Maths

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

Let mid point of PQ is R (h, k)

$\therefore h=\frac{\alpha +\frac{4{\alpha }^2+\alpha +1}{2}}{2}and$

$k=\frac{4{\alpha }^2+1+\frac{4{\alpha }^2+\alpha +1}{2}}{2}$   

Eliminate a from above these two, we get

2 (3x – y)² + (x – 3y) + 2 = 0

$l={\displaystyle \sum _{r=0}^{n-1}f\left(\frac{5r}{n}\right)\frac{1}{n}}\Rightarrow l={\displaystyle \underset{0}{\overset{1}{\int }}f\left(5x\right)dx=}{\displaystyle \underset{0}{\overset{1}{\int }}\left(5x+1\right)dx=\frac{7}{2}}$

$ta{n}^{-1}\left(\frac{2*\frac{3}{5}}{1-\frac{9}{25}}\right)+si{n}^{-1}\frac{5}{13}\Rightarrow ta{n}^{-1}\frac{15}{8}+ta{n}^{-1}\frac{5}{12}=ta{n}^{-1}\frac{\frac{15}{8}+\frac{5}{12}}{1-\frac{15}{8}\frac{5}{12}}=ta{n}^{-1}\frac{220}{21}$

$\therefore tan\left(ta{n}^{-1}\frac{220}{21}\right)=\frac{220}{21}$

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

Kindly consider the following figure

Equation of plane ${\perp }^r$ to line $\frac{x-3}{2}=\frac{x-1}{1}=\frac{x-2}{1}$ and passes through the point

(2, 3, -1) is 2(x – 2) + 1(y – 3) + 1 (z + 1) = 0

=> 2x + y + z – 6 = 0         .(i)

Hence point (1, 2, 2) satisfies equation (i)

$z=\frac{1}{1-cos\theta +2isin\theta }=\frac{1-cos\theta -2isin\theta }{{\left(1-cos\theta \right)}^2-{\left(2isin\theta \right)}^2}=\frac{2si{n}^2\frac{\theta }{2}-2isin\theta }{{\left(1-cos\theta \right)}^2+4si{n}^2\theta }$      

$\therefore Re\left(z\right)=\frac{2si{n}^2\frac{\theta }{2}}{4si{n}^2\frac{\theta }{2}\left(si{n}^2\frac{\theta }{2}+4co{s}^2\frac{\theta }{2}\right)}=\frac{1}{2\left(1+3co{s}^2\frac{\theta }{2}\right)}=\frac{1}{5}\Rightarrow co{s}^2\frac{\theta }{2}=\frac{1}{2}$   

$\Rightarrow \theta =\frac{\pi }{2}\therefore {\displaystyle {\int }_0^{\frac{\pi }{2}}sinxdx=1}$

2ⁿ = 4096 -> n = 12

¹²C₆ = 924 (the greatest coefficient)

$\Rightarrow \frac{6}{3}=\frac{2}{-t}\Rightarrow t=-1$

 ->R (-1,0)

    $P{R}^2+R{Q}^2=20+5=25$          

xf (x) + 2 = $\lambda$ (x – 2) (x – 3) (x – 4) (x – 5)

$x=0\Rightarrow \lambda =\frac{1}{60}$                

x = 10 -> 10f (10) = 26