}{\beta x}$ = Dimensionless

$\beta =\frac{kt}{x}=\frac{\left[M{L}^2{T}^{-2}{k}^{-1}\right]\left[k\right]}{\left[L\right]}$

$\frac{\alpha }{\beta }=$ dimensionless

a = dimensionless of b

a = MLT^-2

According to question, we can write

$d=d_0\left(1+\alpha \Delta T\right)\Rightarrow 6.241=6.230\left(1+1.4\times 10^{-5}\times \Delta T\right)$

$\Rightarrow T=126.18+27=153.18°C$  

This refers to the highest limit of pressure which the body can tolerate in order to regain its original shape after removing the pressure. If the pressure exceeds the elastic limit, the body will never be able to regain its original shape agian. example: rubber stretching.

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$  

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$  

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$  

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$  

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$  

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$  

$\frac{25S}{36}=1+\frac{3/5}{1}$  

$S=\frac{288}{125}$

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According to question, we can write

10 = $\frac{1}{2}a{t}^2...............\left(1\right)and$

$10+x=\frac{1}{2}a{\left(2t\right)}^2................\left(2\right)$

$\Rightarrow X=\frac{1}{2}A\left[{\left(2t\right)}^2-t^2\right]=3\left(\frac{1}{2}a{t}^2\right)=30m$

Kindly go through the solution

B = (I – adjA)⁵

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This is the case of fluids or gases. In such scenarios, the material does not have any limit to its flexibility and can be deformed to any extent. Such materials do not possess any stiffness and cannot regain their original form as they show no resistance towards any kind of pressure.