Maths

f (3x)- f (x) = x

Replace $x\to \frac{x}{3}\Rightarrow f\left(x\right)-f\left(\frac{x}{3}\right)=\frac{x}{3}$

Again replace $x\to \frac{x}{3}\Rightarrow f\left(\frac{x}{3}\right)-f\left(\frac{x}{3^2}\right)-f\left(\frac{x}{3^2}\right)=\frac{x}{3^2}$

$\Rightarrow f\left(3x\right)-f\left(0\right)=\frac{3x}{2}puttingx=\frac{8}{3}\Rightarrow f\left(8\right)-f\left(0\right)=4\therefore f\left(0\right)=3$

Also putting x = $\frac{14}{3}$ in f (3x) – 3 = $\frac{3x}{2}\Rightarrow$ F (14) – 3 = 7 f (14) = 10

 $Letf\left(x\right)=lo{g}_{cosx}cosecx=\frac{logcosecx}{logcosx}$

$f\text{'}\left(x\right)=\frac{logcosx.sinx\left(-cosecxcotx-\left(logcosecx\right)\frac{1}{cosx}.\left(sinx\right)\right)}{{\left(logcosx\right)}^2}$

$Atx=\frac{\pi }{4}f\text{'}\left(\frac{\pi }{4}\right)=\frac{-log\left(\frac{1}{\sqrt[]{2}}\right)+log\sqrt[]{2}}{{\left(log\frac{1}{\sqrt[]{2}}\right)}^2}=\frac{2}{log\sqrt[]{2}}\therefore atx=\frac{\pi }{4},lo{g}_e\left(2f\text{'}\left(x\right)\right)=4$

 $\beta =\frac{\alpha x-\left(e^{3x}-1\right)}{\alpha x\left(e^{3x}-1\right)},\alpha \in R\underset{x\to 0}{lim}\frac{\frac{\alpha }{3}-\left(\frac{e^{3x}-1}{3x}\right)}{\alpha x\left(\frac{e^{3x}-1}{3x}\right)}$

$=\underset{x\to 0}{lim}\frac{1-\frac{\left(1+3x+\frac{9x^2}{2}+..........-1\right)}{3x}}{1+3x+\frac{9x^2}{2}+........-1}=-\frac{1}{2}\therefore \alpha +\beta =\frac{5}{2}$

$f_a\left(x\right)=ta{n}^{-1}2x-3ax+7\Rightarrow f_a\text{'}\left(x\right)=\frac{2}{1+4x^2}-3a\Rightarrow f_a\text{'}\left(x\right)\ge 0\Rightarrow 3a\le \frac{2}{1+4x^2}$

$a_{max}=\frac{2}{3}\left(\frac{1}{1+4*\frac{{\pi }^2}{36}}\right)=\frac{6}{9+{\pi }^2}=\overline{a}\therefore f_a\left(\frac{\pi }{8}\right)=ta{n}^{-1}\frac{2\pi }{8}-3\frac{\pi }{8}\frac{6}{9+{\pi }^2}+7=8-\frac{9\pi }{4\left(9+{\pi }^2\right)}$

Let equation of normal to x² = y at Q (t, t²) is x + 2ty = t + 2t³

It passes through the point (1, -1) so, 2t³ + 3t – 1 = 0

Let f(t) = 2t³ + 3t – 1 f   $\left(\frac{1}{4}\right)f\left(\frac{1}{3}\right)<0\Rightarrow t\in \left(\frac{1}{4},\frac{1}{3}\right)$

Let P(1 – sin q, -1 + cos q) $\therefore$  slope of normal = slope of CP $-\frac{1}{2t}=\frac{cos\theta }{-sin\theta }\Rightarrow 2t$ Þ = tan q according to question $x=1-sin\theta =1-\frac{2t}{\sqrt[]{1+4t^2}}=g\left(t\right)\therefore g\left(t\right)=1-\frac{2t}{\sqrt[]{1+4t^2}}$ ,  

Þ g'(t) < 0 g(t) is decreasing function in  $t\in \left(\frac{1}{4},\frac{1}{3}\right)\Rightarrow g\left(t\right)\in \left(0.440,0.485\right)\in \left(\frac{1}{4},\frac{1}{2}\right)$

$A\text{'}BA=\left[111\right]\left[\begin{array}{ccc}\hfill 9^2\hfill & \hfill -10^2\hfill & \hfill 11^2\hfill \\ \hfill 12^2\hfill & \hfill 13^2\hfill & \hfill -14^2\hfill \\ \hfill -15^2\hfill & \hfill 16^2\hfill & \hfill 17^2\hfill \end{array}\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]=$

$\left[9^2+12^2-15^2-10^2+13^2+16^{2}11^2-14^2+17^2\right]\left[\begin{array}{l}1\\ 1\\ 1\end{array}\right]$

$=\left[9^2+12^2-15^2-10^2+13^2+16^2+11^2-14^2+17^2\right]=\left[539\right]$

$|z-i|=\left|z+5i\right|$ So, z lies on perpendicular bisector of (0, 1) and (0, 5) i.e., line y = 2 as |z| = 2 z = 2i x = 0 and y = 2 so, x + 2y + 4 = 0

Let a and b be the roots of the equation  $x^2+\left(3-a\right)x+1=2a$

Therefore a + b = a – 3, ab = 1 – 2a Þ a² + b² = (a – 3)² – 2 (1 – 2a) = a² – 6a + 9 – 2 + 4a = a² – 2a + 7 = (a – 1)² + 6 Þ So,   ${\alpha }^2+{\beta }^2\ge 6$