Maths

$\underset{x\to 0}{lim}\frac{{\displaystyle \underset{0}{\overset{x^2}{\int }}\left(sin\sqrt[]{t}\right)dt}}{x^3}$     $\left(\frac{0}{0}\right)$  applying L' Hospital rule

$=\underset{x\to 0}{lim}\frac{sin\left(\sqrt[]{x^2}\right).2x}{3x^2}$

$=\frac{2}{3},forx>0$

p + q = 2

 p⁴ + q⁴ = 272

${\left(p^2+q^2\right)}^2-2p^2{q}^2=272$          

  ${\left[{\left(p+q\right)}^2-2pq\right]}^2-2p^2{q}^2=272$          

Let pq = t Þ (4 – 2t)² – 2t² = 272

2t² – 16 t – 256 = 0

->t = pq = 16

Required equation x² – 2x + 16 = 0

A  (A -> B) -> B is a tautology

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{8-sin2x}}=asi{n}^{-1}\left(\frac{sinx+cosx}{b}\right)+c}$

${\displaystyle \int \frac{\left(cosx-sinx\right)dx}{\sqrt[]{9-{\left(sinx+cosx\right)}^2}}=}$  

Put (sin x + cos x) = t Þ (cos x – sin x) dx = dt

$={\displaystyle \int \frac{dt}{\sqrt[]{3^2-t^2}}=si{n}^{-1}\frac{t}{3}+c=si{n}^{-1}\left(\frac{sinx+cosx}{3}\right)+c}$           

= a = 1, b = 3

$\therefore \left(a,b\right)\equiv \left(1,3\right)$       

According to question,

$\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)=\frac{1}{4}$

$\frac{1}{a}+\frac{1}{b}=\frac{1}{2}........(i)$           

Equation of required line is $\frac{x}{a}+\frac{y}{b}=1$  

Obviously B (2, 2) satisfying condition (i)

$\Delta -\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 2\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=\left|\begin{array}{ccc}\hfill 3\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 0\hfill & \hfill -8\hfill & \hfill 0\hfill \\ \hfill 1\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|,\left[R_2-R_1-2R_3\right]$

= -8 (-3 + k)

For inconsistent $\Delta =0\Rightarrow k=3$  

${\Delta }_x=\left|\begin{array}{ccc}\hfill 10\hfill & \hfill -2\hfill & \hfill -k\hfill \\ \hfill 6\hfill & \hfill -4\hfill & \hfill -2\hfill \\ \hfill 5m\hfill & \hfill 2\hfill & \hfill -1\hfill \end{array}\right|=32-40m\ne 0\Rightarrow m\ne \frac{4}{5}$              

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

$\frac{dp}{dt}=0.5p-450andP\left(0\right)=850$        

$\Rightarrow \frac{dp}{P-900}=0.5dt$

${\displaystyle \underset{850}{\overset{0}{\int }}\frac{dp}{P-900}={\displaystyle \underset{0}{\overset{T}{\int }}0.5dt}}$           

  $\Rightarrow ln\left(P-900\right){|}_{805}^0=0.5T$          

  $\frac{T}{2}=ln\left|\frac{900}{50}\right|=ln18$

T = 2 ln 18

$f\left(x\right)=\frac{4x^3-3x^2}{6}-2sinx+\left(2x-1\right)cosx$      

$f\text{'}\left(x\right)=2x^2-x-2cosx+2cosx-\left(2x-1\right)sinx$     

= $\left(2x-1\right)\left(x-sinx\right)\ge 0forx\ge 0$  

  $f\text{'}\left(x\right)\ge 0\forall x\ge 1/2$          

$\therefore f\left(x\right)$ is increasing in $\left[\frac{1}{2},\infty \right)$

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

k = at        ……… (ii)

From (i) & (ii) $\frac{2h}{a}=1+\frac{k^2}{a^2}$  

$\therefore$ required locus of Q is y² = 2a (x – a/2)

Equation of directrix x – a/2 = -a/2-> x = 0